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Quantitative aspects of the composition of substances and mixtures are the subject of this chapter.
06: Composition of Substances and Solutions
Learning Objectives
• Calculate formula masses for covalent and ionic compounds
• Define the amount unit mole and the related quantity Avogadro’s number
• Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another
We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.
Formula Mass
In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance’s formula.
Formula Mass for Covalent Substances
For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl3), a covalent compound once used as a surgical anesthetic and now primarily used in the production of tetrafluoroethylene, the building block for the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure $1$ outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu.
Likewise, the molecular mass of an aspirin molecule, C9H8O4, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure $2$).
Example $1$: Computing Molecular Mass for a Covalent Compound
Ibuprofen, C13H18O2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?
Solution
Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:
Exercise $1$
Acetaminophen, C8H9NO2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?
Answer
151.16 amu
Formula Mass for Ionic Compounds
Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.”
As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na+, and chloride anions, Cl, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (Figure $3$).
Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.
Example $2$: Computing Formula Mass for an Ionic Compound
Aluminum sulfate, Al2(SO4)3, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?
Solution
The formula for this compound indicates it contains Al3+ and SO42 ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al2S3O12. Following the approach outlined above, the formula mass for this compound is calculated as follows:
Exercise $2$
Calcium phosphate, $\ce{Ca3(PO4)2}$, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?
Answer
310.18 amu
The Mole
The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, $\ce{H2O}$, and hydrogen peroxide, $\ce{H2O2}$, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.
The number of entities composing a mole has been experimentally determined to be $6.02214179 \times 10^{23}$, a fundamental constant named Avogadro’s number ($N_A$) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being $6.022 \times 10^{23}/\ce{mol}$.
Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure $4$).
Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12C contains 1 mole of 12C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure $5$).
Table $1$: Mass of one mole of elements
Element Average Atomic Mass (amu) Molar Mass (g/mol) Atoms/Mole
C 12.01 12.01 $6.022 \times 10^{23}$
H 1.008 1.008 $6.022 \times 10^{23}$
O 16.00 16.00 $6.022 \times 10^{23}$
Na 22.99 22.99 $6.022 \times 10^{23}$
Cl 33.45 35.45 $6.022 \times 10^{23}$
While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water after a rainfall. Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.
The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.
Example $3$: Deriving Moles from Grams for an Element
According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?
Solution
The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.
The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):
The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”
$\mathrm{4.7\; \cancel{g} K \left ( \dfrac{mol\; K}{39.10\;\cancel{g}}\right)=0.12\;mol\; K} \nonumber$
The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.
Exercise $3$: Beryllium
Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?
Answer
0.360 mol
Example $4$: Deriving Grams from Moles for an Element
A liter of air contains $9.2 \times 10^{−4}$ mol argon. What is the mass of Ar in a liter of air?
Solution
The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10−3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):
In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):
$\mathrm{9.2 \times10^{-4}\; \cancel{mol} \; Ar \left( \dfrac{39.95\;g}{\cancel{mol}\;Ar} \right)=0.037\;g\; Ar} \nonumber$
The result is in agreement with our expectations, around 0.04 g Ar.
Exercise $4$
What is the mass of 2.561 mol of gold?
Answer
504.4 g
Example $6$: Deriving Number of Atoms from Mass for an Element
Copper is commonly used to fabricate electrical wire (Figure $6$). How many copper atoms are in 5.00 g of copper wire?
Solution
The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (NA) to convert this molar amount to number of Cu atoms:
Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth NA, or approximately 1022 Cu atoms. Carrying out the two-step computation yields:
$\mathrm{5.00\:\cancel{g}\:Cu\left(\dfrac{\cancel{mol}\:Cu}{63.55\:\cancel{g}}\right)\left(\dfrac{6.022\times10^{23}\:atoms}{\cancel{mol}}\right)=4.74\times10^{22}\:atoms\: of\: copper} \nonumber$
The factor-label method yields the desired cancellation of units, and the computed result is on the order of 1022 as expected.
Exercise $6$
A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?
Answer
$4.586 \times 10^{22}\; Au$ atoms
Example $7$: Deriving Moles from Grams for a Compound
Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C2H5O2N. How many moles of glycine molecules are contained in 28.35 g of glycine?
Solution
We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example $6$:
The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C2H5O2N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:
The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:
$\mathrm{28.35\:\cancel{g}\:glycine\left(\dfrac{mol\: glycine}{75.07\:\cancel{g}}\right)=0.378\:mol\: glycine} \nonumber$
This result is consistent with our rough estimate.
Exercise $7$
How many moles of sucrose, $C_{12}H_{22}O_{11}$, are in a 25-g sample of sucrose?
Answer
0.073 mol
Example $8$: Deriving Grams from Moles for a Compound
Vitamin C is a covalent compound with the molecular formula C6H8O6. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10−4 mol. What is the mass of this allowance in grams?
Solution
As for elements, the mass of a compound can be derived from its molar amount as shown:
The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10−4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:
$\mathrm{1.42\times10^{-4}\:\cancel{mol}\:vitamin\: C\left(\dfrac{176.124\:g}{\cancel{mol}\:vitamin\: C}\right)=0.0250\:g\: vitamin\: C} \nonumber$
This is consistent with the anticipated result.
Exercise $8$
What is the mass of 0.443 mol of hydrazine, $N_2H_4$?
Answer
14.2 g
Example $9$: Deriving the Number of Molecules from the Compound Mass
A packet of an artificial sweetener contains 40.0 mg of saccharin (C7H5NO3S), which has the structural formula:
Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?
Solution
The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example $8$, and then multiplying by Avogadro’s number:
Using the provided mass and molar mass for saccharin yields:
$\mathrm{0.0400\:\cancel{g}\:\ce{C7H5NO3S}\left(\dfrac{\cancel{mol}\:\ce{C7H5NO3S}}{183.18\:\cancel{g}\:\ce{C7H5NO3S}}\right)\left(\dfrac{6.022\times10^{23}\:\ce{C7H5NO3S}\:molecules}{1\:\cancel{mol}\:\ce{C7H5NO3S}}\right)}\ =\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\:molecules} \nonumber$
The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:
$\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\: molecules\left(\dfrac{7\:C\: atoms}{1\:\ce{C7H5NO3S}\: molecule}\right)=9.20\times10^{21}\:C\: atoms} \nonumber$
Exercise $9$
How many $C_4H_{10}$ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?
Answer
• $9.545 \times 10^{22}\; \text{molecules}\; C_4H_{10}$
• $9.545 \times 10^{23 }\;\text{atoms}\; H$
Summary
The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be 6.022 × 1023, a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H2O molecule weighs approximately18 amu and 1 mole of H2O molecules weighs approximately 18 g).
Footnotes
1. 1 Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.
Glossary
Avogadro’s number (NA)
experimentally determined value of the number of entities comprising 1 mole of substance, equal to 6.022 × 1023 mol−1
formula mass
sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass
mole
amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of 12C
molar mass
mass in grams of 1 mole of a substance | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/06%3A_Composition_of_Substances_and_Solutions/6.1%3A_Formula_Mass_and_the_Mole_Concept.txt |
Learning Objectives
• Compute the percent composition of a compound
• Determine the empirical formula of a compound
• Determine the molecular formula of a compound
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Percent Composition
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
$\mathrm{\%H=\dfrac{mass\: H}{mass\: compound}\times100\%} \nonumber$
$\mathrm{\%C=\dfrac{mass\: C}{mass\: compound}\times100\%} \nonumber$
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
$\mathrm{\%H=\dfrac{2.5\:g\: H}{10.0\:g\: compound}\times100\%=25\%} \nonumber$
$\mathrm{\%C=\dfrac{7.5\:g\: C}{10.0\:g\: compound}\times100\%=75\%} \nonumber$
Example $1$: Calculation of Percent Composition
Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?
Solution
To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
$\mathrm{\%C=\dfrac{7.34\:g\: C}{12.04\:g\: compound}\times100\%=61.0\%} \nonumber$
$\mathrm{\%H=\dfrac{1.85\:g\: H}{12.04\:g\: compound}\times100\%=15.4\%} \nonumber$
$\mathrm{\%N=\dfrac{2.85\:g\: N}{12.04\:g\: compound}\times100\%=23.7\%} \nonumber$
The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.
Exercise $1$
A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?
Answer
12.1% C, 16.1% O, 71.8% Cl
Determining Percent Composition from Formula Mass
Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:
$\mathrm{\%N=\dfrac{14.01\:amu\: N}{17.03\:amu\:NH_3}\times100\%=82.27\%} \nonumber$
$\mathrm{\%H=\dfrac{3.024\:amu\: N}{17.03\:amu\:NH_3}\times100\%=17.76\%} \nonumber$
This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example $2$. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.
Example $2$: Determining Percent Composition from a Molecular Formula
Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?
Solution
To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:
\begin{align*} \%\ce C&=\mathrm{\dfrac{9\:mol\: C\times molar\: mass\: C}{molar\: mass\:\ce{C9H18O4}}\times100=\dfrac{9\times12.01\:g/mol} \nonumber{180.159\:g/mol}\times100=\dfrac{108.09\:g/mol}{180.159\:g/mol}\times100} \nonumber\ \%\ce C&=\mathrm{60.00\,\%\,C} \nonumber \end{align*} \nonumber
\begin{align*} \%\ce H&=\mathrm{\dfrac{8\:mol\: H\times molar\: mass\: H}{molar\: mass\:\ce{C9H18O4}}\times 100=\dfrac{8\times 1.008\:g/mol} \nonumber {180.159\:g/mol}\times 100=\dfrac{8.064\:g/mol}{180.159\:g/mol}\times 100} \nonumber\ \%\ce H&=4.476\,\%\,\ce H \nonumber \end{align*} \nonumber
\begin{align*} \%\ce O&=\mathrm{\dfrac{4\:mol\: O\times molar\: mass\: O}{molar\: mass\: \ce{C9H18O4}}\times 100=\dfrac{4\times 16.00\:g/mol} \nonumber{180.159\:g/mol}\times 100=\dfrac{64.00\:g/mol}{180.159\:g/mol}\times 100} \nonumber \ \%\ce O&=35.52\% \nonumber \end{align*} \nonumber
Note that these percentages sum to equal 100.00% when appropriately rounded.
Exercise $2$
To three significant digits, what is the mass percentage of iron in the compound $Fe_2O_3$?
Answer
69.9% Fe
Determination of Empirical Formulas
As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:
$\mathrm{1.71\:g\: C\times \dfrac{1\:mol\: C}{12.01\:g\: C}=0.142\:mol\: C} \nonumber$
$\mathrm{0.287\:g\: H\times \dfrac{1\:mol\: H}{1.008\:g\: H}=0.284\:mol\: H} \nonumber$
Thus, we can accurately represent this compound with the formula C0.142H0.284. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:
$\ce C_{\Large{\frac{0.142}{0.142}}}\:\ce H_{\Large{\frac{0.284}{0.142}}}\ce{\:or\:CH2} \nonumber$
(Recall that subscripts of “1” are not written, but rather assumed if no other number is present.)
The empirical formula for this compound is thus CH2. This may or not be the compound’s molecular formula as well; however, we would need additional information to make that determination (as discussed later in this section).
Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:
$\mathrm{Cl_{0.150}O_{0.525}=Cl_{\Large{\frac{0.150}{0.150}}}\: O_{\Large{\frac{0.525}{0.150}}}=ClO_{3.5}} \nonumber$
In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl2O7 as the final empirical formula.
Procedure
In summary, empirical formulas are derived from experimentally measured element masses by:
1. Deriving the number of moles of each element from its mass
2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained
Figure $1$ outlines this procedure in flow chart fashion for a substance containing elements A and X.
Example $3$: Determining an Empirical Formula from Masses of Elements
A sample of the black mineral hematite (Figure $2$), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?
Solution
For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:
\begin{align*} \mathrm{34.97\:g\: Fe\left(\dfrac{mol\: Fe}{55.85\:g}\right)}&=\mathrm{0.6261\:mol\: Fe} \nonumber\ \nonumber\ \mathrm{15.03\:g\: O\left(\dfrac{mol\: O}{16.00\:g}\right)}&=\mathrm{0.9394\:mol\: O} \nonumber \end{align*} \nonumber
Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:
$\mathrm{\dfrac{0.6261}{0.6261}=1.000\:mol\: Fe} \nonumber$
$\mathrm{\dfrac{0.9394}{0.6261}=1.500\:mol\: O} \nonumber$
The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:
$\mathrm{2(Fe_1O_{1.5})=Fe_2O_3} \nonumber$
The empirical formula is $Fe_2O_3$.
Exercise $3$
What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?
Answer
$N_2O_5$
Video $1$: Additional worked examples illustrating the derivation of empirical formulas are presented in the brief video clip.
Deriving Empirical Formulas from Percent Composition
Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.
Example $4$: Determining an Empirical Formula from Percent Composition
The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure $3$). What is the empirical formula for this gas?
Solution
Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:
\begin{align*} 27.29\,\%\,\ce C&=\mathrm{\dfrac{27.29\:g\: C}{100\:g\: compound}} \nonumber \ \nonumber \ 72.71\,\%\,\ce O&=\mathrm{\dfrac{72.71\:g\: O}{100\:g\: compound}} \nonumber \end{align*} \nonumber
The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:
\begin{align*} \mathrm{27.29\:g\: C\left(\dfrac{mol\: C}{12.01\:g}\right)}&=\mathrm{2.272\:mol\: C} \nonumber \ \nonumber \ \mathrm{72.71\:g\: O\left(\dfrac{mol\: O}{16.00\:g}\right)}&=\mathrm{4.544\:mol\: O} \nonumber \end{align*} \nonumber
Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:
$\mathrm{\dfrac{2.272\:mol\: C}{2.272}=1} \nonumber$
$\mathrm{\dfrac{4.544\:mol\: O}{2.272}=2} \nonumber$
Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO2.
Exercise $4$
What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?
Answer
$CH_2O$
Derivation of Molecular Formulas
Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.
Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n:
$\mathrm{\dfrac{molecular\: or\: molar\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}= \mathit n\: formula\: units/molecule} \nonumber$
The molecular formula is then obtained by multiplying each subscript in the empirical formula by n, as shown by the generic empirical formula AxBy:
$\mathrm{(A_xB_y)_n=A_{nx}B_{nx}} \nonumber$
For example, consider a covalent compound whose empirical formula is determined to be CH2O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:
$\mathrm{\dfrac{180\:amu/molecule}{30\:\dfrac{amu}{formula\: unit}}=6\:formula\: units/molecule} \nonumber$
Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:
$\ce{(CH2O)6}=\ce{C6H12O6} \nonumber$
Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.
Example $5$: Determination of the Molecular Formula for Nicotine
Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?
Solution
Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:
\begin{alignat}{2} &\mathrm{(74.02\:g\: C)\left(\dfrac{1\:mol\: C}{12.01\:g\: C}\right)}&&= \:\mathrm{6.163\:mol\: C}\ &\mathrm{(8.710\:g\: H)\left(\dfrac{1\:mol\: H}{1.01\:g\: H}\right)}&&= \:\mathrm{8.624\:mol\: H}\ &\mathrm{(17.27\:g\: N)\left(\dfrac{1\:mol\: N}{14.01\:g\: N}\right)}&&= \:\mathrm{1.233\:mol\: N} \end{alignat} \nonumber
Next, we calculate the molar ratios of these elements relative to the least abundant element, $\ce{N}$.
$6.163 \: \text{mol} \: \ce{C}$ / $1.233 \: \text{mol} \: \ce{N}$ = 5
$8.264 \: \text{mol} \: \ce{H}$ / $1.233 \: \text{mol} \: \ce{N}$ = 7
$1.233 \: \text{mol} \: \ce{N}$ / $1.233\: \text{mol} \: \ce{N}$ = 1
1.233/1.233 = $1.000 \: \text{mol} \: \ce{N}$
6.163/1.233 = $4.998 \: \text{mol} \: \ce{C}$
8.624/1.233 = $6.994 \: \text{mol} \: \ce{H}$
The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C5H7N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.
We calculate the molar mass for nicotine from the given mass and molar amount of compound:
$\mathrm{\dfrac{40.57\:g\: nicotine}{0.2500\:mol\: nicotine}=\dfrac{162.3\:g}{mol}} \nonumber$
Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:
$\mathrm{\dfrac{162.3\:g/mol}{81.13\:\dfrac{g}{formula\: unit}}=2\:formula\: units/molecule} \nonumber$
Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:
$\ce{(C5H7N)2}=\ce{C10H14N2} \nonumber$
Exercise $5$
What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?
Answer
C8H10N4O2
Summary
The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.
Key Equations
• $\mathrm{\%X=\dfrac{mass\: X}{mass\: compound}\times100\%}$
• $\mathrm{\dfrac{molecular\: or\: molar\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}=\mathit n\: formula\: units/molecule}$
• (AxBy)n = AnxBny
Glossary
percent composition
percentage by mass of the various elements in a compound
empirical formula mass
sum of average atomic masses for all atoms represented in an empirical formula | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/06%3A_Composition_of_Substances_and_Solutions/6.2%3A_Determining_Empirical_and_Molecular_Formulas.txt |
Learning Objectives
• Describe the fundamental properties of solutions
• Calculate solution concentrations using molarity
• Perform dilution calculations using the dilution equation
In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (Figure $1$). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.
Solutions
We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions.
The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution.
A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).
Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:
$M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.2}$
Example $1$: Calculating Molar Concentrations
A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?
Solution
Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:
\begin{align*} M &=\dfrac{mol\: solute}{L\: solution} \[4pt] &=\dfrac{0.133\:mol}{355\:mL\times \dfrac{1\:L}{1000\:mL}} \[4pt] &= 0.375\:M \label{3.4.1} \end{align*}
Exercise $1$
A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?
Answer
0.05 M
Example $2$: Deriving Moles and Volumes from Molar Concentrations
How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example $1$?
Solution
In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 3.4.2, 0.375 M:
$M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.3}$
\begin{align*} \mathrm{mol\: solute} &= \mathrm{ M\times L\: solution} \label{3.4.4} \[4pt] \mathrm{mol\: solute} &= \mathrm{0.375\:\dfrac{mol\: sugar}{L}\times \left(10\:mL\times \dfrac{1\:L}{1000\:mL}\right)} &= \mathrm{0.004\:mol\: sugar} \label{3.4.5} \end{align*}
Exercise $2$
What volume (mL) of the sweetened tea described in Example $1$ contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?
Answer
80 mL
Example $3$: Calculating Molar Concentrations from the Mass of Solute
Distilled white vinegar (Figure $2$) is a solution of acetic acid, $CH_3CO_2H$, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?
Solution
As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:
$\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=\dfrac{25.2\: g\: \ce{CH3CO2H}\times \dfrac{1\:mol\: \ce{CH3CO2H}}{60.052\: g\: \ce{CH3CO2H}}}{0.500\: L\: solution}=0.839\: \mathit M} \label{3.4.6}$
$M=\mathrm{\dfrac{0.839\:mol\: solute}{1.00\:L\: solution}} \nonumber$
$\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=0.839\:\mathit M} \label{3.4.7}$
Exercise $3$
Calculate the molarity of 6.52 g of $CoCl_2$ (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.
Answer
0.674 M
Example $4$: Determining the Mass of Solute in a Given Volume of Solution
How many grams of NaCl are contained in 0.250 L of a 5.30-M solution?
Solution
The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example $3$:
$M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.9}$
$\mathrm{mol\: solute= \mathit M\times L\: solution} \label{3.4.10}$
$\mathrm{mol\: solute=5.30\:\dfrac{mol\: NaCl}{L}\times 0.250\:L=1.325\:mol\: NaCl} \label{3.4.11}$
Finally, this molar amount is used to derive the mass of NaCl:
$\mathrm{1.325\: mol\: NaCl\times\dfrac{58.44\:g\: NaCl}{mol\: NaCl}=77.4\:g\: NaCl} \label{3.4.12}$
Exercise $4$
How many grams of $CaCl_2$ (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?
Answer
5.55 g $CaCl_2$
When performing calculations stepwise, as in Example $3$, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example $4$, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.
In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (Example $5$). This eliminates intermediate steps so that only the final result is rounded.
Example $5$: Determining the Volume of Solution
In Example $3$, we found the typical concentration of vinegar to be 0.839 M. What volume of vinegar contains 75.6 g of acetic acid?
Solution
First, use the molar mass to calculate moles of acetic acid from the given mass:
$\mathrm{g\: solute\times\dfrac{mol\: solute}{g\: solute}=mol\: solute} \label{3.4.13}$
Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:
$\mathrm{mol\: solute\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.14}$
Combining these two steps into one yields:
$\mathrm{g\: solute\times \dfrac{mol\: solute}{g\: solute}\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.15}$
$\mathrm{75.6\:g\:\ce{CH3CO2H}\left(\dfrac{mol\:\ce{CH3CO2H}}{60.05\:g}\right)\left(\dfrac{L\: solution}{0.839\:mol\:\ce{CH3CO2H}}\right)=1.50\:L\: solution} \label{3.4.16}$
Exercise $5$:
What volume of a 1.50-M KBr solution contains 66.0 g KBr?
Answer
0.370 L
Dilution of Solutions
Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure $2$).
Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure $3$).
A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters:
$n=ML \nonumber$
Expressions like these may be written for a solution before and after it is diluted:
$n_1=M_1L_1 \nonumber$
$n_2=M_2L_2 \nonumber$
where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution,n1 = n2. Thus, these two equations may be set equal to one another:
$M_1L_1=M_2L_2 \nonumber$
This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:
$C_1V_1=C_2V_2 \nonumber$
where $C$ and $V$ are concentration and volume, respectively.
Example $6$: Determining the Concentration of a Diluted Solution
If 0.850 L of a 5.00-M solution of copper nitrate, Cu(NO3)2, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?
Solution
We are given the volume and concentration of a stock solution, V1 and C1, and the volume of the resultant diluted solution, V2. We need to find the concentration of the diluted solution, C2. We thus rearrange the dilution equation in order to isolate C2:
$C_1V_1=C_2V_2 \nonumber$
$C_2=\dfrac{C_1V_1}{V_2} \nonumber$
Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:
$C_2=\mathrm{\dfrac{0.850\:L\times 5.00\:\dfrac{mol}{L}}{1.80\: L}}=2.36\:M \nonumber$
This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M).
Exercise $6$
What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-M solution of CH3OH to 500.0 mL?
Answer
0.102 M $CH_3OH$
Example $7$: Volume of a Diluted Solution
What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr?
Solution
We are given the volume and concentration of a stock solution, V1 and C1, and the concentration of the resultant diluted solution, C2. We need to find the volume of the diluted solution, V2. We thus rearrange the dilution equation in order to isolate V2:
$C_1V_1=C_2V_2 \nonumber$
$V_2=\dfrac{C_1V_1}{C_2} \nonumber$
Since the diluted concentration (0.12 M) is slightly more than one-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:
$V_2=\dfrac{(0.45\:M)(0.011\: \ce L)}{(0.12\:M)} \nonumber$
$V_2=\mathrm{0.041\:L} \nonumber$
The volume of the 0.12-M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.
Exercise $7$
A laboratory experiment calls for 0.125 M $HNO_3$. What volume of 0.125 M $HNO_3$ can be prepared from 0.250 L of 1.88 M $HNO_3$?
Answer
3.76 L
Example $8$: Volume of a Concentrated Solution Needed for Dilution
What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?
Solution
We are given the concentration of a stock solution, C1, and the volume and concentration of the resultant diluted solution, V2 and C2. We need to find the volume of the stock solution, V1. We thus rearrange the dilution equation in order to isolate V1:
$C_1V_1=C_2V_2 \nonumber$
$V_1=\dfrac{C_2V_2}{C_1} \nonumber$
Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:
$V_1=\dfrac{(0.100\:M)(5.00\:\ce L)}{1.59\:M} \nonumber$
$V_1=0.314\:\ce L \nonumber$
Thus, we would need 0.314 L of the 1.59-M solution to prepare the desired solution. This result is consistent with our rough estimate.
Exercise $8$
What volume of a 0.575-M solution of glucose, C6H12O6, can be prepared from 50.00 mL of a 3.00-M glucose solution?
Answer
0.261
Summary
Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.
Key Equations
• $M=\mathrm{\dfrac{mol\: solute}{L\: solution}}$
• C1V1 = C2V2
Glossary
aqueous solution
solution for which water is the solvent
concentrated
qualitative term for a solution containing solute at a relatively high concentration
concentration
quantitative measure of the relative amounts of solute and solvent present in a solution
dilute
qualitative term for a solution containing solute at a relatively low concentration
dilution
process of adding solvent to a solution in order to lower the concentration of solutes
dissolved
describes the process by which solute components are dispersed in a solvent
molarity (M)
unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution
solute
solution component present in a concentration less than that of the solvent
solvent
solution component present in a concentration that is higher relative to other components | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/06%3A_Composition_of_Substances_and_Solutions/6.3%3A_Molarity.txt |
Learning Objectives
• Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-per-million (ppm), and parts-per-billion (ppb)
• Perform computations relating a solution’s concentration and its components’ volumes and/or masses using these units
In the previous section, we introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. In this section, we will introduce some other units of concentration that are commonly used in various applications, either for convenience or by convention.
Mass Percentage
Earlier in this chapter, we introduced percent composition as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component’s mass to the solution’s mass, expressed as a percentage:
$\text{mass percentage} = \dfrac{\text{mass of component}}{\text{mass of solution}} \times100\% \label{3.5.1}$
We are generally most interested in the mass percentages of solutes, but it is also possible to compute the mass percentage of solvent.
Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).
Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure $1$) cites the concentration of its active ingredient, sodium hypochlorite ($\ce{NaOCl}$), as being 7.4%. A 100.0-g sample of bleach would therefore contain 7.4 g of $\ce{NaOCl}$.
Example $1$: Calculation of Percent by Mass
A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of glucose. What is the percent by mass of glucose in spinal fluid?
Solution
The spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about 0.1%. Substituting the given masses into the equation defining mass percentage yields:
$\mathrm{\%\,glucose=\dfrac{3.75\;mg \;glucose \times \frac{1\;g}{1000\; mg}}{5.0\;g \;spinal\; fluid}=0.075\%} \nonumber$
The computed mass percentage agrees with our rough estimate (it’s a bit less than 0.1%).
Note that while any mass unit may be used to compute a mass percentage (mg, g, kg, oz, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, we converted the units of solute in the numerator from mg to g to match the units in the denominator. We could just as easily have converted the denominator from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.
Exercise $1$
A bottle of a tile cleanser contains 135 g of $\ce{HCl}$ and 775 g of water. What is the percent by mass of $\ce{HCl}$ in this cleanser?
Answer
14.8%
Example $2$: Calculations using Mass Percentage
“Concentrated” hydrochloric acid is an aqueous solution of 37.2% $\ce{HCl}$ that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of $\ce{HCl}$ is contained in 0.500 L of this solution?
Solution
The HCl concentration is near 40%, so a 100-g portion of this solution would contain about 40 g of HCl. Since the solution density isn’t greatly different from that of water (1 g/mL), a reasonable estimate of the HCl mass in 500 g (0.5 L) of the solution is about five times greater than that in a 100 g portion, or $\mathrm{5 \times 40 = 200\: g}$. To derive the mass of solute in a solution from its mass percentage, we need to know the corresponding mass of the solution. Using the solution density given, we can convert the solution’s volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart:
For proper unit cancellation, the 0.500-L volume is converted into 500 mL, and the mass percentage is expressed as a ratio, 37.2 g HCl/g solution:
$\mathrm{500\; mL\; solution \left(\dfrac{1.19\;g \;solution}{mL \;solution}\right) \left(\dfrac{37.2\;g\; HCl}{100\;g \;solution}\right)=221\;g\; HCl} \nonumber$
This mass of HCl is consistent with our rough estimate of approximately 200 g.
Exercise $2$
What volume of concentrated HCl solution contains 125 g of HCl?
Answer
282 mL
Volume Percentage
Liquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, %vol or (v/v)%:
$\text{volume percentage} = \dfrac{\text{volume solute}}{\text{volume solution}} \times100\% \label{3.5.2}$
Example $3$: Calculations using Volume Percentage
Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol?
Solution
Per the definition of volume percentage, the isopropanol volume is 70% of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:
$\text {355 mL solution}(\frac{\text{70 mL isopropyl alcohol}}{\text{100 mL solution}})(\frac{\text{0.785 g isopropyl alcohol}}{\text{1 mL isopropyl alcohol}})=\text{195 g isopropyl alcohol} \nonumber$
Exercise $3$
Wine is approximately 12% ethanol ($\ce{CH_3CH_2OH}$) by volume. Ethanol has a molar mass of 46.06 g/mol and a density 0.789 g/mL. How many moles of ethanol are present in a 750-mL bottle of wine?
Answer
1.5 mol ethanol
Mass-Volume Percentage
“Mixed” percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute’s mass to the solution’s volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of 0.9% mass/volume (m/v), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as “blood sugar”) is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter (100 mL) of blood (Figure $2$).
Parts per Million and Parts per Billion
Very low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage (“part per hundred”) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.
The mass-based definitions of ppm and ppb are given here:
$\text{ppm}=\dfrac{\text{mass solute}}{\text{mass solution}} \times 10^6\; \text{ppm} \label{3.5.3A}$
$\text{ppb}=\dfrac{\text{mass solute}}{\text{mass solution}} \times 10^9\; \text{ppb} \label{3.5.3B}$
Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water (Figure $3$).
Example $4$: Parts per Million and Parts per Billion Concentrations
According to the EPA, when the concentration of lead in tap water reaches 15 ppb, certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead (μg) would be contained in a typical glass of water (300 mL)?
Solution
The definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm. Comparing these two unit definitions shows that ppm is 1000 times greater than ppb (1 ppm = 103 ppb). Thus:
$\mathrm{15\; \cancel{ppb} \times \dfrac{1\; ppm}{10^3\;\cancel{ppb}} =0.015\; ppm} \nonumber$
The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:
$\text{ppb}=\dfrac{\text{mass solute}}{\text{mass solution}} ×10^9\; \text{ppb} \nonumber$
$\text{mass solute} = \dfrac{\text{ppb} \times \text{mass solution}}{10^9\;\text{ppb}} \nonumber$
$\text{mass solute}=\mathrm{\dfrac{15\:ppb×300\:mL×\dfrac{1.00\:g}{mL}}{10^9\:ppb}=4.5 \times 10^{-6}\;g} \nonumber$
Finally, convert this mass to the requested unit of micrograms:
$\mathrm{4.5 \times 10^{−6}\;g \times \dfrac{1\; \mu g}{10^{−6}\;g} =4.5\; \mu g} \nonumber$
Exercise $4$
A 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units.
Answer
9.6 ppm, 9600 ppb
Summary
In addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components’ masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.
Glossary
mass percentage
ratio of solute-to-solution mass expressed as a percentage
mass-volume percent
ratio of solute mass to solution volume, expressed as a percentage
parts per billion (ppb)
ratio of solute-to-solution mass multiplied by 109
parts per million (ppm)
ratio of solute-to-solution mass multiplied by 106
volume percentage
ratio of solute-to-solution volume expressed as a percentage | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/06%3A_Composition_of_Substances_and_Solutions/6.4%3A_Other_Units_for_Solution_Concentrations.txt |
3.1: Formula Mass and the Mole Concept
What is the total mass (amu) of carbon in each of the following molecules?
1. (a) CH4
2. (b) CHCl3
3. (c) C12H10O6
4. (d) CH3CH2CH2CH2CH3
(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu
What is the total mass of hydrogen in each of the molecules?
1. (a) CH4
2. (b) CHCl3
3. (c) C12H10O6
4. (d) CH3CH2CH2CH2CH3
Calculate the molecular or formula mass of each of the following:
(a) P4
(b) H2O
(c) Ca(NO3)2
(d) CH3CO2H (acetic acid)
(e) C12H22O11 (sucrose, cane sugar).
(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu
Determine the molecular mass of the following compounds:
(a)
(b)
(c)
(d)
Determine the molecular mass of the following compounds:
(a)
(b)
(c)
(d)
1. (a) 56.107 amu;
2. (b) 54.091 amu;
3. (c) 199.9976 amu;
4. (d) 97.9950 amu
Which molecule has a molecular mass of 28.05 amu?
(a)
(b)
(c)
Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
Compare 1 mole of H2, 1 mole of O2, and 1 mole of F2.
1. (a) Which has the largest number of molecules? Explain why.
2. (b) Which has the greatest mass? Explain why.
Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2H5OH), 0.60 mol of formic acid (HCO2H), or 1.0 mol of water (H2O)? Explain why.
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C2H5OH), 1 mol of formic acid (HCO2H), or 1 mol of water (H2O)? Explain why.
How are the molecular mass and the molar mass of a compound similar and how are they different?
The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 1023 molecules.
Calculate the molar mass of each of the following compounds:
1. (a) hydrogen fluoride, HF
2. (b) ammonia, NH3
3. (c) nitric acid, HNO3
4. (d) silver sulfate, Ag2SO4
5. (e) boric acid, B(OH)3
Calculate the molar mass of each of the following:
1. (a) S8
2. (b) C5H12
3. (c) Sc2(SO4)3
4. (d) CH3COCH3 (acetone)
5. (e) C6H12O6 (glucose)
(a) 256.528 g/mol; (b) 72.150 g mol−1; (c) 378.103 g mol−1; (d) 58.080 g mol−1; (e) 180.158 g mol−1
Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:
1. (a) limestone, CaCO3
2. (b) halite, NaCl
3. (c) beryl, Be3Al2Si6O18
4. (d) malachite, Cu2(OH)2CO3
5. (e) turquoise, CuAl6(PO4)4(OH)8(H2O)4
Calculate the molar mass of each of the following:
1. (a) the anesthetic halothane, C2HBrClF3
2. (b) the herbicide paraquat, C12H14N2Cl2
3. (c) caffeine, C8H10N4O2
4. (d) urea, CO(NH2)2
5. (e) a typical soap, C17H35CO2Na
(a) 197.382 g mol−1; (b) 257.163 g mol−1; (c) 194.193 g mol−1; (d) 60.056 g mol−1; (e) 306.464 g mol−1
Determine the number of moles of compound and the number of moles of each type of atom in each of the following:
1. (a) 25.0 g of propylene, C3H6
2. (b) 3.06 × 10−3 g of the amino acid glycine, C2H5NO2
3. (c) 25 lb of the herbicide Treflan, C13H16N2O4F (1 lb = 454 g)
4. (d) 0.125 kg of the insecticide Paris Green, Cu4(AsO3)2(CH3CO2)2
5. (e) 325 mg of aspirin, C6H4(CO2H)(CO2CH3)
Determine the mass of each of the following:
1. (a) 0.0146 mol KOH
2. (b) 10.2 mol ethane, C2H6
3. (c) 1.6 × 10−3 mol Na2 SO4
4. (d) 6.854 × 103 mol glucose, C6 H12 O6
5. (e) 2.86 mol Co(NH3)6Cl3
1. (a) 0.819 g;
2. (b) 307 g;
3. (c) 0.23 g;
4. (d) 1.235 × 106 g (1235 kg);
5. (e) 765 g
Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:
1. (a) 2.12 g of potassium bromide, KBr
2. (b) 0.1488 g of phosphoric acid, H3PO4
3. (c) 23 kg of calcium carbonate, CaCO3
4. (d) 78.452 g of aluminum sulfate, Al2(SO4)3
5. (e) 0.1250 mg of caffeine, C8H10N4O2
Determine the mass of each of the following:
1. (a) 2.345 mol LiCl
2. (b) 0.0872 mol acetylene, C2H2
3. (c) 3.3 × 10−2 mol Na2 CO3
4. (d) 1.23 × 103 mol fructose, C6 H12 O6
5. (e) 0.5758 mol FeSO4(H2O)7
1. (a) 99.41 g;
2. (b) 2.27 g;
3. (c) 3.5 g;
4. (d) 222 kg;
5. (e) 160.1 g
The approximate minimum daily dietary requirement of the amino acid leucine, C6H13NO2, is 1.1 g. What is this requirement in moles?
Determine the mass in grams of each of the following:
1. (a) 0.600 mol of oxygen atoms
2. (b) 0.600 mol of oxygen molecules, O2
3. (c) 0.600 mol of ozone molecules, O3
(a) 9.60 g; (b) 19.2 g; (c) 28.8 g
A 55-kg woman has 7.5 × 10−3 mol of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?
Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO4, a semiprecious stone.
zirconium: 2.038 × 1023 atoms; 30.87 g; silicon: 2.038 × 1023 atoms; 9.504 g; oxygen: 8.151 × 1023 atoms; 21.66 g
Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH4, 0.6 mol of C6H6, or 0.4 mol of C3H8.
Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO4, 266 g of Al2Cl6, or 225 g of Al2S3.
AlPO4: 1.000 mol
Al2Cl6: 1.994 mol
Al2S3: 3.00 mol
Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?
The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone?
3.113 × 1025 C atoms
One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?
A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?
0.865 servings, or about 1 serving.
A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na2PO3F) in 100 mL.
1. What mass of fluorine atoms in mg was present?
2. How many fluorine atoms were present?
Which of the following represents the least number of molecules?
1. 20.0 g of H2O (18.02 g/mol)
2. 77.0 g of CH4 (16.06 g/mol)
3. 68.0 g of CaH2 (42.09 g/mol)
4. 100.0 g of N2O (44.02 g/mol)
5. 84.0 g of HF (20.01 g/mol)
20.0 g H2O represents the least number of molecules since it has the least number of moles.
3.2: Determining Empirical and Molecular Formulas
What information do we need to determine the molecular formula of a compound from the empirical formula?
Calculate the following to four significant figures:
1. (a) the percent composition of ammonia, NH3
2. (b) the percent composition of photographic “hypo,” Na2S2O3
3. (c) the percent of calcium ion in Ca3(PO4)2
(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca2+ = 38.76%
Determine the following to four significant figures:
1. the percent composition of hydrazoic acid, HN3
2. the percent composition of TNT, C6H2(CH3)(NO2)3
3. the percent of SO42– in Al2(SO4)3
Determine the percent ammonia, NH3, in Co(NH3)6Cl3, to three significant figures.
% NH3 = 38.2%
Determine the percent water in CuSO4∙5H2O to three significant figures.
Determine the empirical formulas for compounds with the following percent compositions:
(a) 15.8% carbon and 84.2% sulfur
(b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen
(a) CS2
(b) CH2O
Determine the empirical formulas for compounds with the following percent compositions:
(a) 43.6% phosphorus and 56.4% oxygen
(b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O
A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?
C6H6
Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?
Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.
Mg3Si2H3O8 (empirical formula), Mg6Si4H6O16 (molecular formula)
Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:
1. Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O
2. Saran; 24.8% C, 2.0% H, 73.1% Cl
3. polyethylene; 86% C, 14% H
4. polystyrene; 92.3% C, 7.7% H
5. Orlon; 67.9% C, 5.70% H, 26.4% N
A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.
C15H15N3
3.3: Molarity
Questions
Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.
What information do we need to calculate the molarity of a sulfuric acid solution?
We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?
Determine the molarity for each of the following solutions:
1. 0.444 mol of CoCl2 in 0.654 L of solution
2. 98.0 g of phosphoric acid, H3PO4, in 1.00 L of solution
3. 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution
4. 10.5 kg of Na2SO4·10H2O in 18.60 L of solution
5. 7.0 × 10−3 mol of I2 in 100.0 mL of solution
6. 1.8 × 104 mg of HCl in 0.075 L of solution
1. (a) 0.679 M;
2. (b) 1.00 M;
3. (c) 0.06998 M;
4. (d) 1.75 M;
5. (e) 0.070 M;
6. (f) 6.6 M
Determine the molarity of each of the following solutions:
1. 1.457 mol KCl in 1.500 L of solution
2. 0.515 g of H2SO4 in 1.00 L of solution
3. 20.54 g of Al(NO3)3 in 1575 mL of solution
4. 2.76 kg of CuSO4·5H2O in 1.45 L of solution
5. 0.005653 mol of Br2 in 10.00 mL of solution
6. 0.000889 g of glycine, C2H5NO2, in 1.05 mL of solution
Answers:
a.) 0.9713 M
b.) 5.25x10-3 M
c.) 6.122x10-2 M
d.) 7.62 M
e.) 0.5653 M
f.) 1.13x10-2 M
Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C6H12O6, used for intravenous injection?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
Consider this question: What is the mass of solute in 200.0 L of a 1.556-M solution of KBr?
1. (a) Outline the steps necessary to answer the question.
2. (b) Answer the question.
Answer:
(a)
1. Calculate to moles of KBr by multiplying the Molarity by the amount of solution (200.0 L)
2. Find the Molar Mass of KBr and convert moles of solute to grams
(b)
$\dfrac{1.556\:moles\:\ce{KBr}}{1\:\cancel{L}}\times 200.0\:\cancel{L}=311.2\:moles\:\ce{KBr}$
$311.2\:\cancel{moles}\:\ce{KBr}\times\dfrac{119.0\:g\:\ce{KBr}}{1\:\cancel{mole}\:\ce{KBr}}=37,030\:g$
37,030g; 37.03 kg
Calculate the number of moles and the mass of the solute in each of the following solutions:
1. (a) 2.00 L of 18.5 M H2SO4, concentrated sulfuric acid
2. (b) 100.0 mL of 3.8 × 10−5 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum
3. (c) 5.50 L of 13.3 M H2CO, the formaldehyde used to “fix” tissue samples
4. (d) 325 mL of 1.8 × 10−6 M FeSO4, the minimum concentration of iron sulfate detectable by taste in drinking water
(a) 37.0 mol H2SO4;
3.63 × 103 g H2SO4;
(b) 3.8 × 10−6 mol NaCN;
1.9 × 10−4 g NaCN;
(c) 73.2 mol H2CO;
2.20 kg H2CO;
(d) 5.9 × 10−7 mol FeSO4;
8.9 × 10−5 g FeSO4
Calculate the number of moles and the mass of the solute in each of the following solutions:
1. 325 mL of 8.23 × 10−5 M KI, a source of iodine in the diet
2. 75.0 mL of 2.2 × 10−5 M H2SO4, a sample of acid rain
3. 0.2500 L of 0.1135 M K2CrO4, an analytical reagent used in iron assays
4. 10.5 L of 3.716 M (NH4)2SO4, a liquid fertilizer
Answers:
a. 2.67x10-5 moles KI; 4.44x10-3g KI
b. 1.7x10-6 moles H2SO4 ; 1.6x10-4 g H2SO4
c. 2.838x10-2 moles K2CrO4 ; 5.510g K2CrO4
d. 39.0 moles (NH4)2SO4 ; 5,160 g (NH4)2SO4
Consider this question: What is the molarity of KMnO4 in a solution of 0.0908 g of KMnO4 in 0.500 L of solution?
1. (a) Outline the steps necessary to answer the question.
2. (b) Answer the question.
(a) Determine the molar mass of KMnO4; determine the number of moles of KMnO4 in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 × 10−3 M
Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?
1. (a) Outline the steps necessary to answer the question.
2. (b) Answer the question.
Answer:
(a)
1. Convert g of HCl to moles of HCl and convert mL of solution to L of solution
2. Divide moles of HCl by L of solution
(b)
$0.3366\:\cancel{g}\:\ce{HCl}\times\dfrac{1\:mole\:\ce{HCl}}{36.46\:\cancel{g}\:\ce{HCl}}=9.232\times10^{-3}\:moles\:\ce{HCl}$
$35.23\:mL = 0.03523\:L$
$\dfrac{9.232\times10^{-3}\:moles\:\ce{HCl}}{0.03523\:L}=0.2621\:M\:\ce{HCl}$
0.2621 M ;
Calculate the molarity of each of the following solutions:
(a) 0.195 g of cholesterol, C27H46O, in 0.100 L of serum, the average concentration of cholesterol in human serum
(b) 4.25 g of NH3 in 0.500 L of solution, the concentration of NH3 in household ammonia
(c) 1.49 kg of isopropyl alcohol, C3H7OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol
(d) 0.029 g of I2 in 0.100 L of solution, the solubility of I2 in water at 20 °C
(a) 5.04 × 10−3 M;
(b) 0.499 M;
(c) 9.92 M;
(d) 1.1 × 10−3 M
Calculate the molarity of each of the following solutions:
1. 293 g HCl in 666 mL of solution, a concentrated HCl solution
2. 2.026 g FeCl3 in 0.1250 L of a solution used as an unknown in general chemistry laboratories
3. 0.001 mg Cd2+ in 0.100 L, the maximum permissible concentration of cadmium in drinking water
4. 0.0079 g C7H5SNO3 in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.
There is about 1.0 g of calcium, as Ca2+, in 1.0 L of milk. What is the molarity of Ca2+ in milk?
0.025 M
What volume of a 1.00-M Fe(NO3)3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M?
If 0.1718 L of a 0.3556-M C3H7OH solution is diluted to a concentration of 0.1222 M, what is the volume of the resulting solution?
0.5000 L
If 4.12 L of a 0.850 M-H3PO4 solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?
What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?
1.9 mL
What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to 0.105 L?
What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?
1. (a) 1.00 L of a 0.250-M solution of Fe(NO3)3 is diluted to a final volume of 2.00 L
2. (b) 0.5000 L of a 0.1222-M solution of C3H7OH is diluted to a final volume of 1.250 L
3. (c) 2.35 L of a 0.350-M solution of H3PO4 is diluted to a final volume of 4.00 L
4. (d) 22.50 mL of a 0.025-M solution of C12H22O11 is diluted to 100.0 mL
1. (a) 0.125 M;
2. (b) 0.04888 M;
3. (c) 0.206 M;
4. (e) 0.0056 M
What is the final concentration of the solution produced when 225.5 mL of a 0.09988-M solution of Na2CO3 is allowed to evaporate until the solution volume is reduced to 45.00 mL?
A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?
11.9 M
An experiment in a general chemistry laboratory calls for a 2.00-M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?
What volume of a 0.20-M K2SO4 solution contains 57 g of K2SO4?
1.6 L
The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K2Cr2O7), what is the maximum permissible molarity of that substance?
3.4: Other Units for Solution Concentrations
Questions
1. Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO3 by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3 by mass?
1. Outline the steps necessary to answer the question.
2. Answer the question.
2. What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?
3. What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
4. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm–3 and contains 37.21% HCl by mass?
5. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of $\ce{CaCO_3}$, which is equivalent to milligrams of $\ce{CaCO_3}$ per liter of water. What is the molar concentration of Ca2+ ions in a water sample with a hardness count of 175 mg CaCO3/L?
6. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream.
7. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose (C6H12O6) in mg/dL?
8. A throat spray is 1.40% by mass phenol, $\ce{C_6H_5OH}$, in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution.
9. Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass?
10. A cough syrup contains 5.0% ethyl alcohol, C2H5OH, by mass. If the density of the solution is 0.9928 g/mL, determine the molarity of the alcohol in the cough syrup.
11. D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose ($\ce{C_6H_{12}O_6}$) in water. If the density of D5W is 1.029 g/mL, calculate the molarity of dextrose in the solution.
12. Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, $\ce{H_2SO_4}$, for which the density is 1.3057 g/mL.
Solutions
1
• (a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: $\mathrm{\%\,mass_1 \times mass_1=\%\;mass_2 \times mass_2}$ This equation can be rearranged to isolate $\mathrm{mass_1}$ and the given quantities substituted into this equation.
• (b) 58.8 g
3. $\mathrm{114 \;g}$ 5. $1.75 \times 10^{−3} M$ 7 $\mathrm{95\: mg/dL}$ 9 $\mathrm{2.38 \times 10^{−4}\: mol}$ 11 $\mathrm{0.29 mol}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/06%3A_Composition_of_Substances_and_Solutions/6.E%3A_Composition_of_Substances_and_Solutions_%28Exercises%29.txt |
This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions—that is, the reaction stoichiometry.
• 7.0: Prelude to Stoichiometry
This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions—that is, the reaction stoichiometry.
• 7.1: Writing and Balancing Chemical Equations
Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products.
• 7.2: Classifying Chemical Reactions
Chemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and oxidation-reduction (redox). Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for one or more reactant elements.
• 7.3: Reaction Stoichiometry
A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.
• 7.4: Reaction Yields
When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield, which is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (theoretical yield). The extent to which a reaction generates the theoretical amount is expressed as its percent yield.
• 7.5: Quantitative Chemical Analysis
The stoichiometry of chemical reactions may serve as the basis for quantitative chemical analysis methods. Titrations involve measuring the volume of a titrant solution required to completely react with a sample solution. This volume is then used to calculate the concentration of analyte in the sample using the stoichiometry of the titration reaction. Gravimetric analysis involves separating analytes from the sample, determining its mass, and then calculating its concentration.
• 7.E: Stoichiometry of Chemical Reactions (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
07: Stoichiometry of Chemical Reactions
Solid-fuel rockets are a central feature in the world’s space exploration programs, including the new Space Launch System being developed by the National Aeronautics and Space Administration (NASA) to replace the retired Space Shuttle fleet (Figure \(1\)). The engines of these rockets rely on carefully prepared solid mixtures of chemicals combined in precisely measured amounts. Igniting the mixture initiates a vigorous chemical reaction that rapidly generates large amounts of gaseous products. These gases are ejected from the rocket engine through its nozzle, providing the thrust needed to propel heavy payloads into space. Both the nature of this chemical reaction and the relationships between the amounts of the substances being consumed and produced by the reaction are critically important considerations that determine the success of the technology. This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions—that is, the reaction stoichiometry. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.0%3A_Prelude_to_Stoichiometry.txt |
Learning Objectives
• Derive chemical equations from narrative descriptions of chemical reactions.
• Write and balance chemical equations in molecular, total ionic, and net ionic formats.
The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Figure $1$, with space-filling molecular models shown in the lower half of the figure.
This example illustrates the fundamental aspects of any chemical equation:
1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
2. The substances generated by the reaction are called products, and their formulas are placed on the right sight of the equation.
3. Plus signs (+) separate individual reactant and product formulas, and an arrow (⟶) separates the reactant and product (left and right) sides of the equation.
4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.
It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure $2$). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:
• One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
• One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
• One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules.
Balancing Equations
When a chemical equation is balanced it means that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, $\ce{CO2}$ and $\ce{H2O}$, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is
$\left(1\: \cancel{\ce{CO_2} \: \text{molecule}} \times \dfrac{2\: \ce{O} \: \text{atoms}}{ \cancel{\ce{CO_2} \: \text{molecule}}}\right) + \left( \cancel{ \ce{2H_2O} \: \text{molecule} }\times \dfrac{1\: \ce{O}\: \text{atom}}{\cancel{ \ce{H_2O} \: \text{molecule}}}\right)=4\: \ce{O} \: \text{atoms} \nonumber$
The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:
$\ce{CH4 +2O2\rightarrow CO2 +2H2O} \nonumber$
Equation for the reaction $\ce{CH4 +2O2\rightarrow CO2 +2H2O} \nonumber$
Element Reactants Products Balanced?
C 1 × 1 = 1 1 × 1 = 1 1 = 1, yes
H 4 × 1 = 4 2 × 2 = 4 4 = 4, yes
O 2 × 2 = 4 (1 × 2) + (2 × 1) = 4 4 = 4, yes
A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:
$\ce{H2O \rightarrow H2 + O2} \tag{unbalanced}$
Comparing the number of H and O atoms on either side of this equation confirms its imbalance:
Comparisons between H and O atoms
Element Reactants Products Balanced?
H 1 × 2 = 2 1 × 2 = 2 2 = 2, yes
O 1 × 1 = 1 1 × 2 = 2 1 ≠ 2, no
The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H2O to H2O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2.
$\ce{2H2O \rightarrow H2 + O2} \tag{unbalanced}$
O atom balance may be achieved by changing the coefficient for H2O to 2
Element Reactants Products Balanced?
H 2 × 2 = 4 1 × 2 = 2 4 ≠ 2, no
O 2 × 1 = 2 1 × 2 = 2 2 = 2, yes
The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H2 product to 2.
$\ce{2H_2O \rightarrow 2H2 + O2} \tag{balanced}$
H atom balance upset but easily reestablished by changing the coefficient for the H2 product to 2.
Element Reactants Products Balanced?
H 2 × 2 = 4 2 × 2 = 2 4 = 4, yes
O 2 × 1 = 2 1 × 2 = 2 2 = 2, yes
These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:
$\ce{2H_2O \rightarrow 2H_2 + O_2} \nonumber$
Example $1$: Balancing Chemical Equations
Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide.
Solution
First, write the unbalanced equation.
$\ce{N_2 + O_2 \rightarrow N_2O_5} \tag{unbalanced}$
Next, count the number of each type of atom present in the unbalanced equation.
Unbalanced Equation
Element Reactants Products Balanced?
N 1 × 2 = 2 1 × 2 = 2 2 = 2, yes
O 1 × 2 = 2 1 × 5 = 5 2 ≠ 5, no
Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O2 and N2O5 to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).
$\ce{N_2 + 5 O2 \rightarrow 2 N2O5} \tag{unbalanced}$
$\ce{N_2 + 5 O2 \rightarrow 2 N2O5} \tag{unbalanced}$
Element Reactants Products Balanced?
N 1 × 2 = 2 2 × 2 = 4 2 ≠ 4, no
O 5 × 2 = 10 2 × 5 = 10 10 = 10, yes
The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N2 to 2.
$\ce{2N_2 + 5O_2\rightarrow 2N_2O_5} \nonumber$
N atom balance upset but restored by changing the coefficient for the reactant N2 to 2.
Element Reactants Products Balanced?
N 2 × 2 = 4 2 × 2 = 4 4 = 4, yes
O 5 × 2 = 10 2 × 5 = 10 10 = 10, yes
The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.
Exercise $1$
Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)
Answer
$\ce{2NH4NO3 \rightarrow 2N2 + O2 + 4H2O} \nonumber$
Balancing Reactions Which Contain Polyatomics:
It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C2H6) with oxygen to yield H2O and CO2, represented by the unbalanced equation:
$\ce{C_2H_6 + O_2 \rightarrow H_2O + CO_2} \tag{unbalanced}$
Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:
$\ce{C_2H_6 + O_2 \rightarrow 3H_2O + 2CO_2} \tag{unbalanced}$
This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O2 reactant to yield an odd number, so a fractional coefficient, $\ce{7/2}$, is used instead to yield a provisional balanced equation:
$\ce{C2H6 + 7/2 O2\rightarrow 3H2O + 2CO2} \nonumber$
A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:
$\ce{2C2H6 +7O2\rightarrow 6H2O + 4CO2} \nonumber$
Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,
$\ce{3N2 +9H2\rightarrow 6NH3} \nonumber$
the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:
$\ce{N2 + 3H2\rightarrow 2NH3} \nonumber$
Phet Simulation
Use this interactive tutorial for additional practice balancing equations.
Additional Information in Chemical Equations
The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here:
$\ce{2Na (s) + 2H2O (l) \rightarrow 2NaOH (aq) + H2(g)} \nonumber$
This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).
Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow.
$\ce{CaCO3}(s)\xrightarrow{\:\Delta\:} \ce{CaO}(s)+\ce{CO2}(g) \nonumber$
Other examples of these special conditions will be encountered in more depth in later chapters.
Equations for Ionic Reactions
Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of $\ce{CaCl2}$ and $\ce{AgNO3}$ are mixed, a reaction takes place producing aqueous $\ce{Ca(NO3)2}$ and solid $\ce{AgCl}$:
$\ce{CaCl2}(aq)+\ce{2AgNO3}(aq)\rightarrow \ce{Ca(NO3)2}(aq)+\ce{2AgCl}(s) \nonumber$
This balanced equation, derived in the usual fashion, is called a molecular equation because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:
$\ce{CaCl2}(aq)\rightarrow \ce{Ca^2+}(aq)+\ce{2Cl-}(aq) \nonumber$
$\ce{2AgNO3}(aq)\rightarrow \ce{2Ag+}(aq)+\ce{2NO3-}(aq) \nonumber$
$\ce{Ca(NO3)2}(aq)\rightarrow \ce{Ca^2+}(aq)+\ce{2NO3-}(aq) \nonumber$
Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, (s).
Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:
$\ce{Ca^2+}(aq)+\ce{2Cl-}(aq)+\ce{2Ag+}(aq)+\ce{2NO3-}(aq)\rightarrow \ce{Ca^2+}(aq)+\ce{2NO3-}(aq)+\ce{2AgCl}(s) \nonumber$
Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, $\ce{Ca^{2+}(aq)}$ and $\ce{NO3-}(aq)$. These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:
$\cancel{\ce{Ca^2+}(aq)}+\ce{2Cl-}(aq)+\ce{2Ag+}(aq)+\cancel{\ce{2NO3-}(aq)}\rightarrow \cancel{\ce{Ca^2+}(aq)}+\cancel{\ce{2NO3-}(aq)}+\ce{2AgCl}(s) \nonumber$
$\ce{2Cl-}(aq)+\ce{2Ag+}(aq)\rightarrow \ce{2AgCl}(s) \nonumber$
Following the convention of using the smallest possible integers as coefficients, this equation is then written:
$\ce{Cl-}(aq)+\ce{Ag+}(aq)\rightarrow \ce{AgCl}(s) \nonumber$
This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of $\ce{Cl^{−}}$ and $\ce{Ag+}$.
Example $2$: Molecular and Ionic Equations
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.
Solution
Begin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:
$\ce{CO2(aq) + NaOH(aq) \rightarrow Na2CO3(aq) + H2O(l)} \tag{unbalanced}$
Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:
$\ce{CO2(aq)+2NaOH(aq)\rightarrow Na2CO3(aq) + H2O}(l) \nonumber$
The two dissolved ionic compounds, NaOH and Na2CO3, can be represented as dissociated ions to yield the complete ionic equation:
$\ce{CO2 (aq) + 2Na+ (aq) + 2OH- (aq) \rightarrow 2Na+ (aq) + CO3^{2-} (aq) + H2O (l)} \nonumber$
Finally, identify the spectator ion(s), in this case Na+(aq), and remove it from each side of the equation to generate the net ionic equation:
\begin{align*} \ce{CO2}(aq)+\cancel{\ce{2Na+}(aq)}+\ce{2OH-}(aq)&\rightarrow \cancel{\ce{2Na+}(aq)}+\ce{CO3^2-}(aq)+\ce{H2O}(l)\ \ce{CO2}(aq)+\ce{2OH-}(aq)&\rightarrow \ce{CO3^2-}(aq)+\ce{H2O}(l) \end{align*} \nonumber
Exercise $2$
Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:
$\ce{NaCl(aq) + H2O(l) ->[ electricity] NaOH(aq) + H2(g) + Cl2(g)} \nonumber$
Write balanced molecular, complete ionic, and net ionic equations for this process.
Answer
Balanced molecular equation: $\ce{2NaCl(aq) + 2H2O(l) -> 2NaOH(aq) + H2(g) + Cl2(g)} \nonumber$
Balanced ionic equation: $\ce{2Na^{+}(aq) + 2Cl^{-}(aq) + 2H2O (l) -> 2Na^{+}(aq) + 2OH^{-}(aq) + H2(g) + Cl2(g)} \nonumber$
Balanced net ionic equation: $\ce{2Cl^{-}(aq) + 2H2O(l) -> 2OH^{-}(aq) + H2(g) + Cl2(g) } \nonumber$
Key Concepts and Summary
Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations.
Glossary
balanced equation
chemical equation with equal numbers of atoms for each element in the reactant and product
chemical equation
symbolic representation of a chemical reaction
coefficient
number placed in front of symbols or formulas in a chemical equation to indicate their relative amount
complete ionic equation
chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions
molecular equation
chemical equation in which all reactants and products are represented as neutral substances
net ionic equation
chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)
product
substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation
reactant
substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation
spectator ion
ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.1%3A_Writing_and_Balancing_Chemical_Equations.txt |
Learning Objectives
• Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction)
• Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations
• Identify common acids and bases
• Predict the solubility of common inorganic compounds by using solubility rules
• Compute the oxidation states for elements in compounds
Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction.
Precipitation Reactions and Solubility Rules
A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter).
The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table $1$).
Table $1$: Solubilities of Common Ionic Compounds in Water
Soluble compounds contain Exceptions to these solubility rules include
• group 1 metal cations (Li+, Na+, K+, Rb+, and Cs+) and ammonium ion $\left(\ce{NH4+}\right)$
• the halide ions (Cl, Br, and I)
• the acetate $\ce{(C2H3O2- )}$, bicarbonate $\ce{(HCO3- )}$, nitrate $\ce{(NO3- )}$, and chlorate $\ce{(ClO3- )}$ ions
• the sulfate $\ce{(SO4- )}$ ion
• halides of Ag+, $\ce{Hg2^2+}$, and Pb2+
• sulfates of Ag+, Ba2+, Ca2+, $\ce{Hg2^2+}$, Pb2+, and Sr2+
Insoluble compounds contain Exceptions to these insolubility rules include
• carbonate $\ce{(CO3^2- )}$, chromate $\ce{(CrO4^2- )}$, phosphate $\ce{(PO4^3- )}$, and sulfide (S2−) ions
• hydroxide ion (OH)
• compounds of these anions with group 1 metal cations and ammonium ion
• hydroxides of group 1 metal cations and Ba2+
A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:
$\ce{2KI}(aq)+\ce{Pb(NO3)2}(aq)\rightarrow \ce{PbI2}(s)+\ce{2KNO3}(aq) \nonumber$
This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.
The net ionic equation representing this reaction is:
$\ce{Pb^2+}(aq)+\ce{2I-}(aq)\rightarrow \ce{PbI2}(s) \nonumber$
Lead iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow (Figure $1$). The properties of pure PbI2 crystals make them useful for fabrication of X-ray and gamma ray detectors.
The solubility guidelines in Table $1$ may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag+, $\ce{NO3-}$, Na+, and F ions. Aside from the two ionic compounds originally present in the solutions, AgNO3 and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO3 and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:
$\ce{NaF}(aq)+\ce{AgNO3}(aq)\rightarrow \ce{AgF}(s)+\ce{NaNO3}(aq)\hspace{20px}\ce{(molecular)} \nonumber$
$\ce{Ag+}(aq)+\ce{F-}(aq)\rightarrow \ce{AgF}(s)\hspace{20px}\ce{(net\: ionic)} \nonumber$
Example $1$: Predicting Precipitation Reactions
Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.
1. potassium sulfate and barium nitrate
2. lithium chloride and silver acetate
3. lead nitrate and ammonium carbonate
Solution
(a) The two possible products for this combination are KNO3 and BaSO4. The solubility guidelines indicate BaSO4 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is
$\ce{Ba^2+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{BaSO4}(s) \nonumber$
(b) The two possible products for this combination are LiC2H3O2 and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is
$\ce{Ag+}(aq)+\ce{Cl-}(aq)\rightarrow \ce{AgCl}(s) \nonumber$
(c) The two possible products for this combination are PbCO3 and NH4NO3. The solubility guidelines indicate PbCO3 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is
$\ce{Pb^2+}(aq)+\ce{CO3^2-}(aq)\rightarrow \ce{PbCO3}(s) \nonumber$
Exercise $1$
Which solution could be used to precipitate the barium ion, Ba2+, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?
Answer
sodium sulfate, BaSO4
Acid-Base Reactions
An acid-base reaction is one in which a hydrogen ion, H+, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text.
For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, H3O+. As an example, consider the equation shown here:
$\ce{HCl}(aq)+\ce{H2O}(aq)\rightarrow \ce{Cl-}(aq)+\ce{H3O+}(aq) \nonumber$
The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H3O+ ions are produced by a chemical reaction in which H+ ions are transferred from HCl molecules to H2O molecules (Figure $2$).
The nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table $1$). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)\rightleftharpoons \ce{CH3CO2-}(aq)+\ce{H3O+}(aq) \nonumber$
When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, $\ce{CH3CO2-}$ (Figure $3$). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)
Table $2$: Common Strong Acids
Compound Formula Name in Aqueous Solution
HBr hydrobromic acid
HCl hydrochloric acid
HI hydroiodic acid
HNO3 nitric acid
HClO4 perchloric acid
H2SO4 sulfuric acid
A base is a substance that will dissolve in water to yield hydroxide ions, OH. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion—for example, NaOH and Ca(OH)2. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)2 dissolve in water and dissociate completely to produce cations (K+ and Ba2+, respectively) and hydroxide ions, OH. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases.
Consider as an example the dissolution of lye (sodium hydroxide) in water:
$\ce{NaOH}(s)\rightarrow \ce{Na+}(aq)+\ce{OH-}(aq) \nonumber$
This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na+ and OH ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.
Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure $4$). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:
$\ce{NH3}(aq)+\ce{H2O}(l)\rightleftharpoons \ce{NH4+}(aq)+\ce{OH-}(aq) \nonumber$
This is, by definition, an acid-base reaction, in this case involving the transfer of H+ ions from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as $\ce{NH4+}$ ions.
The chemical reactions described in which acids and bases dissolved in water produce hydronium and hydroxide ions, respectively, are, by definition, acid-base reactions. In these reactions, water serves as both a solvent and a reactant. A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base, the products are often a salt and water, and neither reactant is the water itself:
$\mathrm{acid+base\rightarrow salt+water} \nonumber$
To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid Mg(OH)2) is ingested to ease symptoms associated with excess stomach acid (HCl):
$\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{2H2O}(l). \nonumber$
Note that in addition to water, this reaction produces a salt, magnesium chloride.
Example $2$: Writing Equations for Acid-Base Reactions
Write balanced chemical equations for the acid-base reactions described here:
1. the weak acid hydrogen hypochlorite reacts with water
2. a solution of barium hydroxide is neutralized with a solution of nitric acid
Solution
(a) The two reactants are provided, HOCl and H2O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H+ from HOCl to H2O to generate hydronium ions, H3O+ and hypochlorite ions, OCl.
$\ce{HOCl}(aq)+\ce{H2O}(l)\rightleftharpoons \ce{OCl-}(aq)+\ce{H3O+}(aq) \nonumber$
A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.
(b) The two reactants are provided, Ba(OH)2 and HNO3. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba2+) and the anion generated when the acid transfers its hydrogen ion $\ce{(NO3- )}$.
$\ce{Ba(OH)2}(aq)+\ce{2HNO3}(aq)\rightarrow \ce{Ba(NO3)2}(aq)+\ce{2H2O}(l) \nonumber$
Exercise $21$
Write the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acid is dissolved in water.)
Answer
$\ce{H3O+}(aq)+\ce{OH-}(aq)\rightarrow \ce{2H2O}(l) \nonumber$
Explore the microscopic view of strong and weak acids and bases.
Oxidation-Reduction Reactions
Earth’s atmosphere contains about 20% molecular oxygen, O2, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O2, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.
Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:
$\ce{2Na}(s)+\ce{Cl2}(g)\rightarrow \ce{2NaCl}(s) \nonumber$
It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:
\begin{align*} \ce{2Na}(s) &\rightarrow \ce{2Na+}(s)+\ce{2e-} \[4pt] \ce{Cl2}(g)+\ce{2e-} &\rightarrow \ce{2Cl-}(s) \end{align*} \nonumber
These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:
\begin{align} \textbf{oxidation}&=\textrm{loss of electrons}\ \textbf{reduction}&=\textrm{gain of electrons} \end{align}
In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.
\begin{align} \textbf{reducing agent}&=\textrm{species that is oxidized}\ \textbf{oxidizing agent}&=\textrm{species that is reduced} \end{align}
Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding $\ce{NaCl}$:
$\ce{H2}(g)+\ce{Cl2}(g)\rightarrow \ce{2HCl}(g) \nonumber$
The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.
1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion’s charge.
3. Oxidation numbers for common nonmetals are usually assigned as follows:
• Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
• Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, $\ce{O2^2-}$), very rarely $-\dfrac{1}{2}$ (so-called superoxides, $\ce{O2-}$), positive values when combined with F (values vary)
• Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.
Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.
Example $3$: Assigning Oxidation Numbers
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:
1. H2S
2. $\ce{SO3^2-}$
3. Na2SO4
Solution
(a) According to guideline 1, the oxidation number for H is +1.
Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:
$\ce{charge\: on\: H2S}=0=(2\times +1)+(1\times x)$
$x=0-(2\times +1)=-2$
(b) Guideline 3 suggests the oxidation number for oxygen is −2.
Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:
$\ce{charge\: on\: SO3^2-}=-2=(3\times -2)+(1\times x)$
$x=-2-(3\times -2)=+4$
(c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately.
According to guideline 2, the oxidation number for sodium is +1.
Assuming the usual oxidation number for oxygen (−2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:
$\ce{charge\: on\: SO4^2-}=-2=(4\times -2)+(1\times x)$
$x=-2-(4\times -2)=+6$
Exercise $3$
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:
1. KNO3
2. AlH3
3. $\mathrm{\underline{N}H_4^+}$
4. $\mathrm{\sideset{ }{_{\large{4}}^{-}}{H_2\underline{P}O}}$
Answer a
N, +5
Answer b
Al, +3
Answer c
N, −3
Answer d
P, +5
Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist as shown below\). Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:
\begin{align} \textbf{oxidation}&=\textrm{increase in oxidation number}\ \textbf{reduction}&=\textrm{decrease in oxidation number} \end{align} \nonumber
Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in HCl).
Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted below are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:
$\ce{10Al}(s)+\ce{6NH4ClO4}(s)\rightarrow \ce{4Al2O3}(s)+\ce{2AlCl3}(s)+\ce{12H2O}(g)+\ce{3N2}(g) \nonumber$
Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at 3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.
Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:
$\ce{Zn}(s)+\ce{2HCl}(aq)\rightarrow \ce{ZnCl2}(aq)+\ce{H2}(g) \nonumber$
Metallic elements may also be oxidized by solutions of other metal salts; for example:
$\ce{Cu}(s)+\ce{2AgNO3}(aq)\rightarrow \ce{Cu(NO3)2}(aq)+\ce{2Ag}(s) \nonumber$
This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu2+ ions dissolve in the solution to yield a characteristic blue color (Figure $4$).
Example $4$: Describing Redox Reactions
Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.
1. $\ce{ZnCO3}(s)\rightarrow \ce{ZnO}(s)+\ce{CO2}(g)$
2. $\ce{2Ga}(l)+\ce{3Br2}(l)\rightarrow \ce{2GaBr3}(s)$
3. $\ce{2H2O2}(aq)\rightarrow \ce{2H2O}(l)+\ce{O2}(g)$
4. $\ce{BaCl2}(aq)+\ce{K2SO4}(aq)\rightarrow \ce{BaSO4}(s)+\ce{2KCl}(aq)$
5. $\ce{C2H4}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{2H2O}(l)$
Solution
Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.
1. This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
2. This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(l) to +3 in GaBr3(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br2(l) to −1 in GaBr3(s). The oxidizing agent is Br2(l).
3. This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H2O2(aq) to 0 in O2(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H2O2(aq) to −2 in H2O(l). For disproportionation reactions, the same substance functions as an oxidant and a reductant.
4. This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
5. This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C2H4(g) to +4 in CO2(g). The reducing agent (fuel) is C2H4(g). Oxygen is reduced, its oxidation number decreasing from 0 in O2(g) to −2 in H2O(l). The oxidizing agent is O2(g).
Exercise $4$
This equation describes the production of tin(II) chloride:
$\ce{Sn}(s)+\ce{2HCl}(g)\rightarrow \ce{SnCl2}(s)+\ce{H2}(g) \nonumber$
Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.
Answer
Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant.
Balancing Redox Reactions via the Half-Reaction Method
Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:
1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2O molecules.
4. Balance hydrogen atoms by adding H+ ions.
5. Balance charge1 by adding electrons.
6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
• Add OH ions to both sides of the equation in numbers equal to the number of H+ ions.
• On the side of the equation containing both H+ and OH ions, combine these ions to yield water molecules.
• Simplify the equation by removing any redundant water molecules.
9. Finally, check to see that both the number of atoms and the total charges2 are balanced.
Example $5$: Balancing Redox Reactions in Acidic Solution
Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.
$\ce{Cr2O7^2- + Fe^2+ \rightarrow Cr^3+ + Fe^3+} \nonumber$
Solution
Write the two half-reactions.
Each half-reaction will contain one reactant and one product with one element in common.
$\ce{Fe^2+ \rightarrow Fe^3+}$
$\ce{Cr2O7^2- \rightarrow Cr^3+}$
Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.
$\ce{Fe^2+ \rightarrow Fe^3+}$
$\ce{Cr2O7^2- \rightarrow 2Cr^3+}$
Balance oxygen atoms by adding H2O molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.
$\ce{Fe^2+ \rightarrow Fe^3+}$
$\ce{Cr2O7^2- \rightarrow 2Cr^3+ + 7H2O}$
Balance hydrogen atoms by adding H+ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.
$\ce{Fe^2+ \rightarrow Fe^3+}$
$\ce{Cr2O7^2- + 14H+ \rightarrow 2Cr^3+ + 7H2O}$
Balance charge by adding electrons. The iron half-reaction shows a total charge of 2+ on the left side (1 Fe2+ ion) and 3+ on the right side (1 Fe3+ ion). Adding one electron to the right side bring that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved.
The chromium half-reaction shows a total charge of (1 × 2−) + (14 × 1+) = 12+ on the left side ($\ce{1 Cr2O7^2-}$ ion and 14 H+ ions). The total charge on the right side is (2 × 3+) = 6 + (2 Cr3+ ions). Adding six electrons to the left side will bring that side’s total charge to (12+ + 6−) = 6+, and charge balance is achieved.
$\ce{Fe^2+ \rightarrow Fe^3+ + e-}$
$\ce{Cr2O7^2- + 14H+ + 6e- \rightarrow 2Cr^3+ + 7H2O}$
Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6.
$\ce{6Fe^2+ \rightarrow 6Fe^3+ + 6e-}$
$\ce{Cr2O7^2- + 6e- + 14H+ \rightarrow 2Cr^3+ + 7H2O}$
Add the balanced half-reactions and cancel species that appear on both sides of the equation.
$\ce{6Fe^2+ + Cr2O7^2- + 6e- + 14H+ \rightarrow 6Fe^3+ + 6e- + 2Cr^3+ + 7H2O} \nonumber$
Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:
$\ce{6Fe^2+ + Cr2O7^2- + 14H+ \rightarrow 6Fe^3+ + 2Cr^3+ + 7H2O} \nonumber$
A final check of atom and charge balance confirms the equation is balanced.
Final check of atom and charge balance confirms the equation is balanced.
Reactants Products
Fe 6 6
Cr 2 2
O 7 7
H 14 14
charge 24+ 24+
Exercise $5$
In acidic solution, hydrogen peroxide reacts with Fe2+ to produce Fe3+ and H2O. Write a balanced equation for this reaction.
Answer
$\ce{H2O2}(aq)+\ce{2H+}(aq)+\ce{2Fe^2+} \rightarrow \ce{2H2O}(l)+\ce{2Fe^3+} \nonumber$
Summary
Chemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and oxidation-reduction (redox). Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for one or more reactant elements. Writing balanced equations for some redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method.
Footnotes
1. 1 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced.
2. 2 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced.
Glossary
acid
substance that produces H3O+ when dissolved in water
acid-base reaction
reaction involving the transfer of a hydrogen ion between reactant species
base
substance that produces OH when dissolved in water
combustion reaction
vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light
half-reaction
an equation that shows whether each reactant loses or gains electrons in a reaction.
insoluble
of relatively low solubility; dissolving only to a slight extent
neutralization reaction
reaction between an acid and a base to produce salt and water
oxidation
process in which an element’s oxidation number is increased by loss of electrons
oxidation-reduction reaction
(also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements
oxidation number
(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic
oxidizing agent
(also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced
precipitate
insoluble product that forms from reaction of soluble reactants
precipitation reaction
reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double-displacement or metathesis
reduction
process in which an element’s oxidation number is decreased by gain of electrons
reducing agent
(also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized
salt
ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide
single-displacement reaction
(also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species
soluble
of relatively high solubility; dissolving to a relatively large extent
solubility
the extent to which a substance may be dissolved in water, or any solvent
strong acid
acid that reacts completely when dissolved in water to yield hydronium ions
strong base
base that reacts completely when dissolved in water to yield hydroxide ions
weak acid
acid that reacts only to a slight extent when dissolved in water to yield hydronium ions
weak base
base that reacts only to a slight extent when dissolved in water to yield hydroxide ions | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.2%3A_Classifying_Chemical_Reactions.txt |
Learning Objectives
• Explain the concept of stoichiometry as it pertains to chemical reactions
• Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
• Perform stoichiometric calculations involving mass, moles, and solution molarity
A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.
The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Cooking, for example, offers an appropriate comparison. Suppose a recipe for making eight pancakes calls for 1 cup pancake mix, $\dfrac{3}{4}$ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is
$\mathrm{1\:cup\: mix+\dfrac{3}{4}\:cup\: milk+1\: egg \rightarrow 8\: pancakes} \label{4.4.1}$
If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is
$\mathrm{24\: \cancel{pancakes} \times \dfrac{1\: egg}{8\: \cancel{pancakes}}=3\: eggs} \label{4.4.2}$
Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:
$\ce{N2}(g)+\ce{3H2}(g)\rightarrow \ce{2NH3}(g) \label{4.4.3}$
This equation shows that ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:
$\ce{\dfrac{2NH3 \: molecules}{3H2 \: molecules}\: or \: \dfrac{2 \: doz \: NH3\: molecules}{3\: doz\:H2 \:molecules} \: or \: \dfrac{2\: mol\: NH3\: molecules}{3\: mol\: H2\: molecules}} \label{4.4.4}$
These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.
Example $1$: Moles of Reactant Required in a Reaction
How many moles of I2 are required to react with 0.429 mol of Al according to the following equation (see Figure $2$)?
$\ce{2Al + 3I2 \rightarrow 2AlI3} \label{4.4.5}$
Solution
Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is $\ce{\dfrac{3\: mol\: I2}{2\: mol\: Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:
\begin{align*} \mathrm{mol\: I_2} &=\mathrm{0.429\: \cancel{mol\: Al}\times \dfrac{3\: mol\: I_2}{2\:\cancel{mol\: Al}}} \[4pt] &=\mathrm{0.644\: mol\: I_2} \end{align*} \nonumber
Exercise $1$
How many moles of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce Ca3(PO4)2 according to the equation $\ce{3Ca(OH)2 + 2H3PO4 \rightarrow Ca3(PO4)2 + 6H2O}$ ?
Answer
2.04 mol
Example $2$: Number of Product Molecules Generated by a Reaction
How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?
$\ce{C3H8 + 5O2 \rightarrow 3CO2 + 4H2O} \label{4.4.6}$
Solution
The approach here is the same as for Example $1$, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number.
The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:
$\ce{\dfrac{3\: mol\: CO2}{1\: mol\: C3H8}} \label{4.4.7}$
Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,
$\mathrm{0.75\: \cancel{mol\: C_3H_8}\times \dfrac{3\: \cancel{mol\: CO_2}}{1\:\cancel{mol\:C_3H_8}}\times \dfrac{6.022\times 10^{23}\:CO_2\:molecules}{\cancel{mol\:CO_2}}=1.4\times 10^{24}\:CO_2\:molecules} \label{4.4.8}$
Exercise $1$
How many NH3 molecules are produced by the reaction of 4.0 mol of Ca(OH)2 according to the following equation:
$\ce{(NH4)2SO4 + Ca(OH)2 \rightarrow 2NH3 + CaSO4 + 2H2O} \label{4.4.9}$
Answer
4.8 × 1024 NH3 molecules
These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.
Example $3$: Relating Masses of Reactants and Products
What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?
$\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)$
Solution
The approach used previously in Examples $1$ and $2$ is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:
$\mathrm{16\:\cancel{g\: Mg(OH)_2} \times \dfrac{1\:\cancel{mol\: Mg(OH)_2}}{58.3\:\cancel{g\: Mg(OH)_2}}\times \dfrac{2\:\cancel{mol\: NaOH}}{1\:\cancel{mol\: Mg(OH)_2}}\times \dfrac{40.0\: g\: NaOH}{\cancel{mol\: NaOH}}=22\: g\: NaOH} \nonumber$
Exercise $3$
What mass of gallium oxide, Ga2O3, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $\ce{4Ga + 3O2 \rightarrow 2Ga2O3}$.
Answer
39.0 g
Example $4$: Relating Masses of Reactants
What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?
$\ce{2C8H18 + 25O2 \rightarrow 16CO2 + 18H2O} \nonumber$
Solution
The approach required here is the same as for the Example $3$, differing only in that the provided and requested masses are both for reactant species.
$\mathrm{702\:\cancel{g\:\ce{C8H18}}\times \dfrac{1\:\cancel{mol\:\ce{C8H18}}}{114.23\:\cancel{g\:\ce{C8H18}}}\times \dfrac{25\:\cancel{mol\:\ce{O2}}}{2\:\cancel{mol\:\ce{C8H18}}}\times \dfrac{32.00\: g\:\ce{O2}}{\cancel{mol\:\ce{O2}}}=2.46\times 10^3\:g\:\ce{O2}}$
Exercise $4$
What mass of CO is required to react with 25.13 g of Fe2O3 according to the equation $\ce{Fe2O3 + 3CO \rightarrow 2Fe + 3CO2}$?
Answer
13.22 g
These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure $2$ provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.
Airbags
Airbags (Figure $3$) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN3. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN3 to initiate its decomposition:
$\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s) \nonumber$
This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN3 will generate approximately 50 L of N2.
Summary
A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.
Glossary
stoichiometric factor
ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products
stoichiometry
relationships between the amounts of reactants and products of a chemical reaction | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.3%3A_Reaction_Stoichiometry.txt |
Learning Objectives
• Explain the concepts of theoretical yield and limiting reactants/reagents.
• Derive the theoretical yield for a reaction under specified conditions.
• Calculate the percent yield for a reaction.
The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.
Limiting Reactant
Consider another food analogy, making grilled cheese sandwiches (Figure $1$):
$\text{1 slice of cheese} + \text{2 slices of bread} \rightarrow \text{1 sandwich} \label{4.5.A}$
Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.
Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:
$\ce{H2}(g) + \ce{Cl2}(g)\rightarrow \ce{2HCl}(g) \nonumber$
The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation.
For example, imagine combining 6 moles of H2 and 4 moles of Cl2. Identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant.
For the example in the previous paragraph, the complete reaction of the hydrogen would yield
$\mathrm{mol\: HCl\: produced=6\: mol\:H_2\times \dfrac{2\: mol\: HCl}{1\: mol\:H_2}=12\: mol\: HCl} \nonumber$
The complete reaction of the provided chlorine would produce
$\mathrm{mol\: HCl\: produced=4\: mol\:Cl_2\times \dfrac{2\: mol\: HCl}{1\: mol\:Cl_2}=8\: mol\: HCl} \nonumber$
The chlorine will be completely consumed once 8 moles of HCl have been produced. Since enough hydrogen was provided to yield 12 moles of HCl, there will be non-reacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure $2$). To determine the amount of excess reactant that remains, the amount of hydrogen consumed in the reaction can be subtracted from the starting quantity of hydrogen.
The amount of hydrogen consumed is
$\mathrm{mol\: H_2\: produced=8\: mol\:HCl\times \dfrac{1\: mol\: H_2}{2\: mol\:HCl}=4\: mol\: H_2} \nonumber$
Subtract the hydrogen consumed from the starting quantity
$\mathrm{mole\: of\: excess\:H_{2}=6\:mol\:H_{2}\:starting\:-\:4\:mol\:H_{2}\:consumed\:=\:2\:mol\:H_{2}\; excess} \nonumber$
Example $1$: Identifying the Limiting Reactant
Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:
$\ce{3Si}(s)+\ce{2N2}(g)\rightarrow \ce{Si3N4}(s) \nonumber$
Which is the limiting reactant when 2.00 g of Si and 1.50 g of N2 react?
Solution
Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.
$\mathrm{mol\: Si=2.00\:\cancel{g\: Si}\times \dfrac{1\: mol\: Si}{28.09\:\cancel{g\: Si}}=0.0712\: mol\: Si} \nonumber$
$\mathrm{mol\:N_2=1.50\:\cancel{g\:N_2}\times \dfrac{1\: mol\:N_2}{28.02\:\cancel{g\:N_2}}=0.0535\: mol\:N_2} \nonumber$
The provided Si:N2 molar ratio is:
$\mathrm{\dfrac{0.0712\: mol\: Si}{0.0535\: mol\:N_2}=\dfrac{1.33\: mol\: Si}{1\: mol\:N_2}} \nonumber$
The stoichiometric Si:N2 ratio is:
$\mathrm{\dfrac{3\: mol\: Si}{2\: mol\:N_2}=\dfrac{1.5\: mol\: Si}{1\: mol\:N_2}} \nonumber$
Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.
Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield
$\mathrm{mol\:Si_3N_4\:produced=0.0712\: mol\: Si\times \dfrac{1\:mol\:Si_3N_4}{3\: mol\: Si}=0.0237\: mol\:Si_3N_4} \nonumber$
while the 0.0535 moles of nitrogen would produce
$\mathrm{mol\:Si_3N_4\:produced=0.0535\: mol\:N_2\times \dfrac{1\: mol\:Si_3N_4}{2\: mol\:N_2}=0.0268\: mol\:Si_3N_4} \nonumber$
Since silicon yields the lesser amount of product, it is the limiting reactant.
Exercise $1$
Which is the limiting reactant when 5.00 g of H2 and 10.0 g of O2 react and form water?
Answer
O2
Percent Yield
The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this text). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:
$\mathrm{percent\: yield=\dfrac{actual\: yield}{theoretical\: yield}\times 100\%} \nonumber$
Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.
Example $2$: Calculation of Percent Yield
Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:
$\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(s)+\ce{ZnSO4}(aq) \nonumber$
What is the percent yield?
Solution
The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:
$\mathrm{1.274\:\cancel{g\:Cu_SO_4}\times \dfrac{1\:\cancel{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{1\:\cancel{mol\: Cu}}{1\:\cancel{mol\:CuSO_4}}\times \dfrac{63.55\:g\: Cu}{1\:\cancel{mol\: Cu}}=0.5072\: g\: Cu} \nonumber$
Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be
$\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100} \nonumber$
\begin{align*} \mathrm{percent\: yield}&=\mathrm{\left(\dfrac{0.392\: g\: Cu}{0.5072\: g\: Cu}\right)\times 100} \ &=77.3\% \end{align*} \nonumber
Exercise $2$
What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF2Cl2 from 32.9 g of CCl4 and excess HF?
$\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber$
Answer
48.3%
Green Chemistry and Atom Economy
The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry”. One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:
$\mathrm{atom\: economy=\dfrac{mass\: of\: product}{mass\: of\: reactants}\times 100\%} \nonumber$
Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.
The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure $3$). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%. In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997.
Summary
When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.
Key Equations
• $\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100}$
Glossary
actual yield
amount of product formed in a reaction
excess reactant
reactant present in an amount greater than required by the reaction stoichiometry
limiting reactant
reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated
percent yield
measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield
theoretical yield
amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.4%3A_Reaction_Yields.txt |
Learning Objectives
• Describe the fundamental aspects of titrations and gravimetric analysis.
• Perform stoichiometric calculations using typical titration and gravimetric data.
In the 18th century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, K2CO3, which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar.
We now know that the effervescence that occurred during this process was due to reaction with acetic acid, CH3CO2H, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:
$\ce{2CH3CO2H}(aq)+\ce{K2CO3}(s)\rightarrow 2 \ce{KCH3CO3}(aq)+\ce{CO2}(g)+\ce{H2O}(l) \nonumber$
The bubbling was due to the production of CO2.
The test of vinegar with potassium carbonate is one type of quantitative analysis—the determination of the amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.
Titration
The described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret (Figure $1$) to make incremental additions of a solution containing a known concentration of some substance (the titrant) to a sample solution containing the substance whose concentration is to be measured (the analyte). The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration. The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction. The halt of bubble formation in the classic vinegar analysis is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration. Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration’s equivalence point, the volume of titrant actually measured is called the end point. Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.
Example $1$: Titration Analysis
The end point in a titration of a 50.00-mL sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is:
$\ce{HCl}(aq)+\ce{NaOH}(aq)\rightarrow \ce{NaCl}(aq)+\ce{H2O}(l) \nonumber$
What is the molarity of the HCl?
Solution
As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants provided and requested are expressed as solution concentrations.
For this exercise, the calculation will follow the following outlined steps:
The molar amount of HCl is calculated to be:
$\mathrm{35.23\:\cancel{mL\: NaOH}\times \dfrac{1\:\cancel{L}}{1000\:\cancel{mL}}\times \dfrac{0.250\:\cancel{mol\: NaOH}}{1\:\cancel{L}}\times \dfrac{1\: mol\: HCl}{1\:\cancel{mol\: NaOH}}=8.81\times 10^{-3}\:mol\: HCl} \nonumber$
Using the provided volume of HCl solution and the definition of molarity, the HCl concentration is:
\begin{align*} M&=\mathrm{\dfrac{mol\: HCl}{L\: solution}}\ M&=\mathrm{\dfrac{8.81\times 10^{-3}\:mol\: HCl}{50.00\: mL\times \dfrac{1\: L}{1000\: mL}}}\ M&=0.176\:M \end{align*} \nonumber
Note: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:
$M=\mathrm{\dfrac{mol\: solute}{L\: solution}\times \dfrac{\dfrac{10^3\:mmol}{mol}}{\dfrac{10^3\:mL}{L}}=\dfrac{mmol\: solute}{mL\: solution}} \nonumber$
Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors:
$\mathrm{\dfrac{35.23\:mL\: NaOH\times \dfrac{0.250\:mmol\: NaOH}{mL\: NaOH}\times \dfrac{1\:mmol\: HCl}{1\:mmol\: NaOH}}{50.00\:mL\: solution}=0.176\: \mathit M\: HCl} \nonumber$
Exercise $1$
A 20.00-mL sample of aqueous oxalic acid, H2C2O4, was titrated with a 0.09113-M solution of potassium permanganate, KMnO4.
$\ce{2MnO4-}(aq)+\ce{5H2C2O4}(aq)+\ce{6H+}(aq)\rightarrow \ce{10CO2}(g)+\ce{2Mn^2+}(aq)+\ce{8H2O}(l) \nonumber$
A volume of 23.24 mL was required to reach the end point. What is the oxalic acid molarity?
Answer
0.2648 M
Gravimetric Analysis
A gravimetric analysis is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.
The required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water. Also common are gravimetric techniques in which the analyte is subjected to a precipitation reaction of the sort described earlier in this chapter. The precipitate is typically isolated from the reaction mixture by filtration, carefully dried, and then weighed (Figure $2$). The mass of the precipitate may then be used, along with relevant stoichiometric relationships, to calculate analyte concentration.
Example $2$: Gravimetric Analysis
A 0.4550-g solid mixture containing MgSO4 is dissolved in water and treated with an excess of Ba(NO3)2, resulting in the precipitation of 0.6168 g of BaSO4.
$\ce{MgSO4}(aq)+\ce{Ba(NO3)2}(aq)\rightarrow \ce{BaSO4}(s)+\ce{Mg(NO3)2}(aq) \nonumber$
What is the concentration (percent) of MgSO4 in the mixture?
Solution
The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO4 and MgSO4 through their stoichiometric factor. Once the mass of MgSO4 is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.
The mass of MgSO4 that would yield the provided precipitate mass is
$\mathrm{0.6168\:\cancel{g\: BaSO_4}\times \dfrac{1\:\cancel{mol\: BaSO_4}}{233.43\:\cancel{g\: BaSO_4}}\times \dfrac{1\:\cancel{mol\: MgSO_4}}{1\:\cancel{mol\: BaSO_4}}\times \dfrac{120.37\:g\: MgSO_4}{1\:\cancel{mol\: MgSO_4}}=0.3181\:g\: MgSO_4} \nonumber$
The concentration of MgSO4 in the sample mixture is then calculated to be
\begin{align*} \ce{percent\: MgSO4}&=\ce{\dfrac{mass\: MgSO4}{mass\: sample}}\times100\%\ \mathrm{\dfrac{0.3181\: g}{0.4550\: g}}\times100\%&=69.91\% \end{align*} \nonumber
Exercise $2$
What is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess Ag+?
$\ce{Ag+}(aq)+\ce{Cl-}(aq)\rightarrow \ce{AgCl}(s) \nonumber$
Answer
23.76%
Combustion Analysis
The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure $3$). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.
Example $3$: Combustion Analysis
Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO2 and 0.00161 g of H2O. What is the empirical formula of polyethylene?
Solution
The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:
$\mathrm{C_xH_y}(s)+\ce{excess\: O2}(g)\rightarrow x\ce{CO2}(g)+ \dfrac{y}{2} \ce{H2O}(g) \nonumber$
Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.
First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:
$\mathrm{mol\: C=0.00394\:g\: CO_2\times\dfrac{1\:mol\: CO_2}{44.01\: g/mol}\times\dfrac{1\:mol\: C}{1\:mol\: CO_2}=8.95\times10^{-5}\:mol\: C} \nonumber$
$\mathrm{mol\: H=0.00161\:g\: H_2O\times\dfrac{1\:mol\: H_2O}{18.02\:g/mol}\times\dfrac{2\:mol\: H}{1\:mol\: H_2O}=1.79\times10^{-4}\:mol\: H} \nonumber$
The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is
$\mathrm{\dfrac{mol\: H}{mol\: C}=\dfrac{1.79\times10^{-4}\:mol\: H}{8.95\times10^{-5}\:mol\: C}=\dfrac{2\:mol\: H}{1\:mol\: C}} \nonumber$
and the empirical formula for polyethylene is CH2.
Exercise $3$
A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO2 and 0.00148 g of H2O in a combustion analysis. What is the empirical formula for polystyrene?
Answer
CH
Summary
The stoichiometry of chemical reactions may serve as the basis for quantitative chemical analysis methods. Titrations involve measuring the volume of a titrant solution required to completely react with a sample solution. This volume is then used to calculate the concentration of analyte in the sample using the stoichiometry of the titration reaction. Gravimetric analysis involves separating the analyte from the sample by a physical or chemical process, determining its mass, and then calculating its concentration in the sample based on the stoichiometry of the relevant process. Combustion analysis is a gravimetric method used to determine the elemental composition of a compound by collecting and weighing the gaseous products of its combustion.
Glossary
analyte
chemical species of interest
buret
device used for the precise delivery of variable liquid volumes, such as in a titration analysis
combustion analysis
gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products
end point
measured volume of titrant solution that yields the change in sample solution appearance or other property expected for stoichiometric equivalence (see equivalence point)
equivalence point
volume of titrant solution required to react completely with the analyte in a titration analysis; provides a stoichiometric amount of titrant for the sample’s analyte according to the titration reaction
gravimetric analysis
quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and/or sample
indicator
substance added to the sample in a titration analysis to permit visual detection of the end point
quantitative analysis
the determination of the amount or concentration of a substance in a sample
titrant
solution containing a known concentration of substance that will react with the analyte in a titration analysis
titration analysis
quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.5%3A_Quantitative_Chemical_Analysis.txt |
4.1: Writing and Balancing Chemical Equations
Q4.1.1
What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
S4.1.1
An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.
Q4.1.2
Consider molecular, complete ionic, and net ionic equations.
1. What is the difference between these types of equations?
2. In what circumstance would the complete and net ionic equations for a reaction be identical?
Q4.1.3
Balance the following equations:
1. $\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{HCl}(aq)$
2. $\ce{Cu}(s)+\ce{HNO3}(aq)\rightarrow \ce{Cu(NO3)2}(aq)+\ce{H2O}(l)+\ce{NO}(g)$
3. $\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{HI}(s)$
4. $\ce{Fe}(s)+\ce{O2}(g)\rightarrow \ce{Fe2O3}(s)$
5. $\ce{Na}(s)+\ce{H2O}(l)\rightarrow \ce{NaOH}(aq)+\ce{H2}(g)$
6. $\ce{(NH4)2Cr2O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{H2O}(g)$
7. $\ce{P4}(s)+\ce{Cl2}(g)\rightarrow \ce{PCl3}(l)$
8. $\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{Cl2}(g)$
S4.1.3
1. $\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{2HCl}(aq)$;
2. $\ce{3Cu}(s)+\ce{8HNO3}(aq)\rightarrow \ce{3Cu(NO3)2}(aq)+\ce{4H2O}(l)+\ce{2NO}(g)$;
3. $\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{2HI}(s)$ ;
4. $\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)$;
5. $\ce{2Na}(s)+\ce{2H2O}(l)\rightarrow \ce{2NaOH}(aq)+\ce{H2}(g)$;
6. $\ce{(NH4)2Cr52O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{4H2O}(g)$;
7. $\ce{P4}(s)+\ce{6Cl2}(g)\rightarrow \ce{4PCl3}(l)$ ;
8. $\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{2Cl2}(g)$
Q4.1.4
Balance the following equations:
1. $\ce{Ag}(s)+\ce{H2S}(g)+\ce{O2}(g)\rightarrow \ce{Ag2S}(s)+\ce{H2O}(l)$
2. $\ce{P4}(s)+\ce{O2}(g)\rightarrow \ce{P4O10}(s)$
3. $\ce{Pb}(s)+\ce{H2O}(l)+\ce{O2}(g)\rightarrow \ce{Pb(OH)2}(s)$
4. $\ce{Fe}(s)+\ce{H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{H2}(g)$
5. $\ce{Sc2O3}(s)+\ce{SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)$
6. $\ce{Ca3(PO4)2}(aq)+\ce{H3PO4}(aq)\rightarrow \ce{Ca(H2PO4)2}(aq)$
7. $\ce{Al}(s)+\ce{H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{H2}(g)$
8. $\ce{TiCl4}(s)+\ce{H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{HCl}(g)$
S4.1.4
1. $\ce{4Ag}(s)+\ce{2H2S}(g)+\ce{O2}(g)\rightarrow \ce{2Ag2S}(s)+\ce{2H2O}(l)$
2. $\ce{P4}(s)+\ce{5O2}(g)\rightarrow \ce{P4O10}(s)$
3. $\ce{2Pb}(s)+\ce{2H2O}(l)+\ce{O2}(g)\rightarrow \ce{2Pb(OH)2}(s)$
4. $\ce{3Fe}(s)+\ce{4H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{4H2}(g)$
5. $\ce{Sc2O3}(s)+\ce{3SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)$
6. $\ce{Ca3(PO4)2}(aq)+\ce{4H3PO4}(aq)\rightarrow \ce{3Ca(H2PO4)2}(aq)$
7. $\ce{2Al}(s)+\ce{3H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{3H2}(g)$
8. $\ce{TiCl4}(s)+\ce{2H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{4HCl}(g)$
Q4.1.5
Write a balanced molecular equation describing each of the following chemical reactions.
1. Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
2. Gaseous butane, C4H10, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
3. Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
4. Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
S4.1.5
1. $\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(g)$;
2. $\ce{2C4H10}(g)+\ce{13O2}(g)\rightarrow \ce{8CO2}(g)+\ce{10H2O}(g)$;
3. $\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)$;
4. $\ce{2H2O}(g)+\ce{2Na}(s)\rightarrow \ce{2NaOH}(s)+\ce{H2}(g)$
Q4.1.6
Write a balanced equation describing each of the following chemical reactions.
1. Solid potassium chlorate, KClO3, decomposes to form solid potassium chloride and diatomic oxygen gas.
2. Solid aluminum metal reacts with solid diatomic iodine to form solid Al2I6.
3. When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
4. Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.
Q4.1.7
Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.
1. Write the formulas of barium nitrate and potassium chlorate.
2. The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
3. The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
4. Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe3+ ions.)
Q4.1.7
1. Ba(NO3)2, KClO3;
2. $\ce{2KClO3}(s)\rightarrow \ce{2KCl}(s)+\ce{3O2}(g)$;
3. $\ce{2Ba(NO3)2}(s)\rightarrow \ce{2BaO}(s)+\ce{2N2}(g)+\ce{5O2}(g)$;
4. $\ce{2Mg}(s)+\ce{O2}(g)\rightarrow \ce{2MgO}(s)$ ; $\ce{4Al}(s)+\ce{3O2}(g)\rightarrow \ce{2Al2O3}(g)$; $\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)$
Q4.1.8
Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:
Q4.1.9
Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).
1. Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.
2. The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.
S4.1.9
1. $\ce{4HF}(aq)+\ce{SiO2}(s)\rightarrow \ce{SiF4}(g)+\ce{2H2O}(l)$;
2. complete ionic equation: $\ce{2Na+}(aq)+\ce{2F-}(aq)+\ce{Ca^2+}(aq)+\ce{2Cl-}(aq)\rightarrow \ce{CaF2}(s)+\ce{2Na+}(aq)+\ce{2Cl-}(aq)$, net ionic equation: $\ce{2F-}(aq)+\ce{Ca^2+}(aq)\rightarrow \ce{CaF2}(s)$
Q4.1.10
A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.
1. The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
2. The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
3. Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
4. The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
5. Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.
Q4.1.11
From the balanced molecular equations, write the complete ionic and net ionic equations for the following:
1. $\ce{K2C2O4}(aq)+\ce{Ba(OH)2}(aq)\rightarrow \ce{2KOH}(aq)+\ce{BaC2O2}(s)$
2. $\ce{Pb(NO3)2}(aq)+\ce{H2SO4}(aq)\rightarrow \ce{PbSO4}(s)+\ce{2HNO3}(aq)$
3. $\ce{CaCO3}(s)+\ce{H2SO4}(aq)\rightarrow \ce{CaSO4}(s)+\ce{CO2}(g)+\ce{H2O}(l)$
S4.1.11
1. $\ce{2K+}(aq)+\ce{C2O4^2-}(aq)+\ce{Ba^2+}(aq)+\ce{2OH-}(aq)\rightarrow \ce{2K+}(aq)+\ce{2OH-}(aq)+\ce{BaC2O4}(s)\hspace{20px}\ce{(complete)}$ $\ce{Ba^2+}(aq)+\ce{C2O4^2-}(aq)\rightarrow \ce{BaC2O4}(s)\hspace{20px}\ce{(net)}$
2. $\ce{Pb^2+}(aq)+\ce{2NO3-}(aq)+\ce{2H+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{PbSO4}(s)+\ce{2H+}(aq)+\ce{2NO3-}(aq)\hspace{20px}\ce{(complete)}$\ce{Pb^2+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{PbSO4}(s)\hspace{20px}\ce{(net)}$
3. $\ce{CaCO3}(s)+\ce{2H+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{CaSO4}(s)+\ce{CO2}(g)+\ce{H2O}(l)\hspace{20px}\ce{(complete)}$\ce{CaCO3}(s)+\ce{2H+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{CaSO4}(s)+\ce{CO2}(g)+\ce{H2O}(l)\hspace{20px}\ce{(net)}$
4.2: Classifying Chemical Reactions
Q4.2.1
Use the following equations to answer the next five questions:
1. $\ce{H2O}(s)\rightarrow \ce{H2O}(l)$
2. $\ce{Na+}(aq)+\ce{Cl-}(aq)\ce{Ag+}(aq)+\ce{NO3-}(aq) \rightarrow \ce{AgCl}(s)+\ce{Na+}(aq)+\ce{NO3-}(aq)$
3. $\ce{CH3OH}(g)+\ce{O2}(g)\rightarrow \ce{CO2}(g)+\ce{H2O}(g)$
4. $\ce{2H2O}(l)\rightarrow \ce{2H2}(g)+\ce{O2}(g)$
5. $\ce{H+}(aq)+\ce{OH-}(aq)\rightarrow \ce{H2O}(l)$
1. Which equation describes a physical change?
2. Which equation identifies the reactants and products of a combustion reaction?
3. Which equation is not balanced?
4. Which is a net ionic equation?
S4.2.1
a.) i. $H_2O (solid) → H_2O(liquid)$
b.) iii.
c.) iii. $\ce{2CH3OH}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{4H2O}(g)$
d.) v.
Q4.2.2
Indicate what type, or types, of reaction each of the following represents:
1. $\ce{Ca}(s)+\ce{Br2}(l)\rightarrow \ce{CaBr2}(s)$
2. $\ce{Ca(OH)2}(aq)+\ce{2HBr}(aq)\rightarrow \ce{CaBr2}(aq)+\ce{2H2O}(l)$
3. $\ce{C6H12}(l)+\ce{9O2}(g)\rightarrow \ce{6CO2}(g)+\ce{6H2O}(g)$
S4.2.2
oxidation-reduction (addition); acid-base (neutralization); oxidation-reduction (combustion)
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Q4.2.3
Indicate what type, or types, of reaction each of the following represents:
1. $\ce{H2O}(g)+\ce{C}(s)\rightarrow \ce{CO}(g)+\ce{H2}(g)$
2. $\ce{2KClO3}(s)\rightarrow \ce{2KCl}(s)+\ce{3O2}(g)$
3. $\ce{Al(OH)3}(aq)+\ce{3HCl}(aq)\rightarrow \ce{AlBr3}(aq)+\ce{3H2O}(l)$
4. $\ce{Pb(NO3)2}(aq)+\ce{H2SO4}(aq)\rightarrow \ce{PbSO4}(s)+\ce{2HNO3}(aq)$
Q4.2.4
Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.
S4.2.4
It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.
Q4.2.5
Determine the oxidation states of the elements in the following compounds:
1. NaI
2. GdCl3
3. LiNO3
4. H2Se
5. Mg2Si
6. RbO2, rubidium superoxide
7. HF
Q4.2.6
Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
1. H3PO4
2. Al(OH)3
3. SeO2
4. KNO2
5. In2S3
6. P4O6
S4.2.6
H +1, P +5, O −2; Al +3, H +1, O −2; Se +4, O −2; K +1, N +3, O −2; In +3, S −2; P +3, O −2
Q4.2.7
Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
1. H2SO4
2. Ca(OH)2
3. BrOH
4. ClNO2
5. TiCl4
6. NaH
S4.2.7
1. H1+, O2-, S6+
2. H1+, O2-, Ca+2
3. H1+, O2-, Br1+
4. O2-, Cl1-, N5+
5. Cl1-, Ti4+
6. H1+, Na1-
Q4.2.8
Classify the following as acid-base reactions or oxidation-reduction reactions:
1. $\ce{Na2S}(aq)+\ce{2HCl}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{H2S}(g)$
2. $\ce{2Na}(s)+\ce{2HCl}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{H2}(g)$
3. $\ce{Mg}(s)+\ce{Cl2}(g)\rightarrow \ce{MgCl2}(s)$
4. $\ce{MgO}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{H2O}(l)$
5. $\ce{K3P}(s)+\ce{2O2}(g)\rightarrow \ce{K3PO4}(s)$
6. $\ce{3KOH}(aq)+\ce{H3PO4}(aq)\rightarrow \ce{K3PO4}(aq)+\ce{3H2O}(l)$
S4.2.9
acid-base; oxidation-reduction: Na is oxidized, H+ is reduced; oxidation-reduction: Mg is oxidized, Cl2 is reduced; acid-base; oxidation-reduction: P3− is oxidized, O2 is reduced; acid-base
Q4.2.10
Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:
1. $\ce{Mg}(s)+\ce{NiCl2}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{Ni}(s)$
2. $\ce{PCl3}(l)+\ce{Cl2}(g)\rightarrow \ce{PCl5}(s)$
3. $\ce{C2H4}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{2H2O}(g)$
4. $\ce{Zn}(s)+\ce{H2SO4}(aq)\rightarrow \ce{ZnSO4}(aq)+\ce{H2}(g)$
5. $\ce{2K2S2O3}(s)+\ce{I2}(s)\rightarrow \ce{K2S4O6}(s)+\ce{2KI}(s)$
6. $\ce{3Cu}(s)+\ce{8HNO3}(aq)\rightarrow\ce{3Cu(NO3)2}(aq)+\ce{2NO}(g)+\ce{4H2O}(l)$
Q4.2.11
Complete and balance the following acid-base equations:
1. HCl gas reacts with solid Ca(OH)2(s).
2. A solution of Sr(OH)2 is added to a solution of HNO3.
S4.2.11
1. $\ce{2HCl}(g)+\ce{Ca(OH)2}(s)\rightarrow \ce{CaCl2}(s)+\ce{2H2O}(l)$;
2. $\ce{Sr(OH)2}(aq)+\ce{2HNO3}(aq)\rightarrow \ce{Sr(NO3)2}(aq)+\ce{2H2O}(l)$
Q4.2.12
Complete and balance the following acid-base equations:
1. A solution of HClO4 is added to a solution of LiOH.
2. Aqueous H2SO4 reacts with NaOH.
3. Ba(OH)2 reacts with HF gas.
Q4.2.13
Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
1. $\ce{Al}(s)+\ce{F2}(g)\rightarrow$
2. $\ce{Al}(s)+\ce{CuBr2}(aq)\rightarrow$ (single displacement)
3. $\ce{P4}(s)+\ce{O2}(g)\rightarrow$
4. $\ce{Ca}(s)+\ce{H2O}(l)\rightarrow$ (products are a strong base and a diatomic gas)
S4.2.13
1. $\ce{2Al}(s)+\ce{3F2}(g)\rightarrow \ce{2AlF3}(s)$;
2. $\ce{2Al}(s)+\ce{3CuBr2}(aq)\rightarrow \ce{3Cu}(s)+\ce{2AlBr3}(aq)$;
3. $\ce{P4}(s)+\ce{5O2}(g)\rightarrow \ce{P4O10}(s)$; $\ce{Ca}(s)+\ce{2H2O}(l)\rightarrow \ce{Ca(OH)2}(aq)+\ce{H2}(g)$
Q4.2.14
Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
1. $\ce{K}(s)+\ce{H2O}(l)\rightarrow$
2. $\ce{Ba}(s)+\ce{HBr}(aq)\rightarrow$
3. $\ce{Sn}(s)+\ce{I2}(s)\rightarrow$
Q4.2.15
Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.
1. $\ce{Mg(OH)2}(s)+\ce{HClO4}(aq)\rightarrow$
2. $\ce{SO3}(g)+\ce{H2O}(l)\rightarrow$ (assume an excess of water and that the product dissolves)
3. $\ce{SrO}(s)+\ce{H2SO4}(l)\rightarrow$
S4.2.15
1. $\ce{Mg(OH)2}(s)+\ce{2HClO4}(aq)\rightarrow \ce{Mg^2+}(aq)+\ce{2ClO4-}(aq)+\ce{2H2O}(l)$ ;
2. $\ce{SO3}(g)+\ce{2H2O}(l)\rightarrow \ce{H3O+}(aq)+\ce{HSO4-}(aq)$, (a solution of H2SO4);
3. $\ce{SrO}(s)+\ce{H2SO4}(l)\rightarrow \ce{SrSO4}(s)+\ce{H2O}$
Q4.2.16
When heated to 700–800 °C, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction.
Q4.2.17
The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?
S4.2.17
$\ce{H2}(g)+\ce{F2}(g)\rightarrow \ce{2HF}(g)$
Q4.2.18
Write the molecular, total ionic, and net ionic equations for the following reactions:
1. $\ce{Ca(OH)2}(aq)+\ce{HC2H3O2}(aq)\rightarrow$
2. $\ce{H3PO4}(aq)+\ce{CaCl2}(aq)\rightarrow$
Q4.2.19
Great Lakes Chemical Company produces bromine, Br2, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl2.
S4.2.19
$\ce{2NaBr}(aq)+\ce{Cl2}(g)\rightarrow \ce{2NaCl}(aq)+\ce{Br2}(l)$
Q4.2.20
In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of Mg3N2, a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.
Q4.2.21
Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO2. (Hint: Water is one of the products.)
S4.2.21
$\ce{2LiOH}(aq)+\ce{CO2}(g)\rightarrow \ce{Li2CO3}(aq)+\ce{H2O}(l)$
Q4.2.22
Calcium propionate is sometimes added to bread to retard spoilage. This compound can be prepared by the reaction of calcium carbonate, CaCO3, with propionic acid, C2H5CO2H, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate.
Q4.2.23
Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:
1. $\ce{Ca(OH)2}(s)+\ce{H2S}(g) \rightarrow$
2. $\ce{Na2CO3}(aq)+\ce{H2S}(g)\rightarrow$
S4.2.23
1. $\ce{Ca(OH)2}(s)+\ce{H2S}(g)\rightarrow \ce{CaS}(s)+\ce{2H2O}(l)$;
2. $\ce{Na2CO3}(aq)+\ce{H2S}(g)\rightarrow \ce{Na2S}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$
Q4.2.24
Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid hydrogen sulfate as the only product. Write the two equations which represent these reactions.
Q4.2.25
Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.
1. solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)
2. gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction
3. gaseous H2S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)
S4.2.25
1. step 1: $\ce{N2}(g)+\ce{3H2}(g)\rightarrow \ce{2NH3}(g)$, step 2: $\ce{NH3}(g)+\ce{HNO3}(aq)\rightarrow \ce{NH4NO3}(aq)\rightarrow \ce{NH4NO3}(s)\ce{(after\: drying)}$;
2. $\ce{H2}(g)+\ce{Br2}(l)\rightarrow \ce{2HBr}(g)$;
3. $\ce{Zn}(s)+\ce{S}(s)\rightarrow \ce{ZnS}(s)$ and $\ce{ZnS}(s)+\ce{2HCl}(aq)\rightarrow \ce{ZnCl2}(aq)+\ce{H2S}(g)$
Q4.2.26
Calcium cyclamate Ca(C6H11NHSO3)2 is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid C6H11NHSO3H with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions.
Q4.2.27
Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method):
1. $\ce{Sn^4+}(aq)\rightarrow \ce{Sn^2+}(aq)$
2. $\ce{[Ag(NH3)2]+}(aq)\rightarrow \ce{Ag}(s)+\ce{NH3}(aq)$
3. $\ce{Hg2Cl2}(s)\rightarrow \ce{Hg}(l)+\ce{Cl-}(aq)$
4. $\ce{H2O}(l)\rightarrow \ce{O2}(g)\ce{\:(in\: acidic\: solution)}$
5. $\ce{IO3-}(aq)\rightarrow \ce{I2}(s)$
6. $\ce{SO3^2-}(aq)\rightarrow \ce{SO4^2-}(aq)\ce{\:(in\: acidic\: solution)}$
7. $\ce{MnO4-}(aq)\rightarrow \ce{Mn^2+}(aq)\ce{\:(in\: acidic\: solution)}$
8. $\ce{Cl-}(aq)\rightarrow \ce{ClO3-}(aq)\ce{\:(in\: basic\: solution)}$
S4.2.27
1. $\ce{Sn^4+}(aq)+\ce{2e-}\rightarrow \ce{Sn^2+}(aq)$,
2. $\ce{[Ag(NH3)2]+}(aq)+ \ce{e-} \rightarrow \ce{Ag}(s)+\ce{2NH3}(aq)$;
3. $\ce{Hg2Cl2}(s)+ \ce{2e-} \rightarrow \ce{2Hg}(l)+\ce{2Cl-}(aq)$ ;
4. $\ce{2H2O}(l)\rightarrow \ce{O2}(g)+\ce{4H+}(aq)+\ce{4e-}$;
5. $\ce{6H2O}(l)+\ce{2IO3-}(aq)+\ce{10e-}\rightarrow \ce{I2}(s)+\ce{12OH-}(aq)$;
6. $\ce{H2O}(l)+\ce{SO3^2-}(aq)\rightarrow \ce{SO4^2-}(aq)+\ce{2H+}(aq)+\ce{2e-}$;
7. (g) $\ce{8H+}(aq)+\ce{MnO4-}(aq)+\ce{5e-}\rightarrow \ce{Mn^2+}(aq)+\ce{4H2O}(l)$;
8. (h) $\ce{Cl-}(aq)+\ce{6OH-}(aq)\rightarrow \ce{ClO3-}(aq)+\ce{3H2O}(l)+\ce{6e-}$
Q4.2.28
Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method):
1. $\ce{Cr^2+}(aq)\rightarrow \ce{Cr^3+}(aq)$
2. $\ce{Hg}(l)+\ce{Br-}(aq)\rightarrow \ce{HgBr4^2-}(aq)$
3. $\ce{ZnS}(s)\rightarrow \ce{Zn}(s)+\ce{S^2-}(aq)$
4. $\ce{H2}(g)\rightarrow \ce{H2O}(l)\ce{\:(in\: basic\: solution)}$
5. $\ce{H2}(g)\rightarrow \ce{H3O+}(aq)\ce{\:(in\: acidic\: solution)}$
6. $\ce{NO3-}(aq)\rightarrow \ce{HNO2}(aq)\ce{\:(in\: acidic\: solution)}$
7. $\ce{MnO2}(s)\rightarrow \ce{MnO4-}(aq)\ce{\:(in\: basic\: solution)}$
8. $\ce{Cl-}(aq)\rightarrow \ce{ClO3-}(aq)\ce{\:(in\: acidic\: solution)}$
Q4.2.29
Balance each of the following equations according to the half-reaction method:
1. $\ce{Sn^2+}(aq)+\ce{Cu^2+}(aq)\rightarrow \ce{Sn^4+}(aq)+\ce{Cu+}(aq)$
2. $\ce{H2S}(g)+\ce{Hg2^2+}(aq)\rightarrow \ce{Hg}(l)+\ce{S}(s)\ce{\:(in\: acid)}$
3. $\ce{CN-}(aq)+\ce{ClO2}(aq)\rightarrow \ce{CNO-}(aq)+\ce{Cl-}(aq)\ce{\:(in\: acid)}$
4. $\ce{Fe^2+}(aq)+\ce{Ce^4+}(aq)\rightarrow \ce{Fe^3+}(aq)+\ce{Ce^3+}(aq)$
5. $\ce{HBrO}(aq)\rightarrow \ce{Br-}(aq)+\ce{O2}(g)\ce{\:(in\: acid)}$
S4.2.29
1. $\ce{Sn^2+}(aq)+\ce{2Cu^2+}(aq)\rightarrow \ce{Sn^4+}(aq)+\ce{2Cu+}(aq)$;
2. $\ce{H2S}(g)+\ce{Hg2^2+}(aq)+\ce{2H2O}(l)\rightarrow \ce{2Hg}(l)+\ce{S}(s)+\ce{2H3O+}(aq)$;
3. $\ce{5CN-}(aq)+\ce{2ClO2}(aq)+\ce{3H2O}(l)\rightarrow \ce{5CNO-}(aq)+\ce{2Cl-}(aq)+\ce{2H3O+}(aq)$;
4. $\ce{Fe^2+}(aq)+\ce{Ce^4+}(aq)\rightarrow \ce{Fe^3+}(aq)+\ce{Ce^3+}(aq)$;
5. $\ce{2HBrO}(aq)+\ce{2H2O}(l)\rightarrow \ce{2H3O+}(aq)+\ce{2Br-}(aq)+\ce{O2}(g)$
Q4.2.30
Balance each of the following equations according to the half-reaction method:
1. $\ce{Zn}(s)+\ce{NO3-}(aq)\rightarrow \ce{Zn^2+}(aq)+\ce{N2}(g)\ce{\:(in\: acid)}$
2. $\ce{Zn}(s)+\ce{NO3-}(aq)\rightarrow \ce{Zn^2+}(aq)+\ce{NH3}(aq)\ce{\:(in\: base)}$
3. $\ce{CuS}(s)+\ce{NO3-}(aq)\rightarrow \ce{Cu^2+}(aq)+\ce{S}(s)+\ce{NO}(g)\ce{\:(in\: acid)}$
4. $\ce{NH3}(aq)+\ce{O2}(g)\rightarrow \ce{NO2}(g)\ce{\:(gas\: phase)}$
5. $\ce{Cl2}(g)+\ce{OH-}(aq)\rightarrow \ce{Cl-}(aq)+\ce{ClO3-}(aq)\ce{\:(in\: base)}$
6. $\ce{H2O2}(aq)+\ce{MnO4-}(aq)\rightarrow \ce{Mn^2+}(aq)+\ce{O2}(g)\ce{\:(in\: acid)}$
7. $\ce{NO2}(g)\rightarrow \ce{NO3-}(aq)+\ce{NO2-}(aq)\ce{\:(in\: base)}$
8. $\ce{Fe^3+}(aq)+\ce{I-}(aq)\rightarrow \ce{Fe^2+}(aq)+\ce{I2}(aq)$
Q4.2.31
Balance each of the following equations according to the half-reaction method:
1. $\ce{MnO4-}(aq)+\ce{NO2-}(aq)\rightarrow \ce{MnO2}(s)+\ce{NO3-}(aq)\ce{\:(in\: base)}$
2. $\ce{MnO4^2-}(aq)\rightarrow \ce{MnO4-}(aq)+\ce{MnO2}(s)\ce{\:(in\: base)}$
3. $\ce{Br2}(l)+\ce{SO2}(g)\rightarrow \ce{Br-}(aq)+\ce{SO4^2-}(aq)\ce{\:(in\: acid)}$
S4.2.31
1. $\ce{2MnO4-}(aq)+\ce{3NO2-}(aq)+\ce{H2O}(l)\rightarrow \ce{2MnO2}(s)+\ce{3NO3-}(aq)+\ce{2OH-}(aq)$;
2. $\ce{3MnO4^2-}(aq)+\ce{2H2O}(l)\rightarrow \ce{2MnO4-}(aq)+\ce{4OH-}(aq)+\ce{MnO2}(s)\ce{\:(in\: base)}$;
3. $\ce{Br2}(l)+\ce{SO2}(g)+\ce{2H2O}(l)\rightarrow \ce{4H+}(aq)+\ce{2Br-}(aq)+\ce{SO4^2-}(aq)$
4.3: Reaction Stoichiometry
Q4.3.1
Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
1. The number of moles and the mass of chlorine, Cl2, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.
2. The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.
3. The number of moles and the mass of sodium nitrate, NaNO3, required to produce 128 g of oxygen. (NaNO2 is the other product.)
4. The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.
5. The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. (CO2 is the other product.)
Q4.3.2
Determine the number of moles and the mass requested for each reaction in Exercise.
S4.3.2
0.435 mol Na, 0.217 mol Cl2, 15.4 g Cl2; 0.005780 mol HgO, 2.890 × 10−3 mol O2, 9.248 × 10−2 g O2; 8.00 mol NaNO3, 6.8 × 102 g NaNO3; 1665 mol CO2, 73.3 kg CO2; 18.86 mol CuO, 2.330 kg CuCO3; 0.4580 mol C2H4Br2, 86.05 g C2H4Br2
Q4.3.3
Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
1. The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl2 and H2.
2. The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.
3. The number of moles and the mass of magnesium carbonate, MgCO3, required to produce 283 g of carbon dioxide. (MgO is the other product.)
4. The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, C2H2, in an excess of oxygen.
5. The number of moles and the mass of barium peroxide, BaO2, needed to produce 2.500 kg of barium oxide, BaO (O2 is the other product.)
Q4.3.4
Determine the number of moles and the mass requested for each reaction in Exercise.
S4.3.4
0.0686 mol Mg, 1.67 g Mg; 2.701 × 10−3 mol O2, 0.08644 g O2; 6.43 mol MgCO3, 542 g MgCO3 713 mol H2O, 12.8 kg H2O; 16.31 mol BaO2, 2762 g BaO2; 0.207 mol C2H4, 5.81 g C2H4
Q4.3.5
H2 is produced by the reaction of 118.5 mL of a 0.8775-M solution of H3PO4 according to the following equation: $\ce{2Cr + 2H3PO4 \rightarrow 3H2 + 2CrPO4}$.
• Outline the steps necessary to determine the number of moles and mass of H2.
• Perform the calculations outlined.
S4.3.5
a.)
1. Convert mL to L
2. Multiply L by the molarity to determine moles of H3PO4
3. Convert moles of H3PO4 to moles of H2
4. Multiply moles of H2 by the molar mass of H2 to get the answer in grams
b.)
1. $118.5\: mL\times \dfrac{1\: L}{1000\: mL} = 0.1185\: L$
2. $0.1185\: L \times \dfrac{0.8775\: moles\: \ce{H3PO4}}{1\: L} = 0.1040\: moles\: \ce{H3PO4}$
3. $0.1040\: moles\: \ce{H3PO4} \times \dfrac{3\: moles\:\ce{H_2}}{2\: moles\: \ce{H3PO4}} = 0.1560\: moles\: \ce{H2}$
4. $0.1560\: moles\: \ce{H2} \times \dfrac{2.02 g}{1\: mole} = 0.3151g\: \ce{H2}$
Q4.3.6
Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: $\ce{2Ga + 6HCl \rightarrow 2GaCl3 + 3H2}$.
1. Outline the steps necessary to determine the number of moles and mass of gallium chloride.
2. Perform the calculations outlined.
S4.3.6
$\mathrm{volume\: HCl\: solution \rightarrow mol\: HCl \rightarrow mol\: GaCl_3}$; 1.25 mol GaCl3, 2.2 × 102 g GaCl3
Q4.3.7
I2 is produced by the reaction of 0.4235 mol of CuCl2 according to the following equation: $\ce{2CuCl2 + 4KI \rightarrow 2CuI + 4KCl + I2}$.
1. How many molecules of I2 are produced?
2. What mass of I2 is produced?
Q4.3.8
Silver is often extracted from ores as K[Ag(CN)2] and then recovered by the reaction
$\ce{2K[Ag(CN)2]}(aq)+\ce{Zn}(s)\rightarrow \ce{2Ag}(s)+\ce{Zn(CN)2}(aq)+\ce{2KCN}(aq)$
1. How many molecules of Zn(CN)2 are produced by the reaction of 35.27 g of K[Ag(CN)2]?
2. What mass of Zn(CN)2 is produced?
S4.3.8
5.337 × 1022 molecules; 10.41 g Zn(CN)2
Q4.3.9
What mass of silver oxide, Ag2O, is required to produce 25.0 g of silver sulfadiazine, AgC10H9N4SO2, from the reaction of silver oxide and sulfadiazine?
$\ce{2C10H10N4SO2 + Ag2O \rightarrow 2AgC10H9N4SO2 + H2O}$
Q4.3.10
Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO2, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO2 is required to produce 3.00 kg of SiC.
S4.3.10
$\ce{SiO2 + 3C \rightarrow SiC + 2CO}$, 4.50 kg SiO2
Q4.3.11
Automotive air bags inflate when a sample of sodium azide, NaN3, is very rapidly decomposed.
$\ce{2NaN3}(s) \rightarrow \ce{2Na}(s) + \ce{3N2}(g)$
What mass of sodium azide is required to produce 2.6 ft3 (73.6 L) of nitrogen gas with a density of 1.25 g/L?
142g NaN3
Q4.3.12
Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO2 produced by combustion of 1.00×103 kg of carbon followed by the reaction?
$\ce{CO2}(g)+\ce{2NH3}(g)\rightarrow \ce{CO(NH2)2}(s)+\ce{H2O}(l)$
5.00 × 103 kg
Q4.3.13
In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area and CO2 was released by the reaction. Was sufficient Na2CO3 used to neutralize all of the acid?
Q4.3.14
A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).
1.28 × 105 g CO2
Q4.3.15
What volume of a 0.750 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid?
$\ce{NaCl}(s)+\ce{H2SO4}(l)\rightarrow \ce{HCl}(g)+\ce{NaHSO4}(s)$
Q4.3.16
What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M solution of Cu(NO3)2?
$\ce{2Cu(NO3)2 + 4KI \rightarrow 2CuI + I2 + 4KNO3}$
S4.3.16
161.40 mL KI solution
Q4.3.17
A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.
$\ce{2CH3CO2H + Ca(OH)2 \rightarrow Ca(CH3CO2)2 + 2H2O}$
What mass of Ca(OH)2 is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g/mL and containing 58.0% acetic acid by mass?
Q4.3.18
The toxic pigment called white lead, Pb3(OH)2(CO3)2, has been replaced in white paints by rutile, TiO2. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO3) by mass?
$\ce{2FeTiO3 + 4HCl + Cl2 \rightarrow 2FeCl3 + 2TiO2 + 2H2O}$
176 g TiO2
4.4: Reaction Yields
Q4.4.1
The following quantities are placed in a container: 1.5 × 1024 atoms of hydrogen, 1.0 mol of sulfur, and 88.0 g of diatomic oxygen.
1. What is the total mass in grams for the collection of all three elements?
2. What is the total number of moles of atoms for the three elements?
3. If the mixture of the three elements formed a compound with molecules that contain two hydrogen atoms, one sulfur atom, and four oxygen atoms, which substance is consumed first?
4. How many atoms of each remaining element would remain unreacted in the change described in ?
Q4.4.2
What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?
S4.4.2
The limiting reactant is Cl2.
Q4.4.3
Which of the postulates of Dalton's atomic theory explains why we can calculate a theoretical yield for a chemical reaction?
Q4.4.4
A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was his percent yield?
S4.4.4
$\mathrm{Percent\: yield = 31\%}$
Q4.4.5
A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. What is the percent yield for this reaction?
$\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(s)$
Q4.4.6
Freon-12, CCl2F2, is prepared from CCl4 by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl2F2 from 32.9 g of CCl4. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield.
S4.4.6
$\ce{g\: CCl4\rightarrow mol\: CCl4\rightarrow mol\: CCl2F2 \rightarrow g\: CCl2F2}, \mathrm{\:percent\: yield=48.3\%}$
Q4.4.7
Citric acid, C6H8O7, a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The equation representing this reaction is
$\ce{C12H22O11 + H2O + 3O2 \rightarrow 2C6H8O7 + 4H2O}$
What mass of citric acid is produced from exactly 1 metric ton (1.000 × 103 kg) of sucrose if the yield is 92.30%?
Q4.4.8
Toluene, C6H5CH3, is oxidized by air under carefully controlled conditions to benzoic acid, C6H5CO2H, which is used to prepare the food preservative sodium benzoate, C6H5CO2Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?
$\ce{2C6H5CH3 + 3O2 \rightarrow 2C6H5CO2H + 2H2O}$
S4.4.8
$\mathrm{percent\: yield=91.3\%}$
Q4.4.9
In a laboratory experiment, the reaction of 3.0 mol of H2 with 2.0 mol of I2 produced 1.0 mol of HI. Determine the theoretical yield in grams and the percent yield for this reaction.
Q4.4.10
Outline the steps needed to solve the following problem, then do the calculations. Ether, (C2H5)2O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.
2C2H5OH + H2SO4 ⟶ (C2H5)2 + H2SO4·H2O
Q4.4.11
What is the percent yield of ether if 1.17 L (d = 0.7134 g/mL) is isolated from the reaction of 1.500 L of C2H5OH (d = 0.7894 g/mL)?
S4.4.11
Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6%
Q4.4.12
Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.
$\mathrm{percent\: yield=\dfrac{0.8347\:\cancel{g}}{0.9525\:\cancel{g}}\times 100\%=87.6\%}$
Determine the limiting reactant.
Q4.4.13
Outline the steps needed to determine the limiting reactant when 0.50 g of Cr and 0.75 g of H3PO4 react according to the following chemical equation?
$\ce{2Cr + 2H3PO4 \rightarrow 2CrPO4 + 3H2}$
Determine the limiting reactant.
S4.4.13
The conversion needed is $\ce{mol\: Cr \rightarrow mol\: H2PO4}$. Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.
Q4.4.14
What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation?
$\ce{Li + N2 \rightarrow Li3N}$
S4.4.14
$\ce{6Li} + \ce{N2} \rightarrow \: \ce{2Li3N}$
$1.50g\: \ce{Li} \times \dfrac{1\: mole\: \ce{Li}}{6.94g\: \ce{Li}} \times\dfrac{2\: mole\: \ce{Li3N}}{6\:mole\: \ce{Li}} = 0.0720\: moles\: \ce{Li3N}$
$1.50g\: \ce{N2} \times \dfrac{1\: mole\: \ce{N2}}{28.02g\: \ce{N2}} \times\dfrac{2\: mole\: \ce{Li3N}}{1\:mole\: \ce{N2}} = 0.107\: moles\: \ce{Li3N}$
$\ce{Li}$ is the limiting reactant
Q4.4.15
Uranium can be isolated from its ores by dissolving it as UO2(NO3)2, then separating it as solid UO2(C2O4)·3H2O. Addition of 0.4031 g of sodium oxalate, Na2C2O4, to a solution containing 1.481 g of uranyl nitrate, UO2(NO2)2, yields 1.073 g of solid UO2(C2O4)·3H2O.
$\ce{Na2C2O4 + UO2(NO3)2 + 3H2O ⟶ UO2(C2O4)·3H2O + 2NaNO3}$
Determine the limiting reactant and the percent yield of this reaction.
S4.4.15
Na2C2O4 is the limiting reactant. percent yield = 86.6%
Q4.4.16
How many molecules of C2H4Cl2 can be prepared from 15 C2H4 molecules and 8 Cl2 molecules?
Q4.4.17
How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?
S4.4.17
Only four molecules can be made.
Q4.4.18
The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen.
1. What is the limiting reactant when 0.200 mol of P4 and 0.200 mol of O2 react according to $\ce{P4 + 5O2 \rightarrow P4O10}$
2. Calculate the percent yield if 10.0 g of P4O10 is isolated from the reaction.
Q4.4.19
Would you agree to buy 1 trillion (1,000,000,000,000) gold atoms for \$5? Explain why or why not. Find the current price of gold at http://money.cnn.com/data/commodities/ $\mathrm{(1\: troy\: ounce=31.1\: g)}$
S4.4.19
This amount cannot be weighted by ordinary balances and is worthless.
4.5: Quantitative Chemical Analysis
Q4.5.1
What volume of 0.0105-M HBr solution is be required to titrate 125 mL of a 0.0100-M Ca(OH)2 solution?
$\ce{Ca(OH)2}(aq)+\ce{2HBr}(aq) \rightarrow \ce{CaBr2}(aq)+\ce{2H2O}(l)$
Q4.5.2
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?
S4.5.2
3.4 × 10−3 M H2SO4
Q4.5.3
What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point?
$\ce{AgNO3}(aq)+\ce{NaCl}(aq)\rightarrow \ce{AgCl}(s)+\ce{NaNO3}(aq)$
Q4.5.4
In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO3)2 solution.
$\ce{2Cl-}(aq)+\ce{Hg(NO3)2}(aq)\rightarrow \ce{2NO3-}(aq)+\ce{HgCl2}(s)$
What is the Cl concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 5.25 × 10−4 M Hg(NO3)2(aq) to reach the end point?
9.6 × 10−3 M Cl
Q4.5.5
Potatoes can be peeled commercially by soaking them in a 3-M to 6-M solution of sodium hydroxide, then removing the loosened skins by spraying them with water. Does a sodium hydroxide solution have a suitable concentration if titration of 12.00 mL of the solution requires 30.6 mL of 1.65 M HCI to reach the end point?
Q4.5.6
A sample of gallium bromide, GaBr2, weighing 0.165 g was dissolved in water and treated with silver nitrate, AgNO3, resulting in the precipitation of 0.299 g AgBr. Use these data to compute the %Ga (by mass) GaBr2.
22.4%
Q4.5.7
The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO2. Determine its empirical and molecular formulas.
Q4.5.8
A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B2O3. What are the empirical and molecular formulas of the compound.
S4.5.8
The empirical formula is BH3. The molecular formula is B2H6.
Q4.5.9
Sodium bicarbonate (baking soda), NaHCO3, can be purified by dissolving it in hot water (60 °C), filtering to remove insoluble impurities, cooling to 0 °C to precipitate solid NaHCO3, and then filtering to remove the solid, leaving soluble impurities in solution. Any NaHCO3 that remains in solution is not recovered. The solubility of NaHCO3 in hot water of 60 °C is 164 g L. Its solubility in cold water of 0 °C is 69 g/L. What is the percent yield of NaHCO3 when it is purified by this method?
Q4.5.10
What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate?
$\ce{NaHCO3}(aq)+\ce{HCl}(aq)\rightarrow \ce{NaCl}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$
49.6 mL
Q4.5.11
What volume of 0.08892 M HNO3 is required to react completely with 0.2352 g of potassium hydrogen phosphate?
$\ce{2HNO3}(aq)+\ce{K2HPO4}(aq)\rightarrow \ce{H2PO4}(aq)+\ce{2KNO3}(aq)$
Q4.5.12
What volume of a 0.3300-M solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 M oxalic acid?
$\ce{C2O4H2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Na2C2O4}(aq)+\ce{2H2O}(l)$
13.64 mL
Q4.5.13
What volume of a 0.00945-M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a H2SO4 concentration of 1.23 × 10−4 M.
$\ce{H2SO4}(aq)+\ce{2KOH}(aq)\rightarrow \ce{K2SO4}(aq)+\ce{2H2O}(l)$
1.30 mL
Q4.5.14
A sample of solid calcium hydroxide, Ca(OH)2, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 × 10−2 M HCl requires 36.6 mL of the acid to reach the end point.
$\ce{Ca(OH)2}(aq)+\ce{2HCl}(aq)\rightarrow \ce{CaCl2}(aq)+\ce{2H2O}(l)$
What is the molarity?
1.22 M
Q4.5.15
What mass of Ca(OH)2 will react with 25.0 g of propionic acid to form the preservative calcium propionate according to the equation?
Q4.5.16
How many milliliters of a 0.1500-M solution of KOH will be required to titrate 40.00 mL of a 0.0656-M solution of H3PO4?
$\ce{H3PO4}(aq)+\ce{2KOH}(aq)\rightarrow \ce{K2HPO4}(aq)+\ce{2H2O}(l)$
34.99 mL KOH
Q4.5.17
Potassium acid phthalate, KHC6H4O4, or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KHC6H4O4 reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH?
$\ce{KHC6H4O4}(aq)+\ce{NaOH}(aq)\rightarrow \ce{KNaC6H4O4}(aq)+\ce{H2O}(aq)$
Q4.5.18
The reaction of WCl6 with Al at ~400 °C gives black crystals of a compound containing only tungsten and chlorine. A sample of this compound, when reduced with hydrogen, gives 0.2232 g of tungsten metal and hydrogen chloride, which is absorbed in water. Titration of the hydrochloric acid thus produced requires 46.2 mL of 0.1051 M NaOH to reach the end point. What is the empirical formula of the black tungsten chloride?
S4.5.19
The empirical formula is WCl4. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/07%3A_Stoichiometry_of_Chemical_Reactions/7.E%3A_Stoichiometry_of_Chemical_Reactions_%28Exercises%29.txt |
In this chapter, we examine the relationships between gas temperature, pressure, amount, and volume. We will study a simple theoretical model and use it to analyze the experimental behavior of gases. The results of these analyses will show us the limitations of the theory and how to improve on it.
08: Gases
Learning Objectives
• Define the property of pressure
• Define and convert among the units of pressure measurements
• Describe the operation of common tools for measuring gas pressure
• Calculate pressure from manometer data
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure $1$). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.
Pressure is defined as the force exerted on a given area:
$P=\dfrac{F}{A} \label{9.2.1}$
Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either increasing the amount of force or by decreasing the area over which it is applied. Correspondingly, pressure can be decreased by either decreasing the force or increasing the area.
Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure $2$.—the elephant or the figure skater?
A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in2), so the pressure exerted by each foot is about 14 lb/in2:
$\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2}$
The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in2, so the pressure exerted by each blade is about 30 lb/in2:
$\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3}$
Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:
$\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4}$
The SI unit of pressure is the pascal (Pa), with 1 Pa = 1 N/m2, where N is the newton, a unit of force defined as 1 kg m/s2. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch—pounds per square inch (psi)—for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table $1$ provides some information on these and a few other common units for pressure measurements
Table $1$: Pressure Units
Unit Name and Abbreviation Definition or Relation to Other Unit Comment
pascal (Pa) 1 Pa = 1 N/m2 recommended IUPAC unit
kilopascal (kPa) 1 kPa = 1000 Pa
pounds per square inch (psi) air pressure at sea level is ~14.7 psi
atmosphere (atm) 1 atm = 101,325 Pa air pressure at sea level is ~1 atm
bar (bar, or b) 1 bar = 100,000 Pa (exactly) commonly used in meteorology
millibar (mbar, or mb) 1000 mbar = 1 bar
inches of mercury (in. Hg) 1 in. Hg = 3386 Pa used by aviation industry, also some weather reports
torr $\mathrm{1\: torr=\dfrac{1}{760}\:atm}$ named after Evangelista Torricelli, inventor of the barometer
millimeters of mercury (mm Hg) 1 mm Hg ~1 torr
Example $1$: Conversion of Pressure Units
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:
1. torr
2. atm
3. kPa
4. mbar
Solution
This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1.
1. $\mathrm{29.2\cancel{in\: Hg}×\dfrac{25.4\cancel{mm}}{1\cancel{in}} ×\dfrac{1\: torr}{1\cancel{mm\: Hg}} =742\: torr}$
2. $\mathrm{742\cancel{torr}×\dfrac{1\: atm}{760\cancel{torr}}=0.976\: atm}$
3. $\mathrm{742\cancel{torr}×\dfrac{101.325\: kPa}{760\cancel{torr}}=98.9\: kPa}$
4. $\mathrm{98.9\cancel{kPa}×\dfrac{1000\cancel{Pa}}{1\cancel{kPa}} \times \dfrac{1\cancel{bar}}{100,000\cancel{Pa}} \times\dfrac{1000\: mbar}{1\cancel{bar}}=989\: mbar}$
Exercise $1$
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?
Answer
0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar
We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure $3$). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere.
If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be $\dfrac{1}{13.6}$ as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, p:
$p=hρg \label{9.2.5}$
where
• $h$ is the height of the fluid,
• $ρ$ is the density of the fluid, and
• $g$ is acceleration due to gravity.
Example $2$: Calculation of Barometric Pressure
Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = $13.6 \,g/cm^3$.
Solution
The hydrostatic pressure is given by Equation \ref{9.2.5}, with $h = 760 \,mm$, $ρ = 13.6\, g/cm^3$, and $g = 9.81 \,m/s^2$. Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:)
$\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}} \nonumber$
\begin {align*} p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\[4pt] &\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \[4pt] & \mathrm{=1.01×10^5\:Pa} \end {align*} \nonumber
Exercise $2$
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm3.
Answer
10.3 m
A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube (h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure $3$) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.
Example $3$: Calculation of Pressure Using an Open-End Manometer
The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
Solution
The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)
1. In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg
2. $\mathrm{897\cancel{mm Hg}×\dfrac{1\: atm}{760\cancel{mm Hg}}=1.18\: atm}$
3. $\mathrm{1.18\cancel{atm}×\dfrac{101.325\: kPa}{1\cancel{atm}}=1.20×10^2\:kPa}$
Exercise $3$
The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
642 mm Hg
Answer b
0.845 atm
Answer c
85.6 kPa
Application: Measuring Blood Pressure
Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure $5$). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure—the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure—the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg).
Meteorology, Climatology, and Atmospheric Science
Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure $5$) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.
In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.
The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure $7$: the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.
Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.
Summary
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
Key Equations
• $P=\dfrac{F}{A}$
• p = hρg
Glossary
atmosphere (atm)
unit of pressure; 1 atm = 101,325 Pa
bar
(bar or b) unit of pressure; 1 bar = 100,000 Pa
barometer
device used to measure atmospheric pressure
hydrostatic pressure
pressure exerted by a fluid due to gravity
manometer
device used to measure the pressure of a gas trapped in a container
pascal (Pa)
SI unit of pressure; 1 Pa = 1 N/m2
pounds per square inch (psi)
unit of pressure common in the US
pressure
force exerted per unit area
torr
unit of pressure; $\mathrm{1\: torr=\dfrac{1}{760}\,atm}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.1%3A_Gas_Pressure.txt |
Learning Objectives
• Identify the mathematical relationships between the various properties of gases
• Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions
During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure $1$), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas—a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation—the ideal gas law—that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.
Pressure and Temperature: Amontons’s Law
Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure $2$) and the pressure increases.
This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure $3$. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.
Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P-T relationship for gases is known as either Amontons’s law or Gay-Lussac’s law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:
$P∝T\ce{\:or\:}P=\ce{constant}×T\ce{\:or\:}P=k×T \nonumber$
where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.
For a confined, constant volume of gas, the ratio $\dfrac{P}{T}$ is therefore constant (i.e., $\dfrac{P}{T}=k$). If the gas is initially in “Condition 1” (with P = P1 and T = T1), and then changes to “Condition 2” (with P = P2 and T = T2), we have that $\dfrac{P_1}{T_1}=k$ and $\dfrac{P_2}{T_2}=k$, which reduces to $\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$. This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)
Example $1$: Predicting Change in Pressure with Temperature
A can of hair spray is used until it is empty except for the propellant, isobutane gas.
1. On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?
2. The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?
Solution
1. The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)
2. We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P1 and T1 as the initial values, T2 as the temperature where the pressure is unknown and P2 as the unknown pressure, and converting °C to K, we have:
$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\textrm{ which means that }\dfrac{360\:\ce{kPa}}{297\:\ce K}=\dfrac{P_2}{323\:\ce K}$
Rearranging and solving gives:
$P_2=\mathrm{\dfrac{360\:kPa×323\cancel{K}}{297\cancel{K}}=390\:kPa}$
Exercise $1$
A sample of nitrogen, N2, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?
Answer
400 torr
Volume and Temperature: Charles’s Law
If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.
These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure $2$.
The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.
Mathematically, this can be written as:
$VαT\ce{\:or\:}V=\ce{constant}·T\ce{\:or\:}V=k·T \nonumber$
with k being a proportionality constant that depends on the amount and pressure of the gas.
For a confined, constant pressure gas sample, $\dfrac{V}{T}$ is constant (i.e., the ratio = k), and as seen with the P-T relationship, this leads to another form of Charles’s law: $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$.
Example $2$: Predicting Change in Volume with Temperature
A sample of carbon dioxide, CO2, occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?
Solution
Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:
$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\textrm{ which means that }\dfrac{0.300\:\ce L}{283\:\ce K}=\dfrac{V_2}{303\:\ce K}$
Rearranging and solving gives: $V_2=\mathrm{\dfrac{0.300\:L×303\cancel{K}}{283\cancel{K}}=0.321\:L}$
This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).
Exercise $2$
A sample of oxygen, O2, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?
Answer
21.6 mL
Example $3$: Measuring Temperature with a Volume
Change Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm3 when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm3. Find the temperature of boiling ammonia on the kelvin and Celsius scales.
Solution
A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:
$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\textrm{ which means that }\mathrm{\dfrac{150.0\:cm^3}{273.15\:K}}=\dfrac{131.7\:\ce{cm}^3}{T_2} \nonumber$
Rearrangement gives $T_2=\mathrm{\dfrac{131.7\cancel{cm}^3×273.15\:K}{150.0\:cm^3}=239.8\:K}$
Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.
Exercise $3$
What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?
Answer
635 mL
Volume and Pressure: Boyle’s Law
If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure $5$.
Unlike the P-T and V-T relationships, pressure and volume are not directly proportional to each other. Instead, $P$ and $V$ exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:
$P \propto \dfrac{1}{V} \nonumber$
or
$P=k⋅ \dfrac{1}{V} \nonumber$
or
$PV=k \nonumber$
or
$P_1V_1=P_2V_2 \nonumber$
with $k$ being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure $\left(\dfrac{1}{P}\right)$ versus the volume (V), or the inverse of volume $\left(\dfrac{1}{V}\right)$ versus the pressure ($P$). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to “linearize” their data. If we plot P versus V, we obtain a hyperbola (Figure $6$).
The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.
Example $4$: Volume of a Gas Sample
The sample of gas has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:
1. the P-V graph in Figure $\PageIndex{6a}$
2. the $\dfrac{1}{P}$ vs. V graph in Figure $\PageIndex{6b}$
3. the Boyle’s law equation
Comment on the likely accuracy of each method.
Solution
1. Estimating from the P-V graph gives a value for P somewhere around 27 psi.
2. Estimating from the $\dfrac{1}{P}$ versus V graph give a value of about 26 psi.
3. From Boyle’s law, we know that the product of pressure and volume (PV) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P1V1 = k and P2V2 = k which means that P1V1 = P2V2.
Using P1 and V1 as the known values 13.0 psi and 15.0 mL, P2 as the pressure at which the volume is unknown, and V2 as the unknown volume, we have:
$P_1V_1=P_2V_2\mathrm{\:or\:13.0\:psi×15.0\:mL}=P_2×7.5\:\ce{mL} \nonumber$
Solving:
$P_2=\mathrm{\dfrac{13.0\:psi×15.0\cancel{mL}}{7.5\cancel{mL}}=26\:psi} \nonumber$
It was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.
Exercise $4$
The sample of gas has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using:
1. the P-V graph in Figure $\PageIndex{6a}$
2. the $\dfrac{1}{P}$ vs. V graph in Figure $\PageIndex{6b}$
3. the Boyle’s law equation
Comment on the likely accuracy of each method.
Answer a
about 17–18 mL
Answer b
~18 mL
Answer c
17.7 mL; it was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow
Breathing and Boyle’s Law
What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure $7$).
Moles of Gas and Volume: Avogadro’s Law
The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.
In equation form, this is written as:
$V∝n\textrm{ or }V=k×n\textrm{ or }\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2} \nonumber$
Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n versus T.
Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).
The Ideal Gas Law
To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:
• Boyle’s law: PV = constant at constant T and n
• Amontons’s law: $\dfrac{P}{T}$ = constant at constant V and n
• Charles’s law: $\dfrac{V}{T}$ = constant at constant P and n
• Avogadro’s law: $\dfrac{V}{n}$ = constant at constant P and T
Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:
$PV=nRT \nonumber$
where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol–1 K–1 and 8.3145 kPa L mol–1 K–1.
The ideal gas law is easy to remember and apply in solving problems, as long as you use the proper values and units for the gas constant, R.
Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.
The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.
Example $5$: Using the Ideal Gas Law
Methane, CH4, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH4. What is the volume of this much methane at 25 °C and 745 torr?
Exercise $5$
Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.
Answer
350 bar
If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$ using units of atm, L, and K. Both sets of conditions are equal to the product of n × R (where n = the number of moles of the gas and R is the ideal gas law constant).
Example $6$: Using the Combined Gas Law
When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure $8$). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?
Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have:
$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}⟶\mathrm{\dfrac{(153\:atm)(13.2\:L)}{(300\:K)}=\dfrac{(3.13\:atm)(\mathit{V}_2)}{(310\:K)}} \nonumber$
Solving for V2:
$V_2=\mathrm{\dfrac{(153\cancel{atm})(13.2\:L)(310\cancel{K})}{(300\cancel{K})(3.13\cancel{atm})}=667\:L} \nonumber$
(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)
Exercise $6$
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.
Answer
0.193 L
The Interdependence between Ocean Depth and Pressure in Scuba Diving
Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure $9$) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.
Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.
Standard Conditions of Temperature and Pressure
We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K (0.00 °C) and 1 atm (101.325 kPa). At STP, an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure $10$).
Summary
The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law).
The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant.
• PV = nRT
Summary
absolute zero
temperature at which the volume of a gas would be zero according to Charles’s law.
Amontons’s law
(also, Gay-Lussac’s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant
Avogadro’s law
volume of a gas at constant temperature and pressure is proportional to the number of gas molecules
Boyle’s law
volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured
Charles’s law
volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant
ideal gas
hypothetical gas whose physical properties are perfectly described by the gas laws
ideal gas constant (R)
constant derived from the ideal gas equation R = 0.08226 L atm mol–1 K–1 or 8.314 L kPa mol–1 K–1
ideal gas law
relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws
standard conditions of temperature and pressure (STP)
273.15 K (0 °C) and 1 atm (101.325 kPa)
standard molar volume
volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.2%3A_Relating_Pressure_Volume_Amount_and_Temperature%3A_The_Ideal_Gas_Law.txt |
Learning Objectives
• Use the ideal gas law to compute gas densities and molar masses
• Perform stoichiometric calculations involving gaseous substances
• State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it."
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.
Density of a Gas
Recall that the density of a gas is its mass to volume ratio, $ρ=\dfrac{m}{V}$. Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example $1$.
Example $1$: Derivation of a Density Formula from the Ideal Gas Law
Use PV = nRT to derive a formula for the density of gas in g/L
Solution
$PV = nRT \nonumber$
Rearrange to get (mol/L):
$\dfrac{n}{v}=\dfrac{P}{RT} \nonumber$
Multiply each side of the equation by the molar mass, ℳ. When moles are multiplied by ℳ in g/mol, g are obtained:
$(ℳ)\left(\dfrac{n}{V}\right)=\left(\dfrac{P}{RT}\right)(ℳ) \nonumber$
$ℳ/V=ρ=\dfrac{Pℳ}{RT} \nonumber$
Exercise $1$
A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas?
Answer
$ρ=\dfrac{Pℳ}{RT} \nonumber$
$\mathrm{0.0847\:g/L=760\cancel{torr}×\dfrac{1\cancel{atm}}{760\cancel{torr}}×\dfrac{\mathit{ℳ}}{0.0821\: L\cancel{atm}/mol\: K}×290\: K} \nonumber$
ℳ = 2.02 g/mol; therefore, the gas must be hydrogen (H2, 2.02 g/mol)
We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP.
Example $2$: Empirical/Molecular Formula Problems
Using the Ideal Gas Law and Density of a Gas Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?
Solution
Strategy:
First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:
$\mathrm{85.7\: g\: C×\dfrac{1\: mol\: C}{12.01\: g\: C}=7.136\: mol\: C\hspace{20px}\dfrac{7.136}{7.136}=1.00\: mol\: C} \nonumber$
$\mathrm{14.3\: g\: H×\dfrac{1\: mol\: H}{1.01\: g\: H}=14.158\: mol\: H\hspace{20px}\dfrac{14.158}{7.136}=1.98\: mol\: H} \nonumber$
Empirical formula is CH2 [empirical mass (EM) of 14.03 g/empirical unit].
Next, use the density equation related to the ideal gas law to determine the molar mass:
$d=\dfrac{Pℳ}{RT}\hspace{20px}\mathrm{\dfrac{1.56\: g}{1.00\: L}=0.984\: atm×\dfrac{ℳ}{0.0821\: L\: atm/mol\: K}×323\: K} \nonumber$
ℳ = 42.0 g/mol, $\dfrac{ℳ}{Eℳ}=\dfrac{42.0}{14.03}=2.99$, so (3)(CH2) = C3H6 (molecular formula)
Exercise $2$
Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?
Answer
Empirical formula, CH; Molecular formula, C2H2
Molar Mass of a Gas
Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:
$ℳ=\mathrm{\dfrac{grams\: of\: substance}{moles\: of\: substance}}=\dfrac{m}{n} \nonumber$
The ideal gas equation can be rearranged to isolate n:
$n=\dfrac{PV}{RT} \nonumber$
and then combined with the molar mass equation to yield:
$ℳ=\dfrac{mRT}{PV} \nonumber$
This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.
Example $3$: Determining the Molar Mass of a Volatile Liquid
The approximate molar mass of a volatile liquid can be determined by:
1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (Figure $1$)
Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?
Solution
Since
$ℳ=\dfrac{m}{n} \nonumber$
and
$n=\dfrac{PV}{RT} \nonumber$
substituting and rearranging gives
$ℳ=\dfrac{mRT }{PV} \nonumber$
then
$ℳ=\dfrac{mRT}{PV}=\mathrm{\dfrac{(0.494\: g)×0.08206\: L⋅atm/mol\: K×372.8\: K}{0.976\: atm×0.129\: L}=120\:g/mol} \nonumber$
Exercise $3$
A sample of phosphorus that weighs 3.243 × 10−2 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?
Answer
124 g/mol P4
The Pressure of a Mixture of Gases: Dalton’s Law
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container ( Figure $2$). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:
$P_{Total}=P_A+P_B+P_C+...=\sum_iP_i \nonumber$
In the equation PTotal is the total pressure of a mixture of gases, PA is the partial pressure of gas A; PB is the partial pressure of gas B; PC is the partial pressure of gas C; and so on.
The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:
$P_A=X_A×P_{Total}\hspace{20px}\ce{where}\hspace{20px}X_A=\dfrac{n_A}{n_{Total}} \nonumber$
where PA, XA, and nA are the partial pressure, mole fraction, and number of moles of gas A, respectively, and nTotal is the number of moles of all components in the mixture.
Example $2$: The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 × 10−3 mol of H2, 1.00 × 10−3 mol of He, and 3.00 × 10−4 mol of Ne at 35 °C.
1. What are the partial pressures of each of the gases?
2. What is the total pressure in atmospheres?
Solution
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\dfrac{nRT}{V}$:
$P_\mathrm{H_2}=\mathrm{\dfrac{(2.50×10^{−3}\:mol)(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=6.32×10^{−3}\:atm} \nonumber$
$P_\ce{He}=\mathrm{\dfrac{(1.00×10^{−3}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=2.53×10^{−3}\:atm} \nonumber$
$P_\ce{Ne}=\mathrm{\dfrac{(3.00×10^{−4}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=7.58×10^{−4}\:atm} \nonumber$
The total pressure is given by the sum of the partial pressures:
$P_\ce{T}=P_\mathrm{H_2}+P_\ce{He}+P_\ce{Ne}=\mathrm{(0.00632+0.00253+0.00076)\:atm=9.61×10^{−3}\:atm} \nonumber$
Exercise $2$
A 5.73-L flask at 25 °C contains 0.0388 mol of N2, 0.147 mol of CO, and 0.0803 mol of H2. What is the total pressure in the flask in atmospheres?
Answer
1.137 atm
Here is another example of this concept, but dealing with mole fraction calculations.
Example $3$: The Pressure of a Mixture of Gases
A gas mixture used for anesthesia contains 2.83 mol oxygen, O2, and 8.41 mol nitrous oxide, N2O. The total pressure of the mixture is 192 kPa.
1. What are the mole fractions of O2 and N2O?
2. What are the partial pressures of O2 and N2O?
Solution
The mole fraction is given by
$X_A=\dfrac{n_A}{n_{Total}} \nonumber$
and the partial pressure is
$P_A = X_A \times P_{Total} \nonumber$
For O2,
$X_{O_2}=\dfrac{n_{O_2}}{n_{Total}}=\mathrm{\dfrac{2.83 mol}{(2.83+8.41)\:mol}=0.252} \nonumber$
and
$P_{O_2}=X_{O_2}×P_{Total}=\mathrm{0.252×192\: kPa=48.4\: kPa} \nonumber$
For N2O,
$X_{N_2O}=\dfrac{n_{N_2O}}{n_{Total}}=\mathrm{\dfrac{8.41\: mol}{(2.83+8.41)\:mol}=0.748} \nonumber$
and
$P_{N_2O}=X_{N_2O}×P_{Total}=\mathrm{(0.748)×192\: kPa = 143.6 \: kPa} \nonumber$
Exercise $3$
What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?
Answer
1.87 atm
Collection of Gases over Water
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure $3$), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.
However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor.
The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure $4$); more detailed information on the temperature dependence of water vapor can be found in Table $1$, and vapor pressure will be discussed in more detail in the next chapter on liquids.
Table $1$: Vapor Pressure of Ice and Water in Various Temperatures at Sea Level
Temperature (°C) Pressure (torr) Temperature (°C) Pressure (torr) Temperature (°C) Pressure (torr)
–10 1.95 18 15.5 30 31.8
–5 3.0 19 16.5 35 42.2
–2 3.9 20 17.5 40 55.3
0 4.6 21 18.7 50 92.5
2 5.3 22 19.8 60 149.4
4 6.1 23 21.1 70 233.7
6 7.0 24 22.4 80 355.1
8 8.0 25 23.8 90 525.8
10 9.2 26 25.2 95 633.9
12 10.5 27 26.7 99 733.2
14 12.0 28 28.3 100.0 760.0
16 13.6 29 30.0 101.0 787.6
Example $4$: Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure $3$, what is the partial pressure of argon?
Solution
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:
$P_\ce{T}=P_\ce{Ar}+P_\mathrm{H_2O} \nonumber$
Rearranging this equation to solve for the pressure of argon gives:
$P_\ce{Ar}=P_\ce{T}−P_\mathrm{H_2O} \nonumber$
The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so:
$P_\ce{Ar}=\mathrm{750\:torr−25.2\:torr=725\:torr} \nonumber$
Exercise $4$
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?
Answer
0.583 L
Chemical Stoichiometry and Gases
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions. We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.
Avogadro’s Law Revisited
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to
$\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber$
a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in Figure $4$. According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.
Example $5$: Reaction of Gases
Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
Solution
The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:
\begin{align} &\ce{C3H8}(g)+\ce{5O2}(g) ⟶ &&\ce{3CO2}(g)+\ce{4H2O}(l)\ \ce{&1\: volume + 5\: volumes &&3\: volumes + 4\: volumes} \end{align} \nonumber
From the equation, we see that one volume of C3H8 will react with five volumes of O2:
$\mathrm{2.7\cancel{L\:C_3H_8}×\dfrac{5\: L\:\ce{O2}}{1\cancel{L\:C_3H_8}}=13.5\: L\:\ce{O2}} \nonumber$
A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.
Exercise $5$
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene?
$\ce{2C2H2 + 5O2⟶4CO2 + 2H2O} \nonumber$
Answer
3.34 tanks (2.34 × 104 L)
Example $6$: Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?
$\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber$
Solution
Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:
$\mathrm{683\cancel{billion\:ft^3\:NH_3}×\dfrac{3\: billion\:ft^3\:H_2}{2\cancel{billion\:ft^3\:NH_3}}=1.02×10^3\:billion\:ft^3\:H_2} \nonumber$
The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Exercise $6$
What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapor.
Answer
51.0 L
Example $7$: Volume of Gaseous Product
What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?
$\ce{2Ga}(s)+\ce{6HCl}(aq)⟶\ce{2GaCl3}(aq)+\ce{3H2}(g) \nonumber$
Solution
To convert from the mass of gallium to the volume of H2(g), we need to do something like this:
The first two conversions are:
$\mathrm{8.88\cancel{g\: Ga}×\dfrac{1\cancel{mol\: Ga}}{69.723\cancel{g\: Ga}}×\dfrac{3\: mol\:H_2}{2\cancel{mol\: Ga}}=0.191\:mol\: H_2} \nonumber$
Finally, we can use the ideal gas law:
$V_\mathrm{H_2}=\left(\dfrac{nRT}{P}\right)_\mathrm{H_2}=\mathrm{\dfrac{0.191\cancel{mol}×0.08206\: L\cancel{atm\:mol^{−1}\:K^{−1}}×300\: K}{0.951\:atm}=4.94\: L} \nonumber$
Exercise $7$
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?
Answer
1.30 × 103 L
Greenhouse Gases and Climate Change
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost $\dfrac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure $6$).
There is strong evidence from multiple sources that higher atmospheric levels of CO2 are caused by human activity, with fossil fuel burning accounting for about $\dfrac{3}{4}$ of the recent increase in CO2. Reliable data from ice cores reveals that CO2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO2 concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure $7$).
Summary
The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.
Key Equations
• PTotal = PA + PB + PC + … = ƩiPi
• PA = XA PTotal
• $X_A=\dfrac{n_A}{n_{Total}}$
Footnotes
1. “Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, www-history.mcs.st-andrews.ac.../Lagrange.html
Summary
Dalton’s law of partial pressures
total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.
mole fraction (X)
concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
partial pressure
pressure exerted by an individual gas in a mixture
vapor pressure of water
pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.3%3A_Stoichiometry_of_Gaseous_Substances_Mixtures_and_Reactions.txt |
Learning Objectives
• Define and explain effusion and diffusion
• State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure $1$). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure $1$. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:
$\textrm{rate of diffusion}=\dfrac{\textrm{amount of gas passing through an area}}{\textrm{unit of time}} \nonumber$
The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.
A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure $1$). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure $2$). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:
$\textrm{rate of effusion}∝\dfrac{1}{\sqrt{ℳ}} \nonumber$
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
$\dfrac{\textrm{rate of effusion of B}}{\textrm{rate of effusion of A}}=\dfrac{\sqrt{ℳ_\ce{A}}}{\sqrt{ℳ_\ce{B}}} \nonumber$
Example $1$: Applying Graham’s Law to Rates of Effusion
Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.
Solution
From Graham’s law, we have:
$\dfrac{\textrm{rate of effusion of hydrogen}}{\textrm{rate of effusion of oxygen}}=\mathrm{\dfrac{\sqrt{1.43\cancel{g\: L^{−1}}}}{\sqrt{0.0899\cancel{g\: L^{−1}}}}=\dfrac{1.20}{0.300}=\dfrac{4}{1}} \nonumber$
Using molar masses:
$\dfrac{\textrm{rate of effusion of hydrogen}}{\textrm{rate of effusion of oxygen}}=\mathrm{\dfrac{32\cancel{g\: mol^{−1}}}{2\cancel{g\: mol^{−1}}}=\dfrac{\sqrt{16}}{\sqrt{1}}=\dfrac{4}{1}} \nonumber$
Hydrogen effuses four times as rapidly as oxygen.
Exercise $1$
At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide effuse?
Answer
52 mL/s
Here’s another example, making the point about how determining times differs from determining rates.
Example $2$: Effusion Time Calculations
It takes 243 s for 4.46 × 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10−5 mol Ne to effuse?
Solution
It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:
$\textrm{rate of effusion}=\dfrac{\textrm{amount of gas transferred}}{\textrm{time}}\nonumber$
and combine it with Graham’s law:
$\dfrac{\textrm{rate of effusion of gas Xe}}{\textrm{rate of effusion of gas Ne}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}\nonumber$
To get:
$\dfrac{\dfrac{\textrm{amount of Xe transferred}}{\textrm{time for Xe}}}{\dfrac{\textrm{amount of Ne transferred}}{\textrm{time for Ne}}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}\nonumber$
Noting that amount of A = amount of B, and solving for time for Ne:
$\dfrac{\dfrac{\cancel{\textrm{amount of Xe}}}{\textrm{time for Xe}}}{\dfrac{\cancel{\textrm{amount of Ne}}}{\textrm{time for Ne}}}=\dfrac{\textrm{time for Ne}}{\textrm{time for Xe}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}} \nonumber$
and substitute values:
$\mathrm{\dfrac{time\: for\: Ne}{243\:s}=\sqrt{\dfrac{20.2\cancel{g\: mol}}{131.3\cancel{g\: mol}}}=0.392}\nonumber$
Finally, solve for the desired quantity:
$\mathrm{time\: for\: Ne=0.392×243\:s=95.3\:s}\nonumber$
Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for $\ce{Xe}$, which means the time of effusion for Ne will be smaller than that for Xe.
Exercise $2$
A party balloon filled with helium deflates to $\dfrac{2}{3}$ of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to $\dfrac{1}{2}$ of its original volume?
Answer
32 h
Finally, here is one more example showing how to calculate molar mass from effusion rate data.
Example $3$: Determining Molar Mass Using Graham’s Law
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?
Solution
From Graham’s law, we have:
$\mathrm{\dfrac{rate\: of\: effusion\: of\: Unknown}{rate\: of\: effusion\: of\: CO_2}}=\dfrac{\sqrt{ℳ_\mathrm{CO_2}}}{\sqrt{ℳ_{Unknown}}} \nonumber$
Plug in known data:
$\dfrac{1.66}{1}=\dfrac{\sqrt{44.0\:\ce{g/mol}}}{\sqrt{ℳ_{Unknown}}} \nonumber$
Solve:
$ℳ_{Unknown}=\mathrm{\dfrac{44.0\:g/mol}{(1.66)^2}=16.0\:g/mol} \nonumber$
The gas could well be CH4, the only gas with this molar mass.
Exercise
Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Answer
163 g/mol
Application: Use of Diffusion for Uranium Enrichment
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF6, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235UF6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238UF6 molecules. The gas that has passed through the barrier is slightly enriched in 235UF6 and the residual gas is slightly depleted. The small difference in molecular weights between 235UF6 and 238UF6 only about 0.4% enrichment, is achieved in one diffuser (Figure $4$). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
The large scale separation of gaseous 235UF6 from 238UF6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF6.
Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.
Summary
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).
Key Equations
• $\textrm{rate of diffusion}=\dfrac{\textrm{amount of gas passing through an area}}{\textrm{unit of time}}$
• $\dfrac{\textrm{rate of effusion of gas A}}{\textrm{rate of effusion of gas B}}=\dfrac{\sqrt{m_B}}{\sqrt{m_A}}=\dfrac{\sqrt{ℳ_B}}{\sqrt{ℳ_A}}$
Summary
diffusion
movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
effusion
transfer of gaseous atoms or molecules from a container to a vacuum through very small openings
Graham’s law of effusion
rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses
mean free path
average distance a molecule travels between collisions
rate of diffusion
amount of gas diffusing through a given area over a given time | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.4%3A_Effusion_and_Diffusion_of_Gases.txt |
Learning Objectives
• State the postulates of the kinetic-molecular theory
• Use this theory’s postulates to explain the gas laws
The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.
The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)
1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
2. The molecules composing the gas are negligibly small compared to the distances between them.
3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.
The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law.
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I
Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:
• Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure $\PageIndex{1a}$).
• Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.
• Boyle’s law. If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure $\PageIndex{1b}$).
• Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure $\PageIndex{1c}$).
• Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.
Molecular Velocities and Kinetic Energy
The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample.
In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure $2$).
The kinetic energy (KE) of a particle of mass (m) and speed (u) is given by:
$\ce{KE}=\dfrac{1}{2}mu^2 \nonumber$
Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m2 s–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square velocity of a particle, urms, is defined as the square root of the average of the squares of the velocities with n = the number of particles:
$u_\ce{rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u^2_1+u^2_2+u^2_3+u^2_4+…}{n}} \nonumber$
The average kinetic energy, KEavg, is then equal to:
$\mathrm{KE_{avg}}=\dfrac{1}{2}mu^2_\ce{rms} \nonumber$
The KEavg of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:
$\mathrm{KE_{avg}}=\dfrac{3}{2}RT \nonumber$
where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J mol-1K-1 (8.314 kg m2s2mol-1K–1). These two separate equations for KEavg may be combined and rearranged to yield a relation between molecular speed and temperature:
$\dfrac{1}{2}mu^2_\ce{rms}=\dfrac{3}{2}RT \nonumber$
$u_\ce{rms}=\sqrt{\dfrac{3RT}{m}} \label{RMS}$
Example $1$: Calculation of urms
Calculate the root-mean-square velocity for a nitrogen molecule at 30 °C.
Solution
Convert the temperature into Kelvin:
$30°C+273=303\: K \nonumber$
Determine the mass of a nitrogen molecule in kilograms:
$\mathrm{\dfrac{28.0\cancel{g}}{1\: mol}×\dfrac{1\: kg}{1000\cancel{g}}=0.028\:kg/mol} \nonumber$
Replace the variables and constants in the root-mean-square velocity formula (Equation \ref{RMS}), replacing Joules with the equivalent kg m2s–2:
\begin{align*} u_\ce{rms} &= \sqrt{\dfrac{3RT}{m}} \ u_\ce{rms} &=\sqrt{\dfrac{3(8.314\:J/mol\: K)(303\: K)}{(0.028\:kg/mol)}} \ &=\sqrt{2.70 \times 10^5\:m^2s^{−2}} \ &= 519\:m/s \end{align*} \nonumber
Exercise $1$
Calculate the root-mean-square velocity for an oxygen molecule at –23 °C.
Answer
441 m/s
If the temperature of a gas increases, its KEavg increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KEavg decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure $3$.
At a given temperature, all gases have the same KEavg for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher urms, with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower urms, and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure $4$.
The gas simulator may be used to examine the effect of temperature on molecular velocities. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures.
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II
According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.
The rate of effusion of a gas depends directly on the (average) speed of its molecules:
$\textrm{effusion rate} ∝ u_\ce{rms} \nonumber$
Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here:
$u_\ce{rms}=\sqrt{\dfrac{3RT}{m}} \nonumber$
$m=\dfrac{3RT}{u^2_\ce{rms}}=\dfrac{3RT}{\overline{u}^2} \nonumber$
$\mathrm{\dfrac{effusion\: rate\: A}{effusion\: rate\: B}}=\dfrac{u_\mathrm{rms\:A}}{u_\mathrm{rms\:B}}=\dfrac{\sqrt{\dfrac{3RT}{m_\ce{A}}}}{\sqrt{\dfrac{3RT}{m_\ce{B}}}}=\sqrt{\dfrac{m_\ce{B}}{m_\ce{A}}} \nonumber$
The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law.
Summary
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.
Key Equations
• $u_\ce{rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u^2_1+u^2_2+u^2_3+u^2_4+…}{n}}$
• $\mathrm{KE_{avg}}=\dfrac{3}{2}RT$
• $u_\ce{rms}=\sqrt{\dfrac{3RT}{m}}$
Summary
kinetic molecular theory
theory based on simple principles and assumptions that effectively explains ideal gas behavior
root mean square velocity (urms)
measure of average velocity for a group of particles calculated as the square root of the average squared velocity | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.5%3A_The_Kinetic-Molecular_Theory.txt |
Learning Objectives
• Describe the physical factors that lead to deviations from ideal gas behavior
• Explain how these factors are represented in the van der Waals equation
• Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
• Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation
Thus far, the ideal gas law, PV = nRT, has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.
One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, Vm) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with:
$\mathrm{Z=\dfrac{molar\: volume\: of\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}{molar\: volume\: of\: ideal\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}}=\left(\dfrac{PV_m}{RT}\right)_\ce{measured} \nonumber$
Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. Figure $1$ shows plots of Z over a large pressure range for several common gases.
As is apparent from Figure $1$, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.
Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas (Figure $2$). The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law.
At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure $3$). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.
There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.
The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in the ideal gas law is $\dfrac{n^2a}{V^2}$, and the “correction” to the volume is nb. Note that when V is relatively large and n is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, PV = nRT. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table $1$.
Table $1$: Values of van der Waals Constants for Some Common Gases
Gas a (L2 atm/mol2) b (L/mol)
N2 1.39 0.0391
O2 1.36 0.0318
CO2 3.59 0.0427
H2O 5.46 0.0305
He 0.0342 0.0237
CCl4 20.4 0.1383
At low pressures, the correction for intermolecular attraction, a, is more important than the one for molecular volume, b. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by PV = nRT over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in Figure $1$. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing P). At very high pressures, the gas becomes less compressible (Z increases with P), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.
Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case.
Example $1$: Comparison of Ideal Gas Law and van der Waals Equation
A 4.25-L flask contains 3.46 mol CO2 at 229 °C. Calculate the pressure of this sample of CO2:
1. from the ideal gas law
2. from the van der Waals equation
3. Explain the reason(s) for the difference.
Solution
(a) From the ideal gas law:
$P=\dfrac{nRT}{V}=\mathrm{\dfrac{3.46\cancel{mol}×0.08206\cancel{L}atm\cancel{mol^{−1}}\cancel{K^{−1}}×502\cancel{K}}{4.25\cancel{L}}=33.5\:atm}$
(b) From the van der Waals equation:
$\left(P+\dfrac{n^2a}{V^2}\right)×(V−nb)=nRT⟶P=\dfrac{nRT}{(V−nb)}−\dfrac{n^2a}{V^2}$
$P=\mathrm{\dfrac{3.46\:mol×0.08206\:L\:atm\:mol^{−1}\:K^{−1}×502\: K}{(4.25\:L−3.46\:mol×0.0427\:L\:mol^{−1})}−\dfrac{(3.46\:mol)^2×3.59\:L^2\:atm\:mol^2}{(4.25\:L)^2}}$
This finally yields P = 32.4 atm.
(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO2 molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.
Exercise $1$
A 560-mL flask contains 21.3 g N2 at 145 °C. Calculate the pressure of N2:
1. from the ideal gas law
2. from the van der Waals equation
3. Explain the reason(s) for the difference.
Answer a
46.562 atm
Answer b
46.594 atm
Answer c
The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions.
Summary
Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.
Key Equations
• $\mathrm{Z=\dfrac{molar\:volume\: of\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}{molar\: volume\: of\: ideal\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}}=\left(\dfrac{P×V_m}{R×T}\right)_\ce{measured}$
• $\left(P+\dfrac{n^2a}{V^2}\right)×(V−nb)=nRT$
Glossary
compressibility factor (Z)
ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation
van der Waals equation
modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.6%3A_Non-Ideal_Gas_Behavior.txt |
Q9.1.1
Why are sharp knives more effective than dull knives (Hint: think about the definition of pressure)?
S9.1.1
The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively.
Q9.1.2
Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?
Q9.1.3
Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?
S9.1.3
Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice.
Q9.1.4
A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa.
Q9.1.5
A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals?
S9.1.5
0.809 atm; 82.0 kPa
Q9.1.6
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?
Q9.1.7
Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?
2.2 × 102 kPa
Q9.1.8
During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr and kPa?
Q9.1.9
The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
S9.1.9
Earth: 14.7 lb in–2; Venus: 13.1× 103 lb in−2
Q9.1.10
A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. What is the pressure of this gas in atmospheres and torr?
Q9.1.11
Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in., 1013.9 mbar.
1. What was the pressure in kPa?
2. The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Calculate the drop in pressure in torr.
S9.1.11
(a) 101.5 kPa; (b) 51 torr drop
Q9.1.12
Why is it necessary to use a nonvolatile liquid in a barometer or manometer?
Q9.1.13
The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:
1. torr
2. Pa
3. bar
S9.1.13
(a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar
Q9.1.14
The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in. Hg, determine the pressure of the gas in:
1. torr
2. Pa
3. bar
Q9.1.15
The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg, determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
S9.1.15
(a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa
Q9.1.16
The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg, determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
Q9.1.17
How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?
S9.1.17
With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since Pgas = Patm + Pvol liquid.
Q9.2.1
Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
Q9.2.2
Explain how the volume of the bubbles exhausted by a scuba diver (Figure) change as they rise to the surface, assuming that they remain intact.
S9.2.2
As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.
Q9.2.3
One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.” (a) What is the meaning of the term “inversely proportional?” (b) What are the “other things” that must be equal?
Q9.2.4
An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.”
1. What is the meaning of the term “directly proportional?”
2. What are the “other things” that must be equal?
S9.2.4
(a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure
Q9.2.5
How would the graph in Figure change if the number of moles of gas in the sample used to determine the curve were doubled?
The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph’s origin, representing a temperature of absolute zero.
Q9.2.6
How would the graph in Figure change if the number of moles of gas in the sample used to determine the curve were doubled?
When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of $1/P$ vs. $V$ is linear.
S9.2.6
The curve would be farther to the right and higher up, but the same basic shape.
Q9.2.7
In addition to the data found in Figure, what other information do we need to find the mass of the sample of air used to determine the graph?
Q9.2.8
Determine the volume of 1 mol of CH4 gas at 150 K and 1 atm, using Figure.
16.3 to 16.5 L
Q9.2.9
Determine the pressure of the gas in the syringe shown in Figure when its volume is 12.5 mL, using:
1. the appropriate graph
2. Boyle’s law
Q9.2.10
A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?
3.40 × 103 torr
Q9.2.11
What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?
S9.2.11
we must use $\dfrac{P_1V_1}{T_1} =\dfrac{P_2V_2}{T_2}$ and solve for $T_1$
$T_1 = \dfrac{P_1V_1T_2}{P_2V_2}$
Where:
$P_1 = 744\: torr$
$V_1 = 11.2\: L$
$P_2 = 744\: torr$
$V_2 = 13.3\: L$
$T_2 = 328.15°\: K$
$\dfrac{(744\: torr)(11.2\: L)(328.15°\: K)}{(744\: torr)(13.3\: L)} = 276°\: K$
276°K ; 3°C
Q9.2.12
A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
12.1 L
Q9.2.14
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
Q9.2.15
A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
217 L
Q9.2.16
The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa?
Q9.2.17
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF3?
S9.2.17
8.190 × 10–2 mol; 5.553 g
Q9.2.18
Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm?
S9.2.18
1.) Use the equation $PV =nRT$ and solve for $T$
$T= \dfrac{PV}{nR}$
2.) convert grams of I2 to moles of I2 and convert mL to L
$0.292g\: \ce{I2}\times \dfrac{1\: mole\: \ce{I2}}{253.8g\: \ce{I2}} = 1.15 \times10^{-3}\: moles\: \ce{I2}$
$73.3\:mL = 0.0733\:L$
3.) Use these values along with $R= 0.08206\: \dfrac{atm\:L}{mole\:°K}$ to solve for $T$
$T= \dfrac{(0.462\: \cancel{atm})(0.0733\:\cancel{L})}{(1.15\times10^{-3}\: \cancel{moles})(0.08206\: \dfrac{\cancel{atm}\:\cancel{L}}{\cancel{mole}\:°K})} = 359\: °K$
359°K ; 86°C
Q9.2.19
How many grams of gas are present in each of the following cases?
1. 0.100 L of CO2 at 307 torr and 26 °C
2. 8.75 L of C2H4, at 378.3 kPa and 483 K
3. 221 mL of Ar at 0.23 torr and –54 °C
S9.2.19
(a) 7.24 × 10–2 g; (b) 23.1 g; (c) 1.5 × 10–4 g
Q9.2.20
A high altitude balloon is filled with 1.41 × 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
Q9.2.21
A cylinder of medical oxygen has a volume of 35.4 L, and contains O2 at a pressure of 151 atm and a temperature of 25 °C. What volume of O2 does this correspond to at normal body conditions, that is, 1 atm and 37 °C?
5561 L
Q9.2.22
A large scuba tank (Figure) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C?
Q9.2.23
A 20.0-L cylinder containing 11.34 kg of butane, C4H10, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C.
46.4 g
Q9.2.24
While resting, the average 70-kg human male consumes 14 L of pure O2 per hour at 25 °C and 100 kPa. How many moles of O2 are consumed by a 70 kg man while resting for 1.0 h?
Q9.2.25
For a given amount of gas showing ideal behavior, draw labeled graphs of:
1. the variation of P with V
2. the variation of V with T
3. the variation of P with T
4. the variation of $\dfrac{1}{P}$ with V
Q9.2.26
For a gas exhibiting ideal behavior:
Q9.2.27
A liter of methane gas, CH4, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H2, at STP. Using Avogadro’s law as a starting point, explain why.
Q9.2.28
The effect of chlorofluorocarbons (such as CCl2F2) on the depletion of the ozone layer is well known. The use of substitutes, such as CH3CH2F(g), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:
1. CCl2F2(g)
2. CH3CH2F(g)
S9.2.28
(a) 1.85 L CCl2F2; (b) 4.66 L CH3CH2F
Q9.2.29
As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 × 1018 alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 °C?
Q9.2.30
A balloon that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mount Crumpet?
0.644 atm
Q9.2.31
If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
Q9.2.32
If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?
S9.2.32
The pressure decreases by a factor of 3.
Q9.3.1
What is the density of laughing gas, dinitrogen monoxide, N2O, at a temperature of 325 K and a pressure of 113.0 kPa?
S9.3.1
1.) First convert kPa to atm
$113.0\:kPa\times\dfrac{1\:atm}{101.325\:kPa}=1.115\:atm$
2.) The use the equation $d=\dfrac{PM}{RT}$ where d = density in g L-1 and M = molar mass in g mol-1
$d=\dfrac{(1.115\:atm)(44.02\dfrac{g}{\cancel{mol}})}{(0.08206\: \dfrac{\cancel{atm}\:L}{\cancel{mole}\:\cancel{°K}})(325\:\cancel{°K})}=1.84\:\dfrac{g}{L}$
Q9.3.2
Calculate the density of Freon 12, CF2Cl2, at 30.0 °C and 0.954 atm.
4.64 g L−1
Q9.3.3
Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
Q9.3.4
A cylinder of O2(g) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder?
38.8 g
Q9.3.5
What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr?
S9.3.5
1.) convert torr to atm and °C to °K
$307\:torr=0.404atm$
$26°C= 300.°K$
2.) Use the equation $PV=nRT$ and solve for $n$
$n=\dfrac{PV}{RT}$
$n=\dfrac{(0.404\:\cancel{atm})(0.100\:\cancel{L})}{(0.08206\dfrac{\cancel{atm}\:\cancel{L}}{mol\:\cancel{°K}})(300.\cancel{°K})}=0.00165\:moles$
3.) Then divide grams by the number of moles to obtain the molar mass:
$\dfrac{0.0494g}{0.00165\:moles}=30.0\dfrac{g}{mole}$
Q9.3.6
What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr?
72.0 g mol−1
Q9.3.7
How could you show experimentally that the molecular formula of propene is C3H6, not CH2?
Q9.3.8
The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
S9.3.8
88.1 g mol−1; PF3
Q9.3.9
Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies 125 mL with a pressure of 99.5 kPa at 22 °C?
1. Outline the steps necessary to answer the question.
2. Answer the question.
Q9.3.10
A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO2, 805 g O2, and 4,880 g N2. At 25 degrees C, what is the pressure in the cylinder in atmospheres?
141 atm
Q9.3.11
A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2, and the remainder N2 at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
Q9.3.12
A sample of gas isolated from unrefined petroleum contains 90.0% CH4, 8.9% C2H6, and 1.1% C3H8 at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
S9.3.12
CH4: 276 kPa; C2H6: 27 kPa; C3H8: 3.4 kPa
Q9.3.13
A mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.
Q9.3.14
Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O2 is not. If enough O2 is added to a cylinder of H2 at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?
Yes
Q9.3.15
A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 × 10−6 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
Q9.3.16
A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 °C. What is the pressure of the carbon monoxide? (See Table for the vapor pressure of water.)
740 torr
Q9.3.17
In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
Q9.3.18
Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO:
$\ce{2HgO}(s)⟶\ce{2Hg}(l)+\ce{O2}(g)$
1. Outline the steps necessary to answer the following question: What volume of O2 at 23 °C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?
2. Answer the question.
S9.3.18
(a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O2 produced by decomposition of this amount of HgO; and determine the volume of O2 from the moles of O2, temperature, and pressure. (b) 0.308 L
Q9.3.19
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:
$\ce{4H2O}(g)+\ce{3Fe}(s)⟶\ce{Fe3O4}(s)+\ce{4H2}(g)$
1. Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O?
2. Answer the question.
Q9.3.20
The chlorofluorocarbon CCl2F2 can be recycled into a different compound by reaction with hydrogen to produce CH2F2(g), a compound useful in chemical manufacturing:
$\ce{CCl2F2}(g)+\ce{4H2}(g)⟶\ce{CH2F2}(g)+\ce{2HCl}(g)$
1. Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 °C would be required to react with 1 ton (1.000 × 103 kg) of CCl2F2?
2. Answer the question.
S9.3.20
1. Determine the molar mass of CCl2F2. From the balanced equation, calculate the moles of H2 needed for the complete reaction. From the ideal gas law, convert moles of H2 into volume.
2. 3.72 × 103 L
Q9.3.21
Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN3). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
Q9.3.22
Lime, CaO, is produced by heating calcium carbonate, CaCO3; carbon dioxide is the other product.
1. Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875° and 0.966 atm is produced by the decomposition of 1 ton (1.000 × 103 kg) of calcium carbonate?
2. Answer the question.
S9.3.22
(a) Balance the equation. Determine the grams of CO2 produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 × 105 L
Q9.3.23
Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C2H2, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC2, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.
1. Outline the steps necessary to answer the following question: What volume of C2H2 at 1.005 atm and 12.2 °C is formed by the reaction of 15.48 g of CaC2 with water?
2. Answer the question.
Q9.3.24
Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C2H6, to produce carbon dioxide and water, if the volumes of C2H6 and O2 are measured under the same conditions of temperature and pressure.
42.00 L
Q9.3.25
What volume of O2 at STP is required to oxidize 8.0 L of NO at STP to NO2? What volume of NO2 is produced at STP?
Q9.3.26
Consider the following questions:
1. What is the total volume of the CO2(g) and H2O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C2H6(g) measured at STP?
2. What is the partial pressure of H2O in the product gases?
S9.3.26
(a) 18.0 L; (b) 0.533 atm
Q9.3.27
Methanol, CH3OH, is produced industrially by the following reaction:
$\ce{CO}(g)+\ce{2H2}(g)\xrightarrow{\textrm{ copper catalyst 300 °C, 300 atm }}\ce{CH3OH}(g)$
Q9.3.28
Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
Q9.3.29
What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO2 to BaO and O2?
10.57 L O2
Q9.3.30
A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N2 and 1.25 L of O2 at STP. What is the colorless gas?
Q9.3.31
Ethanol, C2H5OH, is produced industrially from ethylene, C2H4, by the following sequence of reactions:
$\ce{3C2H4 + 2H2SO4⟶C2H5HSO4 + (C2H5)2SO4}$
$\ce{C2H5HSO4 + (C2H5)2SO4 + 3H2O⟶3C2H5OH + 2H2SO4}$
What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
5.40 × 105 L
Q9.3.32
One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin?
Q9.3.33
A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
XeF2
Q9.3.34
One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (−NH2) in protein material are allowed to react with nitrous acid, HNO2, to form N2 gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH2(NH2)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N2 collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample?
$\ce{CH2(NH2)CO2H + HNO2⟶CH2(OH)CO2H + H2O + N2}$
Q9.4.1
A balloon filled with helium gas is found to take 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?
4.2 hours
Q9.4.2
Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure.
Q9.4.3
Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.
S9.4.3
Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: $\mathrm{\left(\dfrac{rate\: A}{rate\: B}\right)=\left(\dfrac{molar\: mass\: of\: B}{molar\: mass\: of\: A}\right)^{1/2}}$. Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy:
$\mathrm{KE_A=KE_BKE}=\dfrac{1}{2}mv^2$ Therefore, $\dfrac{1}{2}m_\ce{A}v^2_\ce{A}=\dfrac{1}{2}m_\ce{B}v^2_\ce{B}$ $\dfrac{v^2_\ce{A}}{v^2_\ce{B}}=\dfrac{m_\ce{B}}{m_\ce{A}}$ $\left(\dfrac{v^2_\ce{A}}{v^2_\ce{B}}\right)^{1/2}=\left(\dfrac{m_\ce{B}}{m_\ce{A}}\right)^{1/2}$ $\dfrac{v_\ce{A}}{v_\ce{B}}=\left(\dfrac{m_\ce{B}}{m_\ce{A}}\right)^{1/2}$
Q9.4.4
Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.
Q9.4.5
Which of the following gases diffuse more slowly than oxygen? F2, Ne, N2O, C2H2, NO, Cl2, H2S
S9.4.5
F2, N2O, Cl2, H2S
Q9.4.6
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. Show the calculation that supports this value. The molar mass of 235UF6 = 235.043930 + 6 × 18.998403 = 349.034348 g/mol, and the molar mass of 238UF6 = 238.050788 + 6 × 18.998403 = 352.041206 g/mol.
Q9.4.7
Calculate the relative rate of diffusion of 1H2 (molar mass 2.0 g/mol) compared to that of 2H2 (molar mass 4.0 g/mol) and the relative rate of diffusion of O2 (molar mass 32 g/mol) compared to that of O3 (molar mass 48 g/mol).
1.4; 1.2
Q9.4.8
A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
Q9.4.9
When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.)
$\ce{NH3}(g)+\ce{HCl}(g)⟶\ce{NH4Cl}(s)$
Q9.4.10
At approximately what distance from the ammonia moistened plug does this occur?
51.7 cm
Q9.5.1
Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
Q9.5.2
Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.
S9.5.2
Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature.
Q9.5.3
Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:
1. The pressure of the gas is increased by reducing the volume at constant temperature.
2. The pressure of the gas is increased by increasing the temperature at constant volume.
3. The average velocity of the molecules is increased by a factor of 2.
Q9.5.4
The distribution of molecular velocities in a sample of helium is shown in Figure. If the sample is cooled, will the distribution of velocities look more like that of H2 or of H2O? Explain your answer.
S9.5.4
H2O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier.
Q9.5.5
What is the ratio of the average kinetic energy of a SO2 molecule to that of an O2 molecule in a mixture of two gases? What is the ratio of the root mean square speeds, urms, of the two gases?
Q9.5.6
A 1-L sample of CO initially at STP is heated to 546 °C, and its volume is increased to 2 L.
1. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
2. What is the effect on the average kinetic energy of the molecules?
3. What is the effect on the root mean square speed of the molecules?
S9.5.6
(a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to $\sqrt{2}$ times its initial value; urms is proportional to $\mathrm{KE_{avg}}$.
Q9.5.7
The root mean square speed of H2 molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N2 molecule at 25 °C?
Q9.5.8
Answer the following questions:
1. Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
2. Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
3. At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
4. The average temperature of the gas in a hot air balloon is 1.30 × 102 °F. Calculate its density, assuming the molar mass equals that of dry air.
5. The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
6. An average balloon has a diameter of 60 feet and a volume of 1.1 × 105 ft3. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
7. A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO2 and H2O gas is produced by the combustion of this propane?
8. A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?
S9.5.1
(a) equal; (b) less than; (c) 29.48 g mol−1; (d) 1.0966 g L−1; (e) 0.129 g/L; (f) 4.01 × 105 g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min−1
Q9.5.1
Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, $\dfrac{R_1}{R_2}$, is the same at 0 °C and 100 °C.
Q9.6.1
Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?
S9.6.1
Gases C, E, and F
Q9.6.3
Explain why the plot of PV for CO2 differs from that of an ideal gas.
Q9.6.3
Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.
1. high pressure, small volume
2. high temperature, low pressure
3. low temperature, high pressure
S9.6.3
The gas behavior most like an ideal gas will occur under the conditions in (b). Molecules have high speeds and move through greater distances between collision; they also have shorter contact times and interactions are less likely. Deviations occur with the conditions described in (a) and (c). Under conditions of (a), some gases may liquefy. Under conditions of (c), most gases will liquefy.
Q9.6.4
Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas.
Q9.6.5
For which of the following gases should the correction for the molecular volume be largest: CO, CO2, H2, He, NH3, SF6?
SF6
Q9.6.7
A 0.245-L flask contains 0.467 mol CO2 at 159 °C. Calculate the pressure:
1. using the ideal gas law
2. using the van der Waals equation
3. Explain the reason for the difference.
4. Identify which correction (that for P or V) is dominant and why.
Q9.6.8
Answer the following questions:
1. If XX behaved as an ideal gas, what would its graph of Z vs. P look like?
2. For most of this chapter, we performed calculations treating gases as ideal. Was this justified?
3. What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
4. What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
5. In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?
S9.6.8
(a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures they behave close enough to ideal gases that they are approximated as such, however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the van der Waals equation (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure) (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure. (e) low temperatures
Exercises
Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?
Gases C, E, and F
Explain why the plot of PV for CO2 differs from that of an ideal gas.
Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.
(a) high pressure, small volume
(b) high temperature, low pressure
(c) low temperature, high pressure
The gas behavior most like an ideal gas will occur under the conditions in (b). Molecules have high speeds and move through greater distances between collision; they also have shorter contact times and interactions are less likely. Deviations occur with the conditions described in (a) and (c). Under conditions of (a), some gases may liquefy. Under conditions of (c), most gases will liquefy.
Describe the factors responsible for the deviation of the behavior of real gases from that of an
ideal gas.
For which of the following gases should the correction for the molecular volume be largest:
CO, CO2, H2, He, NH3, SF6?
SF6
A 0.245-L flask contains 0.467 mol CO2 at 159 °C. Calculate the pressure:
(a) using the ideal gas law
(b) using the van der Waals equation
(c) Explain the reason for the difference.
(d) Identify which correction (that for P or V) is dominant and why.
Answer the following questions:
(a) If XX behaved as an ideal gas, what would its graph of Z vs. P look like?
(b) For most of this chapter, we performed calculations treating gases as ideal. Was this justified?
(c) What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
(d) What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
(e) In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?
(a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures they behave close enough to ideal gases that they are approximated as such, however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the van der Waals equation (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure) (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure. (e) low temperatures | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/08%3A_Gases/8.E%3A_Gases_%28Exercises%29.txt |
Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy.
09: Thermochemistry
Chemical reactions, such as those that occur when you light a match, involve changes in energy as well as matter. Societies at all levels of development could not function without the energy released by chemical reactions. In 2012, about 85% of US energy consumption came from the combustion of petroleum products, coal, wood, and garbage. We use this energy to produce electricity (38%); to transport food, raw materials, manufactured goods, and people (27%); for industrial production (21%); and to heat and power our homes and businesses (10%).1 While these combustion reactions help us meet our essential energy needs, they are also recognized by the majority of the scientific community as a major contributor to global climate change.
Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy.
Footnotes
1. US Energy Information Administration, Primary Energy Consumption by Source and Sector, 2012, Total Energy [www.eia.gov]. Data derived from US Energy Information Administration, Monthly Energy Review (January 2014).
9.1: Energy Basics
Learning Objectives
• Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes
• Distinguish the related properties of heat, thermal energy, and temperature
• Define and distinguish specific heat and heat capacity, and describe the physical implications of both
• Perform calculations involving heat, specific heat, and temperature change
Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure $1$). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges.
Over 90% of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world’s energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter.
This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes—an area called thermochemistry. The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science.
Energy
Energy can be defined as the capacity to supply heat or do work. One type of work (w) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire—we move matter (the air in the pump) against the opposing force of the air surrounding the tire.
Like matter, energy comes in different types. One scheme classifies energy into two types: potential energy, the energy an object has because of its relative position, composition, or condition, and kinetic energy, the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure $2$). A battery has potential energy because the chemicals within it can produce electricity that can do work.
Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.)
When one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car’s engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic energy) when they move the cylinders’ pistons.
According to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry. To encompass both chemical and nuclear changes, we combine these laws into one statement: The total quantity of matter and energy in the universe is fixed.
Thermal Energy, Temperature, and Heat
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE, and we say that the object is “cold” (Figure $3$). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.
Interactive Element: PhET
Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure $4$. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.
Heat (q) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance (L). The atoms and molecules in H have a higher average KE than those in L. If we place substance H in contact with substance L, the thermal energy will flow spontaneously from substance H to substance L. The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance L will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure $5$).
Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process—this process also releases energy in the form of light as evidenced by the torch’s flame (Figure $\PageIndex{6a}$). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.
Measuring Energy and Heat Capacity
Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m2/s2, which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules.
We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity (C) of a body of matter is the quantity of heat (q) it absorbs or releases when it experiences a temperature change (ΔT) of 1 degree Celsius (or equivalently, 1 kelvin)
$C=\dfrac{q}{ΔT} \label{5.2.1}$
Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance.
For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,150 J of energy to raise the temperature of the pan by 50.0 °C
$C_{\text{small pan}}=\mathrm{\dfrac{18,140\; J}{50.0\; °C} =363\; J/°C} \label{5.2.2}$
The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change:
$C_{\text{large pan}}=\mathrm{\dfrac{90,700\; J}{50.0\;°C}=1814\; J/°C} \label{5.2.3}$
The specific heat capacity (c) of a substance, commonly called its “specific heat,” is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin):
$c = \dfrac{q}{\mathrm{m\Delta T}} \label{5.2.4}$
Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore:
$c_\ce{iron}=\mathrm{\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C} \label{5.2.5}$
The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron:
$c_\ce{iron}=\mathrm{\dfrac{90,700\; J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C} \label{5.2.6}$
Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure $7$).
Liquid water has a relatively high specific heat (about 4.2 J/g °C); most metals have much lower specific heats (usually less than 1 J/g °C). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table $1$.
Table $1$: Specific Heats of Common Substances at 25 °C and 1 bar
Substance Symbol (state) Specific Heat (J/g °C)
helium He(g) 5.193
water H2O(l) 4.184
ethanol C2H6O(l) 2.376
ice H2O(s) 2.093 (at −10 °C)
water vapor H2O(g) 1.864
nitrogen N2(g) 1.040
air 1.007
oxygen O2(g) 0.918
aluminum Al(s) 0.897
carbon dioxide CO2(g) 0.853
argon Ar(g) 0.522
iron Fe(s) 0.449
copper Cu(s) 0.385
lead Pb(s) 0.130
gold Au(s) 0.129
silicon Si(s) 0.712
If we know the mass of a substance and its specific heat, we can determine the amount of heat, q, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:
\begin{align*} q &= \ce{(specific\: heat)×(mass\: of\: substance)×(temperature\: change)}\label{5.2.7}\q&=c×m×ΔT \[4pt] &=c×m×(T_\ce{final}−T_\ce{initial})\end{align*}
In this equation, $c$ is the specific heat of the substance, m is its mass, and ΔT (which is read “delta T”) is the temperature change, TfinalTinitial. If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, TfinalTinitial has a positive value, and the value of q is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, TfinalTinitial has a negative value, and the value of q is negative.
Example $1$: Measuring Heat
A flask containing $\mathrm{8.0 \times 10^2\; g}$ of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb?
Solution
To answer this question, consider these factors:
• the specific heat of the substance being heated (in this case, water)
• the amount of substance being heated (in this case, 800 g)
• the magnitude of the temperature change (in this case, from 21 °C to 85 °C).
The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C).
This can be summarized using the equation:
\begin{align*} q&=c×m×ΔT \[4pt] &=c×m×(T_\ce{final}−T_\ce{initial}) \[4pt] &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−21)°C}\[4pt] &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(64)°\cancel{C}}\[4pt] &=\mathrm{210,000\: J(=210\: kJ)} \end{align*} \nonumber
Because the temperature increased, the water absorbed heat and $q$ is positive.
Exercise $1$
How much heat, in joules, must be added to a $\mathrm{5.00 \times 10^2 \;g}$ iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C.
Answer
$\mathrm{5.05 \times 10^4\; J}$
Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced.
Example $2$: Determining Other Quantities
A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity).
Solution
Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship:
\begin{align*} q&=c \times m \times \Delta T \[4pt] &=c \times m \times (T_\ce{final}−T_\ce{initial}) \end{align*} \nonumber
Substituting the known values:
$6,640\; \ce J=c \times \mathrm{(348\; g) \times (43.6 − 22.4)\; °C} \nonumber$
Solving:
$c=\mathrm{\dfrac{6,640\; J}{(348\; g) \times (21.2°C)} =0.900\; J/g\; °C} \nonumber$
Comparing this value with the values in Table $1$, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum.
Exercise $2$
A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity.
Answer
$c = \mathrm{0.45 \;J/g \;°C}$; the metal is likely to be iron from checking Table $1$.
Solar Thermal Energy Power Plants
The sunlight that reaches the earth contains thousands of times more energy than we presently capture. Solar thermal systems provide one possible solution to the problem of converting energy from the sun into energy we can use. Large-scale solar thermal plants have different design specifics, but all concentrate sunlight to heat some substance; the heat “stored” in that substance is then converted into electricity.
The Solana Generating Station in Arizona’s Sonora Desert produces 280 megawatts of electrical power. It uses parabolic mirrors that focus sunlight on pipes filled with a heat transfer fluid (HTF) (Figure $8$). The HTF then does two things: It turns water into steam, which spins turbines, which in turn produces electricity, and it melts and heats a mixture of salts, which functions as a thermal energy storage system. After the sun goes down, the molten salt mixture can then release enough of its stored heat to produce steam to run the turbines for 6 hours. Molten salts are used because they possess a number of beneficial properties, including high heat capacities and thermal conductivities.
The 377-megawatt Ivanpah Solar Generating System, located in the Mojave Desert in California, is the largest solar thermal power plant in the world (Figure $9$). Its 170,000 mirrors focus huge amounts of sunlight on three water-filled towers, producing steam at over 538 °C that drives electricity-producing turbines. It produces enough energy to power 140,000 homes. Water is used as the working fluid because of its large heat capacity and heat of vaporization.
Summary
Energy is the capacity to do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics). Matter has thermal energy due to the KE of its molecules and temperature that corresponds to the average KE of its molecules. Heat is energy that is transferred between objects at different temperatures; it flows from a high to a low temperature. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule (J). Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes.
Key Equations
• $q=c×m×ΔT=c×m×(T_\ce{final}−T_\ce{initial})$
Glossary
calorie (cal)
unit of heat or other energy; the amount of energy required to raise 1 gram of water by 1 degree Celsius; 1 cal is defined as 4.184 J
endothermic process
chemical reaction or physical change that absorbs heat
energy
capacity to supply heat or do work
exothermic process
chemical reaction or physical change that releases heat
heat (q)
transfer of thermal energy between two bodies
heat capacity (C)
extensive property of a body of matter that represents the quantity of heat required to increase its temperature by 1 degree Celsius (or 1 kelvin)
joule (J)
SI unit of energy; 1 joule is the kinetic energy of an object with a mass of 2 kilograms moving with a velocity of 1 meter per second, 1 J = 1 kg m2/s and 4.184 J = 1 cal
kinetic energy
energy of a moving body, in joules, equal to $\dfrac{1}{2}mv^2$ (where m = mass and v = velocity)
potential energy
energy of a particle or system of particles derived from relative position, composition, or condition
specific heat capacity (c)
intensive property of a substance that represents the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius (or 1 kelvin)
temperature
intensive property of matter that is a quantitative measure of “hotness” and “coldness”
thermal energy
kinetic energy associated with the random motion of atoms and molecules
thermochemistry
study of measuring the amount of heat absorbed or released during a chemical reaction or a physical change
work (w)
energy transfer due to changes in external, macroscopic variables such as pressure and volume; or causing matter to move against an opposing force | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/09%3A_Thermochemistry/9.0%3A_Prelude_to_Thermochemistry.txt |
Learning Objectives
• Explain the technique of calorimetry
• Calculate and interpret heat and related properties using typical calorimetry data
One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure $1$). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.
By convention, q is given a negative (-) sign when the system releases heat to the surroundings (exothermic); q is given a positive (+) sign when the system absorbs heat from the surroundings (endothermic).
Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment. This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure $2$). These easy-to-use “coffee cup” calorimeters allow more heat exchange with their surroundings, and therefore produce less accurate energy values.
Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure $3$).
Before we practice calorimetry problems involving chemical reactions, consider a simple example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure $4$). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:
$q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{5.3.1}$
This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:
$q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{5.3.2}$
The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that $q_{substance\, M}$ and $q_{substance\, W}$ are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, $q_{substance\, M}$ is a negative value and $q_{substance\, W}$ is positive, since heat is transferred from M to W.
Example $1$: Heat Transfer between Substances at Different Temperatures
A hot 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water is measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = − heat taken in by water, or:
$q_\ce{rebar}=−q_\ce{water} \nonumber$
Since we know how heat is related to other measurable quantities, we have:
$(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \nonumber$
Letting f = final and i = initial, in expanded form, this becomes:
$c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber$
The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:
$\mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=-(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \nonumber$
$\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \nonumber$
Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.
Exercise $\PageIndex{1A}$
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.
Answer
The initial temperature of the copper was 335.6 °C.
Exercise $\PageIndex{1B}$
A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.
Answer
The final temperature (reached by both copper and water) is 38.7 °C.
This method can also be used to determine other quantities, such as the specific heat of an unknown metal.
Example $2$: Identifying a Metal by Measuring Specific Heat
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:
$q_\ce{metal}=−q_\ce{water} \nonumber$
In expanded form, this is:
$c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber$
Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:
$\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \nonumber$
Solving this:
$\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C} \nonumber$
Comparing this with values in Table T4, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.
Exercise $2$
A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).
Answer
$c_{metal}= 0.13 \;J/g\; °C$
This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.
When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), $q_{solution}$, must add up to zero:
$q_\ce{reaction}+q_\ce{solution}=0\ \label{ 5.3.10}$
This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:
$q_\ce{reaction}=−q_\ce{solution} \label{5.3.11}$
This concept lies at the heart of all calorimetry problems and calculations.
Example $3$: Heat Produced by an Exothermic Reaction
When 50.0 mL of 0.10 M HCl(aq) and 50.0 mL of 1.00 M NaOH(aq), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?
$\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l) \nonumber$
Solution
To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.
The heat given off by the reaction is equal to that taken in by the solution. Therefore:
$q_\ce{reaction}=−q_\ce{solution} \nonumber$
(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and its surroundings.)
Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:
$q_\ce{solution}=(c×m×ΔT)_\ce{solution} \nonumber$
To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 × 102 g (two significant figures). The specific heat of water is approximately 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:
$\mathrm{\mathit q_{solution}=(4.184\:J/g\: °C)(1.0×10^2\:g)(28.9°C−22.0°C)=2.89×10^3\:J} \nonumber$
Finally, since we are trying to find the heat of the reaction, we have:
$q_\ce{reaction}=−q_\ce{solution}=−2.89×10^3\:J \nonumber$
The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat.
Exercise $3$
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?
Answer
$1.34 \times 10^3\; J$; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water
Thermochemistry of Hand Warmers
When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure $5$). A common reusable hand warmer contains a supersaturated solution of NaC2H3O2 (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable NaC2H3O2 quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).
The process $\ce{NaC2H3O2}(aq)⟶\ce{NaC2H3O2}(s)$ is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC2H3O2 redissolves and can be reused.
Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is
$\ce{2Fe(s) + 3/2 O2(g) ⟶ Fe2O3(s)}.\ n\nonumber$
Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.
Example $4$: Heat Flow in an Instant Ice Pack
When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (Figure $5$). When 3.21 g of solid NH4NO3 dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature decreases to 20.3 °C.
Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.
Solution
We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:
$q_\ce{rxn}=−q_\ce{soln} \nonumber$
with “rxn” and “soln” used as shorthand for “reaction” and “solution,” respectively.
Assuming also that the specific heat of the solution is the same as that for water, we have:
\begin{align*} q_\ce{rxn} &=−q_\ce{soln}=−(c×m×ΔT)_\ce{soln}\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(20.3°C−24.9°C)]}\ &=\mathrm{−[(4.184J/g\: °C)×(53.2\:g)×(−4.6°C)]}\ &+\mathrm{1.0×10^3\:J=+1.0\:kJ} \end{align*}
The positive sign for q indicates that the dissolution is an endothermic process.
Exercise $4$
When a 3.00-g sample of KCl was added to 3.00 × 102 g of water in a coffee cup calorimeter, the temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the KCl? What assumptions did you make?
Answer
1.33 kJ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water.
If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.
The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter, is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water (Figure $6$). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is trapped in the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q, such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.
Video $1$: Video of view how a bomb calorimeter is prepared for action.
Example $5$: Bomb Calorimetry
When 3.12 g of glucose, C6H12O6, is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample?
Solution
The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)
The heat produced by the reaction is absorbed by the water and the bomb:
\begin{align*} &q_\ce{rxn}=−(q_\ce{water}+q_\ce{bomb})\ &=\mathrm{−[(4.184\:J/g\: °C)×(775\:g)×(35.6°C−23.8°C)+893\:J/°C×(35.6°C−23.8°C)]}\ &=\mathrm{−(38,300\:J+10,500\:J)}\ &=\mathrm{−48,800\: J=−48.8\: kJ} \end{align*} \nonumber
This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.
Exercise $5$
When 0.963 g of benzene, C6H6, is burned in a bomb calorimeter, the temperature of the calorimeter increases by 8.39 °C. The bomb has a heat capacity of 784 J/°C and is submerged in 925 mL of water. How much heat was produced by the combustion of the glucose sample?
Answer
39.0 kJ
Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person.1 These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories (1 kcal), the amount of energy needed to heat 1 kg of water by 1 °C.
Measuring Nutritional Calories
In your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) = 1000 cal, or 1 kcal. (This is approximately the amount of energy needed to heat 1 kg of water by 1 °C.)
The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure $7$).
For the example shown in (b), the total energy per 228-g portion is calculated by:
$\mathrm{(5\:g\: protein×4\:Calories/g)+(31\:g\: carb×4\:Calories/g)+(12\:g\: fat×9\:Calories/g)=252\:Calories} \label{5.3.X}$
So, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freeze-dried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food.
Today, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy.
Summary
Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between the system being studied and its surroundings. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.
Footnotes
1. 1 Francis D. Reardon et al. “The Snellen human calorimeter revisited, re-engineered and upgraded: Design and performance characteristics.” Medical and Biological Engineering and Computing 8 (2006)721–28, The Snellen human calorimeter revisited, re-engineered and upgraded: design and performance characteristics [link.springer.com].
Glossary
bomb calorimeter
device designed to measure the energy change for processes occurring under conditions of constant volume; commonly used for reactions involving solid and gaseous reactants or products
calorimeter
device used to measure the amount of heat absorbed or released in a chemical or physical process
calorimetry
process of measuring the amount of heat involved in a chemical or physical process
nutritional calorie (Calorie)
unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal
surroundings
all matter other than the system being studied
system
portion of matter undergoing a chemical or physical change being studied | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/09%3A_Thermochemistry/9.2%3A_Calorimetry.txt |
Learning Objectives
• State the first law of thermodynamics
• Define enthalpy and explain its classification as a state function
• Write and balance thermochemical equations
• Calculate enthalpy changes for various chemical reactions
• Explain Hess’s law and use it to compute reaction enthalpies
Thermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry in this chapter, we need to consider some widely used concepts of thermodynamics.
Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic energy falls. The total of all possible kinds of energy present in a substance is called the internal energy (U), sometimes symbolized as E.
As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat (q) from the surroundings or when the surroundings do work (w) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire’s temperature. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings.
The relationship between internal energy, heat, and work can be represented by the equation:
$ΔU=q+w \label{5.4.1}$
as shown in Figure $1$. This is one version of the first law of thermodynamics, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive q is heat flow in; negative q is heat flow out) or work done on or by the system. The work, w, is positive if it is done on the system and negative if it is done by the system.
A type of work called expansion work (or pressure-volume work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics.
As discussed, the relationship between internal energy, heat, and work can be represented as ΔU = q + w. Internal energy is a type of quantity known as a state function (or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value does depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m, and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (Figure $2$). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).
Chemists ordinarily use a property known as enthalpy ($H$) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system’s internal energy ($U$) and the mathematical product of its pressure ($P$) and volume ($V$):
$H=U+PV \label{5.4.2}$
Since it is derived from three state functions ($U$, $P$, and $V$), enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change ($ΔH$) is:
$ΔH=ΔU+PΔV\label{5.4.3}$
The mathematical product $PΔV$ represents work ($w$), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of ΔV and w will always be opposite:
$PΔV=−w \label{5.4.4}$
Substituting Equation \ref{5.4.4} and the definition of internal energy (Equation \ref{5.4.1}) into Equation \ref{5.4.3} yields:
\begin{align} ΔH&=ΔU+PΔV \[4pt] &=q_\ce{p}+\cancel{w}−\cancel{w} \[4pt] &=q_\ce{p} \label{5.4.5} \end{align}
where $q_p$ is the heat of reaction under conditions of constant pressure.
And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow ($q_\ce{p}$) and enthalpy change ($ΔH$) for the process are equal.
The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter is not equal to $ΔH$ because the closed, constant-volume metal container prevents expansion work from occurring. Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with $q = ΔH$, which makes enthalpy the most convenient choice for determining heat.
The following conventions apply when we use $ΔH$:
1. Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a ΔH value following the equation for the reaction. This $ΔH$ value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation. For example, consider this equation: $\ce{H2(g) + 1/2 O2(g) ⟶ H2O (l)} \;\; ΔH=\mathrm{−286\:kJ} \label{5.4.6}$ This equation indicates that when 1 mole of hydrogen gas and 12 mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (ΔH is an extensive property).
\begin {align*} &\textrm{(two-fold increase in amounts)}\label{5.4.7}\ &\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(l)\hspace{20px}ΔH=\mathrm{2×(−286\:kJ)=−572\:kJ}\ &\textrm{(two-fold decrease in amounts)}\ &\frac{1}{2}\ce{H2}(g)+\dfrac{1}{4}\ce{O2}(g)⟶\frac{1}{2}\ce{H2O}(l)\hspace{20px}ΔH=\mathrm{\frac{1}{2}×(−286\:kJ)=−143\:kJ} \end {align*} \label{5.4.6B}
1. The enthalpy change of a reaction depends on the physical state of the reactants and products of the reaction (whether we have gases, liquids, solids, or aqueous solutions), so these must be shown. For example, when 1 mole of hydrogen gas and 12 mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.
$\ce{ H2(g) + 1/2 O2(g) ⟶ H2O(g)} \;\;\; ΔH=\ce{−242\:kJ} \label{5.4.7B}$
1. A negative value of an enthalpy change, ΔH, indicates an exothermic reaction; a positive value of ΔH indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its ΔH is changed (a process that is endothermic in one direction is exothermic in the opposite direction).
Example $1$: Measurement of an Enthalpy Change
When 0.0500 mol of HCl(aq) reacts with 0.0500 mol of NaOH(aq) to form 0.0500 mol of NaCl(aq), 2.9 kJ of heat are produced. What is ΔH, the enthalpy change, per mole of acid reacting, for the acid-base reaction run under the conditions described ?
$\ce{HCl (aq) + NaOH(aq) \rightarrow NaCl (aq) + H2O(l)} \nonumber$
Solution
For the reaction of 0.0500 mol acid (HCl), q = −2.9 kJ. This ratio
$\mathrm{\dfrac{−2.9 \; kJ}{0.0500\; mol\; HCl}} \nonumber$
can be used as a conversion factor to find the heat produced when 1 mole of HCl reacts:
$ΔH =\mathrm{1\; \cancel{mol\; HCl} \times \dfrac{ −2.9\; kJ}{0.0500 \;\cancel{ mol\; HCl}} =−58\; kJ} \nonumber$
The enthalpy change when 1 mole of HCl reacts is −58 kJ. Since that is the number of moles in the chemical equation, we write the thermochemical equation as:
$\ce{HCl}_{(aq)}+\ce{NaOH}_{(aq)}⟶\ce{NaCl}_{(aq)}+\ce{H_2O}_{(l)} \;\;\; ΔH=\mathrm{−58\;kJ} \nonumber$
Exercise $1$
When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:
$\ce{Zn}_{(s)}+\ce{2HCl}_{(aq)}⟶\ce{ZnCl}_{(aq)}+\ce{H}_{2(g)} \nonumber$
Answer
ΔH = −153 kJ
Be sure to take both stoichiometry and limiting reactants into account when determining the ΔH for a chemical reaction.
Example $2$: Another Example of the Measurement of an Enthalpy Change
A gummy bear contains 2.67 g sucrose, C12H22O11. When it reacts with 7.19 g potassium chlorate, KClO3, 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction
$\ce{C12H22O11}(aq)+\ce{8KClO3}(aq)⟶\ce{12CO2}(g)+\ce{11H2O}(l)+\ce{8KCl}(aq) \nonumber$
Solution
We have $\mathrm{2.67\:\cancel{g}×\dfrac{1\:mol}{342.3\:\cancel{g}}=0.00780\:mol\:C_{12}H_{22}O_{11}}$ available, and
$\mathrm{7.19\:\cancel{g}×\dfrac{1\:mol}{122.5\:\cancel{g}}=0.0587\:mol\:KClO_3}$ available.
Since
$\mathrm{0.0587\:mol\:KClO_3×\dfrac{1\:mol\:\ce{C12H22O11}}{8\:mol\:KClO_3}=0.00734\:mol\:\ce{C12H22O11}}$
is needed, C12H22O11 is the excess reactant and KClO3 is the limiting reactant.
The reaction uses 8 mol KClO3, and the conversion factor is $\mathrm{\dfrac{−43.7\:kJ}{0.0587\:mol\:KClO_3}}$, so we have $ΔH=\mathrm{8\:mol×\dfrac{−43.7\:kJ}{0.0587\:mol\:KClO_3}=−5960\:kJ}$. The enthalpy change for this reaction is −5960 kJ, and the thermochemical equation is:
$\ce{C12H22O11 + 8KClO3⟶12CO2 + 11H2O + 8KCl}\hspace{20px}ΔH=\ce{−5960\:kJ} \nonumber$
Exercise $2$
When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of $\ce{FeCl}_{2(s)}$ and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of $\ce{FeCl2(s)}$ is produced?
Answer
ΔH = −338 kJ
Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A standard state is a commonly accepted set of conditions used as a reference point for the determination of properties under other different conditions. For chemists, the IUPAC standard state refers to materials under a pressure of 1 bar and solutions at 1 M, and does not specify a temperature (it used too). Many thermochemical tables list values with a standard state of 1 atm. Because the ΔH of a reaction changes very little with such small changes in pressure (1 bar = 0.987 atm), ΔH values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted “o” in the enthalpy change symbol to designate standard state. Since the usual (but not technically standard) temperature is 298.15 K, we will use a subscripted “298” to designate this temperature. Thus, the symbol ($ΔH^\circ_{298}$) is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol ΔH is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions.)
The enthalpy changes for many types of chemical and physical processes are available in the reference literature, including those for combustion reactions, phase transitions, and formation reactions. As we discuss these quantities, it is important to pay attention to the extensive nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.e., as the ΔH for specific amounts of reactants). However, we often find it more useful to divide one extensive property (ΔH) by another (amount of substance), and report a per-amount intensive value of ΔH, often “normalized” to a per-mole basis. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.)
Enthalpy of Combustion
Standard enthalpy of combustion ($ΔH_C^\circ$) is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called “heat of combustion.” For example, the enthalpy of combustion of ethanol, −1366.8 kJ/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 °C and 1 atmosphere pressure, yielding products also at 25 °C and 1 atm.
$\ce{C2H5OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}+\ce{3H2O}(l)\hspace{20px}ΔH_{298}^\circ=\mathrm{−1366.8\: kJ} \label{5.4.8}$
Enthalpies of combustion for many substances have been measured; a few of these are listed in Table $1$. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.
Table $1$: Standard Molar Enthalpies of Combustion
Substance Combustion Reaction Enthalpy of Combustion $ΔH_c^\circ \left(\mathrm{\dfrac{kJ}{mol} \:at\:25°C}\right)$
carbon $\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)$ −393.5
hydrogen $\ce{H2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{H2O}(l)$ −285.8
magnesium $\ce{Mg}(s)+\frac{1}{2}\ce{O2}(g)⟶\ce{MgO}(s)$ −601.6
sulfur $\ce{S}(s)+\ce{O2}(g)⟶\ce{SO2}(g)$ −296.8
carbon monoxide $\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO2}(g)$ −283.0
methane $\ce{CH4}(g)+\ce{2O2}(g)⟶\ce{CO2}(g)+\ce{2H2O}(l)$ −890.8
acetylene $\ce{C2H2}(g)+\dfrac{5}{2}\ce{O2}(g)⟶\ce{2CO2}(g)+\ce{H2O}(l)$ −1301.1
ethanol $\ce{C2H5OH}(l)+\ce{3O2}(g)⟶\ce{CO2}(g)+\ce{3H2O}(l)$ −1366.8
methanol $\ce{CH3OH}(l)+\dfrac{3}{2}\ce{O2}(g)⟶\ce{CO2}(g)+\ce{2H2O}(l)$ −726.1
isooctane $\ce{C8H18}(l)+\dfrac{25}{2}\ce{O2}(g)⟶\ce{8CO2}(g)+\ce{9H2O}(l)$ −5461
Example $3$: Using Enthalpy of Combustion
As Figure $3$ suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g/mL.
Solution
Starting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the necessary conversions. Table $1$ gives this value as −5460 kJ per 1 mole of isooctane (C8H18).
Using these data,
$\mathrm{1.00\:\cancel{L\:\ce{C8H18}}×\dfrac{1000\:\cancel{mL\:\ce{C8H18}}}{1\:\cancel{L\:\ce{C8H18}}}×\dfrac{0.692\:\cancel{g\:\ce{C8H18}}}{1\:\cancel{mL\:\ce{C8H18}}}×\dfrac{1\:\cancel{mol\:\ce{C8H18}}}{114\:\cancel{g\:\ce{C8H18}}}×\dfrac{−5460\:kJ}{1\:\cancel{mol\:\ce{C8H18}}}=−3.31×10^4\:kJ} \nonumber$
The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.)
Note: If you do this calculation one step at a time, you would find:
\begin {align*} &\mathrm{1.00\:L\:\ce{C8H18}⟶1.00×10^3\:mL\:\ce{C8H18}}\ &\mathrm{1.00×10^3\:mL\:\ce{C8H18}⟶692\:g\:\ce{C8H18}}\ &\mathrm{692\:g\:\ce{C8H18}⟶6.07\:mol\:\ce{C8H18}}\ &\mathrm{692\:g\:\ce{C8H18}⟶−3.31×10^4\:kJ} \end {align*}
Exercise $3$
How much heat is produced by the combustion of 125 g of acetylene?
Answer
6.25 × 103 kJ
Emerging Algae-Based Energy Technologies (Biofuels)
As reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. Among the most promising biofuels are those derived from algae (Figure $4$). The species of algae used are nontoxic, biodegradable, and among the world’s fastest growing organisms. About 50% of algal weight is oil, which can be readily converted into fuel such as biodiesel. Algae can yield 26,000 gallons of biofuel per hectare—much more energy per acre than other crops. Some strains of algae can flourish in brackish water that is not usable for growing other crops. Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel.
According to the US Department of Energy, only 39,000 square kilometers (about 0.4% of the land mass of the US or less than $\dfrac{1}{7}$ of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. The cost of algal fuels is becoming more competitive—for instance, the US Air Force is producing jet fuel from algae at a total cost of under \$5 per gallon. The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and CO2 as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (Figure $5$).
Standard Enthalpy of Formation
A standard enthalpy of formation $ΔH^\circ_\ce{f}$ is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law.
The standard enthalpy of formation of CO2(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:
$\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)\hspace{20px}ΔH^\circ_\ce{f}=ΔH^\circ_{298}=−393.5\:\ce{kJ} \label{5.4.9}$
starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO2, also at 1 atm and 25 °C. For nitrogen dioxide, $\ce{NO}_{2(g)}$, $ΔH^\circ_\ce{f}$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:
$\frac{1}{2}\ce{N2}(g)+\ce{O2}(g)⟶\ce{NO2}(g)\hspace{20px}ΔH^\circ_\ce{f}=ΔH^\circ_{298}=+33.2\: \ce{kJ} \label{5.4.10}$
A reaction equation with $\frac{1}{2}$ mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO2(g).
You will find a table of standard enthalpies of formation of many common substances in Tables T1 and T2. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C2H2). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.
Example $4$: Evaluating an Enthalpy of Formation
Ozone, O3(g), forms from oxygen, O2(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, $ΔH^\circ_\ce{f}$ of ozone from the following information:
$\ce{3O2}(g)⟶\ce{2O3}(g)\hspace{20px}ΔH^\circ_{298}=+286\: \ce{kJ} \nonumber$
Solution $ΔH^\circ_\ce{f}$ is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, $ΔH^\circ_\ce{f}$ for O3(g) is the enthalpy change for the reaction:
$\dfrac{3}{2}\ce{O2}(g)⟶\ce{O3}(g) \nonumber$
For the formation of 2 mol of O3(g), $ΔH^\circ_{298}=+286\: \ce{kJ}$. This ratio, $\mathrm{\left(\dfrac{286\:kJ}{2\:mol\:O_3}\right)}$, can be used as a conversion factor to find the heat produced when 1 mole of O3(g) is formed, which is the enthalpy of formation for O3(g):
$ΔH^\circ \ce{\:for\:1\:mole\: of\:O_3}(g)=\mathrm{1\:\cancel{mol\:O_3}×\dfrac{286\:kJ}{2\:\cancel{mol\:O_3}}=143\:kJ} \nonumber$
Therefore, $ΔH^\circ_\ce{f}[\ce{O3}(g)]=\ce{+143\: kJ/mol}$.
Exercise $4$
Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.
Answer
For the reaction
$\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−184.6\:kJ} \nonumber$
Example $5$: Writing Reaction Equations for $ΔH^\circ_\ce{f}$
Write the heat of formation reaction equations for:
1. $\ce{C2H_5OH}_{(l)}$
2. $\ce{Ca_3(PO_4)}_{2(s)}$
Solution
Remembering that $ΔH^\circ_\ce{f}$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:
1. $\ce{2C}(s,\:\ce{graphite})+\ce{3H2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{C2H5OH}(l)$
2. $\ce{3Ca}(s)+\frac{1}{2}\ce{P4}(s)+\ce{4O2}(g)⟶\ce{Ca3(PO4)2}(s)$
Note: The standard state of carbon is graphite, and phosphorus exists as $P_4$.
Exercise $5$
Write the heat of formation reaction equations for:
1. $\ce{C_2H_5OC_2H}_{5(l)}$
2. $\ce{Na_2CO}_{3(s)}$
Answer a
$\ce{4C}(s,\:\ce{graphite})+\ce{5H2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{C2H5OC2H5}(l)$;
Answer b
$\ce{2Na}(s)+\ce{C}(s,\:\ce{graphite})+\dfrac{3}{2}\ce{O2}(g)⟶\ce{Na2CO3}(s)$
Hess’s Law
There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.
This type of calculation usually involves the use of Hess’s law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:
$\ce{C}_{(s)}+\ce{O}_{2(g)}⟶\ce{CO}_{2(g)}\;\;\;ΔH^∘_{298}=\mathrm{−394\;kJ} \label{ 5.4.11}$
In the two-step process, first carbon monoxide is formed:
$\ce{C}_{(s)}+\dfrac{1}{2}\ce{O}_{2(g)}⟶\ce{CO}_{(g)}\;\;\;ΔH^∘_{298}=\mathrm{−111\;kJ} \label{ 5.4.12}$
Then, carbon monoxide reacts further to form carbon dioxide:
$\ce{CO} {(g)}+\dfrac{1}{2}\ce{O2}(g)⟶\ce{CO}_2 {(g)}\;\;\;ΔH^∘_{298}=\mathrm{−283\;kJ} \label{ 5.4.13}$
The equation describing the overall reaction is the sum of these two chemical changes:
\begin {align*} &\textrm{Step 1:} \:\ce{C}(s)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)\ &\underline{\textrm{Step 2:} \:\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO2}(g)}\ &\textrm{Sum:} \:\ce{C}(s)+\frac{1}{2}\ce{O2}(g)+\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)+\ce{CO2}(g) \end {align*} \label{5.4.14}
Because the CO produced in Step 1 is consumed in Step 2, the net change is:
$\ce{C}_{(s)}+\ce{O}_{2(g)}⟶\ce{CO}_{2(g)} \label{5.4.15}$
According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. We can apply the data from the experimental enthalpies of combustion in Table $1$ to find the enthalpy change of the entire reaction from its two steps:
\begin {align*} &\ce{C}(s)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)&&ΔH^\circ_{298}=\mathrm{−111\:kJ}\ &\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO2}(g)&&ΔH^\circ_{298}=\mathrm{−283\:kJ}\ &\overline{\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)\hspace{25px}}&&\overline{ΔH^\circ_{298}=\mathrm{−394\:kJ}} \end {align*} \label{5.4.16}
The result is shown in Figure $6$. We see that ΔH of the overall reaction is the same whether it occurs in one step or two. This finding (overall ΔH for the reaction = sum of ΔH values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.
Before we further practice using Hess’s law, let us recall two important features of ΔH.
1. ΔH is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO2(g) is +33.2 kJ: $\frac{1}{2}\ce{N2}(g)+\ce{O2}(g)⟶\ce{NO2}(g)\hspace{20px}ΔH=\mathrm{+33.2\: kJ} \label{5.4.17}$
When 2 moles of NO2 (twice as much) are formed, the ΔH will be twice as large: $\ce{N2}(g)+\ce{2O2}(g)⟶\ce{2NO2}(g)\hspace{20px}ΔH=\mathrm{+66.4\: kJ} \label{5.4.18}$
In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.
2. ΔH for a reaction in one direction is equal in magnitude and opposite in sign to ΔH for the reaction in the reverse direction. For example, given that: $\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH=\mathrm{−184.6\:kJ} \label{5.4.19}$
Then, for the “reverse” reaction, the enthalpy change is also “reversed”: $\ce{2HCl}(g)⟶\ce{H2}(g)+\ce{Cl2}(g)\hspace{20px}ΔH=\mathrm{+184.6\: kJ} \label{5.4.20}$
Example $6$: Stepwise Calculation of $ΔH^\circ_\ce{f}$
Using Hess’s Law Determine the enthalpy of formation, $ΔH^\circ_\ce{f}$, of FeCl3(s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:
$\ce{Fe}(s)+\ce{Cl2}(g)⟶\ce{FeCl2}(s)\hspace{20px}ΔH°=\mathrm{−341.8\:kJ} \nonumber$
$\ce{FeCl2}(s)+\frac{1}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{20px}ΔH°=\mathrm \nonumber{−57.7\:kJ} \nonumber$
Solution
We are trying to find the standard enthalpy of formation of FeCl3(s), which is equal to ΔH° for the reaction:
$\ce{Fe}(s)+\frac{3}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{20px}ΔH^\circ_\ce{f}=\:? \nonumber$
Looking at the reactions, we see that the reaction for which we want to find ΔH° is the sum of the two reactions with known ΔH values, so we must sum their ΔHs:
$\ce{Fe}(s)+\ce{Cl2}(g)⟶\ce{FeCl2}(s)\hspace{59px}ΔH°=\mathrm{−341.8\:kJ}\ \underline{\ce{FeCl2}(s)+\frac{1}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{20px}ΔH°=\mathrm{−57.7\:kJ}}\ \ce{Fe}(s)+\frac{1}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{43px}ΔH°=\mathrm{−399.5\:kJ} \nonumber$
The enthalpy of formation, $ΔH^\circ_\ce{f}$, of FeCl3(s) is −399.5 kJ/mol.
Exercise $6$
Calculate ΔH for the process:
$\ce{N2}(g)+\ce{2O2}(g)⟶\ce{2NO2}(g) \nonumber$
from the following information:
$\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g)\hspace{20px}ΔH=\mathrm{180.5\:kJ} \nonumber$
$\ce{NO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{NO2}(g)\hspace{20px}ΔH=\mathrm{−57.06\:kJ} \nonumber$
Answer
66.4 kJ
Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of ΔH) if they are difficult to determine experimentally.
Example $7$: A More Challenging Problem
Using Hess’s Law Chlorine monofluoride can react with fluorine to form chlorine trifluoride:
(i) $\ce{ClF}(g)+\ce{F2}(g)⟶\ce{ClF3}(g)\hspace{20px}ΔH°=\:?$
Use the reactions here to determine the ΔH° for reaction (i):
(ii) $\ce{2OF2}(g)⟶\ce{O2}(g)+\ce{2F2}(g)\hspace{20px}ΔH^\circ_{(ii)}=\mathrm{−49.4\:kJ}$
(iii) $\ce{2ClF}(g)+\ce{O2}(g)⟶\ce{Cl2O}(g)+\ce{OF2}(g)\hspace{20px}ΔH^\circ_{(iii)}=\mathrm{+205.6\: kJ}$
(iv) $\ce{ClF3}(g)+\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\dfrac{3}{2}\ce{OF2}(g)\hspace{20px}ΔH^\circ_{(iv)}=\mathrm{+266.7\: kJ}$
Solution
Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that $\ce{ClF}_{(g)}$ is needed as a reactant. This can be obtained by multiplying reaction (iii) by $\frac{1}{2}$, which means that the ΔH° change is also multiplied by $\frac{1}{2}$:
$\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)\hspace{20px} ΔH°=\frac{1}{2}(205.6)=+102.8\: \ce{kJ} \nonumber$
Next, we see that $\ce{F_2}$ is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the ΔH° changes sign and is halved:
$\frac{1}{2}\ce{O2}(g)+\ce{F2}(g)⟶\ce{OF2}(g)\hspace{20px}ΔH°=+24.7\: \ce{kJ} \nonumber$
To get ClF3 as a product, reverse (iv), changing the sign of ΔH°:
$\frac{1}{2}\ce{Cl2O}(g)+\dfrac{3}{2}\ce{OF2}(g)⟶\ce{ClF3}(g)+\ce{O2}(g)\hspace{20px}ΔH°=\mathrm{−266.7\: kJ} \nonumber$
Now check to make sure that these reactions add up to the reaction we want:
\begin {align*} &\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)&&ΔH°=\mathrm{+102.8\: kJ}\ &\frac{1}{2}\ce{O2}(g)+\ce{F2}(g)⟶\ce{OF2}(g)&&ΔH°=\mathrm{+24.7\: kJ}\ &\frac{1}{2}\ce{Cl2O}(g)+\dfrac{3}{2}\ce{OF2}(g)⟶\ce{ClF3}(g)+\ce{O2}(g)&&ΔH°=\mathrm{−266.7\:kJ}\ &\overline{\ce{ClF}(g)+\ce{F2}⟶\ce{ClF3}(g)\hspace{130px}}&&\overline{ΔH°=\mathrm{−139.2\:kJ}} \end {align*} \nonumber
Reactants $\frac{1}{2}\ce{O2}$ and $\frac{1}{2}\ce{O2}$ cancel out product O2; product $\frac{1}{2}\ce{Cl2O}$ cancels reactant $\frac{1}{2}\ce{Cl2O}$; and reactant $\dfrac{3}{2}\ce{OF2}$ is cancelled by products $\frac{1}{2}\ce{OF2}$ and OF2. This leaves only reactants ClF(g) and F2(g) and product ClF3(g), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified ΔH° values will give the desired ΔH°:
$ΔH°=\mathrm{(+102.8\:kJ)+(24.7\:kJ)+(−266.7\:kJ)=−139.2\:kJ}$
Exercise $7$
Aluminum chloride can be formed from its elements:
(i) $\ce{2Al}(s)+\ce{3Cl2}(g)⟶\ce{2AlCl3}(s)\hspace{20px}ΔH°=\:?$
Use the reactions here to determine the ΔH° for reaction (i):
(ii) $\ce{HCl}(g)⟶\ce{HCl}(aq)\hspace{20px}ΔH^\circ_{(ii)}=\mathrm{−74.8\:kJ}$
(iii) $\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH^\circ_{(iii)}=\mathrm{−185\:kJ}$
(iv) $\ce{AlCl3}(aq)⟶\ce{AlCl3}(s)\hspace{20px}ΔH^\circ_{(iv)}=\mathrm{+323\:kJ/mol}$
(v) $\ce{2Al}(s)+\ce{6HCl}(aq)⟶\ce{2AlCl3}(aq)+\ce{3H2}(g)\hspace{20px}ΔH^\circ_{(v)}=\mathrm{−1049\:kJ}$
Answer
−1407 kJ
We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with $\sum$ representing “the sum of” and n standing for the stoichiometric coefficients:
$ΔH^\circ_\ce{reaction}=\sum n×ΔH^\circ_\ce{f}\ce{(products)}−\sum n×ΔH^\circ_\ce{f}\ce{(reactants)} \label{5.4.20B}$
The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.
Example $8$: Using Hess’s Law
What is the standard enthalpy change for the reaction:
$\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g)\hspace{20px}ΔH°=\:? \nonumber$
Solution 1: Using the Equation
Alternatively, we could use the special form of Hess’s law given previously:
$ΔH^\circ_\ce{reaction}=∑n×ΔH^\circ_\ce{f}\ce{(products)}−∑n×ΔH^\circ_\ce{f}\ce{(reactants)} \nonumber$
\begin {align*} &=\mathrm{\left[2\:\cancel{mol\:HNO_3}×\dfrac{−207.4\:kJ}{\cancel{mol\:HNO_3\:(\mathit{aq})}}+1\:\cancel{mol\: NO\:(\mathit{g})}×\dfrac{+90.2\: kJ}{\cancel{mol\: NO\:(\mathit{g})}}\right]}\ &\mathrm{\:−\,\left[3\:\cancel{mol\:NO_2(\mathit{g})}×\dfrac{+33.2\: kJ}{\cancel{mol\:NO_2\:(\mathit{g})}}+1\:\cancel{mol\:H_2O\:(\mathit{l})}×\dfrac{−285.8\:kJ}{\cancel{mol\:H_2O\:(\mathit{l})}}\right]}\ &=\mathrm{2(−207.4\:kJ)+1(+90.2\: kJ)−3(+33.2\: kJ)−1(−285.8\:kJ)}\ &=\mathrm{−138.4\:kJ}\end {align*} \nonumber
Solution 2: Supporting Why the General Equation Is Valid
We can write this reaction as the sum of the decompositions of 3NO2(g) and 1H2O(l) into their constituent elements, and the formation of 2 HNO3(aq) and 1 NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the $ΔH^\circ_\ce{f}$ values for these compounds (from Tables T1 and T2 ), we have:
$\ce{3NO2}(g)⟶ \dfrac{3}{2} \ce{N2}(g)+ 3 \ce{O2}(g)\hspace{20px}ΔH^\circ_{1}=\mathrm{−99.6\:kJ} \nonumber$
$\ce{H2O}(l)⟶\ce{H2}(g)+\frac{1}{2}\ce{O2}(g)\hspace{20px}ΔH^\circ_{2}=+285.8\: \ce{kJ}\:[−1×ΔH^\circ_\ce{f}(\ce{H2O})] \nonumber$
$\ce{H2}(g)+\ce{N2}(g)+ 3 \ce{O2}(g)⟶\ce{2HNO3}(aq)\hspace{20px}ΔH^\circ_{3}=−414.8\:kJ\:[2×ΔH^\circ_\ce{f}(\ce{HNO3 \nonumber})] \nonumber$
$\frac{1}{2}\ce{N2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{NO}(g)\hspace{20px}ΔH^\circ \nonumber_{4}=+90.2\: \ce{kJ}\:[1×(\ce{NO})] \nonumber$
Summing these reaction equations gives the reaction we are interested in:
$\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g) \nonumber$
Summing their enthalpy changes gives the value we want to determine:
\begin {align*} ΔH^\circ_\ce{rxn}&=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3+ΔH^\circ_4=\mathrm{(−99.6\:kJ)+(+285.8\: kJ)+(−414.8\:kJ)+(+90.2\: kJ)}\ &=\mathrm{−138.4\:kJ} \end {align*} \nonumber
So the standard enthalpy change for this reaction is ΔH° = −138.4 kJ.
Note that this result was obtained by:
1. multiplying the $ΔH^\circ_\ce{f}$ of each product by its stoichiometric coefficient and summing those values,
2. multiplying the $ΔH^\circ_\ce{f}$ of each reactant by its stoichiometric coefficient and summing those values, and then
3. subtracting the result found in step 2 from the result found in step 1.
This is also the procedure in using the general equation, as shown.
Exercise $8$
Calculate the heat of combustion of 1 mole of ethanol, C2H5OH(l), when H2O(l) and CO2(g) are formed. Use the following enthalpies of formation: C2H5OH(l), −278 kJ/mol; H2O(l), −286 kJ/mol; and CO2(g), −394 kJ/mol.
Answer
−1368 kJ/mol
Summary
If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, q for the change is called the enthalpy change with the symbol ΔH, or $ΔH^\circ_{298}$ for reactions occurring under standard state conditions. The value of ΔH for a reaction in one direction is equal in magnitude, but opposite in sign, to ΔH for the reaction in the opposite direction, and ΔH is directly proportional to the quantity of reactants and products. Examples of enthalpy changes include enthalpy of combustion, enthalpy of fusion, enthalpy of vaporization, and standard enthalpy of formation. The standard enthalpy of formation, $ΔH^\circ_\ce{f}$, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar (standard state). Many of the processes are carried out at 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.
Key Equations
• $ΔU=q+w$
• $ΔH^\circ_\ce{reaction}=∑n×ΔH^\circ_\ce{f}\ce{(products)}−∑n×ΔH^\circ_\ce{f}\ce{(reactants)}$
Footnotes
1. 1 For more on algal fuel, see www.theguardian.com/environme...n-fuel-problem.
Glossary
chemical thermodynamics
area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes
enthalpy (H)
sum of a system’s internal energy and the mathematical product of its pressure and volume
enthalpy change (ΔH)
heat released or absorbed by a system under constant pressure during a chemical or physical process
expansion work (pressure-volume work)
work done as a system expands or contracts against external pressure
first law of thermodynamics
internal energy of a system changes due to heat flow in or out of the system or work done on or by the system
Hess’s law
if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps
hydrocarbon
compound composed only of hydrogen and carbon; the major component of fossil fuels
internal energy (U)
total of all possible kinds of energy present in a substance or substances
standard enthalpy of combustion ($ΔH^\circ_\ce{c}$)
heat released when one mole of a compound undergoes complete combustion under standard conditions
standard enthalpy of formation ($ΔH^\circ_\ce{f}$)
enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions
standard state
set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations, usually at a temperature of 298.15 K
state function
property depending only on the state of a system, and not the path taken to reach that state | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/09%3A_Thermochemistry/9.3%3A_Enthalpy.txt |
5.1: Energy Basics
Q5.1.1
A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not?
S5.1.1
The temperature of 1 gram of burning wood is approximately the same for both a match and a bonfire. This is an intensive property and depends on the material (wood). However, the overall amount of produced heat depends on the amount of material; this is an extensive property. The amount of wood in a bonfire is much greater than that in a match; the total amount of produced heat is also much greater, which is why we can sit around a bonfire to stay warm, but a match would not provide enough heat to keep us from getting cold.
Q5.1.2
Prepare a table identifying several energy transitions that take place during the typical operation of an automobile.
Q5.1.3
Explain the difference between heat capacity and specific heat of a substance.
S5.1.3
Heat capacity refers to the heat required to raise the temperature of the mass of the substance 1 degree; specific heat refers to the heat required to raise the temperature of 1 gram of the substance 1 degree. Thus, heat capacity is an extensive property, and specific heat is an intensive one.
Q5.1.4
Calculate the heat capacity, in joules and in calories per degree, of the following:
1. 28.4 g of water
2. 1.00 oz of lead
Q5.1.5
Calculate the heat capacity, in joules and in calories per degree, of the following:
1. 45.8 g of nitrogen gas
2. 1.00 pound of aluminum metal
S5.1.5
1. 47.6 J/°C; 11.38 cal °C−1;
2. 407 J/°C; 97.3 cal °C−1
Q5.1.6
How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?
S5.1.6
$q=mCΔ°T$
$q=(75.0g)\times(\dfrac{0.449\:J}{g\:°C})\times(1,510°K) = 50,800J$
50,800J ; 12,200cal
Q5.1.7
How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C?
1310 J; 313 cal
Q5.1.8
How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?
ΔT° = 31.7° C
Q5.1.9
If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?
7.15 °C
Q5.1.10
A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C.
1. What is the specific heat of the substance?
2. If it is one of the substances found in Table, what is its likely identity?
S5.1.10
a.) Solve for the specific heat $C$ and compare the values with the chart
$q=mCΔ°T$
$2110J=(44.7\:g)(C)(66.4°C)$
$C=\dfrac{2110\:J}{2970\:g\:°C}$
$C=\dfrac{0.711\:J}{g\:°C}$
b.) Silicon
Q5.1.11
A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.
1. What is the specific heat of the substance?
2. If it is one of the substances found in Table, what is its likely identity?
S5.1.11
1. 0.390 J/g °C;
2. Copper is a likely candidate.
Q5.1.12
An aluminum kettle weighs 1.05 kg.
1. What is the heat capacity of the kettle?
2. How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C?
3. How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)?
Q5.1.13
Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 × 106 J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)?
S5.1.13
We assume that the density of water is 1.0 g/cm3(1 g/mL) and that it takes as much energy to keep the water at 85 °F as to heat it from 72 °F to 85 °F. We also assume that only the water is going to be heated. Energy required = 7.47 kWh
5.2: Calorimetry
Q5.2.1
A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers.
Q5.2.2
Would the amount of heat measured for the reaction in Example be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.
S5.2.2
lesser; more heat would be lost to the coffee cup and the environment and so ΔT for the water would be lesser and the calculated q would be lesser
Q5.2.3
Would the amount of heat absorbed by the dissolution in Example appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.
Q5.2.4
Would the amount of heat absorbed by the dissolution in Example appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer.
S5.2.4
greater, since taking the calorimeter’s heat capacity into account will compensate for the thermal energy transferred to the solution from the calorimeter; this approach includes the calorimeter itself, along with the solution, as “surroundings”: qrxn = −(qsolution + qcalorimeter); since both qsolution and qcalorimeter are negative, including the latter term (qrxn) will yield a greater value for the heat of the dissolution
Q5.2.5
How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.
Q5.2.6
How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water.
S5.2.6
The temperature of the coffee will drop 1 degree.
Q5.2.7
A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal.
1. What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.
2. The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer.
Q5.2.8
The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C.
S5.2.8
$5.7 \times 10^2\; kJ$
Q5.2.8
A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal?
Q5.2.9
If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure, what is the resulting temperature of the water?
38.5 °C
Q5.2.10
A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter (Figure). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic?
Q5.2.11
Dissolving 3.0 g of CaCl2(s) in 150.0 g of water in a calorimeter (Figure) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic?
S5.2.11
2.2 kJ; The heat produced shows that the reaction is exothermic.
Q5.2.12
When 50.0 g of 0.200 M NaCl(aq) at 24.1 °C is added to 100.0 g of 0.100 M AgNO3(aq) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl(s) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced.
Q5.2.13
The addition of 3.15 g of Ba(OH)2•8H2O to a solution of 1.52 g of NH4SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:
$Ba(OH)_2 \cdot 8H_2O_{(s)} + 2NH_4SCN_{(aq)} \rightarrow Ba(SCN)_{2(aq)} + 2NH_{3(aq)} + 10H_2O_{(l)}$
1.4 kJ
Q5.2.14
The reaction of 50 mL of acid and 50 mL of base described in Example increased the temperature of the solution by 6.9 degrees. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.
Q5.2.15
If the 3.21 g of NH4NO3 in Example were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer.
S5.2.15
22.6. Since the mass and the heat capacity of the solution is approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease of the temperature change.
Q5.2.16
When 1.0 g of fructose, C6H12O6(s), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion?
Q5.2.17
When a 0.740-g sample of trinitrotoluene (TNT), C7H5N2O6, is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?
11.7 kJ
Q5.2.18
One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 × 103 pounds).
Q5.2.19
The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g?
30%
Q5.2.20
A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g?
Q5.2.21
What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g?
0.24 g
Q5.2.22
A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 × 103 Calories if the average number of Calories for fat is 9.1 Calories/g?
Q5.2.23
A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g?
S5.2.23
1.4 × 102 Calories
Q5.2.24
Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs$0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories.
5.3: Enthalpy
Q5.3.1
Explain how the heat measured in [link] differs from the enthalpy change for the exothermic reaction described by the following equation:
$\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)$
The enthalpy change of the indicated reaction is for exactly 1 mol HCL and 1 mol NaOH; the heat in the example is produced by 0.0500 mol HCl and 0.0500 mol NaOH. Using the data in the check your learning section of [link], calculate ΔH in kJ/mol of AgNO3(aq) for the reaction:
$\ce{NaCl}(aq)+\ce{AgNO3}(aq)⟶\ce{AgCl}(s)+\ce{NaNO3}(aq)$
Q5.3.2
Calculate the enthalpy of solution (ΔH for the dissolution) per mole of NH4NO3 under the conditions described in [link].
25 kJ mol−1
Q5.3.3
Calculate ΔH for the reaction described by the equation.
$\ce{Ba(OH)2⋅8H2O}(s)+\ce{2NH4SCN}(aq)⟶\ce{Ba(SCN)2}(aq)+\ce{2NH3}(aq)+\ce{10H2O}(l)$
Q5.3.4
Calculate the enthalpy of solution (ΔH for the dissolution) per mole of CaCl2.
81 kJ mol−1
Q5.3.5
Although the gas used in an oxyacetylene torch is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table. Considering the conditions for which the tabulated data are reported, suggest an explanation.
Q5.3.6
How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?
5204.4 kJ
Q5.3.7
How much heat is produced by combustion of 125 g of methanol under standard state conditions?
Q5.3.8
How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?
1.83 × 10−2 mol
Q5.3.9
What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions?
Q5.3.10
When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?
802 kJ mol−1
Q5.3.11
How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?
$\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\hspace{20px}ΔH^\circ_{298}=\mathrm{−58\:kJ}$
If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation?
Q5.3.12
A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents?
15.5 kJ/ºC
Q5.3.13
Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO2 must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl2F2 (enthalpy of vaporization is 17.4 kJ/mol)?
The vaporization reactions for SO2 and CCl2F2 are
$\ce{SO2}(l)⟶\ce{SO2}(g)$ and $\ce{CCl2F}(l)⟶\ce{CCl2F2}(g)$, respectively.
Q5.3.14
Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C.
7.43 g
Q5.3.15
Which of the enthalpies of combustion in Table the table are also standard enthalpies of formation?
Q5.3.16
Does the standard enthalpy of formation of H2O(g) differ from ΔH° for the reaction $\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g)$?
No.
Q5.3.17
Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO(s) to Hg(l) and O2(g) under standard conditions?
Q5.3.18
How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4(s) at standard state conditions?
459.6 kJ
Q5.3.19
How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe2O3(s) at standard state conditions?
Q5.3.20
The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
$\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−907\:kJ}$
$\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\hspace{20px}ΔH=\mathrm{−113\:kJ}$
$\ce{3NO2}+\ce{H2O}(l)⟶\ce{2HNO2}(aq)+\ce{NO}(g)\hspace{20px}ΔH=\mathrm{−139\:kJ}$
Determine the total energy change for the production of one mole of aqueous nitric acid by this process.
495 kJ/mol
Q5.3.21
Both graphite and diamond burn.
$\ce{C}(s,\:\ce{diamond})+\ce{O2}(g)⟶\ce{CO2}(g)$
For the conversion of graphite to diamond:
$\ce{C}(s,\:\ce{graphite})⟶\ce{C}(s,\:\ce{diamond})\hspace{20px}ΔH^\circ_{298}=\mathrm{1.90\:kJ}$
Which produces more heat, the combustion of graphite or the combustion of diamond?
Q5.3.22
From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: $\ce{H2O}(l)⟶\ce{H2O}(g)$
44.01 kJ/mol
Q5.3.23
Which produces more heat?
$\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(s)$
or
$\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(g)$
for the phase change $\ce{OsO4}(s)⟶\ce{OsO4}(g)\hspace{20px}ΔH=\mathrm{56.4\:kJ}$
Q5.3.24
Calculate $ΔH^\circ_{298}$ for the process
$\ce{Sb}(s)+\dfrac{5}{2}\ce{Cl2}(g)⟶\ce{SbCl5}(g)$
from the following information:
$\ce{Sb}(s)+\dfrac{3}{2}\ce{Cl2}(g)⟶\ce{SbCl3}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−314\:kJ}$
$\ce{SbCl3}(s)+\ce{Cl2}(g)⟶\ce{SbCl5}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−80\:kJ}$
394 kJ
Q5.3.25
Calculate $ΔH^\circ_{298}$ for the process $\ce{Zn}(s)+\ce{S}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)$
from the following information:
$\ce{Zn}(s)+\ce{S}(s)⟶\ce{ZnS}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−206.0\:kJ}$
$\ce{ZnS}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−776.8\:kJ}$
Q5.3.26
Calculate ΔH for the process
$\ce{Hg2Cl2}(s)⟶\ce{2Hg}(l)+\ce{Cl2}(g)$
from the following information:
$\ce{Hg}(l)+\ce{Cl2}(g)⟶\ce{HgCl2}(s)\hspace{20px}ΔH=\mathrm{−224\:kJ}$
$\ce{Hg}(l)+\ce{HgCl2}(s)⟶\ce{Hg2Cl2}(s)\hspace{20px}ΔH=\mathrm{−41.2\:kJ}$
265 kJ
Q5.3.27
Calculate $ΔH^\circ_{298}$ for the process
$\ce{Co3O4}(s)⟶\ce{3Co}(s)+\ce{2O2}(g)$
from the following information:
$\ce{Co}(s)+\dfrac{1}{2}\ce{O2}(g)⟶\ce{CoO}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−237.9\:kJ}$
$\ce{3Co}(s)+\ce{O2}(g)⟶\ce{Co3O4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−177.5\:kJ}$
Q5.3.28
Calculate the standard molar enthalpy of formation of NO(g) from the following data:
$\ce{N2}(g)+\ce{2O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{66.4\:kJ}$
$\ce{2NO}(g)+\ce{O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−114.1\:kJ}$
90.3 mol−1 of NO
Q5.3.29
Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:
1. $\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g)$
2. $\ce{Si}(s)+\ce{2Cl2}(g)⟶\ce{SiCl4}(g)$
3. $\ce{Fe2O3}(s)+\ce{3H2}(g)⟶\ce{2Fe}(s)+\ce{3H2O}(l)$
4. $\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g)$
Q5.3.30
Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:
1. $\ce{Si}(s)+\ce{2F2}(g)⟶\ce{SiF4}(g)$
2. $\ce{2C}(s)+\ce{2H2}(g)+\ce{O2}(g)⟶\ce{CH3CO2H}(l)$
3. $\ce{CH4}(g)+\ce{N2}(g)⟶\ce{HCN}(g)+\ce{NH3}(g)$;
4. $\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g)$
S5.3.30
1. −1615.0 kJ mol−1;
2. −484.3 kJ mol−1;
3. 164.2 kJ;
4. −232.1 kJ
Q5.3.31
The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each.
1. $\ce{2Ag2O}(s)⟶\ce{4Ag}(s)+\ce{O2}(g)$
2. $\ce{SnO}(s)+\ce{CO}(g)⟶\ce{Sn}(s)+\ce{CO2}(g)$
3. $\ce{Cr2O3}(s)+\ce{3H2}(g)⟶\ce{2Cr}(s)+\ce{3H2O}(l)$
4. $\ce{2Al}(s)+\ce{Fe2O3}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)$
Q5.3.32
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions.
$\ce{2H2O2}(l)⟶\ce{2H2O}(g)+\ce{O2}(g)$
−54.04 kJ mol−1
Q5.3.33
Calculate the enthalpy of combustion of propane, C3H8(g), for the formation of H2O(g) and CO2(g). The enthalpy of formation of propane is −104 kJ/mol.
Q5.3.34
Calculate the enthalpy of combustion of butane, C4H10(g) for the formation of H2O(g) and CO2(g). The enthalpy of formation of butane is −126 kJ/mol.
2660 kJ mol−1
Q5.3.35
Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned?
Q5.3.36
The white pigment TiO2 is prepared by the reaction of titanium tetrachloride, TiCl4, with water vapor in the gas phase:
$\ce{TiCl4}(g)+\ce{2H2O}(g)⟶\ce{TiO2}(s)+\ce{4HCl}(g)$.
How much heat is evolved in the production of exactly 1 mole of TiO2(s) under standard state conditions?
67.1 kJ
Q5.3.37
Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon:
$\ce{C}(s)+\ce{H2O}(g)⟶\ce{CO}(g)+\ce{H2}(g)$.
1. Assuming that coke has the same enthalpy of formation as graphite, calculate $ΔH^\circ_{298}$ for this reaction.
2. Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst: $\ce{2H2}(g)+\ce{CO}(g)⟶\ce{CH3OH}(g).$ Under the conditions of the reaction, methanol forms as a gas. Calculate $ΔH^\circ_{298}$ for this reaction and for the condensation of gaseous methanol to liquid methanol.
3. Calculate the heat of combustion of 1 mole of liquid methanol to H2O(g) and CO2(g).
Q5.3.38
In the early days of automobiles, illumination at night was provided by burning acetylene, C2H2. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC2:
$\ce{CaC2}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)+\ce{C2H2}(g)$.
Calculate the standard enthalpy of the reaction. The $ΔH^\circ_\ce{f}$ of CaC2 is −15.14 kcal/mol.
122.8 kJ
Q5.3.39
From the data in Table, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO(g), CH4(g), or C2H2(g).
Q5.3.40
The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 × 105 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane).
3.7 kg
Q5.3.41
Ethanol, C2H5OH, is used as a fuel for motor vehicles, particularly in Brazil.
1. Write the balanced equation for the combustion of ethanol to CO2(g) and H2O(g), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol.
2. The density of ethanol is 0.7893 g/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol.
3. Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C8H18 ($ΔH^\circ_\ce{f}=\mathrm{−208.4\:kJ/mol}$; density = 0.7025 g/mL).
Q5.3.42
Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B2H6, produces B2O3(s) and H2O(g)], methane [CH4, produces CO2(g) and H2O(g)], and hydrazine [N2H4, produces N2(g) and H2O(g)]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The $ΔH^\circ_\ce{f}$ of B2H6(g), CH4(g), and N2H4(l) may be found in Appendix G.
S5.3.42
On the assumption that the best rocket fuel is the one that gives off the most heat, B2H6 is the prime candidate.
Q5.3.43
How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions?
Ethylene, C2H2, a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst.
$\ce{C2H4}(g)+\ce{H2O}(g)⟶\ce{C2H5OH}(l)$
Using the data in the table in Appendix G, calculate ΔH° for the reaction.
88.2 kJ
Q5.3.44
The oxidation of the sugar glucose, C6H12O6, is described by the following equation:
$\ce{C6H12O6}(s)+\ce{6O2}(g)⟶\ce{6CO2}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−2816\:kJ}$
The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.
1. How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?
2. How many Calories can be produced by the metabolism of 1.0 g of glucose?
Q5.3.45
Propane, C3H8, is a hydrocarbon that is commonly used as a fuel.
1. Write a balanced equation for the complete combustion of propane gas.
2. Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O2 by volume. (Hint: we will see how to do this calculation in a later chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O2 per liter.)
3. The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, $ΔH^\circ_\ce{f}$ of propane given that $ΔH^\circ_\ce{f}$ of H2O(l) = −285.8 kJ/mol and $ΔH^\circ_\ce{f}$ of CO2(g) = −393.5 kJ/mol.
4. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.
S5.3.45
1. $\ce{C3H8}(g)+\ce{5O2}(g)⟶\ce{3CO2}(g)+\ce{4H2O}(l)$;
2. 330 L;
3. −104.5 kJ mol−1;
4. 75.4 °C
Q5.3.46
During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).
1. Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was 56 °F; at this temperature and a pressure of 1 atm, natural gas has a density of 0.681 g/L.
2. How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [C3H8: density, 0.5318 g/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO2(g) and H2O(l)] and the furnace used to burn the LPG has the same efficiency as the gas furnace.
3. What mass of carbon dioxide is produced by combustion of the methane used to heat the house?
4. What mass of water is produced by combustion of the methane used to heat the house?
5. What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the month was 1.22 g/L.
6. How many kilowatt–hours (1 kWh = 3.6 × 106 J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house.
7. Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/09%3A_Thermochemistry/9.E%3A_Thermochemistry_%28Exercises%29.txt |
The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent. In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.
• 10.0: Prelude to Liquids and Solids
In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.
• 10.1: Intermolecular Forces
The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient kinetic energy to move past each other. Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature.
• 10.2: Properties of Liquids
The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension. Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for surface wetting and capillary rise.
• 10.3: Phase Transitions
Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened.
• 10.4: Phase Diagrams
The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phase-transition temperatures and pressures. The intersection of all three curves represents the substance’s triple point at which all three phases coexist.
• 10.5: The Solid State of Matter
Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions bet
• 10.6: Lattice Structures in Crystalline Solids
The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest-packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions.
• 10.E: Liquids and Solids (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
10: Liquids and Solids
The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent.
In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.0%3A_Prelude_to_Liquids_and_Solids.txt |
Learning Objectives
• Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)
• Identify the types of intermolecular forces experienced by specific molecules based on their structures
• Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state
As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase “intermolecular attraction” to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.
Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:
• Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.
• Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.
The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles’ KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure \(1\) illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE, of a given substance.
As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure \(2\).
We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, C4H10, is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter’s fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure \(3\).
Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter.
Forces between Molecules
Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure \(4\) illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy—430 kilojoules.
All of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module.
Dispersion Forces
One of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron’s location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species—a so-called dispersion force like that illustrated in Figure \(5\).
Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table \(1\).
Table \(1\): Melting and Boiling Points of the Halogens
Halogen Molar Mass Atomic Radius Melting Point Boiling Point
fluorine, F2 38 g/mol 72 pm 53 K 85 K
chlorine, Cl2 71 g/mol 99 pm 172 K 238 K
bromine, Br2 160 g/mol 114 pm 266 K 332 K
iodine, I2 254 g/mol 133 pm 387 K 457 K
astatine, At2 420 g/mol 150 pm 575 K 610 K
The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.
Example \(1\): London Forces and Their Effects
Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning.
Solution
Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH4 is expected to have the lowest boiling point and SnH4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be
CH4 < SiH4 < GeH4 < SnH4
A graph of the actual boiling points of these compounds versus the period of the group 14 elements shows this prediction to be correct:
Exercise \(1\)
Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10.
Answer
All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore
C2H6 < C3H8 < C4H10.
The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers n-pentane, isopentane, and neopentane (shown in Figure \(6\)) are 36 °C, 27 °C, and 9.5 °C, respectively. Even though these compounds are composed of molecules with the same chemical formula, C5H12, the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for n-pentane and least for neopentane. The elongated shape of n-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip’s contact, the stronger the connection.
Applications: Geckos and Intermolecular Forces
Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way.
Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure \(7\), with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight.
In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications.
Watch this video to learn more about Kellar Autumn’s research that determined that van der Waals forces are responsible for a gecko’s ability to cling and climb.
Dipole-Dipole Attractions
Recall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction—the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure \(8\).
The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F2 molecules. Both HCl and F2 consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F2 molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F2 (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F2 molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances.
Example \(2\): Dipole-Dipole Forces and Their Effects
Predict which will have the higher boiling point: N2 or CO. Explain your reasoning.
Solution
CO and N2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N2 molecules, so CO is expected to have the higher boiling point.
A common method for preparing oxygen is the decomposition
Exercise \(2\)
Predict which will have the higher boiling point: \(\ce{ICl}\) or \(\ce{Br2}\). Explain your reasoning.
Answer
ICl. ICl and Br2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.
Hydrogen Bonding
Nitrosyl fluoride (ONF, molecular mass 49 amu) is a gas at room temperature. Water (H2O, molecular mass 18 amu) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and ONF is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to highly concentrated partial charges with these atoms. Molecules with F-H, O-H, or N-H moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include HF⋯HF, H2O⋯HOH, and H3N⋯HNH2, in which the hydrogen bonds are denoted by dots. Figure \(9\) illustrates hydrogen bonding between water molecules.
Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to 10% as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces.
Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group 15 (NH3, PH3, AsH3, and SbH3), group 16 hydrides (H2O, H2S, H2Se, and H2Te), and group 17 hydrides (HF, HCl, HBr, and HI). The boiling points of the heaviest three hydrides for each group are plotted in Figure \(10\). As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily. For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3, 4, and 5.
If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect NH3 to boil at about −120 °C, H2O to boil at about −80 °C, and HF to boil at about −110 °C. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure \(10\). The stark contrast between our naïve predictions and reality provides compelling evidence for the strength of hydrogen bonding.
Example \(3\): Effect of Hydrogen Bonding on Boiling Points
Consider the compounds dimethylether (CH3OCH3), ethanol (CH3CH2OH), and propane (CH3CH2CH3). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning.
Solution
The VSEPR-predicted shapes of CH3OCH3, CH3CH2OH, and CH3CH2CH3 are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH3CH2CH3 is nonpolar, it may exhibit only dispersion forces. Because CH3OCH3 is polar, it will also experience dipole-dipole attractions. Finally, CH3CH2OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH3CH2CH3 < CH3OCH3 < CH3CH2OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C.
Exercise \(3\)
Ethane (CH3CH3) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH3NH2). Explain your reasoning.
Answer
The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH3CH3 and CH3NH2 are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C.
Hydrogen Bonding and DNA
Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure \(10\).
Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure \(12\)
The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication.
Summary
The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient kinetic energy to move past each other. Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one dipolar molecule for the partial positive end of another. The temporary dipole that results from the motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N.
Glossary
dipole-dipole attraction
intermolecular attraction between two permanent dipoles
dispersion force
(also, London dispersion force) attraction between two rapidly fluctuating, temporary dipoles; significant only when particles are very close together
hydrogen bonding
occurs when exceptionally strong dipoles attract; bonding that exists when hydrogen is bonded to one of the three most electronegative elements: F, O, or N
induced dipole
temporary dipole formed when the electrons of an atom or molecule are distorted by the instantaneous dipole of a neighboring atom or molecule
instantaneous dipole
temporary dipole that occurs for a brief moment in time when the electrons of an atom or molecule are distributed asymmetrically
intermolecular force
noncovalent attractive force between atoms, molecules, and/or ions
polarizability
measure of the ability of a charge to distort a molecule’s charge distribution (electron cloud)
van der Waals force
attractive or repulsive force between molecules, including dipole-dipole, dipole-induced dipole, and London dispersion forces; does not include forces due to covalent or ionic bonding, or the attraction between ions and molecules | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.1%3A_Intermolecular_Forces.txt |
Learning Objectives
• Distinguish between adhesive and cohesive forces
• Define viscosity, surface tension, and capillary rise
• Describe the roles of intermolecular attractive forces in each of these properties/phenomena
When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure $1$, have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly).
The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table $1$ shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases.
Table $1$: Viscosities of Common Substances at 25 °C
Substance Formula Viscosity (mPa·s)
water H2O 0.890
mercury Hg 1.526
ethanol C2H5OH 1.074
octane C8H18 0.508
ethylene glycol CH2(OH)CH2(OH) 16.1
honey variable ~2,000–10,000
motor oil variable ~50–500
The various IMFs between identical molecules of a substance are examples of cohesive forces. The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface—that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure $2$, because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical.
Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table $2$.
Table $2$: Surface Tensions of Common Substances at 25 °C
Substance Formula Surface Tension (mN/m)
water H2O 71.99
mercury Hg 458.48
ethanol C2H5OH 21.97
octane C8H18 21.14
ethylene glycol CH2(OH)CH2(OH) 47.99
Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively “tough skin” that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure $3$, even though they are denser than water, move on its surface because they are supported by the surface tension.
The IMFs of attraction between two different molecules are called adhesive forces. Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not “wet” the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure $4$).
If you place one end of a paper towel in spilled wine, as shown in Figure $5$, the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action—when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity.
Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many −OH groups. Water molecules are attracted to these −OH groups and form hydrogen bonds with them, which draws the H2O molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers.
Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure $6$. If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away.
The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the following equation:
$h=\dfrac{2T\cosθ}{rρg} \label{10.2.1}$
where
• h is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube,
• T is the surface tension of the liquid,
• θ is the contact angle between the liquid and the tube,
• r is the radius of the tube, ρ is the density of the liquid, and
• g is the acceleration due to gravity, 9.8 m/s2.
When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of 0°. This is the situation for water rising in a glass tube.
Example $1$: Capillary Rise
At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm?
For water, T = 71.99 mN/m and ρ = 1.0 g/cm3.
Solution
The liquid will rise to a height h given by Equation $\ref{10.2.1}$ :
$h=\dfrac{2T\cosθ}{rρg} \nonumber$
The Newton is defined as a kg m/s2, and so the provided surface tension is equivalent to 0.07199 kg/s2. The provided density must be converted into units that will cancel appropriately: ρ = 1000 kg/m3. The diameter of the tube in meters is 0.00025 m, so the radius is 0.000125 m. For a glass tube immersed in water, the contact angle is θ = 0°, so cos θ = 1. Finally, acceleration due to gravity on the earth is g = 9.8 m/s2. Substituting these values into the equation, and cancelling units, we have:
$h=\mathrm{\dfrac{2(0.07199\:kg/s^2)}{(0.000125\:m)(1000\:kg/m^3)(9.8\:m/s^2)}=0.12\:m=12\: cm} \nonumber$
Exercise $1$
Water rises in a glass capillary tube to a height of 8.4 cm. What is the diameter of the capillary tube?
Answer
diameter = 0.36 mm
Applications: Capillary Action is Used to Draw Blood
Many medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure $7$. When your finger is pricked, a drop of blood forms and holds together due to surface tension—the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrow-diameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising.
Key Concepts and Summary
The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension (elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise.
Key Equations
• $h=\dfrac{2T\cosθ}{rρg}$
Glossary
adhesive force
force of attraction between molecules of different chemical identities
capillary action
flow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules
cohesive force
force of attraction between identical molecules
surface tension
energy required to increase the area, or length, of a liquid surface by a given amount
viscosity
measure of a liquid’s resistance to flow | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.2%3A_Properties_of_Liquids.txt |
Learning Objectives
• Define phase transitions and phase transition temperatures
• Explain the relation between phase transition temperatures and intermolecular attractive forces
• Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes
We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.
Vaporization and Condensation
When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure $1$, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.
The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.
Example $1$: Explaining Vapor Pressure in Terms of IMFs
Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:
Solution
Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two −OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.
Exercise $1$
At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:
At 20 °C, the vapor pressures of several alcohols
Compound methanol CH3OH ethanol C2H5OH propanol C3H7OH butanol C4H9OH
Vapor Pressure at 20 °C 11.9 kPa 5.95 kPa 2.67 kPa 0.56 kPa
Answer
All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:
$P_{methanol} > P_{ethanol} > P_{propanol} > P_{butanol} \nonumber$
As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure $2$. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.
Boiling Points
When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure $3$ shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure.
Example $2$: A Boiling Point at Reduced Pressure
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure $3$ to determine the boiling point of water at this elevation.
Solution
The graph of the vapor pressure of water versus temperature in Figure $3$ indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.
Exercise $2$
The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure $3$ to determine the approximate atmospheric pressure at the camp.
Answer
Approximately 40 kPa (0.4 atm)
The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation:
$P=Ae^{−ΔH_{vap}/RT} \label{10.4.1}$
where
• $ΔH_{vap}$ is the enthalpy of vaporization for the liquid,
• $R$ is the gas constant, and
• $\ln A$ is a constant whose value depends on the chemical identity of the substance.
Equation $\ref{10.4.1}$ is often rearranged into logarithmic form to yield the linear equation:
$\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A \label{10.4.2}$
This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the examples and exercises that follow. If at temperature $T_1$, the vapor pressure is $P_1$, and at temperature $T_2$, the vapor pressure is $T_2$, the corresponding linear equations are:
$\ln P_1=−\dfrac{ΔH_\ce{vap}}{RT_1}+\ln A \nonumber$
and
$\ln P_2=−\dfrac{ΔH_\ce{vap}}{RT_2}+\ln A \label{10.4.3}$
Since the constant, ln A, is the same, these two equations may be rearranged to isolate $\ln A$ and then set them equal to one another:
$\ln P_1+\dfrac{ΔH_\ce{vap}}{RT_1}=\ln P_2+\dfrac{ΔH_\ce{vap}}{RT_2}\label{10.4.5}$
which can be combined into:
$\ln \left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R} \left( \dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \label{10.4.6}$
Example $3$: Estimating Enthalpy of Vaporization
Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.
Solution
The enthalpy of vaporization, $ΔH_\ce{vap}$, can be determined by using the Clausius-Clapeyron equation (Equation $\ref{10.4.6}$):
$\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber$
Since we have two vapor pressure-temperature values
• $T_1 = 34.0^oC = 307.2\, K$
• $P_1 = 10.0\, kPa$ and
• $T_2 = 98.8 ^oC = 372.0\, K$
• $P_2 = 100\, kPa$
we can substitute them into this equation and solve for $ΔH_{vap}$. Rearranging the Clausius-Clapeyron equation and solving for $ΔH_{vap}$ yields:
\begin{align*} ΔH_\ce{vap} &= \dfrac{R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)} \[4pt] &= \dfrac{(8.3145\:J/mol⋅K)⋅\ln \left(\dfrac{100\: kPa}{10.0\: kPa}\right)}{\left(\dfrac{1}{307.2\:K}−\dfrac{1}{372.0\:K}\right)} \[4pt] &=33,800\, J/mol =33.8\, kJ/mol \end{align*} \nonumber
Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.
Exercise $3$
At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.
Answer
47,782 J/mol = 47.8 kJ/mol
Example $4$: Estimating Temperature (or Vapor Pressure)
For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?
Solution
If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, ΔHvap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation (Equation $\ref{10.4.1}$) :
$\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber$
Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value ($T_1$ = 80.1 °C = 353.3 K, $P_1$ = 101.3 kPa, $ΔH_{vap}$ = 30.8 kJ/mol) and want to find the temperature ($T_2$) that corresponds to vapor pressure P2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for $T_2$. Rearranging the Clausius-Clapeyron equation and solving for $T_2$ yields:
\begin{align*} T_2 &=\left(\dfrac{−R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{ΔH_\ce{vap}}+\dfrac{1}{T_1}\right)^{−1} \[4pt] &=\mathrm{\left(\dfrac{−(8.3145\:J/mol⋅K)⋅\ln\left(\dfrac{83.4\:kPa}{101.3\:kPa}\right)}{30,800\: J/mol}+\dfrac{1}{353.3\:K}\right)^{−1}}\[4pt] &=\mathrm{346.9\: K\: or\:73.8^\circ C} \end{align*} \nonumber
Exercise $4$
For acetone $\ce{(CH3)2CO}$, the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C?
Answer
30.1 kPa
Enthalpy of Vaporization
Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, $ΔH_{vap}$. For example, the vaporization of water at standard temperature is represented by:
$\ce{H2O}(l)⟶\ce{H2O}(g)\hspace{20px}ΔH_\ce{vap}=\mathrm{44.01\: kJ/mol} \nonumber$
As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:
$\ce{H2O}(g)⟶\ce{H2O}(l)\hspace{20px}ΔH_\ce{con}=−ΔH_\ce{vap}=\mathrm{−44.01\:kJ/mol} \nonumber$
Example $5$: Using Enthalpy of Vaporization
One way our body is cooled is by evaporation of the water in sweat (Figure $4$). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); $ΔH_{vap} = 43.46\, kJ/mol$ at 37 °C.
Solution We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:
$\mathrm{1.5\cancel{L}×\dfrac{1000\cancel{g}}{1\cancel{L}}×\dfrac{1\cancel{mol}}{18\cancel{g}}×\dfrac{43.46\:kJ}{1\cancel{mol}}=3.6×10^3\:kJ} \nonumber$
Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.
Exercise $5$: Boiling Ammonia
How much heat is required to evaporate 100.0 g of liquid ammonia, $\ce{NH3}$, at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol?
Answer
28 kJ
Melting and Freezing
When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure $5$.
If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).
The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.
The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process:
$\ce{H2O}_{(s)} \rightarrow \ce{H2O}_{(l)} \;\; ΔH_\ce{fus}=\mathrm{6.01\; kJ/mol} \label{10.4.9}$
The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C:
$\ce{H_2O}_{(l)} \rightarrow \ce{H_2O}_{(s)}\;\; ΔH_\ce{frz}=−ΔH_\ce{fus}=−6.01\;\mathrm{kJ/mol} \label{10.4.10}$
Sublimation and Deposition
Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid CO2) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure $6$). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.
Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:
$\ce{CO2}(s)⟶\ce{CO2}(g)\hspace{20px}ΔH_\ce{sub}=\mathrm{26.1\: kJ/mol} \nonumber$
Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:
$\ce{CO2}(g)⟶\ce{CO2}(s)\hspace{20px}ΔH_\ce{dep}=−ΔH_\ce{sub}=\mathrm{−26.1\:kJ/mol} \nonumber$
Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law.
$\mathrm{solid⟶liquid}\hspace{20px}ΔH_\ce{fus}\\underline{\mathrm{liquid⟶gas}\hspace{20px}ΔH_\ce{vap}}\\mathrm{solid⟶gas}\hspace{20px}ΔH_\ce{sub}=ΔH_\ce{fus}+ΔH_\ce{vap} \nonumber$
Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure $7$. For example:
Heating and Cooling Curves
In the chapter on thermochemistry, the relation between the amount of heat absorbed or related by a substance, q, and its accompanying temperature change, ΔT, was introduced:
$q=mcΔT \nonumber$
where m is the mass of the substance and c is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure $8$ shows a typical heating curve.
Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.
Example $6$: Total Heat Needed to Change Temperature and Phase for a Substance
How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C?
Solution
The transition described involves the following steps:
1. Heat ice from −15 °C to 0 °C
2. Melt ice
3. Heat water from 0 °C to 100 °C
4. Boil water
5. Heat steam from 100 °C to 120 °C
The heat needed to change the temperature of a given substance (with no change in phase) is: q = m × c × ΔT (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by q = n × ΔH.
Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:
\begin{align*} q_\ce{total}&=(m⋅c⋅ΔT)_\ce{ice}+n⋅ΔH_\ce{fus}+(m⋅c⋅ΔT)_\ce{water}+n⋅ΔH_\ce{vap}+(m⋅c⋅ΔT)_\ce{steam}\[7pt] &=\mathrm{(135\: g⋅2.09\: J/g⋅°C⋅15°C)+\left(135⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \right)}\[7pt] &\mathrm{+(135\: g⋅4.18\: J/g⋅°C⋅100°C)+\left(135\: g⋅\dfrac{1\: mol}{18.02\:g}⋅40.67\: kJ/mol\right)}\[7pt] &\mathrm{+(135\: g⋅1.84\: J/g⋅°C⋅20°C)}\[7pt] &=\mathrm{4230\: J+45.0\: kJ+56,500\: J+305\: kJ+4970\: J} \end{align*} \nonumber
Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:
$\mathrm{=4.23\:kJ+45.0\: kJ+56.5\: kJ+305\: kJ+4.97\: kJ=416\: kJ} \nonumber$
Exercise $6$
What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?
Answer
40.5 kJ
Summary
Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.
Key Equations
• $P=Ae^{−ΔH_\ce{vap}/RT}$
• $\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A$
• $\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)$
Glossary
boiling point
temperature at which the vapor pressure of a liquid equals the pressure of the gas above it
Clausius-Clapeyron equation
mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance
condensation
change from a gaseous to a liquid state
deposition
change from a gaseous state directly to a solid state
dynamic equilibrium
state of a system in which reciprocal processes are occurring at equal rates
freezing
change from a liquid state to a solid state
freezing point
temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point
melting
change from a solid state to a liquid state
melting point
temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point
normal boiling point
temperature at which a liquid’s vapor pressure equals 1 atm (760 torr)
sublimation
change from solid state directly to gaseous state
vapor pressure
(also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature
vaporization
change from liquid state to gaseous state | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.3%3A_Phase_Transitions.txt |
Learning Objectives
• Explain the construction and use of a typical phase diagram
• Use phase diagrams to identify stable phases at given temperatures and pressures, and to describe phase transitions resulting from changes in these properties
• Describe the supercritical fluid phase of matter
In the previous module, the variation of a liquid’s equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance’s melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A phase diagram combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in Figure \(1\).
To illustrate the utility of these plots, consider the phase diagram for water shown in Figure \(2\).
We can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of −10 °C correspond to the region of the diagram labeled “ice.” Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state. Note that on the H2O phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here.
The curve BC in Figure \(2\) is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This “liquid-vapor” curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm, the boiling point is 100 °C. Notice that the liquid-vapor curve terminates at a temperature of 374 °C and a pressure of 218 atm, indicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module.
The solid-vapor curve, labeled AB in Figure \(2\), indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in Figure \(2\), we would see that ice has a vapor pressure of about 0.20 kPa at −10 °C. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa, ice will sublime. This is the basis for the “freeze-drying” process often used to preserve foods, such as the ice cream shown in Figure \(3\).
The solid-liquid curve labeled BD shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in Figure \(4\). The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide.
The point of intersection of all three curves is labeled B in Figure \(2\). At the pressure and temperature represented by this point, all three phases of water coexist in equilibrium. This temperature-pressure data pair is called the triple point. At pressures lower than the triple point, water cannot exist as a liquid, regardless of the temperature.
Example \(1\): Determining the State of Water
Using the phase diagram for water given in Figure 10.4.2, determine the state of water at the following temperatures and pressures:
1. −10 °C and 50 kPa
2. 25 °C and 90 kPa
3. 50 °C and 40 kPa
4. 80 °C and 5 kPa
5. −10 °C and 0.3 kPa
6. 50 °C and 0.3 kPa
Solution
Using the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f) gas.
Exercise \(1\)
What phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa? If the pressure is held at 50 kPa?
Answer
At 0.3 kPa: s⟶ g at −58 °C. At 50 kPa: s⟶ l at 0 °C, l ⟶ g at 78 °C
Consider the phase diagram for carbon dioxide shown in Figure \(5\) as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for CO2 increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions. Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous CO2. Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water.
Example \(2\): Determining the State of Carbon Dioxide
Using the phase diagram for carbon dioxide shown in Figure 10.4.5, determine the state of CO2 at the following temperatures and pressures:
1. −30 °C and 2000 kPa
2. −60 °C and 1000 kPa
3. −60 °C and 100 kPa
4. 20 °C and 1500 kPa
5. 0 °C and 100 kPa
6. 20 °C and 100 kPa
Solution
Using the phase diagram for carbon dioxide provided, we can determine that the state of CO2 at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f) gas.
Exercise \(2\)
Determine the phase changes carbon dioxide undergoes when its temperature is varied, thus holding its pressure constant at 1500 kPa? At 500 kPa? At what approximate temperatures do these phase changes occur?
Answer
at 1500 kPa: s⟶ l at −45 °C, l⟶ g at −10 °C; at 500 kPa: s⟶ g at −58 °C
Supercritical Fluids
If we place a sample of water in a sealed container at 25 °C, remove the air, and let the vaporization-condensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm. A distinct boundary between the more dense liquid and the less dense gas is clearly observed. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (Figure \(2\)), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of 374 °C, the vapor pressure has risen to 218 atm, and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a supercritical fluid, and the temperature and pressure above which this phase exists is the critical point (Figure \(5\)). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in Table \(1\).
Table \(1\): Critical Temperatures and Critical Pressures of select substances
Substance Critical Temperature (K) Critical Pressure (atm)
hydrogen 33.2 12.8
nitrogen 126.0 33.5
oxygen 154.3 49.7
carbon dioxide 304.2 73.0
ammonia 405.5 111.5
sulfur dioxide 430.3 77.7
water 647.1 217.7
Like a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus oils. It is nontoxic, relatively inexpensive, and not considered to be a pollutant. After use, the CO2 can be easily recovered by reducing the pressure and collecting the resulting gas.
Example \(3\): The Critical Temperature of Carbon Dioxide
If we shake a carbon dioxide fire extinguisher on a cool day (18 °C), we can hear liquid CO2 sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day (35 °C). Explain these observations.
Solution
On the cool day, the temperature of the CO2 is below the critical temperature of CO2, 304 K or 31 °C (Table \(1\)), so liquid CO2 is present in the cylinder. On the hot day, the temperature of the CO2 is greater than its critical temperature of 31 °C. Above this temperature no amount of pressure can liquefy CO2 so no liquid CO2 exists in the fire extinguisher.
Exercise \(3\)
Ammonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior?
Answer
The critical temperature of ammonia is 405.5 K, which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature.
Decaffeinating Coffee Using Supercritical CO2
Coffee is the world’s second most widely traded commodity, following only petroleum. Across the globe, people love coffee’s aroma and taste. Many of us also depend on one component of coffee—caffeine—to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee’s stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening.
Since the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee’s taste and aroma also dissolve in H2O, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and taste of the decaffeinated coffee. Dichloromethane (CH2Cl2) and ethyl acetate (CH3CO2C2H5) have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee.
Supercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (Figure \(7\)). At temperatures above 304.2 K and pressures above 7376 kPa, CO2 is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes 97−99% of the caffeine, leaving coffee’s flavor and aroma compounds intact. Because CO2 is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs.
Summary
The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phase-transition temperatures and pressures. The point of intersection of all three curves represents the substance’s triple point—the temperature and pressure at which all three phases are in equilibrium. At pressures below the triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance’s critical point, the pressure and temperature above which a liquid phase cannot exist.
Glossary
critical point
temperature and pressure above which a gas cannot be condensed into a liquid
phase diagram
pressure-temperature graph summarizing conditions under which the phases of a substance can exist
supercritical fluid
substance at a temperature and pressure higher than its critical point; exhibits properties intermediate between those of gaseous and liquid states
triple point
temperature and pressure at which the vapor, liquid, and solid phases of a substance are in equilibrium | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.4%3A_Phase_Diagrams.txt |
Learning Objectives
• Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids
• Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids
• Explain the ways in which crystal defects can occur in a solid
When most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure \(1\)).
Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as boron oxide (Figure \(2\)), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.
Crystalline solids are generally classified according the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular.
Ionic Solids
Ionic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure \(3\)). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions—in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.
Metallic Solids
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(4\). The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below 200 °C. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above 1000 °C. These differences reflect differences in strengths of metallic bonding among the metals.
Covalent Network Solids
Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure \(5\). To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above 3500 °C.
Molecular Solids
Molecular solids, such as ice, sucrose (table sugar), and iodine, as shown in Figure \(6\), are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as H2, N2, O2, and F2, have weak attractive forces and form molecular solids with very low melting points (below −200 °C). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar molecules) melt at still higher temperatures. Examples include ice (melting point, 0 °C) and table sugar (melting point, 185 °C).
Properties of Solids
A crystalline solid, like those listed in Table \(1\) has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures.
Table \(1\): Types of Crystalline Solids and Their Properties
Type of Solid Type of Particles Type of Attractions Properties Examples
ionic ions ionic bonds hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points NaCl, Al2O3
metallic atoms of electropositive elements metallic bonds shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature Cu, Fe, Ti, Pb, U
covalent network atoms of electronegative elements covalent bonds very hard, not conductive, very high melting points C (diamond), SiO2, SiC
molecular molecules (or atoms) IMFs variable hardness, variable brittleness, not conductive, low melting points H2O, CO2, I2, C12H22O11
Graphene: Material of the Future
Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure \(7\). You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.
You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure \(8\), is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene.
Crystal Defects
In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure \(9\). Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips.
Summary
Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips.
Glossary
amorphous solid
(also, noncrystalline solid) solid in which the particles lack an ordered internal structure
covalent network solid
solid whose particles are held together by covalent bonds
crystalline solid
solid in which the particles are arranged in a definite repeating pattern
interstitial sites
spaces between the regular particle positions in any array of atoms or ions
ionic solid
solid composed of positive and negative ions held together by strong electrostatic attractions
metallic solid
solid composed of metal atoms
molecular solid
solid composed of neutral molecules held together by intermolecular forces of attraction
vacancy
defect that occurs when a position that should contain an atom or ion is vacant | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.5%3A_The_Solid_State_of_Matter.txt |
Learning Objectives
• Describe the arrangement of atoms and ions in crystalline structures
• Compute ionic radii using unit cell dimensions
• Explain the use of X-ray diffraction measurements in determining crystalline structures
Over 90% of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally.
The Structures of Metals
We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow.
Unit Cells of Metals
The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure $1$.
Let us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure $2$. This arrangement is called simple cubic structure, and the unit cell is called the simple cubic unit cell or primitive cubic unit cell.
In a simple cubic structure, the spheres are not packed as closely as they could be, and they only “fill” about 52% of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure $3$, a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.
In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure $4$. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight “corners,” there is $8×\dfrac{1}{8}=1$ atom within one simple cubic unit cell.
Example $1$: Calculating Atomic Radius and Density for Metals (Part 1)
The edge length of the unit cell of alpha polonium is 336 pm.
1. Determine the radius of a polonium atom.
2. Determine the density of alpha polonium.
Solution
Alpha polonium crystallizes in a simple cubic unit cell:
(a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: $l = 2r$. Therefore, the radius of Po is
$r=\mathrm{\dfrac{l}{2}=\dfrac{336\: pm}{2}=168\: pm}\nonumber$
(b) Density is given by
$\mathrm{density=\dfrac{mass}{volume}}.\nonumber$
The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom.
The mass of a Po unit cell can be found by:
$\mathrm{1\: Po\: unit\: cell×\dfrac{1\: Po\: atom}{1\: Po\: unit\: cell}×\dfrac{1\: mol\: Po}{6.022\times 10^{23}\:Po\: atoms}×\dfrac{208.998\:g}{1\: mol\: Po}=3.47\times 10^{−22}\:g}\nonumber$
The volume of a Po unit cell can be found by:
$V=l^3=\mathrm{(336\times 10^{−10}\:cm)^3=3.79\times 10^{−23}\:cm^3}\nonumber$
(Note that the edge length was converted from pm to cm to get the usual volume units for density.)
Therefore, the density of
$\mathrm{Po=\dfrac{3.471\times 10^{−22}\:g}{3.79\times 10^{−23}\:cm^3}=9.16\: g/cm^3}\nonumber$
Exercise $1$
The edge length of the unit cell for nickel is 0.3524 nm. The density of Ni is 8.90 g/cm3. Does nickel crystallize in a simple cubic structure? Explain.
Answer
No. If Ni were simple cubic, its density would be given by:
$\mathrm{1\: Ni\: atom×\dfrac{1\: mol\: Ni}{6.022\times 10^{23}\:Ni\: atoms}×\dfrac{58.693\:g}{1\: mol\: Ni}=9.746\times 10^{−23}\:g}\nonumber$
$V=l^3=\mathrm{(3.524\times 10^{−8}\:cm)^3=4.376\times 10^{−23}\:cm^3}\nonumber$
Then the density of Ni would be
$(\mathrm{=\dfrac{9.746\times 10^{−23}\:g}{4.376\times 10^{−23}\:cm^3}=2.23\: g/cm^3}\nonumber$
Since the actual density of Ni is not close to this, Ni does not form a simple cubic structure.
Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell, and face-centered cubic unit cell—all of which are illustrated in Figure $5$. (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)
Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure $6$. This is called a body-centered cubic (BCC) solid. Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners ($8×\dfrac{1}{8}=1$ atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight.
Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about 68% of the total volume. Isomorphous metals with a BCC structure include K, Ba, Cr, Mo, W, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous.)
Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure $7$. This arrangement is called a face-centered cubic (FCC) solid. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners ($8×\dfrac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces ($6×\dfrac{1}{2}=3$ atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.
Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying 74% of the volume. This structure is also called cubic closest packing (CCP). In CCP, there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12. The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure $8$.
Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in Figure $9$. Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB⋯). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type A, type B, and type C close-packed layers (i.e., ABCABCABC⋯). About two–thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include Cd, Co, Li, Mg, Na, and Zn, and metals that crystallize in a CCP structure include Ag, Al, Ca, Cu, Ni, Pb, and Pt.
Example $2$: Calculating Atomic Radius and Density for Metals (Part 2)
Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm.
1. What is the atomic radius of Ca in this structure?
2. Calculate the density of Ca.
Solution
(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4r).
Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:
\begin{align*} a^2+a^2 &=d^2 \[4pt] \mathrm{(558.8\:pm)^2+(558.5\:pm)^2} &=(4r)^2 \end{align*} \nonumber
Solving this gives
$r=\mathrm{\sqrt{\dfrac{(558.8\:pm)^2+(558.5\:pm)^2}{16}}}=\textrm{197.6 pmg for a Ca radius}. \nonumber$
(b) Density is given by $\mathrm{density=\dfrac{mass}{volume}}$. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face-centered Ca unit cell has one-eighth of an atom at each of the eight corners ($8 \times \dfrac{1}{8}=1$atom) and one-half of an atom on each of the six faces $6×\dfrac{1}{2}=3$ atoms), for a total of four atoms in the unit cell.
The mass of the unit cell can be found by:
$\mathrm{1\: Ca\: unit\: cell×\dfrac{4\: Ca\: atoms}{1\: Ca\: unit\: cell}×\dfrac{1\: mol\: Ca}{6.022\times 10^{23}\:Ca\: atoms}×\dfrac{40.078\:g}{1\: mol\: Ca}=2.662\times 10^{−22}\:g} \nonumber$
The volume of a Ca unit cell can be found by:
$V=a^3=\mathrm{(558.8\times 10^{−10}\:cm)^3=1.745\times 10^{−22}\:cm^3} \nonumber$
(Note that the edge length was converted from pm to cm to get the usual volume units for density.)
Then, the density of polonium:
$\mathrm{Po=\dfrac{2.662\times 10^{−22}\:g}{1.745\times 10^{−22}\:cm^3}=1.53\: g/cm^3} \nonumber$
Exercise $2$
Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm.
1. What is the atomic radius of Ag in this structure?
2. Calculate the density of Ag.
Answer a
144 pm
Answer b
10.5 g/cm3
In general, a unit cell is defined by the lengths of three axes (a, b, and c) and the angles (α, β, and γ) between them, as illustrated in Figure $10$. The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments.
There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure $11$.
The Structures of Ionic Crystals
Ionic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size. Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are surrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.
In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole. The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole. Figure $12$ illustrates both of these types of holes.
Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure $13$. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing.
There are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of 2:1; all of the tetrahedral holes are filled at this ratio. Examples include Li2O, Na2O, Li2S, and Na2S. Compounds with a ratio of less than 2:1 may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant.
Example $3$: Occupancy of Tetrahedral Holes
Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide?
Solution
Because there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be $\dfrac{1}{2}×2$, or 1, zinc ion per sulfide ion. Thus, the formula is ZnS.
Exercise $3$: Lithium selenide
Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide?
Answer
$\ce{Li2Se}$
The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO, MnS, NaCl, and KH, for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty.
Example $4$: Stoichiometry of Ionic Compounds Sapphire
Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?
Solution
Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be $\dfrac{2}{3}$:1, which would give $\mathrm{Al_{2/3}O}$. The simplest whole number ratio is 2:3, so the formula is Al2O3.
Exercise $4$
The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide?
Answer
$\ce{TiO2}$
In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl, and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in SrH2, UO2, SrCl2, and CaF2.
Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar.
Unit Cells of Ionic Compounds
Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures. When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl, (Figure $14$) is an example of this, with Cs+ and Cl having radii of 174 pm and 181 pm, respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by Cs+ ions overlapping unit cells formed by Cl ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion.
We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures.
When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure $15$. Sodium chloride, NaCl, is an example of this, with Na+ and Cl having radii of 102 pm and 181 pm, respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl.
The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure $16$. This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about 40% of the radius of a sulfide ion, so these small Zn2+ ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS.
A calcium fluoride unit cell, like that shown in Figure $17$, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, CaF2. Close examination of Figure $17$ will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs.
Calculation of Ionic Radii
If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.
Example $5$: Calculation of Ionic Radii
The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, calculate the ionic radius for the chloride ion. Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10−10 m.
Solution
On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:
Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so $d = 4r$. From the Pythagorean theorem, we have:
$a^2+a^2=d^2 \nonumber$
which yields:
$\mathrm{(0.514\:nm)^2+(0.514\:nm)^2}=(4r)^2=16r^2 \nonumber$
Solving this gives:
$r=\mathrm{\sqrt{\dfrac{(0.514\:nm)^2+(0.514\:nm)^2}{16}}=0.182\: nm\:(1.82\: Å)\:for\: a\: Cl^−\: radius.} \nonumber$
Exercise $6$
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å.
Answer
The radius of the potassium ion is 1.33 Å.
It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations.
X-Ray Crystallography
The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few Å).
When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference, a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves’ maxima are separated (Figure $18$).
When X-rays of a certain wavelength, λ, are scattered by atoms in adjacent crystal planes separated by a distance, d, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, n, of the wavelength. This condition is satisfied when the angle of the diffracted beam, θ, is related to the wavelength and interatomic distance by the equation:
$nλ=2d\sin \theta \label{Eq1}$
This relation is known as the Bragg equation in honor of W. H. Bragg, the English physicist who first explained this phenomenon. Figure $18$ illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference.
An X-ray diffractometer, such as the one illustrated in Figure $20$, may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise.
Example $6$: Using the Bragg Equation
In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction (n = 1) occurred at an angle θ = 25.25°. Determine the spacing between the diffracting planes in copper.
Solution
The distance between the planes is found by solving the Bragg equation (Equation $\ref{Eq1}$) for d.
This gives
$d=\dfrac{nλ}{2\sinθ}=\mathrm{\dfrac{1(0.1315\:nm)}{2\sin(25.25°)}=0.154\: nm}\nonumber$
Exercise $6$
A crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm. What is the angle for the first order diffraction?
Answer
21.9°
X-ray Crystallographer Rosalind Franklin
The discovery of the structure of DNA in 1953 by Francis Crick and James Watson is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice Wilkins, who provided experimental proof of DNA’s structure. British chemist Rosalind Franklin made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin’s research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type “B”) and a short, wide fiber formed when dried (type “A”). Her X-ray diffraction images of DNA provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the RNA that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued to work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo.
Key Concepts and Summary
The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest-packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements.
Glossary
body-centered cubic (BCC) solid
crystalline structure that has a cubic unit cell with lattice points at the corners and in the center of the cell
body-centered cubic unit cell
simplest repeating unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube
Bragg equation
equation that relates the angles at which X-rays are diffracted by the atoms within a crystal
coordination number
number of atoms closest to any given atom in a crystal or to the central metal atom in a complex
cubic closest packing (CCP)
crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations (ABC)
diffraction
redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions
face-centered cubic (FCC) solid
crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face
face-centered cubic unit cell
simplest repeating unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face
hexagonal closest packing (HCP)
crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB)
hole
(also, interstice) space between atoms within a crystal
isomorphous
possessing the same crystalline structure
octahedral hole
open space in a crystal at the center of six particles located at the corners of an octahedron
simple cubic unit cell
(also, primitive cubic unit cell) unit cell in the simple cubic structure
simple cubic structure
crystalline structure with a cubic unit cell with lattice points only at the corners
space lattice
all points within a crystal that have identical environments
tetrahedral hole
tetrahedral space formed by four atoms or ions in a crystal
unit cell
smallest portion of a space lattice that is repeated in three dimensions to form the entire lattice
X-ray crystallography
experimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]). | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.6%3A_Lattice_Structures_in_Crystalline_Solids.txt |
10.1: Intermolecular Forces
Q10.1.1
In terms of their bulk properties, how do liquids and solids differ? How are they similar?
S10.1.1
Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid.
Q10.1.2
In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids?
Q10.1.3
In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases?
S10.1.3
They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast, a gas will expand without limit to fill the space into which it is placed.
Q10.1.4
Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape.
Q10.1.5
What is the evidence that all neutral atoms and molecules exert attractive forces on each other?
S10.1.5
All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature.
Q10.1.6
Open the PhET States of Matter Simulation to answer the following questions:
1. Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?
2. For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.
3. Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain.
Q10.1.7
Define the following and give an example of each:
1. dispersion force
2. dipole-dipole attraction
3. hydrogen bond
S10.1.7
1. Dispersion forces occur as an atom develops a temporary dipole moment when its electrons are distributed asymmetrically about the nucleus. This structure is more prevalent in large atoms such as argon or radon. A second atom can then be distorted by the appearance of the dipole in the first atom. The electrons of the second atom are attracted toward the positive end of the first atom, which sets up a dipole in the second atom. The net result is rapidly fluctuating, temporary dipoles that attract one another (example: Ar).
2. A dipole-dipole attraction is a force that results from an electrostatic attraction of the positive end of one polar molecule for the negative end of another polar molecule (example: ICI molecules attract one another by dipole-dipole interaction).
3. Hydrogen bonds form whenever a hydrogen atom is bonded to one of the more electronegative atoms, such as a fluorine, oxygen, nitrogen, or chlorine atom. The electrostatic attraction between the partially positive hydrogen atom in one molecule and the partially negative atom in another molecule gives rise to a strong dipole-dipole interaction called a hydrogen bond (example: $\mathrm{HF⋯HF}$).
Q10.1.8
The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid?
Q10.1.9
Why do the boiling points of the noble gases increase in the order He < Ne < Ar < Kr < Xe?
S10.1.9
The London forces typically increase as the number of electrons increase.
Q10.1.10
Neon and HF have approximately the same molecular masses.
1. Explain why the boiling points of Neon and HF differ.
2. Compare the change in the boiling points of Ne, Ar, Kr, and Xe with the change of the boiling points of HF, HCl, HBr, and HI, and explain the difference between the changes with increasing atomic or molecular mass.
Q10.1.11
Arrange each of the following sets of compounds in order of increasing boiling point temperature:
1. HCl, H2O, SiH4
2. F2, Cl2, Br2
3. CH4, C2H6, C3H8
4. O2, NO, N2
S10.1.11
(a) SiH4 < HCl < H2O; (b) F2 < Cl2 < Br2; (c) CH4 < C2H6 < C3H8; (d) N2 < O2 < NO
Q10.1.12
The molecular mass of butanol, C4H9OH, is 74.14; that of ethylene glycol, CH2(OH)CH2OH, is 62.08, yet their boiling points are 117.2 °C and 174 °C, respectively. Explain the reason for the difference.
Q10.1.13
On the basis of intermolecular attractions, explain the differences in the boiling points of n–butane (−1 °C) and chloroethane (12 °C), which have similar molar masses.
S10.1.13
Only rather small dipole-dipole interactions from C-H bonds are available to hold n-butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the Cl-C bond; the interaction is therefore stronger, leading to a higher boiling point.
Q10.1.14
On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses.
Q10.1.15
The melting point of H2O(s) is 0 °C. Would you expect the melting point of H2S(s) to be −85 °C, 0 °C, or 185 °C? Explain your answer.
S10.1.15
−85 °C. Water has stronger hydrogen bonds so it melts at a higher temperature.
Q10.1.16
Silane (SiH4), phosphine (PH3), and hydrogen sulfide (H2S) melt at −185 °C, −133 °C, and −85 °C, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds?
Q10.1.17
Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules.
S10.1.17
The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of F is greater than that of O. Consequently, the partial negative charge on F is greater than that on O. The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O.
Q10.1.18
Under certain conditions, molecules of acetic acid, CH3COOH, form “dimers,” pairs of acetic acid molecules held together by strong intermolecular attractions:
Draw a dimer of acetic acid, showing how two CH3COOH molecules are held together, and stating the type of IMF that is responsible.
Q10.1.19
Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix. What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together:
S10.1.19
H-bonding is the principle IMF holding the DNA strands together. The H-bonding is between the $\mathrm{N−H}$ and $\mathrm{C=O}$.
Q10.1.20
The density of liquid NH3 is 0.64 g/mL; the density of gaseous NH3 at STP is 0.0007 g/mL. Explain the difference between the densities of these two phases.
Q10.1.21
Identify the intermolecular forces present in the following solids:
1. CH3CH2OH
2. CH3CH2CH3
3. CH3CH2Cl
S10.1.21
(a) hydrogen bonding and dispersion forces; (b) dispersion forces; (c) dipole-dipole attraction and dispersion forces
10.2: Properties of Liquids
Q10.2.1
The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration. Rank the motor oils in order of increasing viscosity, and explain your reasoning:
Q10.2.2
Although steel is denser than water, a steel needle or paper clip placed carefully lengthwise on the surface of still water can be made to float. Explain at a molecular level how this is possible:
(credit: Cory Zanker)
S10.2.2
The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of “skin” at its surface. This skin can support a bug or paper clip if gently placed on the water.
Q10.2.3
The surface tension and viscosity values for diethyl ether, acetone, ethanol, and ethylene glycol are shown here.
1. Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.
2. Explain their differences in surface tension in terms of the size and shape of their molecules and their IMFs:
Q10.2.4
You may have heard someone use the figure of speech “slower than molasses in winter” to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature.
S10.2.4
Temperature has an effect on intermolecular forces: the higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid; the lower the temperature, the lesser the intermolecular forces are overcome, and so the less viscous the liquid.
Q10.2.5
It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this.
Q10.2.6
The surface tension and viscosity of water at several different temperatures are given in this table.
Water Surface Tension (mN/m) Viscosity (mPa s)
0 °C 75.6 1.79
20 °C 72.8 1.00
60 °C 66.2 0.47
100 °C 58.9 0.28
1. As temperature increases, what happens to the surface tension of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
2. As temperature increases, what happens to the viscosity of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
S10.2.6
(a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason.
Q10.2.7
At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm? Refer to Example for the required information.
Q10.2.8
Water rises in a glass capillary tube to a height of 17 cm. What is the diameter of the capillary tube?
9.5 × 10−5 m
10.3: Phase Transitions
Q10.3.1
Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change?
Q10.3.2
Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change?
S10.3.2
The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise.
Q10.3.3
What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container?
Q10.3.4
Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate?
S10.3.4
We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids.
Q10.3.5
Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime?
Q10.3.6
What is the relationship between the intermolecular forces in a liquid and its vapor pressure?
S10.3.7
The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases.
Q10.3.7
What is the relationship between the intermolecular forces in a solid and its melting temperature?
Q10.3.8
Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day?
S10.3.8
As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures.
Q10.3.9
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl4 is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl4.
Q10.3.10
When is the boiling point of a liquid equal to its normal boiling point?
S10.3.10
When the pressure of gas above the liquid is exactly 1 atm
Q10.3.11
How does the boiling of a liquid differ from its evaporation?
Q10.3.12
Use the information in Figure to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa.
S10.3.12
approximately 95 °C
Q10.3.13
A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced?
Q10.3.14
Explain the following observations:
1. It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level).
2. Perspiring is a mechanism for cooling the body.
S10.3.14
(a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water.
Q10.3.15
The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.
Q10.3.16
Explain why the molar enthalpies of vaporization of the following substances increase in the order CH4 < C2H6 < C3H8, even though all three substances experience the same dispersion forces when in the liquid state.
S10.3.16
Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids.
Q10.3.17
Explain why the enthalpies of vaporization of the following substances increase in the order CH4 < NH3 < H2O, even though all three substances have approximately the same molar mass.
Q10.3.18
The enthalpy of vaporization of CO2(l) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS2(l) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers.
S10.3.18
The boiling point of CS2 is higher than that of CO2 partially because of the higher molecular weight of CS2; consequently, the attractive forces are stronger in CS2. It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO2. A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible.
Q10.3.19
The hydrogen fluoride molecule, HF, is more polar than a water molecule, H2O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.
Q10.3.20
Ethyl chloride (boiling point, 13 °C) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride.
S10.3.20
The thermal energy (heat) needed to evaporate the liquid is removed from the skin.
Q10.3.21
Which contains the compounds listed correctly in order of increasing boiling points?
1. N2 < CS2 < H2O < KCl
2. H2O < N2 < CS2 < KCl
3. N2 < KCl < CS2 < H2O
4. CS2 < N2 < KCl < H2O
5. KCl < H2O < CS2 < N2
Q10.3.22
How much heat is required to convert 422 g of liquid H2O at 23.5 °C into steam at 150 °C?
1130 kJ
Q10.3.22
Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)
Q10.3.24
Titanium tetrachloride, TiCl4, has a melting point of −23.2 °C and has a ΔH fusion = 9.37 kJ/mol.
1. How much energy is required to melt 263.1 g TiCl4?
2. For TiCl4, which will likely have the larger magnitude: ΔH fusion or ΔH vaporization? Explain your reasoning.
S10.3.24
(a) 13.0 kJ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them.
10.4: Phase Diagrams
Q10.4.1
From the phase diagram for water, determine the state of water at:
1. 35 °C and 85 kPa
2. −15 °C and 40 kPa
3. −15 °C and 0.1 kPa
4. 75 °C and 3 kPa
5. 40 °C and 0.1 kPa
6. 60 °C and 50 kPa
Q10.4.2
What phase changes will take place when water is subjected to varying pressure at a constant temperature of 0.005 °C? At 40 °C? At −40 °C?
S10.4.2
At low pressures and 0.005 °C, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At 40 °C, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At −40 °C, water goes from a gas to a solid as the pressure increases above very low values.
Q10.4.3
Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm. What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning.
Q10.4.4
From the phase diagram for carbon dioxide, determine the state of CO2 at:
1. 20 °C and 1000 kPa
2. 10 °C and 2000 kPa
3. 10 °C and 100 kPa
4. −40 °C and 500 kPa
5. −80 °C and 1500 kPa
6. −80 °C and 10 kPa
The pressure and temperature axes on this phase diagram of carbon dioxide are not drawn to constant scale in order to illustrate several important properties.
S10.4.4
(a) liquid; (b) solid; (c) gas; (d) gas; (e) gas; (f) gas
Q10.4.5
Determine the phase changes that carbon dioxide undergoes as the pressure changes if the temperature is held at −50 °C? If the temperature is held at −40 °C? At 20 °C?
The pressure and temperature axes on this phase diagram of carbon dioxide are not drawn to constant scale in order to illustrate several important properties.
Q10.4.6
Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of 65 atm and a temperature of 20 °C. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature.
Q10.4.7
Dry ice, CO2(s), does not melt at atmospheric pressure. It sublimes at a temperature of −78 °C. What is the lowest pressure at which CO2(s) will melt to give CO2(l)? At approximately what temperature will this occur? (See Figure for the phase diagram.)
S10.4.7
Dry ice, CO2(s), will melt to give CO2(l) at 5.11 atm at −56.6 °C, the triple point of carbon dioxide.
Q10.4.8
If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer.
S10.4.8
Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry.
Q10.4.9
Is it possible to liquefy nitrogen at room temperature (about 25 °C)? Is it possible to liquefy sulfur dioxide at room temperature? Explain your answers.
Q10.4.10
Elemental carbon has one gas phase, one liquid phase, and three different solid phases, as shown in the phase diagram:
1. On the phase diagram, label the gas and liquid regions.
2. Graphite is the most stable phase of carbon at normal conditions. On the phase diagram, label the graphite phase.
3. If graphite at normal conditions is heated to 2500 K while the pressure is increased to 1010 Pa, it is converted into diamond. Label the diamond phase.
4. Circle each triple point on the phase diagram.
5. In what phase does carbon exist at 5000 K and 108 Pa?
6. If the temperature of a sample of carbon increases from 3000 K to 5000 K at a constant pressure of 106 Pa, which phase transition occurs, if any?
10.5: The Solid State of Matter
Q10.5.1
What types of liquids typically form amorphous solids?
S10.5.1
Amorphous solids lack an ordered internal structure. Liquid materials that contain large, cumbersome molecules that cannot move readily into ordered positions generally form such solids.
Q10.5.2
At very low temperatures oxygen, O2, freezes and forms a crystalline solid. Which best describes these crystals?
1. ionic
2. covalent network
3. metallic
4. amorphous
5. molecular crystals
S10.5.3
(e) molecular crystals
Q10.5.4
As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid?
1. ionic
2. covalent network
3. metallic
4. amorphous
5. molecular crystals
(d) amorphous
Q10.5.5
Explain why ice, which is a crystalline solid, has a melting temperature of 0 °C, whereas butter, which is an amorphous solid, softens over a range of temperatures.
S10.5.6
Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range.
Q10.5.7
Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:
1. SiO2
2. KCl
3. Cu
4. CO2
5. C (diamond)
6. BaSO4
7. NH3
8. NH4F
9. C2H5OH
S10.5.7
(a) SiO2, covalent network; (b) KCl, ionic; (c) Cu, metallic; (d) CO, molecular; (e) C (diamond), covalent network; (f) BaSO4, ionic; (g) NH3, molecular; (h) NH4F, ionic; (i) C2H5OH, molecular
Q10.5.8
Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:
1. CaCl2
2. SiC
3. N2
4. Fe
5. C (graphite)
6. CH3CH2CH2CH3
7. HCl
8. NH4NO3
9. K3PO4
S10.5.8
(a) CaCl2, ionic; (b) SiC, covalent network; (c) N2, molecular; (d) Fe, metallic; (e) C (graphite), covalent network; (f) CH3CH2CH2CH3, molecular; (g) HCl, molecular; (h) NH4NO3, ionic; (i) K3PO4, ionic
Q10.5.9
Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:
Substance Appearance Melting Point Electrical Conductivity Solubility in Water
X lustrous, malleable 1500 °C high insoluble
Y soft, yellow 113 °C none insoluble
Z hard, white 800 °C only if melted/dissolved soluble
S10.5.9
X = metallic; Y = covalent network; Z = ionic
Q10.5.10
Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:
Substance Appearance Melting Point Electrical Conductivity Solubility in Water
X brittle, white 800 °C only if melted/dissolved soluble
Y shiny, malleable 1100 °C high insoluble
Z hard, colorless 3550 °C none insoluble
S10.5.10
X = ionic; Y = metallic; Z = covalent network
Q10.5.11
Identify the following substances as ionic, metallic, covalent network, or molecular solids:
Substance A is malleable, ductile, conducts electricity well, and has a melting point of 1135 °C. Substance B is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072 °C. Substance C is very hard, does not conduct electricity, and has a melting point of 3440 °C. Substance D is soft, does not conduct electricity, and has a melting point of 185 °C.
S10.5.11
A = metallic; B = ionic; C = covalent network; D = molecular
Q10.5.12
Substance A is shiny, conducts electricity well, and melts at 975 °C. Substance A is likely a(n):
1. ionic solid
2. metallic solid
3. molecular solid
4. covalent network solid
S10.5.12
(b) metallic solid
Q10.5.13
Substance B is hard, does not conduct electricity, and melts at 1200 °C. Substance B is likely a(n):
1. ionic solid
2. metallic solid
3. molecular solid
4. covalent network solid
S10.5.13
(d) covalent network solid
10.6: Lattice Structures
Q10.6.1
Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell.
S10.6.1
The structure of this low-temperature form of iron (below 910 °C) is body-centered cubic. There is one-eighth atom at each of the eight corners of the cube and one atom in the center of the cube.
Q10.6.2
Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell.
Q10.6.3
What is the coordination number of a chromium atom in the body-centered cubic structure of chromium?
eight
Q10.6.4
What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum?
Q10.6.5
Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom?
12
Q10.6.6
Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom?
Q10.6.7
Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 Å.
1. What is the atomic radius of tungsten in this structure?
2. Calculate the density of tungsten.
S10.6.7
(a) 1.370 Å; (b) 19.26 g/cm
Q10.6.8
Platinum (atomic radius = 1.38 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum.
Q10.6.9
Barium crystallizes in a body-centered cubic unit cell with an edge length of 5.025 Å
1. What is the atomic radius of barium in this structure?
2. Calculate the density of barium.
S10.6.9
(a) 2.176 Å; (b) 3.595 g/cm3
Q10.6.10
Aluminum (atomic radius = 1.43 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum.
Q10.6.11
The density of aluminum is 2.7 g/cm3; that of silicon is 2.3 g/cm3. Explain why Si has the lower density even though it has heavier atoms.
S10.6.11
The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12).
Q10.6.12
The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space?
Q10.6.13
Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer.
S10.6.13
In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS.
Q10.6.14
A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer.
Q10.6.15
What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions?
Co3O4
Q10.6.16
A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound?
Q10.6.17
A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? Explain your answer.
S10.6.17
In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is TlI.
Q10.6.18
Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closest-packed array with all of the octahedral holes occupied: Li, Na, Be, Ca, or Al?
Q10.6.19
What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium?
S10.6.19
59.95%; The oxidation number of titanium is +4.
Q10.6.20
Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure.
Q10.6.21
As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation.
S10.6.21
Both ions are close in size: Mg, 0.65; Li, 0.60. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of Si4+ for Al3+.
Q10.6.22
Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. What is the formula of the compound?
Q10.6.23
One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound?
Mn2O3
Q10.6.24
NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is 4.880 Å.
1. Calculate the ionic radius of H. (The ionic radius of Li+ is 0.0.95 Å.)
2. Calculate the density of NaH.
Q10.6.25
Thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å. Calculate the ionic radius of TI+. (The ionic radius of I is 2.16 Å.)
1.48 Å
Q10.6.26
A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge.
1. What is the empirical formula of this compound? Explain your answer.
2. What is the coordination number of the Mn3+ ion?
3. Calculate the edge length of the unit cell if the radius of a Mn3+ ion is 0.65 A.
4. Calculate the density of the compound.
Q10.6.27
What is the spacing between crystal planes that diffract X-rays with a wavelength of 1.541 nm at an angle θ of 15.55° (first order reflection)?
2.874 Å
Q10.6.28
A diffractometer using X-rays with a wavelength of 0.2287 nm produced first-order diffraction peak for a crystal angle θ = 16.21°. Determine the spacing between the diffracting planes in this crystal.
Q10.6.29
A metal with spacing between planes equal to 0.4164 nm diffracts X-rays with a wavelength of 0.2879 nm. What is the diffraction angle for the first order diffraction peak?
20.2°
Q10.6.30
Gold crystallizes in a face-centered cubic unit cell. The second-order reflection (n = 2) of X-rays for the planes that make up the tops and bottoms of the unit cells is at θ = 22.20°. The wavelength of the X-rays is 1.54 Å. What is the density of metallic gold?
Q10.6.31
When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted. These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64 Å. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first-order diffraction?
1.74 × 104 eV | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.E%3A_Liquids_and_Solids_%28Exercises%29.txt |
In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions.
11: Solutions and Colloids
Coral reefs are home to about 25% of all marine species. They are being threatened by climate change, oceanic acidification, and water pollution, all of which change the composition of the solution we know as seawater. Dissolved oxygen in seawater is critical for sea creatures, but as the oceans warm, oxygen becomes less soluble. As the concentration of carbon dioxide in the atmosphere increases, the concentration of carbon dioxide in the oceans increases, contributing to oceanic acidification. Coral reefs are particularly sensitive to the acidification of the ocean, since the exoskeletons of the coral polyps are soluble in acidic solutions. Humans contribute to the changing of seawater composition by allowing agricultural runoff and other forms of pollution to affect our oceans.
Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions.
11.1: The Dissolution Process
Learning Objectives
• Describe the basic properties of solutions and how they form.
• Predict whether a given mixture will yield a solution based on molecular properties of its components.
• Explain why some solutions either produce or absorb heat when they form.
An earlier chapter of this text introduced solutions, defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, $\ce{C12H22O11}$. When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:
$\ce{C12H22O11 (s) ⟶ C12H22O11 (aq) } \label{Eq1}$
The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time.
Potassium dichromate, $\ce{K_2Cr_2O_7}$, is an ionic compound composed of colorless potassium ions, $\mathrm{K^+}$, and orange dichromate ions, $\ce{Cr_2O_7^{2−}}$. When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure $1$), as indicated in this equation:
$\ce{K2Cr2O7(s) ⟶ 2K^{+} (aq) + Cr2O7^{2-} (aq)} \label{Eq2}$
As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules.
Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table $1$ gives examples of several different solutions and the phases of the solutes and solvents.
Table $1$: Different Types of Solutions
Solution Solute Solvent
air O2(g) N2(g)
soft drinks CO2(g) H2O(l)
hydrogen in palladium H2(g) Pd(s)
rubbing alcohol H2O(l) C3H8O(l) (2-propanol)
saltwater NaCl(s) H2O(l)
brass Zn(s) Cu(s)
Solutions exhibit these defining traits:
• They are homogeneous; that is, after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).
• The physical state of a solution—solid, liquid, or gas—is typically the same as that of the solvent, as demonstrated by the examples in Table $1$.
• The components of a solution are dispersed on a molecular scale; that is, they consist of a mixture of separated molecules, atoms, and/or ions.
• The dissolved solute in a solution will not settle out or separate from the solvent.
• The composition of a solution, or the concentrations of its components, can be varied continuously, within limits.
The Formation of Solutions
The formation of a solution is an example of a spontaneous process, a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes we stir a mixture to speed up the dissolution process, but this is not necessary; a homogeneous solution would form if we waited long enough. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that favor, but do not guarantee, the spontaneous formation of a solution:
1. a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)
2. an increase in the disorder in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics)
In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in disorder always results when a solution forms.
When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution. A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions.
When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure $2$). The formation of this solution clearly involves an increase in disorder, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.
Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH3OH) and ethanol (C2H5OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, $\ce{C5H12}$, and hexane, $\ce{C6H14}$. Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure $2$ will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how diffusion alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution.
Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent-solvent, and solute-solvent. As illustrated in Figure $3$, the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation.
For example, cooking oils and water will not mix to any appreciable extent to yield solutions (Figure $4$). Hydrogen bonding is the dominant intermolecular attractive force present in liquid water; the nonpolar hydrocarbon molecules of cooking oils are not capable of hydrogen bonding, instead being held together by dispersion forces. Forming an oil-water solution would require overcoming the very strong hydrogen bonding in water, as well as the significantly strong dispersion forces between the relatively large oil molecules. And, since the polar water molecules and nonpolar oil molecules would not experience very strong intermolecular attraction, very little energy would be released by solvation.
On the other hand, a mixture of ethanol and water will mix in any proportions to yield a solution. In this case, both substances are capable of hydrogen bonding, and so the solvation process is sufficiently exothermic to compensate for the endothermic separations of solute and solvent molecules.
As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH4NO3) is one such example and is used to make instant cold packs for treating injuries like the one pictured in Figure $5$. A thin-walled plastic bag of water is sealed inside a larger bag with solid NH4NO3. When the smaller bag is broken, a solution of NH4NO3 forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution.
Video $1$: Watch this brief video illustrating endothermic and exothermic dissolution processes.
Summary
A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy.
Footnotes
1. If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution.
Glossary
alloy
solid mixture of a metallic element and one or more additional elements
ideal solution
solution that forms with no accompanying energy change
solvation
exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established
spontaneous process
physical or chemical change that occurs without the addition of energy from an external source | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.0%3A_Prelude_to_Solutions_and_Colloids.txt |
Learning Objectives
• Define and give examples of electrolytes
• Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes
• Relate electrolyte strength to solute-solvent attractive forces
When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.
Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure $1$).
Ionic Electrolytes
Water and other polar molecules are attracted to ions, as shown in Figure $2$. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes.
Let us consider what happens at the microscopic level when we add solid KCl to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules penetrate between individual K+ and Cl ions and surround them, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure shows. The reduction of the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution, resulting in an increase in the disorder of the system as the ions change from their fixed and ordered positions in the crystal to mobile and much more disordered states in solution. This increased disorder is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat.
In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in disorder cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust).
Covalent Electrolytes
Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton to another molecule of water, yielding hydronium and hydroxide ions.
$\ce{H_2O (l)+ H_2O (l) \rightleftharpoons H_3O^{+} (aq) + OH^{−} (aq)} \label{11.3.2}$
In some cases, we find that solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, when we dissolve hydrogen chloride in water, we find that the solution is a very good conductor. The water molecules play an essential part in forming ions: Solutions of hydrogen chloride in many other solvents, such as benzene, do not conduct electricity and do not contain ions.
Hydrogen chloride is an acid, and so its molecules react with water, transferring H+ ions to form hydronium ions ($H_3O^+$) and chloride ions (Cl):
This reaction is essentially 100% complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry.
Summary
Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes, while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water.
Glossary
dissociation
physical process accompanying the dissolution of an ionic compound in which the compound’s constituent ions are solvated and dispersed throughout the solution
electrolyte
substance that produces ions when dissolved in water
ion-dipole attraction
electrostatic attraction between an ion and a polar molecule
nonelectrolyte
substance that does not produce ions when dissolved in water
strong electrolyte
substance that dissociates or ionizes completely when dissolved in water
weak electrolyte
substance that ionizes only partially when dissolved in water | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.2%3A_Electrolytes.txt |
Learning Objectives
• Describe the effects of temperature and pressure on solubility
• State Henry’s law and use it in calculations involving the solubility of a gas in a liquid
• Explain the degrees of solubility possible for liquid-liquid solutions
Imagine adding a small amount of salt to a glass of water, stirring until all the salt has dissolved, and then adding a bit more. You can repeat this process until the salt concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent-solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved salt remains. The concentration of salt in the solution at this point is known as its solubility.
The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium. Referring to the example of salt in water:
$\ce{NaCl}(s)⇌\ce{Na+}(aq)+\ce{Cl-}(aq) \label{11.4.1}$
When a solute’s concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute’s concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.
If we add more salt to a saturated solution of salt, we see it fall to the bottom and no more seems to dissolve. In fact, the added salt does dissolve, as represented by the forward direction of the dissolution equation. Accompanying this process, dissolved salt will precipitate, as depicted by the reverse direction of the equation. The system is said to be at equilibrium when these two reciprocal processes are occurring at equal rates, and so the amount of undissolved and dissolved salt remains constant. Support for the simultaneous occurrence of the dissolution and precipitation processes is provided by noting that the number and sizes of the undissolved salt crystals will change over time, though their combined mass will remain the same.
Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states. For example, the carbonated beverage in an open container that has not yet “gone flat” is supersaturated with carbon dioxide gas; given time, the CO2 concentration will decrease until it reaches its equilibrium value.
Solutions of Gases in Liquids
In an earlier module of this chapter, the effect of intermolecular attractive forces on solution formation was discussed. The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl3. Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C6H14, is approximately 20 times greater than it is in water.
Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure $1$). This is one of the major impacts resulting from the thermal pollution of natural bodies of water.
When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure $2$).
The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO2 at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure $3$). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.”
For many gaseous solutes, the relation between solubility, $C_g$, and partial pressure, $P_g$, is a proportional one:
$C_\ce{g}=kP_\ce{g} \nonumber$
where $k$ is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry’s law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.
Example $1$: Application of Henry’s Law
At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10−3 mol L−1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere.
Solution
According to Henry’s law, for an ideal solution the solubility, Cg, of a gas (1.38 × 10−3 mol L−1, in this case) is directly proportional to the pressure, Pg, of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both Cg and Pg, we can rearrange this expression to solve for k.
\begin{align*} C_\ce{g}&=kP_\ce{g}\[4pt] k&=\dfrac{C_\ce{g}}{P_\ce{g}}\[4pt] &=\mathrm{\dfrac{1.38×10^{−3}\:mol\:L^{−1}}{101.3\:kPa}}\[4pt] &=\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}}\[4pt] &\hspace{15px}\mathrm{(1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1})} \end{align*} \nonumber
Now we can use k to find the solubility at the lower pressure.
$C_\ce{g}=kP_\ce{g} \nonumber$
$\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}×20.7\:kPa\[4pt] (or\:1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1}×155\:torr)\[4pt] =2.82×10^{−4}\:mol\:L^{−1}} \nonumber$
Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.
Exercise $1$
Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10−3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr).
Answer
7.25 × 10−3 g in 100.0 mL or 0.0725 g/L
Case Study: Decompression Sickness (“The Bends”)
Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law.
As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure $4$).
Deviations from Henry’s law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.
Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure $5$), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO2 were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.
Solutions of Liquids in Liquids
We know that some liquids mix with each other in all proportions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in Figure $6$) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline.
Liquids that mix with water in all proportions are usually polar substances or substances that form hydrogen bonds. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solvent-solvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom “like dissolves like.”
Two liquids that do not mix to an appreciable extent are called immiscible. Layers are formed when we pour immiscible liquids into the same container. Gasoline, oil (Figure $7$), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. The attraction between the molecules of such nonpolar liquids and polar water molecules is ineffectively weak. The only strong attractions in such a mixture are between the water molecules, so they effectively squeeze out the molecules of the nonpolar liquid. The distinction between immiscibility and miscibility is really one of degrees, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility.
Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure $8$).
Solutions of Solids in Liquids
The dependence of solubility on temperature for a number of inorganic solids in water is shown by the solubility curves in Figure $9$. Reviewing these data indicate a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate.
The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure $10$, take advantage of this behavior.
Video $2$: This video shows the crystallization process occurring in a hand warmer.
Summary
The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances’ atoms, ions, or molecules. This tendency to dissolve is quantified as substance’s solubility, its maximum concentration in a solution at equilibrium under specified conditions. A saturated solution contains solute at a concentration equal to its solubility. A supersaturated solution is one in which a solute’s concentration exceeds its solubility—a nonequilibrium (unstable) condition that will result in solute precipitation when the solution is appropriately perturbed. Miscible liquids are soluble in all proportions, and immiscible liquids exhibit very low mutual solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature. The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry’s law.
Key Equations
• $C_\ce{g}=kP_\ce{g}$
Glossary
Henry’s law
law stating the proportional relationship between the concentration of dissolved gas in a solution and the partial pressure of the gas in contact with the solution
immiscible
of negligible mutual solubility; typically refers to liquid substances
miscible
mutually soluble in all proportions; typically refers to liquid substances
partially miscible
of moderate mutual solubility; typically refers to liquid substances
saturated
of concentration equal to solubility; containing the maximum concentration of solute possible for a given temperature and pressure
solubility
extent to which a solute may be dissolved in water, or any solvent
supersaturated
of concentration that exceeds solubility; a nonequilibrium state
unsaturated
of concentration less than solubility | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.3%3A_Solubility.txt |
Learning Objectives
• Express concentrations of solution components using mole fraction and molality
• Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
• Perform calculations using the mathematical equations that describe these various colligative effects
• Describe the process of distillation and its practical applications
• Explain the process of osmosis and describe how it is applied industrially and in nature
The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.
Mole Fraction and Molality
Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:
$M=\dfrac{\text{mol solute}}{\text{L solution}} \label{11.5.1}$
Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.
The mole fraction, $\chi$, of a component is the ratio of its molar amount to the total number of moles of all solution components:
$\chi_\ce{A}=\dfrac{\text{mol A}}{\text{total mol of all components}} \label{11.5.2}$
Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:
$m=\dfrac{\text{mol solute}}{\text{kg solvent}} \label{11.5.3}$
Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.
Example $1$: Calculating Mole Fraction and Molality
The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C2H4(OH)2, in a solution prepared from $\mathrm{2.22 \times 10^3 \;g}$ of ethylene glycol and $\mathrm{2.00 \times 10^3\; g}$ of water (approximately 2 L of glycol and 2 L of water)?
Solution
(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.
$\mathrm{mol\:C_2H_4(OH)_2=2220\:g×\dfrac{1\:mol\:C_2H_4(OH)_2}{62.07\:g\:C_2H_4(OH)_2}=35.8\:mol\:C_2H_4(OH)_2}$
$\mathrm{mol\:H_2O=2000\:g×\dfrac{1\:mol\:H_2O}{18.02\:g\:H_2O}=111\:mol\:H_2O}$
$\chi_\mathrm{ethylene\:glycol}=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{(35.8+111)\:mol\: total}=0.245}$
Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:
$\mathrm{2220\:g\:C_2H_4(OH)_2\left(\dfrac{mol\:C_2H_2(OH)_2}{62.07\:g}\right)=35.8\:mol\:C_2H_4(OH)_2} \nonumber$
Then, convert the mass of the water from grams to kilograms:
$\mathrm{2000\: g\:H_2O\left(\dfrac{1\:kg}{1000\:g}\right)=2\: kg\:H_2O} \nonumber$
Finally, calculate molarity per its definition:
\begin{align*} \ce{molality}&=\mathrm{\dfrac{mol\: solute}{kg\: solvent}}\ \ce{molality}&=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{2\:kg\:H_2O}}\ \ce{molality}&=17.9\:m \end{align*} \nonumber
Exercise $1$
What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH3, dissolved in 125 g of water?
Answer
7.14 × 10−3; 0.399 m
Example $2$: Converting Mole Fraction and Molal Concentrations
Calculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.
Solution
Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:
$\mathrm{\dfrac{3.0\;mol\; NaCl}{1.0\; kg\; H_2O}} \nonumber$
The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg
$\mathrm{1.0\:kg\:H_2O\left(\dfrac{1000\:g}{1\:kg}\right)\left(\dfrac{mol\:H_2O}{18.02\:g}\right)=55\:mol\:H_2O} \nonumber$
and then substituting these molar amounts into the definition for mole fraction.
\begin{align*} X_\mathrm{H_2O}&=\mathrm{\dfrac{mol\:H_2O}{mol\: NaCl + mol\:H_2O}}\ X_\mathrm{H_2O}&=\mathrm{\dfrac{55\:mol\:H_2O}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\ X_\mathrm{H_2O}&=0.95\ X_\mathrm{NaCl}&=\mathrm{\dfrac{mol\: NaCl}{mol\: NaCl+mol\:H_2O}}\ X_\mathrm{NaCl}&=\mathrm{\dfrac{3.0\:mol\:NaCl}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\ X_\mathrm{NaCl}&=0.052 \end{align*} \nonumber
Exercise $2$
The mole fraction of iodine, $\ce{I_2}$, dissolved in dichloromethane, $\ce{CH_2Cl_2}$, is 0.115. What is the molal concentration, m, of iodine in this solution?
Answer
1.50 m
Vapor Pressure Lowering
As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:
$\text{liquid} \rightleftharpoons \text{gas} \label{11.5.4}$
Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure $1$). While this kinetic interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the greater entropy of a solution in comparison to its separate solvent and solute serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module.
The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
$P_\ce{A}=X_\ce{A}P^\circ_\ce{A} \label{11.5.5}$
where PA is the partial pressure exerted by component A in the solution, $P^\circ_\ce{A}$ is the vapor pressure of pure A, and XA is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.)
Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing i components is
$P_\ce{solution}=\sum_{i}P_i=\sum_{i}X_iP^\circ_i \label{11.5.6}$
A nonvolatile substance is one whose vapor pressure is negligible (P° ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:
$P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent} \label{11.5.7}$
Example $3$: Calculation of a Vapor Pressure
Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C3H5(OH)3, and 184.4 g of ethanol, C2H5OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature.
Solution
Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as:
$P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent}$
First, calculate the molar amounts of each solution component using the provided mass data.
$\mathrm{92.1\cancel{g\:C_3H_5(OH)_3}×\dfrac{1\:mol\:C_3H_5(OH)_3}{92.094\cancel{g\:C_3H_5(OH)_3}}=1.00\:mol\:C_3H_5(OH)_3}$
$\mathrm{184.4\cancel{g\:C_2H_5OH}×\dfrac{1\:mol\:C_2H_5OH}{46.069\cancel{g\:C_2H_5OH}}=4.000\:mol\:C_2H_5OH}$
Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure.
$X_\mathrm{C_2H_5OH}=\mathrm{\dfrac{4.000\:mol}{(1.00\:mol+4.000\:mol)}=0.800}$
$P_\ce{solv}=X_\ce{solv}P^\circ_\ce{solv}=\mathrm{0.800×0.178\:atm=0.142\:atm}$
Exercise $3$
A solution contains 5.00 g of urea, CO(NH2)2 (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
Answer
23.4 torr
Elevation of the Boiling Point of a Solvent
As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, $ΔT_b$, is called boiling point elevation and is directly proportional to the molal concentration of solute species:
$ΔT_b=K_bm \label{11.5.8}$
where
• $K_\ce{b}$ is the boiling point elevation constant, or the ebullioscopic constant and
• $m$ is the molal concentration (molality) of all solute species.
Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in Table $1$.
Table $1$: Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents
Solvent Boiling Point (°C at 1 atm) Kb (Cm−1) Freezing Point (°C at 1 atm) Kf (Cm−1)
water 100.0 0.512 0.0 1.86
hydrogen acetate 118.1 3.07 16.6 3.9
benzene 80.1 2.53 5.5 5.12
chloroform 61.26 3.63 −63.5 4.68
nitrobenzene 210.9 5.24 5.67 8.1
The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose (342 g/mol) and a 1 m aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.
Example $4$: Calculating the Boiling Point of a Solution
What is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?
Solution
Use the equation relating boiling point elevation to solute molality to solve this problem in two steps.
1. Calculate the change in boiling point.
$ΔT_\ce{b}=K_\ce{b}m=2.53\:°\ce C\:m^{−1}×0.33\:m=0.83\:°\ce C$
• Add the boiling point elevation to the pure solvent’s boiling point.
$\mathrm{Boiling\: temperature=80.1\:°C+0.83\:°C=80.9\:°C}$
Exercise $4$
What is the boiling point of the antifreeze described in Example $4$?
Answer
109.2 °C
Example $5$: The Boiling Point of an Iodine Solution
Find the boiling point of a solution of 92.1 g of iodine, $\ce{I2}$, in 800.0 g of chloroform, $\ce{CHCl3}$, assuming that the iodine is nonvolatile and that the solution is ideal.
Solution
We can solve this problem using four steps.
1. Convert from grams to moles of $\ce{I2}$ using the molar mass of $\ce{I2}$ in the unit conversion factor.
Result: 0.363 mol
Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.
Result: 0.454 m
Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.
Result: 1.65 °C
Determine the new boiling point from the boiling point of the pure solvent and the change.
Result: 62.91 °C
Check each result as a self-assessment.
Exercise $5$: glycerin:Water Solution
What is the boiling point of a solution of 1.0 g of glycerin, $\ce{C3H5(OH)3}$, in 47.8 g of water? Assume an ideal solution.
Answer
100.12 °C
Distillation of Solutions
Distillation is a technique for separating the components of mixtures that is widely applied in both in the laboratory and in industrial settings. It is used to refine petroleum, to isolate fermentation products, and to purify water. This separation technique involves the controlled heating of a sample mixture to selectively vaporize, condense, and collect one or more components of interest. A typical apparatus for laboratory-scale distillations is shown in Figure $2$.
Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure $3$.
Depression of the Freezing Point of a Solvent
Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure $4$), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).
The decrease in freezing point of a dilute solution compared to that of the pure solvent, ΔTf, is called the freezing point depression and is directly proportional to the molal concentration of the solute
$ΔT_\ce{f}=K_\ce{f}m \label{11.5.9}$
where
• $m$ is the molal concentration of the solute in the solvent and
• $K_f$ is called the freezing point depression constant (or cryoscopic constant).
Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of Kf for several solvents are listed in Table $1$.
Example $5$: Calculation of the Freezing Point of a Solution
What is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example $4$?
Solution
Use the equation relating freezing point depression to solute molality to solve this problem in two steps.
1. Calculate the change in freezing point. $ΔT_\ce{f}=K_\ce{f}m=5.12\:°\ce C\:m^{−1}×0.33\:m=1.7\:°\ce C \nonumber$
2. Subtract the freezing point change observed from the pure solvent’s freezing point. $\mathrm{Freezing\: Temperature=5.5\:°C−1.7\:°C=3.8\:°C} \nonumber$
Exercise $6$
What is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?
Answer
−9.3 °C
Colligative Properties and De-Icing
Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.
Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Video $1$).
Video $1$: Freezing point depression is exploited to remove ice from the control surfaces of aircraft.
Phase Diagram for an Aqueous Solution of a Nonelectrolyte
The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid. Phase diagrams for water and an aqueous solution are shown in Figure $5$.
The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, ΔP, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTf, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed solvent only, and so transitions between these phases are not subject to colligative effects.
Osmosis and Osmotic Pressure of Solutions
A number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes.
Consider the apparatus illustrated in Figure $6$, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.
When osmosis is carried out in an apparatus like that shown in Figure $6$, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure ($\Pi$) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, M, and absolute temperature, T, according to the equation
$Π=MRT \label{11.5.10}$
where $R$ is the universal gas constant.
Example $7$: Calculation of Osmotic Pressure
What is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C?
Solution
We can find the osmotic pressure, $Π$, using Equation \ref{11.5.10}, where T is on the Kelvin scale (310 K) and the value of R is expressed in appropriate units (0.08206 L atm/mol K).
\begin{align*} Π&=MRT\ &=\mathrm{0.03\:mol/L×0.08206\: L\: atm/mol\: K×310\: K}\ &=\mathrm{7.6\:atm} \end{align*} \nonumber
Exercise $7$
What is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, CH3OH, in water at 37 °C?
Answer
5.3 atm
If a solution is placed in an apparatus like the one shown in Figure $7$, applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking.
Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation (Figure 11.5.8).
Determination of Molar Masses
Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.
Example $8$: Determining Molar Mass from Freezing Point Depression
A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?
Solution
We can solve this problem using the following steps.
1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table 11.5.1).
$ΔT_\ce{f}=\mathrm{5.5\:°C−2.32\:°C=3.2\:°C}$
1. Determine the molal concentration from Kf, the freezing point depression constant for benzene (Table 11.5.1), and ΔTf.
$ΔT_\ce{f}=K_\ce{f}m$
$m=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:m$
1. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
$\mathrm{Moles\: of\: solute=\dfrac{0.62\:mol\: solute}{1.00\cancel{kg\: solvent}}×0.0550\cancel{kg\: solvent}=0.035\:mol}$
2. Determine the molar mass from the mass of the solute and the number of moles in that mass.
$\mathrm{Molar\: mass=\dfrac{4.00\:g}{0.034\:mol}=1.2×10^2\:g/mol}$
Exercise $8$
A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?
Answer
1.8 × 102 g/mol
Example $9$: Determination of a Molar Mass from Osmotic Pressure
A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?
Solution
Here is one set of steps that can be used to solve the problem:
1. $\Pi=\mathrm{\dfrac{5.9\:torr×1\:atm}{760\:torr}=7.8×10^{−3}\:atm} \nonumber$
$\Pi=MRT \nonumber$
$M=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:L\: atm/mol\: K)(295\:K)}=3.2×10^{−4}\:M}$
1. $\mathrm{moles\: of\: hemoglobin=\dfrac{3.2×10^{−4}\:mol}{1\cancel{L\: solution}}×0.500\cancel{L\: solution}=1.6×10^{−4}\:mol}$
2. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
$\mathrm{molar\: mass=\dfrac{10.0\:g}{1.6×10^{−4}\:mol}=6.2×10^4\:g/mol}$
Exercise $9$
What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?
Answer
2.7 × 104 g/mol
Colligative Properties of Electrolytes
As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.
Example $10$: The Freezing Point of a Solution of an Electrolyte
The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater).
Solution
We can solve this problem using the following series of steps.
• Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Result: 0.072 mol NaCl
• Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl). Result: 0.14 mol ions
• Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms. Result: 1.1 m
• Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes. Result: 2.0 °C
• Determine the new freezing point from the freezing point of the pure solvent and the change. Result: −2.0 °C
Check each result as a self-assessment.
Exercise $10$
Assume that each of the ions in calcium chloride, CaCl2, has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl2 in 175 g of water.
Answer
−0.208 °C
Assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl) per each kilogram of water, and its freezing point depression is expected to be
$ΔT_\ce{f}=\mathrm{2.0\:mol\: ions/kg\: water×1.86\:°C\: kg\: water/mol\: ion=3.7\:°C.} \label{11.5.11}$
When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.
To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The van’t Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:
$i=\dfrac{\textrm{moles of particles in solution}}{\textrm{moles of formula units dissolved}} \label{11.5.12}$
Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table $2$.
Table $2$: Expected and Observed van’t Hoff Factors for Several 0.050 m Aqueous Electrolyte Solutions
Electrolyte Particles in Solution i (Predicted) i (Measured)
HCl H+, Cl 2 1.9
NaCl Na+, Cl 2 1.9
MgSO4 Mg2+, $\ce{SO4^2-}$ 2 1.3
MgCl2 Mg2+, 2Cl 3 2.7
FeCl3 Fe3+, 3Cl 4 3.4
glucose (a non-electrolyte) C12H22O11 1 1.0
In 1923, the chemists Peter Debye and Erich Hückel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure $9$). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table $2$ are for 0.05 m solutions, at which concentration the value of i for NaCl is 1.9, as opposed to an ideal value of 2.
Summary
Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.
Key Equations
• $\left(P_\ce{A}=X_\ce{A}P^\circ_\ce{A}\right)$
• $P_\ce{solution}=\sum_{i}P_i=\sum_{i}X_iP^\circ_i$
• $P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent}$
• ΔTb = Kbm
• ΔTf = Kfm
• Π = MRT
Footnotes
1. A nonelectrolyte shown for comparison.
Glossary
boiling point elevation
elevation of the boiling point of a liquid by addition of a solute
boiling point elevation constant
the proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant
colligative property
property of a solution that depends only on the concentration of a solute species
crenation
process whereby biological cells become shriveled due to loss of water by osmosis
freezing point depression
lowering of the freezing point of a liquid by addition of a solute
freezing point depression constant
(also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality
hemolysis
rupture of red blood cells due to the accumulation of excess water by osmosis
hypertonic
of greater osmotic pressure
hypotonic
of less osmotic pressure
ion pair
solvated anion/cation pair held together by moderate electrostatic attraction
isotonic
of equal osmotic pressure
molality (m)
a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
osmosis
diffusion of solvent molecules through a semipermeable membrane
osmotic pressure (Π)
opposing pressure required to prevent bulk transfer of solvent molecules through a semipermeable membrane
Raoult’s law
the partial pressure exerted by a solution component is equal to the product of the component’s mole fraction in the solution and its equilibrium vapor pressure in the pure state
semipermeable membrane
a membrane that selectively permits passage of certain ions or molecules
van’t Hoff factor (i)
the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.4%3A_Colligative_Properties.txt |
Learning Objectives
• Describe the composition and properties of colloidal dispersions
• List and explain several technological applications of colloids
As a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, when we make a solution, we prepare a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. A group of mixtures called colloids (or colloidal dispersions) exhibit properties intermediate between those of suspensions and solutions (Figure $1$). The particles in a colloid are larger than most simple molecules; however, colloidal particles are small enough that they do not settle out upon standing.
The particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect. This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure $2$. Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out.
The term “colloid”—from the Greek words kolla, meaning “glue,” and eidos, meaning “like”—was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units.
Analogous to the identification of solution components as “solute” and “solvent,” the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the dispersed phase and the substance or solution throughout which the particulate is dispersed is called the dispersion medium. Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table $1$.
Table $2$: Examples of Colloidal Systems
Dispersed Phase Dispersion Medium Common Examples Name
solid gas smoke, dust
solid liquid starch in water, some inks, paints, milk of magnesia sol
solid solid some colored gems, some alloys
liquid gas clouds, fogs, mists, sprays aerosol
liquid liquid milk, mayonnaise, butter emulsion
liquid solid jellies, gels, pearl, opal (H2O in SiO2) gel
gas liquid foams, whipped cream, beaten egg whites foam
gas solid pumice, floating soaps
Preparation of Colloidal Systems
We can prepare a colloidal system by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods:
1. Dispersion methods: that is, by breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills.
2. Condensation methods: that is, growth from smaller units, such as molecules or ions. For example, clouds form when water molecules condense and form very small droplets.
A few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air.
We can prepare an emulsion by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an emulsifying agent, a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents.
Condensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. We can prepare a red colloidal suspension of iron(III) hydroxide by mixing a concentrated solution of iron(III) chloride with hot water:
$\mathrm{Fe^{3+}}_{(aq)}+\mathrm{3Cl^-}_{(aq)}+\mathrm{6H_2O}_{(l)}⟶\mathrm{Fe(OH)}_{3(s)}+\mathrm{H_3O^+}_{(aq)}+\mathrm{3Cl^-}_{(aq)} \label{11.6.1}$
A colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate:
$\ce{Au}^{3+}+ \ce{3e}^− \rightarrow \ce{Au} \label{11.6.2}$
Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids.
Soaps and Detergents
Pioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, $\ce{K_2CO_3}$, from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula $\ce{C_{17}H_{35}CO_2Na}$ and contains an uncharged nonpolar hydrocarbon chain, the $\mathrm{C_{17}H_{35}-}$ unit, and an ionic carboxylate group, the $-\mathrm{\sideset{ }{_{2}^{-}}{CO}}$ unit (Figure $3$).
Detergents (soap substitutes) also contain nonpolar hydrocarbon chains, such as $\mathrm{C_{12}H_{25}—}$, and an ionic group, such as a sulfate $—\mathrm{\sideset{ }{_{3}^{-}}{OSO}}$, or a sulfonate $—\mathrm{\sideset{ }{_{3}^{-}}{SO}}$ (Figure $4$). Soaps form insoluble calcium and magnesium compounds in hard water; detergents form water-soluble products—a definite advantage for detergents.
The cleaning action of soaps and detergents can be explained in terms of the structures of the molecules involved. The hydrocarbon (nonpolar) end of a soap or detergent molecule dissolves in, or is attracted to, nonpolar substances such as oil, grease, or dirt particles. The ionic end is attracted by water (polar), illustrated in Figure $5$. As a result, the soap or detergent molecules become oriented at the interface between the dirt particles and the water so they act as a kind of bridge between two different kinds of matter, nonpolar and polar. Molecules such as this are termed amphiphilic since they have both a hydrophobic (“water-fearing”) part and a hydrophilic (“water-loving”) part. As a consequence, dirt particles become suspended as colloidal particles and are readily washed away.
Deepwater Horizon Oil Spill
The blowout of the Deepwater Horizon oil drilling rig on April 20, 2010, in the Gulf of Mexico near Mississippi began the largest marine oil spill in the history of the petroleum. In the 87 days following the blowout, an estimated 4.9 million barrels (210 million gallons) of oil flowed from the ruptured well 5000 feet below the water’s surface. The well was finally declared sealed on September 19, 2010.
Crude oil is immiscible with and less dense than water, so the spilled oil rose to the surface of the water. Floating booms, skimmer ships, and controlled burns were used to remove oil from the water’s surface in an attempt to protect beaches and wetlands along the Gulf coast. In addition to removal of the oil, attempts were also made to lessen its environmental impact by rendering it “soluble” (in the loose sense of the term) and thus allowing it to be diluted to hopefully less harmful levels by the vast volume of ocean water. This approach used 1.84 million gallons of the oil dispersant Corexit 9527, most of which was injected underwater at the site of the leak, with small amounts being sprayed on top of the spill. Corexit 9527 contains 2-butoxyethanol (C6H14O2), an amphiphilic molecule whose polar and nonpolar ends are useful for emulsifying oil into small droplets, increasing the surface area of the oil and making it more available to marine bacteria for digestion (Figure $6$). While this approach avoids many of the immediate hazards that bulk oil poses to marine and coastal ecosystems, it introduces the possibility of long-term effects resulting from the introduction of the complex and potential toxic components of petroleum into the ocean’s food chain. A number of organizations are involved in monitoring the extended impact of this oil spill, including the National Oceanic and Atmospheric Administration.
Electrical Properties of Colloidal Particles
Dispersed colloidal particles are often electrically charged. A colloidal particle of iron(III) hydroxide, for example, does not contain enough hydroxide ions to compensate exactly for the positive charges on the iron(III) ions. Thus, each individual colloidal particle bears a positive charge, and the colloidal dispersion consists of charged colloidal particles and some free hydroxide ions, which keep the dispersion electrically neutral. Most metal hydroxide colloids have positive charges, whereas most metals and metal sulfides form negatively charged dispersions. All colloidal particles in any one system have charges of the same sign. This helps keep them dispersed because particles containing like charges repel each other.
We can take advantage of the charge on colloidal particles to remove them from a variety of mixtures. If we place a colloidal dispersion in a container with charged electrodes, positively charged particles, such as iron(III) hydroxide particles, would move to the negative electrode. There, the colloidal particles lose their charge and coagulate as a precipitate.
The carbon and dust particles in smoke are often colloidally dispersed and electrically charged. The charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust (Figure $7$. This is one of the important methods used to clean up the smoke from a variety of industrial processes. The process is also important in the recovery of valuable products from the smoke and flue dust of smelters, furnaces, and kilns. There are also ionic air filters designed for home use to improve indoor air quality.
Gels
When we make gelatin, such as Jell-O, we are making a type of colloid (Figure $8$). Gelatin sets on cooling because the hot aqueous mixture of gelatin coagulates as it cools and the whole mass, including the liquid, sets to an extremely viscous body known as a gel, a colloid in which the dispersing medium is a solid and the dispersed phase is a liquid. It appears that the fibers of the dispersing medium form a complex three-dimensional network, the interstices being filled with the liquid medium or a dilute solution of the dispersing medium. Because the formation of a gel is accompanied by the taking up of water or some other solvent, the gel is said to be hydrated or solvated.
Pectin, a carbohydrate from fruit juices, is a gel-forming substance important in jelly making. Silica gel, a colloidal dispersion of hydrated silicon dioxide, is formed when dilute hydrochloric acid is added to a dilute solution of sodium silicate. Canned Heat is a gel made by mixing alcohol and a saturated aqueous solution of calcium acetate.
Summary
Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity, and they are often electrically charged. Colloids are widespread in nature and are involved in many technological applications.
Glossary
amphiphilic
molecules possessing both hydrophobic (nonpolar) and a hydrophilic (polar) parts
colloid
(also, colloidal dispersion) mixture in which relatively large solid or liquid particles are dispersed uniformly throughout a gas, liquid, or solid
dispersion medium
solid, liquid, or gas in which colloidal particles are dispersed
dispersed phase
substance present as relatively large solid or liquid particles in a colloid
emulsifying agent
amphiphilic substance used to stabilize the particles of some emulsions
emulsion
colloid formed from immiscible liquids
gel
colloidal dispersion of a liquid in a solid
Tyndall effect
scattering of visible light by a colloidal dispersion | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.5%3A_Colloids.txt |
11.2: The Dissolution Process
Q11.2.1
How do solutions differ from compounds? From other mixtures?
S11.2.1
A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous.
Q11.2.2
Which of the principal characteristics of solutions can we see in the solutions of $\ce{K2Cr2O7}$ shown in
Figure: When potassium dichromate ($\ce{K2Cr2O7}$) is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott)
S11.2.2
The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K2Cr2O7 in water can vary.
Q11.2.3
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally.
1. Is the dissolution of KNO3 an endothermic or an exothermic process?
2. What conclusions can you draw about the intermolecular attractions involved in the process?
3. Is the resulting solution an ideal solution?
S11.2.3
(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and $\ce{NO3-}$ ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption.
Q11.2.4
Give an example of each of the following types of solutions:
1. a gas in a liquid
2. a gas in a gas
3. a solid in a solid
S11.2.4
(a) CO2 in water; (b) O2 in N2 (air); (c) bronze (solution of tin or other metals in copper)
Q11.2.5
Indicate the most important types of intermolecular attractions in each of the following solutions:
1. The solution in Figure.
2. NO(l) in CO(l)
3. Cl2(g) in Br2(l)
4. HCl(aq) in benzene C6H6(l)
5. Methanol CH3OH(l) in H2O(l)
S11.2.5
(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding
Q11.2.5
Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C7H16, nonpolar solvent):
1. vegetable oil (nonpolar)
2. isopropyl alcohol (polar)
3. potassium bromide (ionic)
S11.2.5
(a) heptane; (b) water; (c) water
Q11.2.6
Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.
S11.2.6
Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat.
Q11.2.7
Solutions of hydrogen in palladium may be formed by exposing Pd metal to H2 gas. The concentration of hydrogen in the palladium depends on the pressure of H2 gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal.
1. Determine the molarity of this solution (solution density = 1.8 g/cm3).
2. Determine the molality of this solution (solution density = 1.8 g/cm3).
3. Determine the percent by mass of hydrogen atoms in this solution (solution density = 1.8 g/cm3).
S11.2.7
http://cnx.org/contents/mH6aqegx@2/The-Dissolution-Process
11.3: Electrolytes
Q11.3.1
Explain why the ions Na+ and Cl are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
S11.3.2
Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals.
Q11.3.2
Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.
S11.3.2
HBr is an acid and so its molecules react with water molecules to form H3O+ and Br ions that provide conductivity to the solution. Though HBr is soluble in benzene, it does not react chemically but remains dissolved as neutral HBr molecules. With no ions present in the benzene solution, it is electrically nonconductive.
Q11.3.3
Consider the solutions presented:
(a) Which of the following sketches best represents the ions in a solution of Fe(NO3)3(aq)?
(b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO3)3.
S11.3.3
(a) Fe(NO3)3 is a strong electrolyte, thus it should completely dissociate into Fe3+ and $\ce{(NO3- )}$ ions. Therefore, (z) best represents the solution. (b) $\ce{Fe(NO3)3}(s)⟶\ce{Fe^3+}(aq)+\ce{3NO3- }(aq)$
Q11.3.4
Compare the processes that occur when methanol (CH3OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
S11.3.3
Methanol, $CH_3OH$, dissolves in water in all proportions, interacting via hydrogen bonding.
Methanol:
$CH_3OH_{(l)}+H_2O_{(l)}⟶CH_3OH_{(aq)}$
Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions.
Hydrogen chloride:
$HCl{(g)}+H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)}$
Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively.
Sodium hydroxide:
$NaOH_{(s)} \rightarrow Na^+_{(aq)} + OH^−_{(aq)}$
Q11.3.5
What is the expected electrical conductivity of the following solutions?
1. NaOH(aq)
2. HCl(aq)
3. C6H12O6(aq) (glucose)
4. NH3(l)
S11.3.5
(a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved)
Q11.3.6
Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain.
S11.3.6
A medium must contain freely mobile, charged entities to be electrically conductive. The ions present in a typical ionic solid are immobilized in a crystalline lattice and so the solid is not able to support an electrical current. When the ions are mobilized, either by melting the solid or dissolving it in water to dissociate the ions, current may flow and these forms of the ionic compound are conductive.
Q11.3.7
Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:
1. the solutions in Figure
2. methanol, CH3OH, dissolved in ethanol, C2H5OH
3. methane, CH4, dissolved in benzene, C6H6
4. the polar halocarbon CF2Cl2 dissolved in the polar halocarbon CF2ClCFCl2
5. O2(l) in N2(l)
S11.3.7
(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces
11.4: Solubility
Q11.4.1
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3. How could you determine whether the solution is unsaturated, saturated, or supersaturated?
S11.4.1
Add a small crystal of $Na_2S_2O_3$. It will dissolve in an unsaturated solution, remain apparently unchanged in a saturated solution, or initiate precipitation in a supersaturated solution.
Q11.4.2
Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures.
S11.4.2
The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.
Q11.4.3
Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.
S11.4.3
The hydrogen bonds between water and C2H5OH are much stronger than the intermolecular attractions between water and C2H5SH.
Q11.4.4
Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C using the following figure for useful data, and report the computed percentage to one significant digit.
This graph shows how the solubility of several solids changes with temperature.
S11.4.4
At 10 °C, the solubility of KBr in water is approximately 60 g per 100 g of water.
$\%\; KBr =\dfrac{60\; g\; KBr}{(60+100)\;g\; solution} = 40\%$
Q11.4.5
Which of the following gases is expected to be most soluble in water? Explain your reasoning.
1. CH4
2. CCl4
3. CHCl3
S11.4.5
(c) CHCl3 is expected to be most soluble in water. Of the three gases, only this one is polar and thus capable of experiencing relatively strong dipole-dipole attraction to water molecules.
Q11.4.6
At 0 °C and 1.00 atm, as much as 0.70 g of O2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O2 dissolve in 1 L of water?
S11.4.6
This problem requires the application of Henry’s law. The governing equation is $C_g = kP_g$.
$k=\dfrac{C_g}{P_g}=\dfrac{0.70\;g}{1.00\; atm} =0.70\;g\; atm^{−1}$
Under the new conditions, $C_g=0.70\;g\;atm^{−1} \times 4.00\; atm = 2.80\; g$.
Q11.4.7
Refer to following figure for the following three questions:
1. How did the concentration of dissolved CO2 in the beverage change when the bottle was opened?
2. What caused this change?
3. Is the beverage unsaturated, saturated, or supersaturated with CO2?
S11.4.7
(a) It decreased as some of the CO2 gas left the solution (evidenced by effervescence). (b) Opening the bottle released the high-pressure CO2 gas above the beverage. The reduced CO2 gas pressure, per Henry’s law, lowers the solubility for CO2. (c) The dissolved CO2 concentration will continue to slowly decrease until equilibrium is reestablished between the beverage and the very low CO2 gas pressure in the opened bottle. Immediately after opening, the beverage, therefore, contains dissolved CO2 at a concentration greater than its solubility, a nonequilibrium condition, and is said to be supersaturated.
Q11.4.8
The Henry’s law constant for CO2 is $3.4 \times 10^{−2}\; M/atm$ at 25 °C. What pressure of carbon dioxide is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?
S11.4.8
$P_g=\dfrac{C_g}{k}=\dfrac{0.10\; M}{3.4 \times 10^{−2}\;M/atm} =2.9\; atm$
Q11.4.9
The Henry’s law constant for O2 is $1.3\times 10^{−3}\; M/atm$ at 25 °C. What mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O2 is 0.21 atm?
S11.4.9
Start with Henry's law
$C_g=kP_g$
and apply it to $O_2$
$C(O_2)=(1.3 \times 10^{−3}\; M/atm) (0.21\;atm)=2.7 \times 10^{−4}\;mol/L$
The total amount is $(2.7 \times 10^{−4}\; mol/L)(40\;L=1.08 \times 10^{−2} \;mol\] The mass of oxygen is \((1.08 \times 10^{−2}\; mol)(32.0\; g/mol)=0.346\;g$
or, using two significant figures, $0.35\; g$.
Q11.4.10
How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20-M solution of hydrochloric acid?
S11.4.10
First, calculate the moles of HCl needed. Then use the ideal gas law to find the volume required.
M = mol L−1
3.20M=xmol1.25L
x = 4.00 mol HCl
Before using the ideal gas law, change pressure to atmospheres and convert temperature from °C to kelvin.
\[1\;atmx=760torr745torr
x = 0.9803 atm
V=nRTP=(4.000molHCl)(0.08206LatmK−1mol−1)(303.15K)0.9803atm=102 L HCl
102 L HCl
2488fW6W@2/Solubility
11.5: Colligative Properties
Q11.5.1
Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion?
Q11.5.2
What is the microscopic explanation for the macroscopic behavior illustrated in [link]?
S11.5.2
The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region.
Q11.5.3
Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume.
Q11.5.4
A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C3H5(OH)3), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical?
S11.5.4
Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions.
Q11.5.5
What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?
Q11.5.6
What are the mole fractions of HNO3 and water in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
S11.5.6
1. Find number of moles of HNO3 and H2O in 100 g of the solution. Find the mole fractions for the components.
2. The mole fraction of HNO3 is 0.378. The mole fraction of H2O is 0.622.
Q11.5.7
Calculate the mole fraction of each solute and solvent:
1. 583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
2. 0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
3. 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
4. 25 g of I2 in 125 g of ethanol, C2H5OH
S11.5.7
a. $583\:g\:\ce{H2SO4}\times\dfrac{1\:mole\:\ce{H2SO4}}{98.08\:g\:\ce{H2SO4}}=5.94\:mole\:\ce{H2SO4}$ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ $1.50\:kg\:\ce{H2O}\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mole\:\ce{H2O}}{18.02\:g\:\ce{H2O}}=83.2\:moles\:\ce{H2O}$
Q11.5.8
Calculate the mole fraction of each solute and solvent:
1. 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0 °C
2. 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
3. 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
4. 0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3
S11.5.8
1. $X_\mathrm{Na_2CO_3}=0.0119$; $X_\mathrm{H_2O}=0.988$;
2. $X_\mathrm{NH_4NO_3}=0.9927$; $X_\mathrm{H_2O}=0.907$;
3. $X_\mathrm{Cl_2}=0.192$; $X_\mathrm{CH_2CI_2}=0.808$;
4. $X_\mathrm{C_5H_9N}=0.00426$; $X_\mathrm{CHCl_3}=0.997$
Q11.5.9
Calculate the mole fractions of methanol, CH3OH; ethanol, C2H5OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.)
Q11.5.10
What is the difference between a 1 M solution and a 1 m solution?
S11.5.10
In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.
Q11.5.11
What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?
Q11.5.12
What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
S11.5.12
(a) Determine the molar mass of HNO3. Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m
Q11.5.13
Calculate the molality of each of the following solutions:
1. 583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
2. 0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
3. 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
4. 25 g of I2 in 125 g of ethanol, C2H5OH
Q11.5.14
Calculate the molality of each of the following solutions:
1. 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C
2. 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
3. 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
4. 0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3
S11.5.14
(a) 6.70 × 10−1 m; (b) 5.67 m; (c) 2.8 m; (d) 0.0358 m
Q11.5.15
The concentration of glucose, C6H12O6, in normal spinal fluid is $\mathrm{\dfrac{75\:mg}{100\:g}}$. What is the molality of the solution?
Q11.5.16
A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.
1.08 m
Q11.5.17
1. Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin?
2. What is the boiling point of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water?
S11.5.17
1. Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point.
2. 100.5 °C
Q11.5.18
What is the boiling point of a solution of 9.04 g of I2 in 75.5 g of benzene, assuming the I2 is nonvolatile?
Q11.5.19
What is the freezing temperature of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water, which freezes at 0.0 °C when pure?
S11.5.19
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C
Q11.5.20
What is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?
Q11.5.21
What is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO3)2 in water at 25 °C? The volume of the solution is 275 mL.
S11.5.21
(a) Determine the molar mass of Ca(NO3)2; determine the number of moles of Ca(NO3)2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm
Q11.5.22
What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol−1) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin?
Q11.5.23
What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; Kb = 5.02 °C/m) that boils at 81.5 °C at 1 atm?
S11.5.24
(a) Determine the molal concentration from the change in boiling point and Kb; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 × 102 g mol−1
Q11.5.25
A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound.
Q11.5.26
A 1.0 m solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain.
S11.5.26
No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by ΔTf = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is ΔTf = (1.0 m)(5.14 °C/m) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized.
Q11.5.27
A solution contains 5.00 g of urea, CO(NH2)2, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
Q11.5.28
A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance.
144 g mol−1
Q11.5.29
Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na3PO4, 0.1 m C2H5OH, 0.01 m CO2, 0.15 m NaCl, and 0.2 m CaCl2.
Q11.5.30
Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na2SO4, and 0.030 mol of MgCl2, assuming complete dissociation of these electrolytes.
0.870 °C
Q11.5.31
How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? What is the freezing point of this solution?
Q11.5.32
A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS2 (Kb = 2.43 °C/m). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide?
S8
Q11.5.33
In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI2?
Q11.5.34
Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 × 10−3 atm at 25 °C. What is the molar mass of lysozyme?
S11.5.34
1.39 × 104 g mol−1
Q11.5.35
The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. What is the molar mass of insulin?
Q11.5.36
The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C6H12O6, is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C?
54 g
Q11.5.37
What is the freezing point of a solution of dibromobenzene, C6H4Br2, in 0.250 kg of benzene, if the solution boils at 83.5 °C?
Q11.5.38
What is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C?
100.26 °C
Q11.5.39
The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and Kb for ethanol is 1.20 °C/m. What is the molecular formula of fructose?
Q11.5.40
The vapor pressure of methanol, CH3OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C2H5OH, is 44 torr at the same temperature.
1. Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol.
2. Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at 20 °C.
3. Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.
S11.5.40
(a) $X_\mathrm{CH_3OH}=0.590$; $X_\mathrm{C_2H_5OH}=0.410$; (b) Vapor pressures are: CH3OH: 55 torr; C2H5OH: 18 torr; (c) CH3OH: 0.75; C2H5OH: 0.25
Q11.5.41
The triple point of air-free water is defined as 273.15 K. Why is it important that the water be free of air?
Q11.5.42
Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature?
S11.5.42
The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C.
Q11.5.43
An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. Kf for camphor is 37.7 °C/m. What is the molecular formula of the solute? Show your calculations.
Q11.5.44
A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C2H5OH (Kb = 1.20 °C/m). The boiling point elevation of the solution is 1.27 °C. Is HgCl2 an electrolyte in ethanol? Show your calculations.
S11.5.44
$\mathrm{Δbp}=K_\ce{b}m=(1.20\:°\ce C/m)\mathrm{\left(\dfrac{9.41\:g×\dfrac{1\:mol\: HgCl_2}{271.496\:g}}{0.03275\:kg}\right)=1.27\:°\ce C}$
The observed change equals the theoretical change; therefore, no dissociation occurs.
Q11.5.45
A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations.
11.6: Colloids
Q11.6.1
Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby.
S11.6.1
Colloidal System Dispersed Phase Dispersion Medium
starch dispersion starch water
smoke solid particles air
fog water air
pearl water calcium carbonate (CaCO3)
whipped cream air cream
floating soap air soap
jelly fruit juice pectin gel
milk butterfat water
ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3)
Q11.6.2
Distinguish between dispersion methods and condensation methods for preparing colloidal systems.
S11.6.2
Dispersion methods use a grinding device or some other means to bring about the subdivision of larger particles. Condensation methods bring smaller units together to form a larger unit. For example, water molecules in the vapor state come together to form very small droplets that we see as clouds.
Q11.6.3
How do colloids differ from solutions with regard to dispersed particle size and homogeneity?
S11.6.3
Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.
Q11.6.4
Explain the cleansing action of soap.
S11.6.4
Soap molecules have both a hydrophobic and a hydrophilic end. The charged (hydrophilic) end, which is usually associated with an alkali metal ion, ensures water solubility The hydrophobic end permits attraction to oil, grease, and other similar nonpolar substances that normally do not dissolve in water but are pulled into the solution by the soap molecules.
Q11.6.5
How can it be demonstrated that colloidal particles are electrically charged?
S11.6.5
If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.E%3A_Solutions_and_Colloids_%28Exercises%29.txt |
Among the many capabilities of chemistry is its ability to predict if a process will occur under specified conditions. Thermodynamics, the study of relationships between the energy and work associated with chemical and physical processes, provides this predictive ability. Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions.
12: Thermodynamics
Skills to Develop
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous processes
In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur by force. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions and how relatively quickly or slowly that natural change proceeds. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.
Spontaneous and Nonspontaneous Processes
Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. Iron exposed to the earth’s atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.
The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure $1$).
As another example, consider the conversion of diamond into graphite (Figure $2$).
$\ce{C}_{(s,\textrm{ diamond})}⟶\ce{C}_{(s,\textrm{ graphite})} \label{Eq1}$
The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.
Figure $2$:The conversion of carbon from the diamond allotrope to the graphite allotrope is spontaneous at ambient pressure, but its rate is immeasurably slow at low to moderate temperatures. This process is known as graphitization, and its rate can be increased to easily measurable values at temperatures in the 1000–2000 K range. (credit "diamond" photo: modification of work by "Fancy Diamonds"/Flickr; credit "graphite" photo: modificaton of work by images-of-elements.com/carbon.php)
Dispersal of Matter and Energy
As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure $3$). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.
$w=−PΔV=0 \;\;\; \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2}$
Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process.
$ΔU=q+w=0+0=0 \label{Eq3}$
The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the driving force appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).
Figure $3$:An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a second flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks.
Now consider two objects at different temperatures: object X at temperature TX and object Y at temperature TY, with TX > TY (Figure $4$). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.
$q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4}$
From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.
Figure $4$:When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object.
As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.
Example $1$: Redistribution of Matter during a Spontaneous Process
Describe how matter is redistributed when the following spontaneous processes take place:
1. A solid sublimes.
2. A gas condenses.
3. A drop of food coloring added to a glass of water forms a solution with uniform color.
Solution
Figure $5$:(credit a: modification of work by Jenny Downing; credit b: modification of work by “Fuzzy Gerdes”/Flickr; credit c: modification of work by Sahar Atwa)
1. (a) Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition.
2. (b) Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after the solid-to-gas transition.
3. (c) The process in question is dilution. The food dye molecules initially occupy a much smaller volume (the drop of dye solution) than they occupy once the process is complete (in the full glass of water). The process therefore entails a greater dispersal of matter. The process may also yield a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop, zero in the water), and the final state of the system contains a single dye concentration throughout.
Exercise $1$
Describe how matter and/or energy is redistributed when you empty a canister of compressed air into a room.
Answer:
This is also a dilution process, analogous to example (c). It entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room.
Summary
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system.
Glossary
nonspontaneous process
process that requires continual input of energy from an external source
spontaneous change
process that takes place without a continuous input of energy from an external source | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/12%3A_Thermodynamics/12.1%3A_Spontaneity.txt |
Learning Objectives
• Define entropy
• Explain the relationship between entropy and the number of microstates
• Predict the sign of the entropy change for chemical and physical processes
In 1824, at the age of 28, Nicolas Léonard Sadi Carnot (Figure $2$) published the results of an extensive study regarding the efficiency of steam heat engines. In a later review of Carnot’s findings, Rudolf Clausius introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the reversible heat (qrev) and the kelvin temperature (T). The term reversible process refers to a process that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be “reversed”) by an infinitesimally small change is some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as irreversible.
Similar to other thermodynamic properties, this new quantity is a state function, and so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property entropy (S) and defined its change for any process as the following:
$ΔS=\dfrac{q_\ce{rev}}{T} \label{Eq1}$
The entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states.
Entropy and Microstates
Following the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates possible for the system. A microstate ($\Omega$) is a specific configuration of the locations and energies of the atoms or molecules that comprise a system like the following:
$S=k \ln \Omega \label{Eq2}$
Here k is the Boltzmann constant and has a value of $1.38 \times 10^{−23}\, J/K$.
As for other state functions, the change in entropy for a process is the difference between its final (Sf) and initial (Si) values:
\begin{align} ΔS &=S_\ce{f}−S_\ce{i} \nonumber \[4pt] &=k \ln \Omega_\ce{f} − k \ln \Omega_\ce{i} \nonumber \[4pt] &=k \ln\dfrac{\Omega_\ce{f}}{\Omega_\ce{i}} \label{Eq2a} \end{align}
For processes involving an increase in the number of microstates of the system, $\Omega_f > \Omega_i$, the entropy of the system increases, $ΔS > 0$. Conversely, processes that reduce the number of microstates in the system, $\Omega_f < \Omega_i$, yield a decrease in system entropy, $ΔS < 0$. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs.
Consider the general case of a system comprised of N particles distributed among n boxes. The number of microstates possible for such a system is nN. For example, distributing four particles among two boxes will result in 24 = 16 different microstates as illustrated in Figure $2$. Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions (sometimes called macrostates or configurations). The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates, the most probable distribution is therefore the one of greatest entropy.
For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is
$\dfrac{6}{16} = \dfrac{3}{8} \nonumber$
The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of
$\dfrac{1}{16} \nonumber$
The probability of finding all particles in only one box (either the left box or right box) is then
$\left(\dfrac{1}{16}+\dfrac{1}{16}\right)=\dfrac{2}{16} = \dfrac{1}{8} \nonumber$
As you add more particles to the system, the number of possible microstates increases exponentially (2N). A macroscopic (laboratory-sized) system would typically consist of moles of particles (N ~ 1023), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations.
The most probable distribution is therefore the one of greatest entropy.
The previous description of an ideal gas expanding into a vacuum is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. The spontaneous process whereby the gas contained initially in one flask expands to fill both flasks equally therefore yields an increase in entropy for the system.
A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of energy (represented as “*”) in Figure $3$. The hot object is comprised of particles A and B and initially contains both energy units. The cold object is comprised of particles C and D, which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. And so, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is $\frac{3}{10}$. More likely is the flow of heat to yield one of the other two distribution, the combined probability being $\frac{7}{10}$. The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being $\frac{4}{10}$. As for the previous example of matter dispersal, extrapolating this treatment to macroscopic collections of particles dramatically increases the probability of the uniform distribution relative to the other distributions. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects’ temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy.
Example $1$: Determination of ΔS
Consider the system shown here. What is the change in entropy for a process that converts the system from distribution (a) to (c)?
Solution
We are interested in the following change:
The initial number of microstates is one, the final six:
\begin{align} ΔS &=k \ln\dfrac{\Omega_\ce{c}}{\Omega_\ce{a}} \nonumber \[4pt] &= 1.38×10^{−23}\:J/K × \ln\dfrac{6}{1} \nonumber \[4pt] &= 2.47×10^{−23}\:J/K \nonumber \end{align} \nonumber
The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive.
Exercise $1$
Consider the system shown in Figure $3$. What is the change in entropy for the process where all the energy is transferred from the hot object (AB) to the cold object (CD)?
Answer
0 J/K
Predicting the Sign of ΔS
The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure $4$. In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system’s component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, Sliquid > Ssolid and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, ΔS > 0. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, ΔS < 0.
Now consider the vapor or gas phase. The atoms or molecules occupy a much greater volume than in the liquid phase; therefore each atom or molecule can be found in many more locations than in the liquid (or solid) phase. Consequently, for any substance, Sgas > Sliquid > Ssolid, and the processes of vaporization and sublimation likewise involve increases in entropy, ΔS > 0. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, ΔS < 0.
According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure $5$ ).
The entropy of a substance is influenced by structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle’s mass and the spacing of quantized translational energy levels (which is a topic beyond the scope of our treatment). For molecules, greater numbers of atoms (regardless of their masses) increase the ways in which the molecules can vibrate and thus the number of possible microstates and the system entropy.
Finally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, ΔS > 0.
Considering the various factors that affect entropy allows us to make informed predictions of the sign of ΔS for various chemical and physical processes as illustrated in Example .
Example $2$: Predicting the Sign of ∆S
Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions.
1. One mole liquid water at room temperature $⟶$ one mole liquid water at 50 °C
2. $\ce{Ag+}(aq)+\ce{Cl-}(aq)⟶\ce{AgCl}(s)$
3. $\ce{C6H6}(l)+\dfrac{15}{2}\ce{O2}(g)⟶\ce{6CO2}(g)+\ce{3H2O}(l)$
4. $\ce{NH3}(s)⟶\ce{NH3}(l)$
Solution
1. positive, temperature increases
2. negative, reduction in the number of ions (particles) in solution, decreased dispersal of matter
3. negative, net decrease in the amount of gaseous species
4. positive, phase transition from solid to liquid, net increase in dispersal of matter
Exercise $2$
Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.
1. $\ce{NaNO3}(s)⟶\ce{Na+}(aq)+\ce{NO3-}(aq)$
2. the freezing of liquid water
3. $\ce{CO2}(s)⟶\ce{CO2}(g)$
4. $\ce{CaCO}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)$
Answer a
Positive; The solid dissolves to give an increase of mobile ions in solution.
Answer b
Negative; The liquid becomes a more ordered solid.
Answer c
Positive; The relatively ordered solid becomes a gas
Answer d
Positive; There is a net production of one mole of gas.
Summary
Entropy ($S$) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system. For a given substance, $S_{solid} < S_{liquid} \ll S_{gas}$ in a given physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions may be reliably predicted.
Key Equations
• $ΔS=\dfrac{q_\ce{rev}}{T}$
• S = k ln W
• $ΔS=k\ln\dfrac{W_\ce{f}}{W_\ce{i}}$
Glossary
entropy (S)
state function that is a measure of the matter and/or energy dispersal within a system, determined by the number of system microstates often described as a measure of the disorder of the system
microstate (W)
possible configuration or arrangement of matter and energy within a system
reversible process
process that takes place so slowly as to be capable of reversing direction in response to an infinitesimally small change in conditions; hypothetical construct that can only be approximated by real processes removed | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/12%3A_Thermodynamics/12.2%3A_Entropy.txt |
Learning Objectives
• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
The Second Law of Thermodynamics
In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the systemS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:
$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{1}$
To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following: $ΔS_\ce{sys}=\dfrac{−q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{q_\ce{rev}}{T_\ce{surr}} \label{$2$}$ The arithmetic signs of qrev denote the loss of heat by the system and the gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of ΔSsys and ΔSsurr will yield a positive value for ΔSuniv. This process involves an increase in the entropy of the universe.
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following: $ΔS_\ce{sys}=\dfrac{q_\ce{rev}}{T_\ce{sys}}\hspace{20px}\ce{and}\hspace{20px}ΔS_\ce{surr}=\dfrac{−q_\ce{rev}}{T_\ce{surr}} \label{$3$}$ The arithmetic signs of qrev denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The temperature difference between the objects is infinitesimally small, TsysTsurr, and so the heat flow is thermodynamically reversible. See the previous section’s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.
These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table $1$.
Table $1$: The Second Law of Thermodynamics
ΔSuniv > 0 spontaneous
ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction)
ΔSuniv = 0 reversible (system is at equilibrium)
Definition: The Second Law of Thermodynamics
All spontaneous changes cause an increase in the entropy of the universe.
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, $q_{surr}$ is a good approximation of $q_{rev}$, and the second law may be stated as the following:
$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4}$
We may use this equation to predict the spontaneity of a process as illustrated in Example $1$.
Example $1$: Will Ice Spontaneously Melt?
The entropy change for the process
$\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber$
is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?
Solution
We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.
At −10.00 °C (263.15 K), the following is true:
\begin{align*} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \ &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}=−0.7\:J/K} \end{align*} \nonumber
$S_{univ} < 0$, so melting is nonspontaneous (not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
\begin{align*} ΔS_\ce{univ} &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \[4pt] &=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K \end{align*} \nonumber
$S_{univ} > 0$, so melting is spontaneous at 10.00 °C.
Exercise $1$
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv?
Answer
Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
The Third Law of Thermodynamics
The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (W = 1). According to the Boltzmann equation, the entropy of this system is zero.
$S=k\ln W=k\ln(1)=0 \label{5}$
This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.
Definition: Third Law of Thermodynamics
The entropy of a pure, perfect crystalline substance at 0 K is zero.
We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies are given the label $S^\circ_{298}$ for values determined for one mole of substance, isolated in its pure form in its own container, at a pressure of 1 bar and a temperature of 298 K.
Definition: Term
The thermodynamic standard state of a substance refers to an isolated sample of that substance, in its own container, at 1.000 bar (0.9869 atm) pressure. If the substance is a solute, the most common standard state is one in which the concentration of the solute is 1.000 molal (sometimes approximated with 1.000 M). There is no defined temperature for the standard state, but most discussions about standard state assume that the temperature is 298.15 K (25ºC) unless otherwise noted.
This may seem like a strange definition, because it requires that each of the reactants and each of the products of a reaction are kept separate from one another, unmixed. The entropy of mixing must be determined separately.
The standard entropy change (ΔS°) for any process may be computed from the standard entropies of its reactant and product species like the following:
$ΔS°=\sum νS^\circ_{298}(\ce{products})−\sum νS^\circ_{298}(\ce{reactants}) \label{$6$}$
Here, ν represents stoichiometric coefficients in the balanced equation representing the process. For example, ΔS° for the following reaction at room temperature
$m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D} \label{$7$}$
is computed as the following:
$=[xS^\circ_{298}(\ce{C})+yS^\circ_{298}(\ce{D})]−[mS^\circ_{298}(\ce{A})+nS^\circ_{298}(\ce{B})] \label{$8$}$
Table $2$ lists some standard entropies at 298.15 K. You can find additional standard entropies in Tables T1 or T2.
Table $2$: Standard Entropies (at 298.15 K, 1 atm)
Substance $S^\circ_{298} \, \dfrac{J}{mol \, K}$
carbon
C(s, graphite) 5.740
C(s, diamond) 2.38
CO(g) 197.7
CO2(g) 213.8
CH4(g) 186.3
C2H4(g) 219.5
C2H6(g) 229.5
CH3OH(l) 126.8
C2H5OH(l) 160.7
hydrogen
H2(g) 130.57
H(g) 114.6
H2O(g) 188.71
H2O(l) 69.91
HCI(g) 186.8
H2S(g) 205.7
oxygen
O2(g) 205.03
Example $2$: Determination of ΔS°
Calculate the standard entropy change for the following process:
$\ce{H2O}(g)⟶\ce{H2O}(l) \nonumber$
Solution
The value of the standard entropy change at room temperature, $ΔS^\circ_{298}$, is the difference between the standard entropy of the product, H2O(l), and the standard entropy of the reactant, H2O(g).
\begin{align*} ΔS^\circ_{298}&=S^\circ_{298}(\ce{H2O}(l))−S^\circ_{298}(\ce{H2O}(g))\[4pt] &= (70.0\: J\:mol^{−1}K^{−1})−(188.8\: Jmol^{−1}K^{−1})=−118.8\:J\:mol^{−1}K^{−1} \end{align*} \nonumber
The value for $ΔS^o_{298}$ is negative, as expected for this phase transition (condensation), which the previous section discussed.
Exercise $2$
Calculate the standard entropy change for the following process:
$\ce{H2}(g)+\ce{C2H4}(g)⟶\ce{C2H6}(g) \nonumber$
Answer
−120.6 J mol−1 K−1
Example $3$: Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH at room temperature:
$\ce{2CH3OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{4H2O}(l) \nonumber$
Solution
The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients.
\begin{align*} ΔS^\circ &=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants}) \[4pt] &=[2S^\circ_{298}(\ce{CO2}(g))+4S^\circ_{298}(\ce{H2O}(l))]−[2S^\circ_{298}(\ce{CH3OH}(l))+3S^\circ_{298}(\ce{O2}(g))] \[4pt] &=\{[2(213.8)+4×70.0]−[2(126.8)+3(205.03)]\}=−161.1\:J/mol⋅K \end{align*} \nonumber
Exercise $3$
Calculate the standard entropy change for the following reaction:
$\ce{Ca(OH)2}(s)⟶\ce{CaO}(s)+\ce{H2O}(l) \nonumber$
Answer
24.7 J/mol•K
Summary
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.
Key Equations
• $ΔS^\circ=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants})$
• $ΔS=\dfrac{q_\ce{rev}}{T}$
• ΔSuniv = ΔSsys + ΔSsurr
• $ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}$
Glossary
second law of thermodynamics
entropy of the universe increases for a spontaneous process
standard entropy (S°)
entropy for a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K and denoted $S^\circ_{298}$
standard entropy change (ΔS°)
change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted $ΔS^\circ_{298}$
third law of thermodynamics
entropy of a perfect crystal at absolute zero (0 K) is zero | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/12%3A_Thermodynamics/12.3%3A_The_Second_and_Third_Laws_of_Thermodynamics.txt |
Learning Objectives
• Define Gibbs free energy, and describe its relation to spontaneity
• Calculate the standard free energy change for a process using standard free energies of formation for its reactants and products
• Calculate standard free energy change for a process using senthalpies of formation and the entropies for its reactants and products
• Explain how temperature affects the spontaneity of some processes
• Relate standard free energy changes to equilibrium constants
One of the challenges of using the second law of thermodynamics to determine if a chemical reaction is spontaneous is that we must determine the entropy change for the system and the entropy change for the surroundings. A second challenge when working with a chemical reaction is that we need to take into account the mixing of the substances, an issue that does not occur when observing the phase change if a pure substance. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy ($G$) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:
$G=H−TS \nonumber$
Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following:
$ΔG^º_\ce{sys}=ΔH^º_\ce{sys}−TΔS^º_\ce{sys} \nonumber$
(For simplicity’s sake, the subscript “sys” will be omitted henceforth.) We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression:
$ΔS_\ce{univ}=ΔS+\dfrac{q_\ce{surr}}{T} \nonumber$
The first law requires that $q_{surr} = −q_{sys}$, and at constant pressure $q_{sys} = ΔH$, and so this expression may be rewritten as the following:
$ΔS_\ce{univ}=ΔS−\dfrac{ΔH}{T} \nonumber$
ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:
$−TΔS_\ce{univ}=ΔH−TΔS \nonumber$
Comparing this equation to the previous one for free energy change shows the following relation:
$ΔG=−TΔS_\ce{univ} \label{6}$
The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, $ΔS_{univ}$. Table $1$ summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.
Table $1$: Relation between Process Spontaneity and Signs of Thermodynamic Properties
$ΔS_{univ} > 0$ ΔG < 0 moves spontaneously in the forward direction, as written, to reach equilibrium
$ΔS_{univ} < 0$ ΔG > 0
nonspontaneous in the forward direction, as written, but moves spontaneously in the reverse direction, as written, to reach equilibrium
$ΔS_{univ} = 0$ ΔG = 0 reversible (at equilibrium)
Calculating Free Energy Change
Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system that have undergone some change. A convenient and common approach to the calculation of free energy changes for physical changes and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in Example $1$.
$ΔG°=ΔH°−TΔS° \label{7}$
It is important to understand that for phase changes, $\Delta G^º$ tells you if the phase change is spontaneous or not; will it happen, or not happen. For chemical reactions, $\Delta G^º$ tells you the extent of a reaction. In other words, $\Delta G^º$ for a reaction tells you how much product will be present at equilibrium. A reaction with $\Delta G^º$ < 0 is considered product-favored at equilibrium; there will be more products than reactants when the reaction reaches equilibrium. A reaction with $\Delta G^º$ > 0 is considered reactant-favored at equilibrium; there will be more reactants than products when the reaction reaches equilibrium.
Example $1$: Evaluation of ΔG° for a Phase Change of a Pure Substance
Use standard enthalpy and entropy data from Tables T1 or T2 to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this physical change for a pure substance?
Solution
The process of interest is the following:
$\ce{H2O}(l)⟶\ce{H2O}(g) \label{$8$}$
The standard change in free energy may be calculated using the following equation:
$ΔG^\circ_{298}=ΔH°−TΔS° \label{$9$}$
From Tables T1 or T2, here are the data:
Data from Tables T1 or T2
Substance $ΔH^\circ_\ce{f}\ce{(kJ/mol)}$ $S^\circ_{298}\textrm{(J/K⋅mol)}$
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8
Combining at 298 K:
\begin{align*} ΔH°&=ΔH^\circ_{298}=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \[4pt] &=[−241.82\: kJ−(−285.83)]\:kJ/mol \[4pt] &=44.01\: kJ/mol \[4pt] ΔS° &=ΔS^\circ_{298}=S^\circ_{298}(\ce{H2O}(g))−S^\circ_{298}(\ce{H2O}(l)) \[4pt] &=188.8\:J/mol⋅K−70.0\:J/K \[4pt] &=118.8\:J/mol⋅K \end{align*} \nonumber
then use Equation \ref{7}:
$ΔG°=ΔH°−TΔS° \nonumber$
Converting everything into kJ and combining at 298 K:
\begin{align*}ΔG^\circ_{298} &=ΔH°−TΔS° \[4pt] &=44.01\: kJ/mol−(298\: K×118.8\:J/mol⋅K)×\dfrac{1\: kJ}{1000\: J} \end{align*} \nonumber
$\mathrm{44.01\: kJ/mol−35.4\: kJ/mol=8.6\: kJ/mol} \nonumber$
At 298 K (25 °C) $ΔG^\circ_{298}>0$, and so boiling is nonspontaneous (not spontaneous) at 298 K.
Exercise $1$: Evaluation of ΔG° for a Chemical Reaction
Use standard enthalpy and entropy data from Tables T1 or T2 to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the extent of this reaction at 298 K?
$\ce{C2H6}(g)⟶\ce{H2}(g)+\ce{C2H4}(g) \nonumber$
Answer
$ΔG^\circ_{298}=\mathrm{102.0\: kJ/mol}$; the reaction is reactant-favored at equilibrium at 25 °C. There will be more $\ce{C2H6}(g)$ than $\ce{H2}(g)$ and $\ce{C2H4}(g)$ at equilibrium
Free energy changes may also use the standard free energy of formation $(ΔG^\circ_\ce{f})$, for each of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of formation, $(ΔG^\circ_\ce{f})$ is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
$m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D}, \nonumber$
the standard free energy change at room temperature may be calculated as
\begin{align} ΔG^\circ_{298}&=ΔG° \[4pt] &=∑νΔG^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants})\[4pt] &=[xΔG^\circ_\ce{f}(\ce{C})+yΔG^\circ_\ce{f}(\ce{D})]−[mΔG^\circ_\ce{f}(\ce{A})+nΔG^\circ_\ce{f}(\ce{B})]. \end{align} \nonumber
Example $2$: Calculation of $ΔG^\circ_{298}$
Consider the decomposition of yellow mercury(II) oxide.
$\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+ \ce{ 1/2 O2(g)} \nonumber$
Calculate the standard free energy change at room temperature, $ΔG^\circ_{298}$, using:
1. standard free energies of formation and
2. standard enthalpies of formation and standard entropies.
Do the results indicate the reaction to be product-favored or reactant-favored at equilibrium?
Solution
The required data are available in Tables T1 or T2 and are shown here.
Required data for Tables T1 or T2
Compound $ΔG^\circ_\ce{f}\:\mathrm{(kJ/mol)}$ $ΔH^\circ_\ce{f}\:\mathrm{(kJ/mol)}$ $S^\circ_{298}\:\textrm{(J/K⋅mol)}$
HgO (s, yellow) −58.43 −90.46 71.13
Hg(l) 0 0 75.9
O2(g) 0 0 205.2
(a) Using free energies of formation:
\begin{align*} ΔG^\circ_{298}&=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) \[4pt] &=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \[4pt] & \mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} \end{align*} \nonumber
(b) Using enthalpies and entropies of formation:
\begin{align*}ΔH^\circ_{298}&=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) \[4pt] &=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \[4pt] &\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} \[4pt] ΔS^\circ_{298} &=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) \[4pt] &=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \[4pt] & \mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} \end{align*} \nonumber
then we can use Equation \ref7} directly:
\begin{align*}ΔG°&=ΔH°−TΔS°\[4pt] &=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} \[4pt] &=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} \end{align*} \nonumber
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is reactant-favored at equilibrium at room temperature.
Exercise $2$
Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Tables T1 or T2). Do the results indicate the reaction to be product-favored or reactant-favored at equilibrium at 25 °C?
$\ce{C2H4}(g)⟶\ce{H2}(g)+\ce{C2H2}(g) \nonumber$
Answer
141.5 kJ/mol, reactant-favored at equilibrium
Temperature Dependence of Spontaneity and Extent of Reaction
As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. In a similar, but not identical fashion, some chemical reactions can switch from being product-favored at equilibrium, to being reactant-favored at equilibrium, depending on the temperature.
Note
The numerical value of $\Delta G^º$ is always dependent on the temperature. In this section we are determining whether or not the sign of $\Delta G^º$ is dependent on the temperature.
To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
$ΔG^º=ΔH^º−TΔS^º \nonumber$
The extent of a process, as reflected in the arithmetic sign of its standard free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (Kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:
1. Both ΔHº and ΔSº are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is greater than ΔHº. If the TΔSº term is less than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at high temperatures and reactant-favored at equilibrium at low temperatures.
2. Both ΔHº and ΔSº are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is less than ΔHº. If the TΔSº term’s magnitude is greater than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at low temperatures and reactant-favored at equilibrium at high temperatures.
3. ΔHº is positive and ΔSº is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔGº will be positive regardless of the temperature. Such a process is reactant-favored at equilibrium at all temperatures.
4. ΔHº is negative and ΔSº is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔGº will be negative regardless of the temperature. Such a process is product-favored at equilibrium at all temperatures.
These four scenarios are summarized in Table $1$
Table $1$
Sign of $\Delta H^o$
Sign of $\Delta S^o$
Sign of $\Delta G^o$
Temperature Dependence of $\Delta G^o$
- + - The sign of $\Delta G^o$ does not depend on the temperature.The reaction is product-favored at equilibrium at all temperatures.
+ - + The sign of $\Delta G^o$ does not depend on the temperature.The reaction is reactant-favored at equilibrium at all temperatures.
- - - or + The sign of $\Delta G^o$ does depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
+ + - or +
The sign of $\Delta G^o$ does depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
Example $3$: Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:
$\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber$
Does the sign of $\Delta G^º$ of this process depend upon temperature?
Solution
Combustion processes are exothermic ($ΔH^º < 0$). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, $ΔS^º > 0$). The reaction is therefore product-favored at equilibrium ($ΔG^º < 0$) at all temperatures.
Exercise $3$
Popular chemical hand warmers generate heat by the air-oxidation of iron:
$\ce{4Fe}(s)+\ce{3O2}(g)⟶\ce{2Fe2O3}(s) \nonumber$
Does the sign of $\Delta G^o$ of this process depend upon temperature?
Answer
ΔHº and ΔSº are both negative; the reaction is product-favored at equilibrium at low temperatures.
When considering the conclusions drawn regarding the temperature dependence of the sign of ΔGº, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is reactant-favored at equilibrium at one temperature but product-favored at equilibrium at another temperature will necessarily undergo a change in “extent” (as reflected by its ΔGº) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔGº is plotted on the y axis versus T on the x axis:
$ΔG^º=ΔH^º−TΔS^º \nonumber$
$y=b+mx \nonumber$
Such a plot is shown in Figure $2$. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependence for the sign of ΔGº as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔGº) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔGº is zero:
$ΔG^º=0=ΔH^º−TΔS^º \nonumber$
$T=\dfrac{ΔH^º}{ΔS^º} \nonumber$
Thus, saying a process is product-favored at equilibrium at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔGº for the process is zero.
Note
In this discussion, we have used two different descriptions for the meaning of the sign of ΔGº. You should be aware of the meaning of each description.
a) Extent of Reaction: This description is used to predict the ratio of the product and reactant concentrations at equilibrium. In this description, we use the thermodynamic term ΔGº to tell us the same information as the equilibrium constant, K. When ΔGº < 0, K > 1, and the reaction will be product-favored at equilibrium. When ΔGº > 0, K< 1, and the reaction is reactant-favored at equilibrium. When ΔGº = 0, K =1, and the reaction will have roughly equal amounts of products and reactants at equilibrium. In all cases, the reaction will form a mixture of products and reactants at equilibrium. We use the sign and magnitude of ΔGº to tell us how much product will be made if the reaction is allowed to reach equilibrium.
b) Spontaneity: This description is much more complicated because it involves two different interpretations of how a reaction at standard state occurs. One interpretation involves the hypothetical process in which the reaction proceeds from a starting point of pure reactants to a finishing point of pure products, with all substances isolated in their own containers under standard state conditions. In the second, more realistic interpretation, the reaction starts with all reactants and all products in their standard state in one container. We then allow this specific mixture to react an infinitesimally small amount so that we can obtain a rate of change in free energy with respect to the extent of reaction when all reactants and products are mixed and (essentially) in their standard states. Although each interpretation describes a different reaction scenario, the value of the difference in free energy and the value of the rate of change in free energy are the same number. If ΔGº < 0, we say that the reaction is spontaneous, meaning that the reaction would proceed in the forward direction, as written, to form pure products in their standard state. If ΔGº > 0, we say that the reaction is nonspontaneous, meaning that the reaction would proceed in the reverse direction, as written, to form pure reactants in their standard state. If ΔGº = 0, we say that the neither the reactants nor the products are favored to be formed.
A detailed treatment of the meaning of ΔGº can be found in the paper, "Free Energy versus Extent of Reaction" by Richard S. Treptow, Journal of Chemical Education, 1996, Volume 73 (1), 51-54.
Example $4$: Equilibrium Temperature for a Phase Transition
As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Tables T1 or T2 to estimate the boiling point of water.
Solution
The process of interest is the following phase change:
$\ce{H2O}(l)⟶\ce{H2O}(g) \nonumber$
When this process is at equilibrium, ΔG = 0, so the following is true:
$0=ΔH°−TΔS°\hspace{40px}\ce{or}\hspace{40px}T=\dfrac{ΔH°}{ΔS°} \nonumber$
Using the standard thermodynamic data from Tables T1 or T2,
\begin{align*} ΔH°&=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\ &=\mathrm{−241.82\: kJ/mol−(−285.83\: kJ/mol)=44.01\: kJ/mol} \nonumber \end{align*} \nonumber
\begin{align*} ΔS°&=ΔS^\circ_{298}(\ce{H2O}(g))−ΔS^\circ_{298}(\ce{H2O}(l)) \nonumber\ &=\mathrm{188.8\: J/K⋅mol−70.0\: J/K⋅mol=118.8\: J/K⋅mol} \nonumber \end{align*} \nonumber
$T=\dfrac{ΔH°}{ΔS°}=\mathrm{\dfrac{44.01×10^3\:J/mol}{118.8\:J/K⋅mol}=370.5\:K=97.3\:°C} \nonumber$
The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Tables T1 or T2.). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.
Exercise $4$
Use the information in Tables T1 or T2 to estimate the boiling point of CS2.
Answer
313 K (accepted value 319 K).
Free Energy and Equilibrium
The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).
In the chapter on equilibrium the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.
The free energy change for a process taking place with reactants and products present under nonstandard conditions, ΔG, is related to the standard free energy change, ΔG°, according to this equation:
$ΔG=ΔG°+RT\ln Q \label{eq10A}$
R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in Example $1$.
Example $5$: Calculating ΔG under Nonstandard Conditions
What is the free energy change for the process shown here under the specified conditions?
T = 25 °C, $P_{\ce{N2}}=\mathrm{0.870\: atm}$, $P_{\ce{H2}}=\mathrm{0.250\: atm}$, and $P_{\ce{NH3}}=\mathrm{12.9\: atm}$
$\ce{2NH3}(g)⟶\ce{3H2}(g)+\ce{N2}(g) \hspace{20px} ΔG°=\mathrm{33.0\: kJ/mol} \nonumber$
Solution
Equation \ref{eq10A} relates free energy change to standard free energy change and reaction quotient and may be used directly:
\begin{align*} ΔG&=ΔG°+RT\ln Q \[4pt] &=\mathrm{33.0\:\dfrac{kJ}{mol}+\left(8.314\:\dfrac{J}{mol\: K}×298\: K×\ln\dfrac{(0.250^3)×0.870}{12.9^2}\right)}\[4pt] &=\mathrm{9680\:\dfrac{J}{mol}\:or\: 9.68\: kJ/mol} \end{align*} \nonumber
Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions. The reaction will proceed in the reverse direction to reach equilibrium.
Exercise $5$
Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?
Answer
ΔG = −47 kJ; yes, the reaction proceeds in the forward direction, as written, to reach equilibrium.
For a system at equilibrium, Q = K and ΔG = 0, and the Equation \ref{eq10A} may be written as
$\underbrace{0=ΔG°+RT\ln K}_{\text{at equilibrium}} \nonumber$
$ΔG°=−RT\ln K \label{eq4A}$
or alternatively
$K=e^{−\frac{ΔG°}{RT}} \label{eq4B}$
This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table $1$.
Table $1$: Relations between Standard Free Energy Changes and Equilibrium Constants
K ΔG° Comments
< 1 > 0 Reactants are more abundant at equilibrium.
= 1 = 0 Reactants and products are equally abundant at equilibrium.
> 1 < 0 Products are more abundant at equilibrium.
Example $6$: Equilibrium Constant using Standard Free Energy Change
Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.
Solution
The reaction of interest is the following:
$\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ag+][Cl- ]} \nonumber$
The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:
\begin{align*} ΔG° =ΔG^\circ_{298} &=[ΔG^\circ_\ce{f}(\ce{Ag+}(aq))+ΔG^\circ_\ce{f}(\ce{Cl-}(aq))]−[ΔG^\circ_\ce{f}(\ce{AgCl}(s))] \[4pt] &=[77.1\: kJ/mol−131.2\: kJ/mol]−[−109.8\: kJ/mol] \[4pt] &=55.7\: kJ/mol \end{align*} \nonumber
The equilibrium constant for the reaction may then be derived from its standard free energy change via Equation \ref{eq4B}:
\begin{align*} K_\ce{sp}&=e^{−\dfrac{ΔG°}{RT}}=\exp\left(−\dfrac{ΔG°}{RT}\right) \[4pt] &=\mathrm{\exp\left(−\dfrac{55.7×10^3\:J/mol}{8.314\:J/mol⋅K×298.15\:K}\right)}\&=\mathrm{\exp(−22.470)=e^{−22.470}=1.74×10^{−10}} \end{align*} \nonumber
Exercise $6$: dissociation of dinitrogen tetroxide
Use the thermodynamic data provided in Tables T1 or T2 to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
$\ce{NO}_{2(g)} \rightleftharpoons \ce{N_2O}_{4(g)} \nonumber$
Answer
K = 6.9
To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure $3$). If a system is present with reactants and products present in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.
Summary
Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates that the process will proceed in the forward direction to reach equilibrium; a positive ΔG indicates that the process will proceed in the reverse direction to reach equilibrium ; and a ΔG of zero indicates that the system is at equilibrium. A negative value for ΔGº means that the reaction is product-favored at equilibrium. A positive value for ΔGº means that the reaction is reactant-favored at equilibrium. If ΔGº equals 0 (a rare occurrence), the reaction has roughly equal amounts of reactants and products at equilibrium.A number of approaches to the computation of free energy changes are possible.
Key Equations
• ΔG = ΔHTΔS
• ΔG = ΔG° + RT ln Q
• ΔG° = −RT ln K
Glossary
Gibbs free energy change (G)
thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G
standard free energy change (ΔG°)
change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)
standard free energy of formation $(ΔG^\circ_\ce{f})$
change in free energy accompanying the formation of one mole of substance from its elements in their standard states | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/12%3A_Thermodynamics/12.4%3A_Gibbs_Energy.txt |
16.1: Spontaneity Exercises
Q16.1.1
What is a spontaneous reaction?
S16.1.1
A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.
Q16.1.2
What is a nonspontaneous reaction?
Q16.1.3
Indicate whether the following processes are spontaneous or nonspontaneous.
1. Liquid water freezing at a temperature below its freezing point
2. Liquid water freezing at a temperature above its freezing point
3. The combustion of gasoline
4. A ball thrown into the air
5. A raindrop falling to the ground
6. Iron rusting in a moist atmosphere
S16.1.2
spontaneous; nonspontaneous; spontaneous; nonspontaneous; spontaneous; spontaneous
Q16.1.4
A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
Q16.1.5
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
S16.1.5
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
16.2: Entropy Exercises
Q16.2.1
In the below Figure all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
Q16.2.2
In Figure all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, for the system when it is converted from distribution to distribution (d).
S16.2.2
There are four initial microstates and four final microstates.
$ΔS=k\ln\dfrac{W_\ce{f}}{W_\ce{i}}=\mathrm{1.38×10^{−23}\:J/K×\ln\dfrac{4}{4}=0}$
Q16.2.3
How does the process described in the previous item relate to the system shown in [link]?
Q16.2.4
Consider a system similar to the one below, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to $\dfrac{1}{8}$. What does this comparison tell us about even larger systems?
S16.2.4
The probability for all the particles to be on one side is $\dfrac{1}{32}$. This probability is noticeably lower than the $\dfrac{1}{8}$ result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
Q16.2.5
Consider the system shown in Figure. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?
Q16.2.6
Consider the system shown in Figure. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
S16.2.6
There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.
$ΔS=k\ln\left(\dfrac{W_\ce{f}}{W_\ce{i}}\right)=\mathrm{1.38×10^{−23}\:J/K×\ln\left(\dfrac{4}{1}\right)=1.91×10^{−23}\:J/K}$
Q16.2.7
Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.
1. H2(g), HBrO4(g), HBr(g)
2. H2O(l), H2O(g), H2O(s)
3. He(g), Cl2(g), P4(g)
Q16.2.8
At room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.
S16.2.8
The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I2 is a solid, Br2 is a liquid, and Cl2 is a gas.
Q16.2.9
Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature but somewhat higher pressure).
$\ce{I2}(s)⟶\ce{I2}(g)$
$\ce{I2}(s)⟶\ce{I2}(l)$
Is ΔS positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?
Q16.2.11
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
1. C2H5OH(l) or C3H7OH(l)
2. C2H5OH(l) or C2H5OH(g)
3. 2Hor H(g)
S16.2.11
C3H7OH(l) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. C2H5OH(g) as it is in the gaseous state. 2H(g), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
Q16.2.11
Predict the sign of the entropy change for the following processes:
1. An ice cube is warmed to near its melting point.
2. Exhaled breath forms fog on a cold morning.
3. Snow melts.
Q16.2.12
Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.
1. $\ce{Pb^2+}(aq)+\ce{S^2-}(aq)⟶\ce{PbS}(s)$
2. $\ce{2Fe}(s)+\ce{3O2}(g)⟶\ce{Fe2O3}(s)$
3. $\ce{2C6H14}(l)+\ce{19O2}(g)⟶\ce{14H2O}(g)+\ce{12CO2}(g)$
S16.2.12
Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. Negative. There is a net loss of three moles of gas from reactants to products. Positive. There is a net increase of seven moles of gas from reactants to products.
Q16.2.13
Write the balanced chemical equation for the combustion of methane, CH4(g), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether ΔS is positive or negative for this chemical reaction.
Q16.2.14
Write the balanced chemical equation for the combustion of benzene, C6H6(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process?
S16.2.14
$\ce{C6H6}(l)+7.5\ce{O2}(g)⟶\ce{3H2O}(g)+\ce{6CO2}(g)$
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and ΔS is positive.
16.3: The Second and Third law
Q16.3.0
What is the difference between ΔS, ΔS°, and $ΔS^\circ_{298}$ for a chemical change?
Q16.3.1
Calculate $ΔS^\circ_{298}$ for the following changes.
1. $\ce{SnCl4}(l)⟶\ce{SnCl4}(g)$
2. $\ce{CS2}(g)⟶\ce{CS2}(l)$
3. $\ce{Cu}(s)⟶\ce{Cu}(g)$
4. $\ce{H2O}(l)⟶\ce{H2O}(g)$
5. $\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(l)$
6. $\ce{2HCl}(g)+\ce{Pb}(s)⟶\ce{PbCl2}(s)+\ce{H2}(g)$
7. $\ce{Zn}(s)+\ce{CuSO4}(s)⟶\ce{Cu}(s)+\ce{ZnSO4}(s)$
S16.3.1
107 J/K; −86.4 J/K; 133.2 J/K; 118.8 J/K; −326.6 J/K; −171.9 J/K; (g) −7.2 J/K
Q16.3.2
Determine the entropy change for the combustion of liquid ethanol, C2H5OH, under standard state conditions to give gaseous carbon dioxide and liquid water.
Q16.3.3
Determine the entropy change for the combustion of gaseous propane, C3H8, under standard state conditions to give gaseous carbon dioxide and water.
100.6 J/K
Q16.3.4
“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is:
$\ce{Fe2O3}(s)+\ce{2Al}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)$
Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
Q16.3.5
Using the relevant $S^\circ_{298}$ values listed in Appendix G, calculate $S^\circ_{298}$ for the following changes:
1. $\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g)$
2. $\ce{N2}(g)+\dfrac{5}{2}\ce{O2}(g)⟶\ce{N2O5}(g)$
S16.3.5
−198.1 J/K; −348.9 J/K
Q16.3.6
From the following information, determine $ΔS^\circ_{298}$ for the following:
1. $\ce{N}(g)+\ce{O}(g)⟶\ce{NO}(g) \hspace{20px} ΔS^\circ_{298}=\,?$
2. $\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{24.8\: J/K}$
3. $\ce{N2}(g)⟶\ce{2N}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{115.0\: J/K}$
4. $\ce{O2}(g)⟶\ce{2O}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{117.0\: J/K}$
Q16.3.7
By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
$S^\circ_{\ce{NaCl}(s)}=\mathrm{72.11\:\dfrac{J}{mol⋅K}}\hspace{40px} S^\circ_{\ce{NaCl}(l)}=\mathrm{95.06\:\dfrac{J}{mol⋅K}}\hspace{40px ΔH^\circ_\ce{fusion}=\mathrm{27.95\: kJ/mol}$
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
S16.3.7
As ΔSuniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
Q16.3.8
Use the standard entropy data in Appendix G to determine the change in entropy for each of the reactions listed in [link]. All are run under standard state conditions and 25 °C.
Q16.3.8
2.86 J/K; 24.8 J/K; −113.2 J/K; −24.7 J/K; 15.5 J/K; 290.0 J/K
16.4: Free Energy
Q16.4.1
What is the difference between ΔG, ΔG°, and $ΔG^\circ_{298}$ for a chemical change?
Q16.4.2
A reactions has $ΔH^\circ_{298}$ = 100 kJ/mol and $ΔS^\circ_{298}=\textrm{250 J/mol⋅K}$. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
S16.4.2
The reaction is nonspontaneous at room temperature. Above 400 K, ΔG will become negative, and the reaction will become spontaneous.
Q16.4.3
Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
1. $\ce{MnO2}(s)⟶\ce{Mn}(s)+\ce{O2}(g)$
2. $\ce{H2}(g)+\ce{Br2}(l)⟶\ce{2HBr}(g)$
3. $\ce{Cu}(s)+\ce{S}(g)⟶\ce{CuS}(s)$
4. $\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g)$
5. $\ce{CH4}(g)+\ce{O2}(g)⟶\ce{C}(s,\,\ce{graphite})+\ce{2H2O}(g)$
6. $\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g)$
S16.4.3
465.1 kJ nonspontaneous; −106.86 kJ spontaneous; −53.6 kJ spontaneous; −83.4 kJ spontaneous; −406.7 kJ spontaneous; −30.0 kJ spontaneous
Q16.4.4
Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
1. $\ce{C}(s,\, \ce{graphite})+\ce{O2}(g)⟶\ce{CO2}(g)$
2. $\ce{O2}(g)+\ce{N2}(g)⟶\ce{2NO}(g)$
3. $\ce{2Cu}(s)+\ce{S}(g)⟶\ce{Cu2S}(s)$
4. $\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)$
5. $\ce{Fe2O3}(s)+\ce{3CO}(g)⟶\ce{2Fe}(s)+\ce{3CO2}(g)$
6. $\ce{CaSO4⋅2H2O}(s)⟶\ce{CaSO4}(s)+\ce{2H2O}(g)$
Given:
$\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s) \hspace{20px} ΔG^\circ_{298}=\mathrm{−2697.0\: kJ/mol}$
$\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{−457.18\: kJ/mol}$
$\ce{6H2O}(g)+\ce{P4O10}(g)⟶\ce{4H3PO4}(l) \hspace{20px} ΔG^\circ_{298}=\mathrm{−428.66\: kJ/mol}$
Q16.4.5
1. Determine the standard free energy of formation, $ΔG^\circ_\ce{f}$, for phosphoric acid.
2. How does your calculated result compare to the value in Appendix G? Explain.
S16.4.5
−1124.3 kJ/mol for the standard free energy change. The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.
Q16.4.6
Is the formation of ozone (O3(g)) from oxygen (O2(g)) spontaneous at room temperature under standard state conditions?
Q16.4.7
Consider the decomposition of red mercury(II) oxide under standard state conditions.
$\ce{2HgO}(s,\,\ce{red})⟶\ce{2Hg}(l)+\ce{O2}(g)$
1. Is the decomposition spontaneous under standard state conditions?
2. Above what temperature does the reaction become spontaneous?
S16.4.7
The reaction is nonspontaneous; Above 566 °C the process is spontaneous.
Q16.4.8
Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
1. Ammonia: $\ce{2NH3}(g)⟶\ce{N2}(g)+\ce{3H2}(g)$
2. Diborane: $\ce{B2H6}(g)⟶\ce{2B}(g)+\ce{3H2}(g)$
3. Hydrazine: $\ce{N2H4}(g)⟶\ce{N2}(g)+\ce{2H2}(g)$
4. Hydrogen peroxide: $\ce{H2O2}(l)⟶\ce{H2O}(g)+\dfrac{1}{2}\ce{O2}(g)$
Q16.4.9
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
1. $\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} \mathrm{T=2000\:°C} \hspace{20px} K_p=4.1×10^{−4}$
2. $\ce{H2}(g)+\ce{I2}(g)⟶\ce{2HI}(g) \hspace{20px} \mathrm{T=400\:°C} \hspace{20px} K_p=50.0$
3. $\ce{CO2}(g)+\ce{H2}(g)⟶\ce{CO}(g)+\ce{H2O}(g) \hspace{20px} \mathrm{T=980\:°C} \hspace{20px} K_p=1.67$
4. $\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g) \hspace{20px} \mathrm{T=900\:°C} \hspace{20px} K_p=1.04$
5. $\ce{HF}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{F-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=7.2×10^{−4}$
6. $\ce{AgBr}(s)⟶\ce{Ag+}(aq)+\ce{Br-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=3.3×10^{−13}$
S16.4.9
1.5 × 102 kJ; −21.9 kJ; −5.34 kJ; −0.383 kJ; 18 kJ; 71 kJ
Q16.4.10
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
1. $\ce{Cl2}(g)+\ce{Br2}(g)⟶\ce{2BrCl}(g) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=4.7×10^{−2}$
2. $\ce{2SO2}(g)+\ce{O2}(g)⇌\ce{2SO3}(g) \hspace{20px} \mathrm{T=500\:°C} \hspace{20px} K_p=48.2$
3. $\ce{H2O}(l)⇌\ce{H2O}(g) \hspace{20px} \mathrm{T=60\:°C} \hspace{20px} K_p=\mathrm{0.196\: atm}$
4. $\ce{CoO}(s)+\ce{CO}(g)⇌\ce{Co}(s)+\ce{CO2}(g) \hspace{20px} \mathrm{T=550\:°C} \hspace{20px} K_p=4.90×10^2$
5. $\ce{CH3NH2}(aq)+\ce{H2O}(l)⟶\ce{CH3NH3+}(aq)+\ce{OH-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=4.4×10^{−4}$
6. $\ce{PbI2}(s)⟶\ce{Pb^2+}(aq)+\ce{2I-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=8.7×10^{−9}$
Q16.4.11
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
1. $\ce{O2}(g)+\ce{2F2}(g)⟶\ce{2OF2}(g) \hspace{20px} ΔG°=\mathrm{−9.2\: kJ}$
2. $\ce{I2}(s)+\ce{Br2}(l)⟶\ce{2IBr}(g) \hspace{20px} ΔG°=\mathrm{7.3\: kJ}$
3. $\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g) \hspace{20px} ΔG°=\mathrm{−79\: kJ}$
4. $\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g) \hspace{20px} ΔG°=\mathrm{−1.6\: kJ}$
5. $\ce{SnCl4}(l)⟶\ce{SnCl4}(l) \hspace{20px} ΔG°=\mathrm{8.0\: kJ}$
S16.4.11
K = 41; K = 0.053; K = 6.9 × 1013; K = 1.9; K = 0.04
Q16.4.2
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
1. $\ce{I2}(s)+\ce{Cl2}(g)⟶\ce{2ICl}(g) \hspace{20px} ΔG°=\mathrm{−10.88\: kJ}$
2. $\ce{H2}(g)+\ce{I2}(s)⟶\ce{2HI}(g) \hspace{20px} ΔG°=\mathrm{3.4\: kJ}$
3. $\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g) \hspace{20px} ΔG°=\mathrm{−39\: kJ}$
4. $\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} ΔG°=\mathrm{−141.82\: kJ}$
5. $\ce{CS2}(g)⟶\ce{CS2}(l) \hspace{20px} ΔG°=\mathrm{−1.88\: kJ}$
Q16.4.13
Calculate the equilibrium constant at the temperature given.
1. (a) $\ce{O2}(g)+\ce{2F2}(g)⟶\ce{2F2O}(g) \hspace{20px} \mathrm{(T=100\:°C)}$
2. $\ce{I2}(s)+\ce{Br2}(l)⟶\ce{2IBr}(g) \hspace{20px} \mathrm{(T=0.0\:°C)}$
3. $\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g) \hspace{20px} \mathrm{(T=575\:°C)}$
4. $\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g) \hspace{20px} \mathrm{(T=−10.0\:°C)}$
5. $\ce{SnCl4}(l)⟶\ce{SnCl4}(g) \hspace{20px} \mathrm{(T=200\:°C)}$
S16.4.13
In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG° from the values ΔH° and ΔS° and then calculate ΔG from the relation ΔG° = ΔH° − TΔS°.
1. K = 1.29
2. K = 2.51 × 10−3
3. K = 4.83 × 103
4. K = 0.219
5. K = 16.1
Q16.4.14
Calculate the equilibrium constant at the temperature given.
1. (a) $\ce{I2}(s)+\ce{Cl2}(g)⟶\ce{2ICl}(g) \hspace{20px} \mathrm{(T=100\:°C)}$
2. $\ce{H2}(g)+\ce{I2}(s)⟶\ce{2HI}(g) \hspace{20px} \mathrm{(T=0.0\:°C)}$
3. $\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g) \hspace{20px} \mathrm{(T=125\:°C)}$
4. $\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} \mathrm{(T=675\:°C)}$
5. $\ce{CS2}(g)⟶\ce{CS2}(l) \hspace{20px} \mathrm{(T=90\:°C)}$
Q16.4.15
Consider the following reaction at 298 K:
$\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_P=0.142$
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
S16.4.16
The standard free energy change is $ΔG^\circ_{298}=−RT\ln K=\mathrm{4.84\: kJ/mol}$. When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and $ΔG_{298}$ becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.
Q16.4.17
Determine the normal boiling point (in kelvin) of dichloroethane, CH2Cl2. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
Q16.4.18
Under what conditions is $\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g)$ spontaneous?
S16.4.18
The reaction will be spontaneous at temperatures greater than 287 K.
Q16.4.19
At room temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.00 × 10−14. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
Q16.4.20
Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction $\ce{2H2S}(g)+\ce{SO2}(g)⇌\dfrac{3}{8}\ce{S8}(s,\,\ce{rhombic})+\ce{2H2O}(l)$. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
S16.4.20
K = 5.35 × 1015
The process is exothermic.
Q16.4.21
Consider the decomposition of CaCO3(s) into CaO(s) and CO2(g). What is the equilibrium partial pressure of CO2 at room temperature?
Q16.4.22
In the laboratory, hydrogen chloride (HCl(g)) and ammonia (NH3(g)) often escape from bottles of their solutions and react to form the ammonium chloride (NH4Cl(s)), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH3 in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
S16.4.22
1.0 × 10−8 atm. This is the maximum pressure of the gases under the stated conditions.
Q16.4.23
Benzene can be prepared from acetylene. $\ce{3C2H2}(g)⇌\ce{C6H6}(g)$. Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
Q16.4.24
Carbon dioxide decomposes into CO and O2 at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO2 was 1.15 atm?
$x=\mathrm{1.29×10^{−5}\:atm}=P_{\ce{O2}}$
Q16.4.25
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
$\ce{CH4}(g)+\ce{4Cl2}(g)⟶\ce{CCl4}(g)+\ce{4HCl}(g)$
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
Q16.4.25B
Acetic acid, CH3CO2H, can form a dimer, (CH3CO2H)2, in the gas phase.
$\ce{2CH3CO2H}(g)⟶\ce{(CH3CO2H)2}(g)$
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
At 25 °C, the equilibrium constant for the dimerization is 1.3 × 103 (pressure in atm). What is ΔS° for the reaction?
−0.16 kJ
Q16.4.26
Nitric acid, HNO3, can be prepared by the following sequence of reactions:
$\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g)$
$\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)$
$\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(l)+\ce{NO}(g)$
How much heat is evolved when 1 mol of NH3(g) is converted to HNO3(l)? Assume standard states at 25 °C.
Q16.4.27A
Determine ΔG for the following reactions.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
$\ce{SbCl5}(g)⟶\ce{SbCl3}(g)+\ce{Cl2}(g)$
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
Chlorine molecules dissociate according to this reaction:
$\ce{Cl2}(g)⟶\ce{2Cl}(g)$
1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.
S16.4.27A
1. (a) −22.1 kJ;
2. 61.6 kJ/mol
Q16.4.27
Given that the $ΔG^\circ_\ce{f}$ for Pb2+(aq) and Cl(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).
Q16.4.28
Determine the standard free energy change, $ΔG^\circ_\ce{f}$, for the formation of S2−(aq) given that the $ΔG^\circ_\ce{f}$ for Ag+(aq) and Ag2S(s) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 × 10−51.
90 kJ/mol
Q16.4.29
Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
Q16.4.30
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
$\ce{H2O}(l)⇌\ce{H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{8.58\: kJ}$
1. (a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
2. Determine the equilibrium constant, KP, for this physical process.
3. By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, $P_{\ce{H2O}}$, is 0.011 atm.
4. If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of $P_{\ce{H2O}}$ in the air?
S16.4.30
(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; Kp = 0.031; The evaporation of water is spontaneous; $P_{\ce{H2O}}$ must always be less than Kp or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
Q16.4.31
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
$\mathrm{Glu + ATP ⟶ G6P + ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−17\: kJ}$
In this process, ATP becomes ADP summarized by the following equation:
$\mathrm{ATP⟶ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−30\: kJ}$
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
$\mathrm{Glu⟶G6P} \hspace{20px} ΔG^\circ_{298}=\:?$
Q16.4.32
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
$\mathrm{G6P⇌F6P} \hspace{20px} ΔG^\circ_{298}=\mathrm{1.7\: kJ}$
1. (a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
2. Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
S16.4.32
(a) Nonspontaneous as $ΔG^\circ_{298}>0$; $ΔG^\circ_{298}=−RT\ln K,$ $ΔG = 1.7×10^3 + \left(8.314 × 335 × \ln\dfrac{28}{128}\right) = \mathrm{−2.5\: kJ}$. The forward reaction to produce F6P is spontaneous under these conditions.
Q16.4.33
Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
1. (a) $\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g)$
2. $\ce{HCl}(g)+\ce{NH3}(g)⟶\ce{NH4Cl}(s)$
3. $\ce{(NH4)2Cr2O7}(s)⟶\ce{Cr2O3}(s)+\ce{4H2O}(g)+\ce{N2}(g)$
4. $\ce{2Fe}(s)+\ce{3O2}(g)⟶\ce{Fe2O3}(s)$
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
S16.4.33
ΔG is negative as the process is spontaneous. ΔH is positive as with the solution becoming cold, the dissolving must be endothermic. ΔS must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
Q16.4.34
An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the following equation:
$\ce{Cu2S}(s)⟶\ce{Cu}(s)+\ce{S}(s)$
1. (a) Determine $ΔG^\circ_{298}$ for the decomposition of Cu2S(s).
2. The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine $ΔG^\circ_{298}$ for the process.
3. The production of copper from chalcocite is performed by roasting the Cu2S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
Q16.4.35
What happens to $ΔG^\circ_{298}$ (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
1. (a) $\ce{S}(s)+\ce{O2}(g)⟶\ce{SO2}(g)$
2. $\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{SO3}(g)$
3. $\ce{HgO}(s)⟶\ce{Hg}(l)+\ce{O2}(g)$
S16.4.35
1. (a) Increasing $P_{\ce{O2}}$ will shift the equilibrium toward the products, which increases the value of K. $ΔG^\circ_{298}$ therefore becomes more negative.
2. Increasing $P_{\ce{O2}}$ will shift the equilibrium toward the products, which increases the value of K. $ΔG^\circ_{298}$ therefore becomes more negative.
3. Increasing $P_{\ce{O2}}$ will shift the equilibrium the reactants, which decreases the value of K. $ΔG^\circ_{298}$ therefore becomes more positive. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/12%3A_Thermodynamics/12.E%3A_Thermodynamics_%28Exercises%29.txt |
In this chapter, you will learn how to predict the position of the balance and the yield of a product of a reaction under specific conditions, how to change a reaction's conditions to increase or reduce yield, and how to evaluate an equilibrium system's reaction to disturbances.
13: Fundamental Equilibrium Concepts
Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot and want to cool off, they head into the surf to swim. As the swimmers tire, they head to the beach to rest. If these two rates of transfer (sunbathers entering the water, swimmers leaving the water) are equal, the number of sunbathers and swimmers would be constant, or at equilibrium, although the identities of the people are constantly changing from sunbather to swimmer and back. An analogous situation occurs in chemical reactions. Reactions can occur in both directions simultaneously (reactants to products and products to reactants) and eventually reach a state of balance.
These balanced two-way reactions occur all around and even in us. For example, they occur in our blood, where the reaction between carbon dioxide and water forms carbonic acid \(\ce{(HCO3- )}\) (Figure \(1\)). Human physiology is adapted to the amount of ionized products produced by this reaction (\(\ce{HCO3-}\) and H+). In this chapter, you will learn how to predict the position of the balance and the yield of a product of a reaction under specific conditions, how to change a reaction's conditions to increase or reduce yield, and how to evaluate an equilibrium system's reaction to disturbances.
13.1: Chemical Equilibria
Learning Objectives
• Describe the nature of equilibrium systems
• Explain the dynamic nature of a chemical equilibrium
A chemical reaction is usually written in a way that suggests it proceeds in one direction, the direction in which we read, but all chemical reactions are reversible, and both the forward and reverse reaction occur to one degree or another depending on conditions. In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. If we run a reaction in a closed system so that the products cannot escape, we often find the reaction does not give a 100% yield of products. Instead, some reactants remain after the concentrations stop changing. At this point, when there is no further change in concentrations of reactants and products, we say the reaction is at equilibrium. A mixture of reactants and products is found at equilibrium.
For example, when we place a sample of dinitrogen tetroxide ($N_2O_4$, a colorless gas) in a glass tube, it forms nitrogen dioxide ($\ce{NO2}$, a brown gas) by the reaction
$\ce{ N2O4(g) \rightleftharpoons 2NO2(g)} \label{13.2.1}$
The color becomes darker as $\ce{N2O4}$ is converted to $\ce{NO2}$. When the system reaches equilibrium, both $\ce{N2O4}$ and $\ce{NO2}$ are present (Figure $1$).
The formation of $\ce{NO2}$ from $\ce{N2O4}$ is a reversible reaction, which is identified by the equilibrium arrow ($\rightleftharpoons$). All reactions are reversible, but many reactions, for all practical purposes, proceed in one direction until the reactants are exhausted and will reverse only under certain conditions. Such reactions are often depicted with a one-way arrow from reactants to products. Many other reactions, such as the formation of $\ce{NO2}$ from $\ce{N2O4}$, are reversible under more easily obtainable conditions and, therefore, are named as such. In a reversible reaction, the reactants can combine to form products and the products can react to form the reactants. Thus, not only can $\ce{N2O4}$ decompose to form $\ce{NO2}$, but the $\ce{NO2}$ produced can react to form $\ce{N2O4}$. As soon as the forward reaction produces any $\ce{NO2}$, the reverse reaction begins and $\ce{NO2}$ starts to react to form $\ce{N2O4}$. At equilibrium, the concentrations of $\ce{N2O4}$ and $\ce{NO2}$ no longer change because the rate of formation of $\ce{NO2}$ is exactly equal to the rate of consumption of $\ce{NO2}$, and the rate of formation of $\ce{N2O4}$ is exactly equal to the rate of consumption of $\ce{N2O4}$. Chemical equilibrium is a dynamic process: As with the swimmers and the sunbathers, the numbers of each remain constant, yet there is a flux back and forth between them (Figure $2$).
In a chemical equilibrium, the forward and reverse reactions do not stop, rather they continue to occur at the same rate, leading to constant concentrations of the reactants and the products. Plots showing how the reaction rates and concentrations change with respect to time are shown in Figure $1$.
We can detect a state of equilibrium because the concentrations of reactants and products do not appear to change. However, it is important that we verify that the absence of change is due to equilibrium and not to a reaction rate that is so slow that changes in concentration are difficult to detect.
We use a double arrow when writing an equation for a reversible reaction. Such a reaction may or may not be at equilibrium. For example, Figure $1$ shows the reaction:
$\ce{N2O4(g) \rightleftharpoons 2NO2(g)} \label{13.2.2}$
When we wish to speak about one particular component of a reversible reaction, we use a single arrow. For example, in the equilibrium shown in Figure $1$, the rate of the forward reaction
$\ce{2NO2(g) \rightarrow N2O4(g)} \label{13.2.3}$
is equal to the rate of the backward reaction
$\ce{N2O4(g) \rightarrow 2NO2(g)} \label{13.2.4}$
Equilibrium and Soft Drinks
The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804; mostly known today for his role in the discovery and identification of oxygen) discovered a method of infusing water with carbon dioxide to make carbonated water. In 1772, Priestly published a paper entitled “Impregnating Water with Fixed Air.” The paper describes dripping oil of vitriol (today we call this sulfuric acid, but what a great way to describe sulfuric acid: “oil of vitriol” literally means “liquid nastiness”) onto chalk (calcium carbonate). The resulting $CO_2$ falls into the container of water beneath the vessel in which the initial reaction takes place; agitation helps the gaseous $CO_2$ mix into the liquid water.
$\ce{H2SO4(l) + CaCO3(s) \rightarrow CO2(g) + H2O (l) + CaSO4 (aq)} \nonumber$
Carbon dioxide is slightly soluble in water. There is an equilibrium reaction that occurs as the carbon dioxide reacts with the water to form carbonic acid ($H_2CO_3$). Since carbonic acid is a weak acid, it can dissociate into protons ($H^+$) and hydrogen carbonate ions ($HCO_3^−$).
$\ce{ CO2 (aq) + H2O(l) \rightleftharpoons H2CO3 (aq) \rightleftharpoons HCO3^{-} (aq) + H^{+} (aq)} \nonumber$
Today, $\ce{CO_2}$ can be pressurized into soft drinks, establishing the equilibrium shown above. Once you open the beverage container, however, a cascade of equilibrium shifts occurs. First, the $\ce{CO_2}$ gas in the air space on top of the bottle escapes, causing the equilibrium between gas-phase $\ce{CO_2}$ and dissolved or aqueous $\ce{CO_2}$ to shift, lowering the concentration of $\ce{CO_2}$ in the soft drink. Less $\ce{CO_2}$ dissolved in the liquid leads to carbonic acid decomposing to dissolved $\ce{CO_2}$ and H2O. The lowered carbonic acid concentration causes a shift of the final equilibrium. As long as the soft drink is in an open container, the $\ce{CO_2}$ bubbles up out of the beverage, releasing the gas into the air (Figure $3$). With the lid off the bottle, the $\ce{CO_2}$ reactions are no longer at equilibrium and will continue until no more of the reactants remain. This results in a soft drink with a much lowered $\ce{CO_2}$ concentration, often referred to as “flat.”
Sublimation of Bromine
Let us consider the evaporation of bromine as a second example of a system at equilibrium.
$\ce{Br2(l) \rightleftharpoons Br2(g)} \nonumber$
An equilibrium can be established for a physical change—like this liquid to gas transition—as well as for a chemical reaction. Figure $4$ shows a sample of liquid bromine at equilibrium with bromine vapor in a closed container. When we pour liquid bromine into an empty bottle in which there is no bromine vapor, some liquid evaporates, the amount of liquid decreases, and the amount of vapor increases. If we cap the bottle so no vapor escapes, the amount of liquid and vapor will eventually stop changing and an equilibrium between the liquid and the vapor will be established. If the bottle were not capped, the bromine vapor would escape and no equilibrium would be reached.
Summary
A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process, meaning the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.
Glossary
equilibrium
in chemical reactions, the state in which the conversion of reactants into products and the conversion of products back into reactants occur simultaneously at the same rate; state of balance
reversible reaction
chemical reaction that can proceed in both the forward and reverse directions under given conditions | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/13%3A_Fundamental_Equilibrium_Concepts/13.0%3A_Prelude_to_Equilibrium.txt |
Learning Objectives
• Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
• Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
• Relate the magnitude of an equilibrium constant to properties of the chemical system
Now that we have a symbol ($\rightleftharpoons$) to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:
$m\ce{A}+n\ce{B}+ \rightleftharpoons x\ce{C}+y\ce{D} \label{13.3.1}$
We can write the reaction quotient ($Q$) for this equation. When evaluated using concentrations, it is called $Q_c$. We use brackets to indicate molar concentrations of reactants and products.
$Q_c=\dfrac{[\ce{C}]^x[\ce{D}]^y}{[\ce{A}]^m[\ce{B}]^n} \label{13.3.2}$
The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction
$\ce{2NO}_{2(g)} \rightleftharpoons \ce{N_2O}_{4(g)} \label{13.3.3}$
is given by this expression:
$Q_c=\ce{\dfrac{[N_2O_4]}{[NO_2]^2}} \label{13.3.4}$
Example $1$: Writing Reaction Quotient Expressions
Write the expression for the reaction quotient for each of the following reactions:
1. $\ce{3O}_{2(g)} \rightleftharpoons \ce{2O}_{3(g)}$
2. $\ce{N}_{2(g)}+\ce{3H}_{2(g)} \rightleftharpoons \ce{2NH}_{3(g)}$
3. $\ce{4NH}_{3(g)}+\ce{7O}_{2(g)} \rightleftharpoons \ce{4NO}_{2(g)}+\ce{6H_2O}_{(g)}$
Solution
1. $Q_c=\dfrac{[\ce{O3}]^2}{[\ce{O2}]^3}$
2. $Q_c=\dfrac{[\ce{NH3}]^2}{\ce{[N2][H2]}^3}$
3. $Q_c=\dfrac{\ce{[NO2]^4[H2O]^6}}{\ce{[NH3]^4[O2]^7}}$
Exercise $1$
Write the expression for the reaction quotient for each of the following reactions:
1. $\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g)$
2. $\ce{C4H8}(g) \rightleftharpoons \ce{2C2H4}(g)$
3. $\ce{2C4H10}(g)+\ce{13O2}(g) \rightleftharpoons \ce{8CO2}(g)+\ce{10H2O}(g)$
Answer a
$Q_c=\dfrac{[\ce{SO3}]^2}{\ce{[SO2]^2[O2]}}$
Answer b
$Q_c=\dfrac{[\ce{C2H4}]^2}{[\ce{C4H8}]}$
Answer c
$Q_c=\dfrac{\ce{[CO2]^8[H2O]^{10}}}{\ce{[C4H10]^2[O2]^{13}}}$
The numeric value of $Q_c$ for a given reaction varies; it depends on the concentrations of products and reactants present at the time when $Q_c$ is determined. When pure reactants are mixed, $Q_c$ is initially zero because there are no products present at that point. As the reaction proceeds, the value of $Q_c$ increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease (Figure $1$). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.
When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant ($K$) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as $K_c$.
That a reaction quotient always assumes the same value at equilibrium can be expressed as:
$Q_c \textrm{ at equilibrium}=K_c=\dfrac{[\ce C]^x[\ce D]^y...}{[\ce A]^m[\ce B]^n...} \label{13.3.5}$
This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.
Example $2$: Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
$\ce{2NO}_{2(g)} \rightleftharpoons \ce{N_2O}_{4(g)} \nonumber$
When 0.10 mol $\ce{NO2}$ is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] = 0.016 M and [N2O4] = 0.042 M.
1. What is the value of the reaction quotient before any reaction occurs?
2. What is the value of the equilibrium constant for the reaction?
Solution
1. Before any product is formed, $\mathrm{[NO_2]=\dfrac{0.10\:mol}{1.0\:L}}=0.10\:M$, and [N2O4] = 0 M. Thus, $Q_c=\ce{\dfrac{[N2O4]}{[NO2]^2}}=\dfrac{0}{0.10^2}=0 \nonumber$
2. At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, $K_c=Q_c=\ce{\dfrac{[N2O4]}{[NO2]^2}}=\dfrac{0.042}{0.016^2}=1.6\times 10^2. \nonumber$
The equilibrium constant is 1.6 × 102.
Note that dimensional analysis would suggest the unit for this $K_c$ value should be M−1. However, it is common practice to omit units for $K_c$ values computed as described here, since it is the magnitude of an equilibrium constant that relays useful information. As will be discussed later in this module, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from activities instead of molar concentrations, and so $K_c$ values are truly unitless.
Exercise $2$
For the reaction
$\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g) \nonumber$
the concentrations at equilibrium are [SO2] = 0.90 M, [O2] = 0.35 M, and [SO3] = 1.1 M. What is the value of the equilibrium constant, Kc?
Answer
Kc = 4.3
The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for $K_c$ indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of $K_c$—much less than 1—indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.
Once a value of $K_c$ is known for a reaction, it can be used to predict directional shifts when compared to the value of $Q_c$. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure $2$ illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of $\ce{CO}$, $\ce{H_2O}$, $\ce{CO_2}$, and $\ce{H_2}$ react to reach compositions adhering to the same equilibrium (the value of $Q_c$ changes until it equals the value of Kc). This value is 0.640, the equilibrium constant for the reaction under these conditions.
$\ce{CO}(g)+\ce{H2O}(g) \rightleftharpoons \ce{CO2}(g)+\ce{H2}(g) \hspace{20px} K_c=0.640 \hspace{20px} \mathrm{T=800°C} \label{13.3.6}$
It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure $2$ when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.
Example $3$: Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
$\ce{CO}(g)+\ce{H2O}(g) \rightleftharpoons \ce{CO2}(g)+\ce{H2}(g) \nonumber$
with $K_c=0.64$. Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
Reactant/Products for Three Experiments
Reactants/Products Experiment 1 Experiment 2 Experiment 3
[CO]i 0.0203 M 0.011 M 0.0094 M
[H2O]i 0.0203 M 0.0011 M 0.0025 M
[CO2]i 0.0040 M 0.037 M 0.0015 M
[H2]i 0.0040 M 0.046 M 0.0076 M
Solution
Experiment 1:
$Q_c=\ce{\dfrac{[CO2][H2]}{[CO][H2O]}}=\dfrac{(0.0040)(0.0040)}{(0.0203)(0.0203)}=0.039. \nonumber$
Qc < $K_c$ (0.039 < 0.64)
The reaction will shift to the right.
Experiment 2:
$Q_c=\ce{\dfrac{[CO2][H2]}{[CO][H2O]}}=\dfrac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber$
Qc > $K_c$ (140 > 0.64)
The reaction will shift to the left.
Experiment 3:
$Q_c=\ce{\dfrac{[CO2][H2]}{[CO][H2O]}}=\dfrac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber$
Qc < $K_c$ (0.48 < 0.64)
The reaction will shift to the right.
Exercise $3$
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
(a) A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:
$\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g)\hspace{20px}K_c=4.6\times 10^4 \nonumber$
(b) A 5.0-L flask containing 17 g of NH3, 14 g of N2, and 12 g of H2:
$\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)\hspace{20px}K_c=0.060 \nonumber$
(c) A 2.00-L flask containing 230 g of SO3(g):
$\ce{2SO3}(g)⇌\ce{2SO2}(g)+\ce{O2}(g)\hspace{20px}K_c=0.230 \nonumber$
Answer a
$Q_c$ = 6.45 × 103, shifts right.
Answer b
$Q_c$ = 0.12, shifts left.
Answer c
$Q_c$ = 0, shifts right
In Example $2$, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are not constant) at high solution concentrations. This may be avoided by computing $K_c$ values using the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:
• Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
• For relatively dilute solutions, a solute's activity and its molar concentration are roughly equal.
• Activities for pure condensed phases (solids and liquids) are equal to 1.
• Activities for solvents in dilute solutions are equal to 1.
As a consequence of this last two aspects, $Q_c$ and $K_c$ expressions do not contain terms for solids or liquids or solvents in dilute solutions (being numerically equal to 1, these terms have no effect on the expression's value). Several examples of equilibria yielding such expressions will be encountered in this section.
Homogeneous Equilibria
A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here:
Example 1
$\ce{C2H2}(aq)+\ce{2Br2}(aq) \rightleftharpoons \ce{C2H2Br4}(aq)\hspace{20px} \label{13.3.7a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[C2H2Br4]}{[C2H2][Br2]^2}} \label{13.3.7b}$
Example 2
$\ce{I2}(aq)+\ce{I-}(aq) \rightleftharpoons \ce{I3-}(aq) \label{13.3.8b}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[I3- ]}{[I2][I- ]}} \label{13.3.8c}$
Example 3
$\ce{Hg2^2+}(aq)+\ce{NO3-}(aq)+\ce{3H3O+}(aq) \rightleftharpoons \ce{2Hg^2+}(aq)+\ce{HNO2}(aq)+\ce{4H2O}(l) \label{13.3.9a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[Hg^2+]^2[HNO2]}{[Hg2^2+][NO3- ][H3O+]^3}} \label{13.3.9b}$
Example 4
$\ce{HF}(aq)+\ce{H2O}(l) \rightleftharpoons \ce{H3O+}(aq)+\ce{F-}(aq) \label{13.3.10a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[H3O+][F- ]}{[HF]}} \label{13.3.10b}$
Example 5
$\ce{NH3}(aq)+\ce{H2O}(l) \rightleftharpoons \ce{NH4+}(aq)+\ce{OH-}(aq) \label{13.3.11a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[NH4+][OH- ]}{[NH3]}} \label{13.3.11b}$
In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H2O(l) is the solvent for these solutions, it is assigned an activity of 1, and thus does not appear explicitly as a term in the $K_c$ expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.
Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well:
Example 1
$\ce{C2H6}(g) \rightleftharpoons \ce{C2H4}(g)+\ce{H2}(g) \label{13.3.12a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[C2H4][H2]}{[C2H6]}} \label{13.3.12b}$
Example 2
$\ce{3O2}(g) \rightleftharpoons \ce{2O3}(g) \label{13.3.13a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[O3]^2}{[O2]^3}} \label{13.3.13b}$
Example 3
$\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g) \label{13.3.14a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[NH3]^2}{[N2][H2]^3}} \label{13.3.14b}$
Example 4
$\ce{C3H8}(g)+\ce{5O2}(g) \rightleftharpoons \ce{3CO2}(g)+\ce{4H2O}(g)\label{13.3.15a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[CO2]^3[H2O]^4}{[C3H8][O2]^5}}\label{13.3.15b}$
Note that the concentration of $\ce{H_2O}_{(g)}$ has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.
Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, $\dfrac{n}{V}$.
\begin{align} PV&=nRT \label{13.3.16} \[4pt] P &=\left(\dfrac{n}{V}\right)RT \label{13.3.17} \[4pt] &=MRT \label{13.3.18} \end{align}
Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration. Using the partial pressures of the gases, we can write the reaction quotient for the system
$\ce{C2H6}(g) \rightleftharpoons \ce{C2H4}(g)+\ce{H2}(g) \label{13.3.19}$
by following the same guidelines for deriving concentration-based expressions:
$Q_P=\dfrac{P_{\ce{C2H4}}P_{\ce{H2}}}{P_{\ce{C2H6}}} \label{13.3.20}$
In this equation we use QP to indicate a reaction quotient written with partial pressures: $P_{\ce{C2H6}}$ is the partial pressure of C2H6; $P_{\ce{H2}}$, the partial pressure of H2; and $P_{\ce{C2H6}}$, the partial pressure of C2H4. At equilibrium:
$K_P=Q_P=\dfrac{P_{\ce{C2H4}}P_{\ce{H2}}}{P_{\ce{C2H6}}} \label{13.3.21}$
The subscript $P$ in the symbol $K_P$ designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, KP, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.
Conversion between a value for $K_c$, an equilibrium constant expressed in terms of concentrations, and a value for $K_P$, an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).
The equation relating $K_c$ and $K_P$ is derived as follows. For the gas-phase reaction:
$m\ce{A}+n\ce{B} \rightleftharpoons x\ce{C}+y\ce{D} \label{13.3.22}$
with
\begin{align} K_P &=\dfrac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n} \label{13.3.23} \[4pt] &=\dfrac{([\ce C]×RT)^x([\ce D]×RT)^y}{([\ce A]×RT)^m([\ce B]×RT)^n} \label{13.3.24} \[4pt] &=\dfrac{[\ce C]^x[\ce D]^y}{[\ce A]^m[\ce B]^n}×\dfrac{(RT)^{x+y}}{(RT)^{m+n}} \label{13.3.25} \[4pt] &=K_c(RT)^{(x+y)−(m+n)} \label{13.3.26} \[4pt] &=K_c(RT)^{Δn} \label{13.3.27} \end{align}
The relationship between $K_c$ and $K_P$ is
$\color{red} K_P=K_c(RT)^{Δn} \label{13.3.28}$
In this equation, Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction $m\ce{A}+n\ce{B} \rightleftharpoons x\ce{C}+y\ce{D}$, we have
$Δn=(x+y)−(m+n) \label{13.3.29}$
Example $4$: Calculation of KP
Write the equations for the conversion of $K_c$ to KP for each of the following reactions:
1. $\ce{C2H6}(g) \rightleftharpoons \ce{C2H4}(g)+\ce{H2}(g)$
2. $\ce{CO}(g)+\ce{H2O}(g) \rightleftharpoons \ce{CO2}(g)+\ce{H2}(g)$
3. $\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g)$
4. Kc is equal to 0.28 for the following reaction at 900 °C:
$\ce{CS2}(g)+\ce{4H2}(g) \rightleftharpoons \ce{CH4}(g)+\ce{2H2S}(g) \nonumber$
What is KP at this temperature?
Solution
(a) Δn = (2) − (1) = 1
KP = $K_c$ (RT)Δn = $K_c$ (RT)1 = $K_c$ (RT)
(b) Δn = (2) − (2) = 0
KP = $K_c$ (RT)Δn = $K_c$ (RT)0 = Kc
(c) Δn = (2) − (1 + 3) = −2
KP = $K_c$ (RT)Δn = $K_c$ (RT)−2 = $\dfrac{K_c}{(RT)^2}$
d) KP = $K_c$ (RT)Δn = (0.28)[(0.0821)(1173)]−2 = 3.0 × 10−5
Exercise $4$
Write the equations for the conversion of $K_c$ to KP for each of the following reactions, which occur in the gas phase:
1. $\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g)$
2. $\ce{N2O4}(g) \rightleftharpoons \ce{2NO2}(g)$
3. $\ce{C3H8}(g)+\ce{5O2}(g) \rightleftharpoons \ce{3CO2}(g)+\ce{4H2O}(g)$
4. At 227 °C, the following reaction has $K_c$ = 0.0952:
$\ce{CH3OH}(g) \rightleftharpoons \ce{CO}(g)+\ce{2H2}(g) \nonumber$
What would be the value of KP at this temperature?
Answer a
KP = $K_c$ (RT)−1
Answer b
KP = $K_c$ (RT)
Answer c
KP = $K_c$ (RT);
Answer d
160 or 1.6 × 102
Heterogeneous Equilibria
A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).
Some heterogeneous equilibria involve chemical changes:
Example 1
$\ce{PbCl2}(s) \rightleftharpoons \ce{Pb^2+}(aq)+\ce{2Cl-}(aq) \label{13.3.30a}$
with associated equilibrium constant
$K_c=\ce{[Pb^2+][Cl- ]^2} \label{13.3.30b}$
Example 1
$\ce{CaO}(s)+\ce{CO2}(g) \rightleftharpoons \ce{CaCO3}(s) \label{13.3.31a}$
with associated equilibrium constant
$K_c=\dfrac{1}{[\ce{CO2}]} \label{13.3.31b}$
Example 1
$\ce{C}(s)+\ce{2S}(g) \rightleftharpoons \ce{CS2}(g) \label{13.3.32a}$
with associated equilibrium constant
$K_c=\ce{\dfrac{[CS2]}{[S]^2}} \label{13.3.32b}$
Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:
$\ce{Br2}(l) \rightleftharpoons \ce{Br2}(g) \label{13.3.33a}$
with associated equilibrium constant
$K_c=[\ce{Br2}] \label{13.3.33b}$
We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:
$\ce{CaO}(s)+\ce{CO2}(g) \rightleftharpoons \ce{CaCO3}(s)\label{13.3.34a}$
with associated equilibrium constant
$K_P=\dfrac{1}{P_{\ce{CO2}}} \label{13.3.34b}$
$\ce{C}(s)+\ce{2S}(g) \rightleftharpoons \ce{CS2}(g)\label{13.3.35a}$
with associated equilibrium constant
$K_P=\dfrac{P_{\ce{CS2}}}{(P_{\ce S})^2} \label{13.3.35b}$
Summary
For any reaction that is at equilibrium, the reaction quotient Q is equal to the equilibrium constant K for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant. Their particular values may vary depending on conditions, but the value of the reaction quotient will always equal K (Kc when using concentrations or KP when using partial pressures).
A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases. We can decide whether a reaction is at equilibrium by comparing the reaction quotient with the equilibrium constant for the reaction.
Key Equations
• $Q=\dfrac{[\ce C]^x[\ce D]^y}{[\ce A]^m[\ce B]^n}\hspace{20px}\textrm{where }m\ce A+n\ce B⇌x\ce C+y\ce D$
• $Q_P=\dfrac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n}\hspace{20px}\textrm{where }m\ce A+n\ce B⇌x\ce C+y\ce D$
• P = MRT
• KP = $K_c$ (RT)Δn
Glossary
equilibrium constant (K)
value of the reaction quotient for a system at equilibrium
heterogeneous equilibria
equilibria between reactants and products in different phases
homogeneous equilibria
equilibria within a single phase
Kc
equilibrium constant for reactions based on concentrations of reactants and products
KP
equilibrium constant for gas-phase reactions based on partial pressures of reactants and products
law of mass action
when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant
reaction quotient (Q)
ratio of the product of molar concentrations (or pressures) of the products to that of the reactants, each concentration (or pressure) being raised to the power equal to the coefficient in the equation | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/13%3A_Fundamental_Equilibrium_Concepts/13.2%3A_Equilibrium_Constants.txt |
Learning Objectives
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient ($Q$) is equal to the equilibrium constant (K). We next address what happens when a system at equilibrium is disturbed so that $Q$ is no longer equal to $K$. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of $Q$ will no longer equal the value of $K$. To re-establish equilibrium, the system will either shift toward the products (if $Q < K$) or the reactants (if $Q > K$) until $Q$ returns to the same value as $K$.
This process is described by Le Chatelier's principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in $Q$; the reaction will shift to re-establish $Q = K$.
Predicting the Direction of a Reversible Reaction
Le Chatelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of Q and K for the system to predict the changes.
A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.
The stress on the system in Figure $1$ is the reduction of the equilibrium concentration of SCN (lowering the concentration of one of the reactants would cause Q to be larger than K). As a consequence, Le Chatelier's principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.
The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:
$\ce{H}_{2(g)}+\ce{I}_{2(g)} \rightleftharpoons \ce{2HI}_{(g)} \label{13.4.1a}$
$K_c=\mathrm{50.0 \; at\; 400°C} \label{13.4.1b}$
The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with $\mathrm{[H_2] = [I_2]} = 0.221\; M$ and $\ce{[HI]} = 1.563 \;M$ is at equilibrium; for this mixture, $Q_c = K_c = 50.0$. If $\ce{H_2}$ is introduced into the system so quickly that its concentration doubles before it begins to react (new $\ce{[H_2]} = 0.442\; M$), the reaction will shift so that a new equilibrium is reached, at which
• $\ce{[H_2]} = 0.374\; M$,
• $\ce{[I_2]} = 0.153\; M$, and
• $\ce{[HI]} = 1.692\; M$.
This gives:
$Q_c=\mathrm{\dfrac{[HI]^2}{[H_2][I_2]}}=\dfrac{(1.692)^2}{(0.374)(0.153)}=50.0=K_c \label{13.4.2}$
We have stressed this system by introducing additional $\ce{H_2}$. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess $\ce{H_2}$, reducing the amount of uncombined $\ce{I_2}$, and forming additional $\ce{HI}$.
Effect of Change in Pressure on Equilibrium
Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for $K_c$) or partial pressure (for $K_P$). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.
As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.
Consider what happens when we increase the pressure on a system in which $\ce{NO}$, $\ce{O_2}$, and $\ce{NO_2}$ are at equilibrium:
$\ce{2NO (g) + O2(g) \rightleftharpoons 2NO2(g)} \label{13.4.3}$
The formation of additional amounts of $\ce{NO2}$ decreases the total number of molecules in the system because each time two molecules of $\ce{NO_2}$ form, a total of three molecules of $\ce{NO}$ and $\ce{O_2}$ are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of $\ce{NO_2}$ into $\ce{NO}$ and $\ce{O_2}$, which tends to restore the pressure.
Now consider this reaction:
$\ce{N2 (g) + O2 (g) \rightleftharpoons 2NO (g)} \label{13.4.4}$
Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.
Effect of Change in Temperature on Equilibrium
Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle.
When hydrogen reacts with gaseous iodine, heat is evolved.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g) } \;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{13.4.5}$
Because this reaction is exothermic, we can write it with heat as a product.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} + \text{heat} \label{13.4.6}$
Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of $\ce{H2}$ and $\ce{I2}$ and a reduction in the concentration of $\ce{HI}$. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.
When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the $\ce{HI}$ system increases the equilibrium constant: At the new equilibrium the concentration of $\ce{HI}$ has increased and the concentrations of $\ce{H2}$ and $\ce{I2}$ decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.
Temperature affects the equilibrium between $\ce{NO_2}$ and $\ce{N_2O_4}$ in this reaction
$\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{13.4.7}$
The positive ΔH value tells us that the reaction is endothermic and could be written
$\text{heat}+\ce{N2O4(g) \rightleftharpoons 2NO2(g)} \label{13.4.8}$
At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown $\ce{NO_2}$ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless $\ce{N_2O_4}$ increases, and the concentration of brown $\ce{NO_2}$ decreases, causing the brown color to fade.
Catalysts Do Not Affect Equilibrium
As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.
The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{13.4.9}$
A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year. Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.
Fritz Haber
Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures.
For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate.
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \nonumber$
Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008. The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen ($\ce{N_2}$) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).
In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.”1 Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.
Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.
It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood.
To be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system in which N2, H2, and NH3 are at equilibrium or are coming to equilibrium.
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{13.4.10}$
The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.
Although increasing the pressure of a mixture of N2, H2, and NH3 will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of N2 and H2, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:
$\ce{N2(g) + 3H2(g) \rightarrow 2NH3(g)} \;\;\; ΔH=\mathrm{−92.2\; kJ} \label{13.4.11}$
Thus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature.
Part of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly. In the commercial production of ammonia, conditions of about 500 °C, 150–900 atm, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure $2$).
Summary
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.
Table $1$: Effects of Disturbances of Equilibrium and K
Disturbance Observed Change as Equilibrium is Restored Direction of Shift Effect on K
reactant added added reactant is partially consumed toward products none
product added added product is partially consumed toward reactants none
decrease in volume/increase in gas pressure pressure decreases toward side with fewer moles of gas none
increase in volume/decrease in gas pressure pressure increases toward side with more moles of gas none
temperature increase heat is absorbed toward products for endothermic, toward reactants for exothermic changes
temperature decrease heat is given off toward reactants for endothermic, toward products for exothermic changes
Footnotes
1. Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764.
Glossary
Le Chatelier's principle
when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance
position of equilibrium
concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance)
stress
change to a reaction's conditions that may cause a shift in the equilibrium | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/13%3A_Fundamental_Equilibrium_Concepts/13.3%3A_Shifting_Equilibria%3A_Le_Chateliers_Principle.txt |
Learning Objectives
• Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
• Use algebra to perform various types of equilibrium calculations
We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.
Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.
On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:
$\ce{2NH3}(g)⇌\ce{N2}(g)+\ce{3H2}(g) \nonumber$
If a sample of ammonia decomposes in a closed system and the concentration of N2 increases by 0.11 M, the change in the N2 concentration, Δ[N2], the final concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of N2 increases.
The change in the H2 concentration, Δ[H2], is also positive—the concentration of H2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H2 is three times the change in the concentration of N2 because for each mole of N2 produced, 3 moles of H2 are produced.
\begin{align*} \ce{Δ[H2]} &=3×\ce{Δ[N2]} \[4pt] &=3×(0.11\:M) \[4pt] &=0.33\:M \end{align*} \nonumber
The change in concentration of NH3, Δ[NH3], is twice that of Δ[N2]; the equation indicates that 2 moles of NH3 must decompose for each mole of N2 formed. However, the change in the NH3 concentration is negative because the concentration of ammonia decreases as it decomposes.
\begin{align*} Δ[\ce{NH3}] &=−2×Δ[\ce{N2}] \[4pt] &=−2×(0.11\:M) \[4pt] &=−0.22\:M \end{align*} \nonumber
We can relate these relationships directly to the coefficients in the equation
\begin{align} &\phantom{Δ[NH3}\ce{2NH3}(g) &&⇌ &&\phantom{Δ[N2}\ce{N2}(g) &&+ &&\phantom{Δ[H2]}\ce{3H2}(g)\ &Δ[\ce{NH3}]=−2×Δ[\ce{N2}]&& && Δ[\ce{N2}]=0.11\:M && && Δ[\ce{H2}]=3×Δ[\ce{N2}] \end{align} \nonumber
Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.
If we did not know the magnitude of the change in the concentration of N2, we could represent it by the symbol x.
$Δ[\ce{N2}]=x$
The changes in the other concentrations would then be represented as:
$Δ[\ce{H2}]=3×Δ[\ce{N2}]=3x$
$Δ[\ce{NH3}]=−2×Δ[\ce{N2}]=−2x$
The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.
\begin{alignat}{3} &\ce{2NH3}(g)⇌\:&&\ce{N2}(g)+\:&&\ce{3H2}(g)\ &−2x &&x &&3x \end{alignat}
The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.
Example $1$: Determining Relative Changes in Concentration
Complete the changes in concentrations for each of the following reactions.
(a) \begin{alignat}{3} &\ce{C2H2}(g)+\:&&\ce{2Br2}(g)⇌\:&&\ce{C2H2Br4}(g)\ &x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
(b) \begin{alignat}{3} &\ce{I2}(aq)+\:&&\ce{I-}(aq)⇌\:&&\ce{I3-}(aq)\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&x \end{alignat}
(c) \begin{alignat}{3} &\ce{C3H8}(g)+\:&&\ce{5O2}(g)⇌\:&&\ce{3CO2}(g)+\:&&\ce{4H2O}(g)\ &x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
Solution
(a) \begin{alignat}{3} &\ce{C2H2}(g)+\:&&\ce{2Br2}(g)⇌\:&&\ce{C2H2Br4}(g)\ &x &&2x &&-x \end{alignat}
(b) \begin{alignat}{3} &\ce{I2}(aq)+\:&&\ce{I-}(aq)⇌\:&&\ce{I3-}(aq)\ &-x &&-x &&x \end{alignat}
(c) \begin{alignat}{3} &\ce{C3H8}(g)+\:&&\ce{5O2}(g)⇌\:&&\ce{3CO2}(g)+\:&&\ce{4H2O}(g)\ &x &&5x &&-3x &&-4x \end{alignat}
Exercise $1$
Complete the changes in concentrations for each of the following reactions:
(a) \begin{alignat}{3} &\ce{2SO2}(g)+\:&&\ce{O2}(g)⇌\:&&\ce{2SO3}(g)\ &\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}} \end{alignat}
(b) \begin{alignat}{3} &\ce{C4H8}(g)⇌\:&&\ce{2C2H4}(g)\ &\underline{\hspace{40px}} &&-2x \end{alignat}
(c) \begin{alignat}{3} &\ce{4NH3}(g)+\:&&\ce{7H2O}(g)⇌\:&&\ce{4NO2}(g)+\:&&\ce{6H2O}(g)\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
Answer a
2x, x, −2x
Answer b
x, −2x
Answer c
4x, 7x, −4x, −6x or −4x, −7x, 4x, 6x
Calculations Involving Equilibrium Concentrations
Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Qc (i.e., the law of mass action) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Qc = Kc (at equilibrium) in all of these situations and that there are only three basic types of calculations:
1. Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
2. Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.
3. Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.
A similar list could be generated using QP, KP, and partial pressure. We will look at solving each of these cases in sequence.
Calculation of an Equilibrium Constant
Since the law of mass action is the only equation we have to describe the relationship between Kc and the concentrations of reactants and products, any problem that requires us to solve for Kc must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for Kc, as it will be the only unknown.
Example $1$ showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE table—for Initial, Change, and Equilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.
Example $2$: Calculation of an Equilibrium Constant
Iodine molecules react reversibly with iodide ions to produce triiodide ions.
$\ce{I2}(aq)+\ce{I-}(aq)⇌\ce{I3-}(aq) \nonumber$
If a solution with the concentrations of I2 and I both equal to 1.000 × 10−3 M before reaction gives an equilibrium concentration of I2 of 6.61 × 10−4 M, what is the equilibrium constant for the reaction?
Solution
We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −x as the change in concentration of I2.
Since the equilibrium concentration of I2 is given, we can solve for x. At equilibrium the concentration of I2 is 6.61 × 10−4 M so that
$1.000×10^{−3}−x=6.61×10^{−4}$
$x=1.000×10^{−3}−6.61×10^{−4}$
$=3.39×10^{−4}\:M$
Now we can fill in the table with the concentrations at equilibrium.
We now calculate the value of the equilibrium constant.
$K_c=Q_c=\ce{\dfrac{[I3- ]}{[I2][I- ]}}$
$=\dfrac{3.39×10^{−4}\:M}{(6.61×10^{−4}\:M)(6.61×10^{−4}\:M)}=776$
Exercise $2$
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.
$\ce{C2H5OH + CH3CO2H ⇌ CH3CO2C2H5 + H2O}$
When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $\dfrac{1}{3}$ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)
Answer
Kc = 4
Calculation of a Missing Equilibrium Concentration
If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.
Example $3$: Calculation of a Missing Equilibrium Concentration
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, $\ce{N2}(g)+\ce{O2}(g)⇌\ce{2NO}(g)$, is 4.1 × 10−4. Find the concentration of NO(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N2] = 0.036 mol/L and [O2] 0.0089 mol/L.
Solution
We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.
$K_c=Q_c=\ce{\dfrac{[NO]^2}{[N2][O2]}}$
$\ce{[NO]^2}=K_c\ce{[N2][O2]}$
$\ce{[NO]}=\sqrt{K_c\ce{[N2][O2]}}$
$=\sqrt{(4.1×10^{−4})(0.036)(0.0089)}$
$=\sqrt{1.31×10^{−7}}$
$=3.6×10^{−4}$
Thus [NO] is 3.6 × 10−4 mol/L at equilibrium under these conditions.
We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.
\begin{align*} Q_c=\ce{\dfrac{[NO]^2}{[N2][O2]}} \[4pt] &=\dfrac{(3.6×10^{−4})^2}{(0.036)(0.0089)} \[4pt] &=4.0×10^{−4}=K_c \end{align*} \nonumber
The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.
Exercise $3$
The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10−2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.
Answer
1.53 mol/L
Calculation of Changes in Concentration
If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.
1. Determine the direction the reaction proceeds to come to equilibrium.
• Write a balanced chemical equation for the reaction.
• If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Qc from the initial concentrations and compare to Kc to determine the direction of change.
2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
• Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
• Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
3. Solve for the change and the equilibrium concentrations.
• Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and check any assumptions used to find x.
• Calculate the equilibrium concentrations.
4. Check the arithmetic.
• Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.
In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.
Example $4$: Calculation of Concentration as a Reaction Equilibrates
Under certain conditions, the equilibrium constant for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of PCl5, PCl3, and Cl2 if the initial concentration of PCl5 was 1.00 M?
Solution
Use the stepwise process described earlier.
1. The balanced equation for the decomposition of PCl5 is
$\ce{PCl5}(g)⇌\ce{PCl3}(g)+\ce{Cl2}(g)$
Because we have no products initially, Qc = 0 and the reaction will proceed to the right.
• Let us represent the increase in concentration of PCl3 by the symbol x. The other changes may be written in terms of x by considering the coefficients in the chemical equation.
\begin{alignat}{3} &\ce{PCl5}(g)⇌\:&&\ce{PCl3}(g)+\:&&\ce{Cl2}(g)\ &-x &&x &&x \end{alignat}
• The changes in concentration and the expressions for the equilibrium concentrations are:
• Substituting the equilibrium concentrations into the equilibrium constant equation gives
$K_c=\ce{\dfrac{[PCl3][Cl2]}{[PCl5]}}=0.0211$
$=\dfrac{(x)(x)}{(1.00−x)}$
This equation contains only one variable, x, the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.
$0.0211=\dfrac{(x)(x)}{(1.00−x)}$
$0.0211(1.00−x)=x^2$
$x^2+0.0211x−0.0211=0$
An equation of the form ax2 + bx + c = 0 can be rearranged to solve for x:
$x=\dfrac{−b±\sqrt{b^2−4ac}}{2a}$
In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:
$x=\dfrac{−0.0211±\sqrt{(0.0211)^2−4(1)(−0.0211)}}{2(1)}$
$=\dfrac{−0.0211±\sqrt{(4.45×10^{−4})+(8.44×10^{−2})}}{2}$
$=\dfrac{−0.0211±0.291}{2}$
Hence
$x=\dfrac{−0.0211+0.291}{2}=0.135$
or
$x=\dfrac{−0.0211−0.291}{2}=−0.156$
Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M.
The equilibrium concentrations are
$\ce{[PCl5]}=1.00−0.135=0.87\:M$
$\ce{[PCl3]}=x=0.135\:M$
$\ce{[Cl2]}=x=0.135\:M$
• Substitution into the expression for Kc (to check the calculation) gives
$K_c=\ce{\dfrac{[PCl3][Cl2]}{[PCl5]}}=\dfrac{(0.135)(0.135)}{0.87}=0.021$
The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.
Exercise $\PageIndex{4A}$
Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5.
$\ce{CH3CO2H + C2H5OH ⇌ CH3CO2C2H5 + H2O} \nonumber$
The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O are mixed in enough dioxane to make 1.0 L of solution?
Answer
[CH3CO2H] = 0.36 M, [C2H5OH] = 0.36 M, [CH3CO2C2H5] = 0.17 M, [H2O] = 0.17 M
Exercise $\PageIndex{4B}$
A 1.00-L flask is filled with 1.00 moles of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2, I2, and HI in moles/L?
$\ce{H2}(g)+\ce{I2}(g)\rightleftharpoons\ce{2HI}(g) \nonumber$
Answer
[H2] = 0.06 M, [I2] = 1.06 M, [HI] = 1.88 M
Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products.
Consider the ionization of 0.150 M HA, a weak acid.
$\ce{HA}(aq)⇌\ce{H+}(aq)+\ce{A-}(aq) \hspace{20px} K_c=6.80×10^{−4}$
The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.
Setting up and solving the quadratic equation gives
$K_c=\ce{\dfrac{[H+][A- ]}{[HA]}}=\dfrac{(x)(x)}{(0.150−x)}=6.80×10^{−4}$
$x^2+6.80×10^{−4}x−1.02×10^{−4}=0$
$x={−6.80×10^{−4}±\sqrt{(6.80×10^{−4})^2−(4)(1)(−1.02×10^{−4})}}{(2)(1)}$
$x=0.00977\:M\textrm{ or }−0.0104\:M$
Using the positive (physical) root, the equilibrium concentrations are
$\ce{[HA]}=0.150−x=0.140\:M$
$\ce{[H+]}=\ce{[A- ]}=x=0.00977\:M$
A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could be called the “all product” starting point. Assuming all of the HA ionizes gives
$\ce{[HA]}=0.150−0.150=0\:M$
$\ce{[H+]}=0+0.150=0.150\:M$
$\ce{[A- ]}=0+0.150=0.150\:M$
Using these as initial concentrations and “y” to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.
Setting up and solving the quadratic equation gives
$K_c=\ce{\dfrac{[H+][A- ]}{[HA]}}=\dfrac{(0.150−y)(0.150−y)}{(y)}=6.80×10^{−4}$
$6.80×10^{−4}y=0.0225−0.300y+y^2$
Retain a few extra significant figures to minimize rounding problems.
$y^2−0.30068y+0.022500=0$
$y=\dfrac{0.30068±\sqrt{(0.30068)^2−(4)(1)(0.022500)}}{(2)(1)}$
$y=\dfrac{0.30068±0.020210}{2}$
Rounding each solution to three significant figures gives
$y=0.160\:M\textrm{ or }y=0.140\:M$
Using the physically significant root (0.140 M) gives the equilibrium concentrations as
$\ce{[HA]}=y=0.140\:M$
$\ce{[H+]}=0.150−y=0.010\:M$
$\ce{[A- ]}=0.150−y=0.010\:M$
Thus, the two approaches give the same results (to three decimal places), and show that both starting points lead to the same equilibrium conditions. The “all reactant” starting point resulted in a relatively small change (x) because the system was close to equilibrium, while the “all product” starting point had a relatively large change (y) that was nearly the size of the initial concentrations. It can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions (x) to reach equilibrium.
Recall that a small Kc means that very little of the reactants form products and a large Kc means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change (x) is small compared to any initial concentrations, it can be neglected. Small is usually defined as resulting in an error of less than 5%. The following two examples demonstrate this.
Example $5$: Approximate Solution Starting Close to Equilibrium
What are the concentrations at equilibrium of a 0.15 M solution of HCN?
$\ce{HCN}(aq)⇌\ce{H+}(aq)+\ce{CN-}(aq) \hspace{20px} K_c=4.9×10^{−10} \nonumber$
Solution
Using “x” to represent the concentration of each product at equilibrium gives this ICE table.
The exact solution may be obtained using the quadratic formula with
$K_c=\dfrac{(x)(x)}{0.15−x} \nonumber$
solving
$x^2+4.9×10^{−10}−7.35×10^{−11}=0 \nonumber$
$x=8.56×10^{−6}\:M\textrm{ (3 sig. figs.)}=8.6×10^{−6}\:M\textrm{ (2 sig. figs.)} \nonumber$
Thus [H+] = [CN] = x = 8.6 × 10–6 M and [HCN] = 0.15 – x = 0.15 M.
In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, x must be small compared to 0.15 M. More formally, if $x≪0.15$, then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x
$K_c=\dfrac{(x)(x)}{0.15−x}≈\dfrac{x^2}{0.15} \nonumber$
$4.9×10^{−10}=\dfrac{x^2}{0.15} \nonumber$
$x^2=(0.15)(4.9×10^{−10})=7.4×10^{−11} \nonumber$
$x=\sqrt{7.4×10^{−11}}=8.6×10^{−6}\:M \nonumber$
In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to IF (0.15 – x) ≈ 0.15 M, so if
$\dfrac{x}{0.15}×100\%=\dfrac{8.6×10^{−6}}{0.15}×100\%=0.006\% \nonumber$
is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.
Exercise $5$
What are the equilibrium concentrations in a 0.25 M NH3 solution?
$\ce{NH3}(aq)+\ce{H2O}(l)⇌\ce{NH4+}(aq)+\ce{OH-}(aq) \hspace{20px} K_c=1.8×10^{−5}$
Assume that x is much less than 0.25 M and calculate the error in your assumption.
Answer
$\ce{[OH- ]}=\ce{[NH4+]}=0.0021\:M$; [NH3] = 0.25 M, error = 0.84%
The second example requires that the original information be processed a bit, but it still can be solved using a small x approximation.
Example $6$: Approximate Solution After Shifting Starting Concentration
Copper(II) ions form a complex ion in the presence of ammonia
$\ce{Cu^2+}(aq)+\ce{4NH3}(aq)⇌\ce{Cu(NH3)4^2+}(aq) \hspace{20px} K_c=5.0×10^{13}=\ce{\dfrac{[Cu(NH3)4^2+]}{[Cu^2+(aq)][NH3]^4}} \nonumber$
If 0.010 mol Cu2+ is added to 1.00 L of a solution that is 1.00 M NH3 what are the concentrations when the system comes to equilibrium?
Solution The initial concentration of copper(II) is 0.010 M. The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants.” Note that Cu2+ is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would be
$\ce{[Cu^2+]}=0.010−0.010=0\:M \nonumber$
$\ce{[Cu(NH3)4^2+]}=0.010\:M \nonumber$
$\ce{[NH3]}=1.00−4×0.010=0.96\:M \nonumber$
Using these “shifted” values as initial concentrations with x as the free copper(II) ion concentration at equilibrium gives this ICE table.
Since we are starting close to equilibrium, x should be small so that
$0.96+4x≈0.96\:M$
$0.010−x≈0.010\:M$
$K_c=\dfrac{(0.010−x)}{x(0.96−4x)^4}≈\dfrac{(0.010)}{x(0.96)^4}=5.0×10^{13}$
$x=\dfrac{(0.010)}{K_c(0.96)^4}=2.4×10^{−16}\:M$
Select the smallest concentration for the 5% rule.
$\dfrac{2.4×10^{−16}}{0.010}×100\%=2×10^{−12}\%$
This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are
$\ce{[Cu^2+]}=x=2.4×10^{−16}\:M$
$\ce{[NH3]}=0.96−4x=0.96\:M$
$\ce{[Cu(NH3)4^2+]}=0.010−x=0.010\:M$
By starting with the maximum amount of product, this system was near equilibrium and the change (x) was very small. With only a small change required to get to equilibrium, the equation for x was greatly simplified and gave a valid result well within the 5% error maximum.
Exercise $6$
What are the equilibrium concentrations when 0.25 mol Ni2+ is added to 1.00 L of 2.00 M NH3 solution?
$\ce{Ni^2+}(aq)+\ce{6NH3}(aq)⇌\ce{Ni(NH3)6^2+}(aq) \nonumber$
with $K_c=5.5×10^8$.
With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount (x) of the product shifts left. Calculate the error in your assumption.
Answer
$\ce{[Ni(NH3)6^2+]}=0.25\:M$, [NH3] = 0.50 M, [Ni2+] = 2.9 × 10–8 M, error = 1.2 × 10–5%
Summary
The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant; when given the equilibrium constant and some of the concentrations involved, we can solve for the missing concentrations; and when given the equilibrium constant and the initial concentrations, we can solve for the concentrations at equilibrium. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/13%3A_Fundamental_Equilibrium_Concepts/13.4%3A_Equilibrium_Calculations.txt |
13.1: Chemical Equilibria Exercises
Q13.1.1
What does it mean to describe a reaction as “reversible”?
S13.1.1
The reaction can proceed in both the forward and reverse directions.
Q13.1.2
When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?
Q13.1.3
If a reaction is reversible, when can it be said to have reached equilibrium?
S13.1.3
When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the reactions continue to occur, but at equivalent rates.
Q13.1.4
Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
Q13.1.5
If the concentrations of products and reactants are equal, is the system at equilibrium?
S13.1.5
The concept of equilibrium does not imply equal concentrations, though it is possible.
13.2: Equilibrium Constant Exercises
Q13.2.1
Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
Q13.2.2
Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown below:
S13.2.2
Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.
Q13.2.3
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4?
$\ce{2NO2}(g) \rightleftharpoons \ce{N2O4}(g)$
Q13.2.4
Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2, AgCl, PbCl2, and CuCl.
Q13.2.5
1. (a) Write the expression for the equilibrium constant for the reaction represented by the equation $\ce{AgCl}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{Cl-}(aq)$. Is Kc > 1, < 1, or ≈ 1? Explain your answer.
2. (b) Write the expression for the equilibrium constant for the reaction represented by the equation $\ce{Pb^2+}(aq)+\ce{2Cl-}(aq) \rightleftharpoons \ce{PbCl2}(s)$. Is Kc > 1, < 1, or ≈ 1? Explain your answer.
S13.2.5
(a) Kc = [Ag+][Cl] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M; (b) $K_c=\ce{\dfrac{1}{[Pb^2+][Cl- ]^2}}$ > 1 because PbCl2 is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).
Q13.2.6
Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.
1. Write the expression for the equilibrium constant for the reaction represented by the equation $\ce{CaCO3}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{CO3-}(aq)$. Is Kc > 1, < 1, or ≈ 1? Explain your answer.
2. Write the expression for the equilibrium constant for the reaction represented by the equation $\ce{3Ba^2+}(aq)+\ce{2PO4^3-}(aq) \rightleftharpoons \ce{Ba3(PO4)2}(s)$. Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Q13.2.7
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: $\ce{3C2H2}(g)⟶\ce{C6H6}(g)$. Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer.
S13.2.7
Since $K_c=\ce{\dfrac{[C6H6]}{[C2H2]^3}}$, a value of Kc ≈ 10 means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
Q13.2.8
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $\ce{KI}(aq)+\ce{I2}(aq) \rightleftharpoons \ce{KI3}(aq)$ give the same expression for the reaction quotient. KI3 is composed of the ions K+ and I3.
Q13.2.9
For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction?
Kc > 1
Q13.2.10
For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is Kc > 1, < 1, or ≈ 1 for a useful precipitation reaction?
Q13.2.11
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
1. $\ce{CH4}(g)+\ce{Cl2}(g) \rightleftharpoons \ce{CH3Cl}(g)+\ce{HCl}(g)$
2. $\ce{N2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2NO}(g)$
3. $\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g)$
4. $\ce{BaSO3}(s) \rightleftharpoons \ce{BaO}(s)+\ce{SO2}(g)$
5. $\ce{P4}(g)+\ce{5O2}(g) \rightleftharpoons \ce{P4O10}(s)$
6. $\ce{Br2}(g) \rightleftharpoons \ce{2Br}(g)$
7. $\ce{CH4}(g)+\ce{2O2}(g) \rightleftharpoons \ce{CO2}(g)+\ce{2H2O}(l)$
8. $\ce{CuSO4⋅5H2O}(s) \rightleftharpoons \ce{CuSO4}(s)+\ce{5H2O}(g)$
S13.2.11
(a) $Q_c=\ce{\dfrac{[CH3Cl][HCl]}{[CH4][Cl2]}}$; (b) $Q_c=\ce{\dfrac{[NO]^2}{[N2][O2]}}$; (c) $Q_c=\ce{\dfrac{[SO3]^2}{[SO2]^2[O2]}}$; (d) $Q_c$ = [SO2]; (e) $Q_c=\ce{\dfrac{1}{[P4][O2]^5}}$; (f) $Q_c=\ce{\dfrac{[Br]^2}{[Br2]}}$; (g) $Q_c=\ce{\dfrac{[CO2]}{[CH4][O2]^2}}$; (h) $Q_c$ = [H2O]5
Q13.2.12
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
1. $\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g)$
2. $\ce{4NH3}(g)+\ce{5O2}(g) \rightleftharpoons \ce{4NO}(g)+\ce{6H2O}(g)$
3. $\ce{N2O4}(g) \rightleftharpoons \ce{2NO2}(g)$
4. $\ce{CO2}(g)+\ce{H2}(g) \rightleftharpoons \ce{CO}(g)+\ce{H2O}(g)$
5. $\ce{NH4Cl}(s) \rightleftharpoons \ce{NH3}(g)+\ce{HCl}(g)$
6. $\ce{2Pb(NO3)2}(s) \rightleftharpoons \ce{2PbO}(s)+\ce{4NO2}(g)+\ce{O2}(g)$
7. $\ce{2H2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2H2O}(l)$
8. $\ce{S8}(g) \rightleftharpoons \ce{8S}(g)$
S13.2.12
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
1. $\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_c=17$; [NH3] = 0.20 M, [N2] = 1.00 M, [H2] = 1.00 M
2. $\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_P=6.8×10^4$; initial pressures: NH3 = 3.0 atm, N2 = 2.0 atm, H2 = 1.0 atm
3. $\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_c=0.230$; [SO3] = 0.00 M, [SO2] = 1.00 M, [O2] = 1.00 M
4. $\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_P=16.5$; initial pressures: SO3 = 1.00 atm, SO2 = 1.00 atm, O2 = 1.00 atm
5. $\ce{2NO}(g)+\ce{Cl2}(g) \rightleftharpoons \ce{2NOCl}(g) \hspace{20px} K_c=4.6×10^4$; [NO] = 1.00 M, [Cl2] = 1.00 M, [NOCl] = 0 M
6. $\ce{N2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2NO}(g) \hspace{20px} K_P=0.050$; initial pressures: NO = 10.0 atm, N2 = O2 = 5 atm
S13.2.13
(a) $Q_c$ 25 proceeds left; (b) QP 0.22 proceeds right; (c) $Q_c$ undefined proceeds left; (d) QP 1.00 proceeds right; (e) QP 0 proceeds right; (f) $Q_c$ 4 proceeds left
Q13.2.14
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
1. $\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_c=17$; [NH3] = 0.50 M, [N2] = 0.15 M, [H2] = 0.12 M
2. $\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_P=6.8×10^4$; initial pressures: NH3 = 2.00 atm, N2 = 10.00 atm, H2 = 10.00 atm
3. $\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_c=0.230$; [SO3] = 2.00 M, [SO2] = 2.00 M, [O2] = 2.00 M
4. $\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_P=\mathrm{6.5\:atm}$; initial pressures: SO2 = 1.00 atm, O2 = 1.130 atm, SO3 = 0 atm
5. $\ce{2NO}(g)+\ce{Cl2}(g) \rightleftharpoons \ce{2NOCl}(g) \hspace{20px} K_P=2.5×10^3$; initial pressures: NO = 1.00 atm, Cl2 = 1.00 atm, NOCl = 0 atm
6. $\ce{N2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2NO}(g) \hspace{20px} K_c=0.050$; [N2] = 0.100 M, [O2] = 0.200 M, [NO] = 1.00 M
Q13.2.15
The following reaction has KP = 4.50 × 10−5 at 720 K.
$\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g)$
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH3) = 93 atm, P(N2) = 48 atm, and P(H2) = 52
S13.2.15
The system will shift toward the reactants to reach equilibrium.
Q13.2.16
Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
$\ce{SO2Cl2}(g) \rightleftharpoons \ce{SO2}(g)+\ce{Cl2}(g)$
[SO2Cl2] = 0.12 M, [Cl2] = 0.16 M and [SO2] = 0.050 M. Kc for the reaction is 0.078.
Q13.2.17
Which of the systems described in Exercise give homogeneous equilibria? Which give heterogeneous equilibria?
S13.2.17
(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous
Q13.2.18
Which of the systems described in Exercise give homogeneous equilibria? Which give heterogeneous equilibria?
Q13.2.19
For which of the reactions in Exercise does Kc (calculated using concentrations) equal KP (calculated using pressures)?
S13.2.19
This situation occurs in (a) and (b).
Q13.2.19
For which of the reactions in Exercise does Kc (calculated using concentrations) equal KP (calculated using pressures)?
Q13.2.20
Convert the values of Kc to values of KP or the values of KP to values of Kc.
1. $\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g) \hspace{20px} K_c=\textrm{0.50 at 400°C}$
2. $\ce{H2 + I2 \rightleftharpoons 2HI} \hspace{20px} K_c=\textrm{50.2 at 448°C}$
3. $\ce{Na2SO4⋅10H2O}(s) \rightleftharpoons \ce{Na2SO4}(s)+\ce{10H2O}(g) \hspace{20px} K_P=4.08×10^{−25}\textrm{ at 25°C}$
4. $\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g) \hspace{20px} K_P=\textrm{0.122 at 50°C}$
S13.2.20
(a) KP = 1.6 × 10−4; (b) KP = 50.2; (c) Kc = 5.31 × 10−39; (d) Kc = 4.60 × 10−3
Q13.2.21
Convert the values of Kc to values of KP or the values of KP to values of Kc.
1. $\ce{Cl2}(g)+\ce{Br2}(g) \rightleftharpoons \ce{2BrCl}(g) \hspace{20px} K_c=4.7×10^{−2}\textrm{ at 25°C}$
2. $\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g) \hspace{20px} K_P=\textrm{48.2 at 500°C}$
3. $\ce{CaCl2⋅6H2O}(s) \rightleftharpoons \ce{CaCl2}(s)+\ce{6H2O}(g) \hspace{20px} K_P=5.09×10^{−44}\textrm{ at 25°C}$
4. $\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g) \hspace{20px} K_P=\textrm{0.196 at 60°C}$
Q13.2.22
What is the value of the equilibrium constant expression for the change $\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g)$ at 30 °C?
S13.2.22
$K_P=P_{\ce{H2O}}=0.042.$
Q13.2.23
Write the expression of the reaction quotient for the ionization of HOCN in water.
Q13.2.24
Write the reaction quotient expression for the ionization of NH3 in water.
S13.2.24
$Q_c=\ce{\dfrac{[NH4+][OH- ]}{[HN3]}}$
Q13.2.25
What is the approximate value of the equilibrium constant KP for the change $\ce{C2H5OC2H5}(l) \rightleftharpoons \ce{C2H5OC2H5}(g)$ at 25 °C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)
13.3: Shifting Equilbria Exercises
Q13.3.1
The following equation represents a reversible decomposition:
$\ce{CaCO3}(s)\rightleftharpoons\ce{CaO}(s)+\ce{CO2}(g)$
Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains?
S13.3.1
The amount of CaCO3 must be so small that $P_{\ce{CO2}}$ is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full $P_{\ce{CO2}}$ required for equilibrium.
Q13.3.2
Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.
Q13.3.3
What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
S13.3.3
The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants' side.
Q13.3.4
What would happen to the color of the solution in part (b) of Figure if a small amount of NaOH were added and Fe(OH)3 precipitated? Explain your answer.
Q13.3.5
The following reaction occurs when a burner on a gas stove is lit:
$\ce{CH4}(g)+\ce{2O2}(g)\rightleftharpoons\ce{CO2}(g)+\ce{2H2O}(g)$
Is an equilibrium among CH4, O2, CO2, and H2O established under these conditions? Explain your answer.
S13.3.5
No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.
Q13.3.6
A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of $\ce{SO3 }$is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures.
$\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)$
1. (a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
2. (b) Is the reaction endothermic or exothermic?
Q13.3.7a
Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation:
$\ce{N2}(g)+\ce{2H2}(g)\rightleftharpoons\ce{N2H4}(g) \hspace{20px} ΔH=\ce{95\:kJ}$
S13.3.7a
Add N2; add H2; decrease the container volume; heat the mixture.
Q13.3.7b
Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
$\ce{P4}(g)+\ce{6H2}(g)\rightleftharpoons\ce{4PH3}(g) \hspace{20px} ΔH=\mathrm{110.5\:kJ}$
Q13.3.8
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
1. $\ce{2NH3}(g)\rightleftharpoons\ce{N2}(g)+\ce{3H2}(g) \hspace{20px} ΔH=\mathrm{92\:kJ}$
2. $\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g) \hspace{20px} ΔH=\mathrm{181\:kJ}$
3. $\ce{2O3}(g)\rightleftharpoons\ce{3O2}(g) \hspace{20px} ΔH=\mathrm{−285\:kJ}$
4. $\ce{CaO}(s)+\ce{CO2}(g)\rightleftharpoons\ce{CaCO3}(s) \hspace{20px} ΔH=\mathrm{-176\:kJ}$
S13.3.8
(a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.
Q13.3.9
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
1. $\ce{2H2O}(g)\rightleftharpoons\ce{2H2}(g)+\ce{O2}(g) \hspace{20px} ΔH=\ce{484\:kJ}$
2. $\ce{N2}(g)+\ce{3H2}(g)\rightleftharpoons\ce{2NH3}(g) \hspace{20px} ΔH=\mathrm{-92.2\:kJ}$
3. $\ce{2Br}(g)\rightleftharpoons\ce{Br2}(g) \hspace{20px} ΔH=\mathrm{-224\:kJ}$
4. $\ce{H2}(g)+\ce{I2}(s)\rightleftharpoons\ce{2HI}(g) \hspace{20px} ΔH=\ce{53\:kJ}$
Q13.3.10
Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction:
$\ce{H2O}(g)+\ce{C}(s)\rightleftharpoons\ce{H2}(g)+\ce{CO}(g).$
Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
1. Write the expression for the equilibrium constant ($K_c$) for the reversible reaction $\ce{2H2}(g)+\ce{CO}(g)\rightleftharpoons\ce{CH3OH}(g) \hspace{20px} ΔH=\mathrm{-90.2\:kJ}$
2. What will happen to the concentrations of $\ce{H2}$, $\ce{CO}$, and $\ce{CH3OH}$ at equilibrium if more H2 is added?
3. What will happen to the concentrations of H$\ce{H2}$, $\ce{CO}$, and $\ce{CH3OH}$ at equilibrium if CO is removed?
4. What will happen to the concentrations of $\ce{H2}$, $\ce{CO}$, and $\ce{CH3OH}$ at equilibrium if CH3OH is added?
5. What will happen to the concentrations of H$\ce{H2}$, $\ce{CO}$, and $\ce{CH3OH}$ at equilibrium if the temperature of the system is increased?
6. What will happen to the concentrations of $\ce{H2}$, $\ce{CO}$, and $\ce{CH3OH}$ at equilibrium if more catalyst is added?
S13.3.10
1. $K_c=\ce{\dfrac{[CH3OH]}{[H2]^2[CO]}}$;
2. [H2] increases, [CO] decreases, [CH3OH] increases;
3. [H2] increases, [CO] decreases, [CH3OH] decreases;
4. [H2] increases, [CO] increases, [CH3OH] increases;
5. [H2] increases, [CO] increases, [CH3OH] decreases;
6. no changes.
Q13.3.11
Nitrogen and oxygen react at high temperatures.
1. Write the expression for the equilibrium constant (Kc) for the reversible reaction $\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g)\hspace{20px}ΔH=\ce{181\:kJ}$
2. What will happen to the concentrations of N2, O2, and NO at equilibrium if more O2 is added?
3. What will happen to the concentrations of N2, O2, and NO at equilibrium if N2 is removed?
4. What will happen to the concentrations of N2, O2, and NO at equilibrium if NO is added?
5. What will happen to the concentrations of N2, O2, and NO at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
6. What will happen to the concentrations of N2, O2, and NO at equilibrium if the temperature of the system is increased?
7. What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added?
Q13.3.12
Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
1. Write the expression for the equilibrium constant for the reversible reaction $\ce{C}(s)+\ce{H2O}(g)\rightleftharpoons\ce{CO}(g)+\ce{H2}(g)\hspace{20px}ΔH=\mathrm{131.30\:kJ}$
2. What will happen to the concentration of each reactant and product at equilibrium if more C is added?
3. What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
4. What will happen to the concentration of each reactant and product at equilibrium if CO is added?
5. What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
S13.3.12
(a) $K_c=\ce{\dfrac{[CO][H2]}{[H2O]}}$; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.
Q13.3.13
Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
1. Write the expression for the equilibrium constant (Kc) for the reversible reaction $\ce{Fe2O3}(s)+\ce{3H2}(g)\rightleftharpoons\ce{2Fe}(s)+\ce{3H2O}(g) \hspace{20px} ΔH=\mathrm{98.7\:kJ}$
2. What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
3. What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
4. What will happen to the concentration of each reactant and product at equilibrium if H2 is added?
5. What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
6. What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
Q13.3.14
Ammonia is a weak base that reacts with water according to this equation:
$\ce{NH3}(aq)+\ce{H2O}(l)\rightleftharpoons\ce{NH4+}(aq)+\ce{OH-}(aq)$
Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water and why?
1. Addition of NaOH
2. Addition of HCl
3. Addition of NH4Cl
Only (b)
Q13.3.15
Acetic acid is a weak acid that reacts with water according to this equation:
$\ce{CH3CO2H}(aq)+\ce{H2O}(aq)\rightleftharpoons\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)$
Will any of the following increase the percent of acetic acid that reacts and produces $\ce{CH3CO2-}$ ion?
1. Addition of HCl
2. Addition of NaOH
3. Addition of NaCH3CO2
Q13.3.16
Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl, Ag+, and $\ce{NO3-}$, in contact with solid AgCl.
$\ce{Na+}(aq)+\ce{Cl-}(aq)+\ce{Ag+}(aq)+\ce{NO3-}(aq)\rightleftharpoons\ce{AgCl}(s)+\ce{Na+}(aq)+\ce{NO3-}(aq)$
$ΔH=\mathrm{−65.9\:kJ}$
S13.3.16
Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).
Q13.3.17
How can the pressure of water vapor be increased in the following equilibrium?
$\ce{H2O}(l)\rightleftharpoons\ce{H2O}(g) \hspace{20px} ΔH=\ce{41\:kJ}$
Q13.3.18
Additional solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and sulfate ion at equilibrium with solid silver sulfate.
$\ce{2Ag+}(aq)+\ce{SO4^2-}(aq)\rightleftharpoons\ce{Ag2SO4}(s)$
Which of the following will occur?
1. Ag+ or $\ce{SO4^2-}$ concentrations will not change.
2. The added silver sulfate will dissolve.
3. Additional silver sulfate will form and precipitate from solution as Ag+ ions and $\ce{SO4^2-}$ ions combine.
4. The Ag+ ion concentration will increase and the $\ce{SO4^2-}$ ion concentration will decrease.
(a)
Q13.3.19
The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium constant for ionization $\ce{(HX \rightleftharpoons H+ + X- )}$?
13.4: Equilibrium Calculations Exercises
Q13.4.1
A reaction is represented by this equation: $\ce{A}(aq)+\ce{2B}(aq)⇌\ce{2C}(aq) \hspace{20px} K_c=1×10^3$
1. Write the mathematical expression for the equilibrium constant.
2. Using concentrations ≤1 M, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
S13.4.1
$K_c=\ce{\dfrac{[C]^2}{[A][B]^2}}$. [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.
Q13.4.2
A reaction is represented by this equation: $\ce{2W}(aq)⇌\ce{X}(aq)+\ce{2Y}(aq) \hspace{20px} K_c=5×10^{−4}$
1. Write the mathematical expression for the equilibrium constant.
2. Using concentrations of ≤1 M, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.
Q13.4.3
What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation?
$\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)$
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 × 10−1 M NH3.
Kc = 6.00 × 10−2
Q13.4.4
Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.
$\ce{CH4}(g)+\ce{H2O}(g)⇌\ce{3H2}(g)+\ce{CO}(g)$
What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH4, 0.126 M; H2O, 0.242 M; CO, 0.126 M; H2 1.15 M, at a temperature of 760 °C?
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature.
Kc = 0.50
Q13.4.5
At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 3.3 × 10−3% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction.
$\ce{2NO2}(g)⇌\ce{2NO}(g)+\ce{O2}(g)$
Q13.4.6
Calculate the value of the equilibrium constant KP for the reaction $\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g)$ from these equilibrium pressures: NO, 0.050 atm; Cl2, 0.30 atm; NOCl, 1.2 atm.
S13.4.6
The equilibrium equation is KP = 1.9 × 103
Q13.4.7
When heated, iodine vapor dissociates according to this equation:
$\ce{I2}(g)⇌\ce{2I}(g)$
At 1274 K, a sample exhibits a partial pressure of I2 of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, KP, for the decomposition at 1274 K.
Q13.4.8
A sample of ammonium chloride was heated in a closed container.
$\ce{NH4Cl}(s)⇌\ce{NH3}(g)+\ce{HCl}(g)$
At equilibrium, the pressure of NH3(g) was found to be 1.75 atm. What is the value of the equilibrium constant KP for the decomposition at this temperature?
KP = 3.06
Q13.4.9
At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant KP for the transformation at 60 °C?
$\ce{H2O}(l)⇌\ce{H2O}(g)$
Q13.4.10
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
(a)
\begin{alignat}{3} &\ce{2SO3}(g)⇌\:&&\ce{2SO2}(g)+\:&&\ce{O2}(g)\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&+x\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&0.125\:M \end{alignat}
(b)
\begin{alignat}{3} &\ce{4NH3}(g)+\:&&\ce{3O2}(g)⇌\:&&\ce{2N2}(g)+\:&&\ce{6H2O}(g)\ &\underline{\hspace{40px}} &&3x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\ &\underline{\hspace{40px}} &&0.24\:M &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
(c) Change in pressure:
\begin{alignat}{3} &\ce{2CH4}(g)⇌\:&&\ce{C2H2}(g)+\:&&\ce{3H2}(g)\ &\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}}\ &\underline{\hspace{40px}} &&\textrm{25 torr} &&\underline{\hspace{40px}} \end{alignat}
(d) Change in pressure:
\begin{alignat}{3} &\ce{CH4}(g)+\:&&\ce{H2O}(g)⇌\:&&\ce{CO}(g)+\:&&\ce{3H2}(g)\ &\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\ &\underline{\hspace{40px}} &&\textrm{5 atm} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
(e)
\begin{alignat}{3} &\ce{NH4Cl}(s)⇌\:&&\ce{NH3}(g)+\:&&\ce{HCl}(g)\ & &&x &&\underline{\hspace{40px}}\ & &&1.03×10^{−4}\:M &&\underline{\hspace{40px}} \end{alignat}
(f) change in pressure:
\begin{alignat}{3} &\ce{Ni}(s)+\:&&\ce{4CO}(g)⇌\:&&\ce{Ni(CO)4}(g)\ & &&4x &&\underline{\hspace{40px}}\ & &&\textrm{0.40 atm} &&\underline{\hspace{40px}} \end{alignat}
S13.4.10
1. −2x, 2x, −0.250 M, 0.250 M;
2. 4x, −2x, −6x, 0.32 M, −0.16 M, −0.48 M;
3. −2x, 3x, −50 torr, 75 torr;
4. x, − x, −3x, 5 atm, −5 atm, −15 atm;
5. x, 1.03 × 10−4 M; (f) x, 0.1 atm.
Q13.4.11
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
(a)
\begin{alignat}{3} &\ce{2H2}(g)+\:&&\ce{O2}(g)⇌\:&&\ce{2H2O}(g)\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&+2x\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&1.50\:M \end{alignat}
(b)
\begin{alignat}{3} &\ce{CS2}(g)+\:&&\ce{4H2}(g)⇌\:&&\ce{CH4}(g)+\:&&\ce{2H2S}(g)\ &x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\ &0.020\:M &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
(c) Change in pressure:
\begin{alignat}{3} &\ce{H2}(g)+\:&&\ce{Cl2}(g)⇌\:&&\ce{2HCl}(g)\ &x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\ &\textrm{1.50 atm} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} \end{alignat}
(d) Change in pressure:
\begin{alignat}{3} &\ce{2NH3}(g)+\:&&\ce{2O2}(g)⇌\:&&\ce{N2O}(g)+\:&&\ce{3H2O}(g)\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&x\ &\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\textrm{60.6 torr} \end{alignat}
(e)
\begin{alignat}{3} &\ce{NH4HS}(s)⇌\:&&\ce{NH3}(g)+\:&&\ce{H2S}(g)\ & &&x &&\underline{\hspace{40px}}\ & &&9.8×10^{−6}\:M &&\underline{\hspace{40px}} \end{alignat}
(f) Change in pressure:
\begin{alignat}{3} &\ce{Fe}(s)+\:&&\ce{5CO}(g)⇌\:&&\ce{Fe(CO)4}(g)\ & &&\underline{\hspace{40px}} &&x\ & &&\underline{\hspace{40px}} &&\textrm{0.012 atm} \end{alignat}
Q13.4.12
Why are there no changes specified for Ni in Exercise, part (f)? What property of Ni does change?
S13.4.12
Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.
Q13.4.13
Why are there no changes specified for NH4HS in Exercise, part (e)? What property of NH4HS does change?
Q13.4.14
Analysis of the gases in a sealed reaction vessel containing NH3, N2, and H2 at equilibrium at 400 °C established the concentration of N2 to be 1.2 M and the concentration of H2 to be 0.24 M.
$\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g) \hspace{20px} K_c=\textrm{0.50 at 400 °C}$
Calculate the equilibrium molar concentration of NH3.
S13.4.14
[NH3] = 9.1 × 10−2 M
Q13.4.16
Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00−L flask at 448 °C.
$\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}$
Q13.4.17
What is the pressure of BrCl in an equilibrium mixture of Cl2, Br2, and BrCl if the pressure of Cl2 in the mixture is 0.115 atm and the pressure of Br2 in the mixture is 0.450 atm?
$\ce{Cl2}(g)+\ce{Br2}(g)⇌\ce{2BrCl}(g) \hspace{20px} K_P=4.7×10^{−2}$
S13.4.17
PBrCl = 4.9 × 10−2 atm
Q13.4.18
What is the pressure of CO2 in a mixture at equilibrium that contains 0.50 atm H2, 2.0 atm of H2O, and 1.0 atm of CO at 990 °C?
$\ce{H2}(g)+\ce{CO2}(g)⇌\ce{H2O}(g)+\ce{CO}(g) \hspace{20px} K_P=\textrm{1.6 at 990 °C}$
Q13.4.12
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
$\ce{CoO}(s)+\ce{CO}(g)⇌\ce{Co}(s)+\ce{CO2}(g) \hspace{20px} K_c=4.90×10^2\textrm{ at 550 °C}$
What concentration of CO remains in an equilibrium mixture with [CO2] = 0.100 M?
S13.4.12
[CO] = 2.0 × 10−4 M
Q13.4.13
Carbon reacts with water vapor at elevated temperatures.
$\ce{C}(s)+\ce{H2O}(g)⇌\ce{CO}(g)+\ce{H2}(g) \hspace{20px} K_c=\textrm{0.2 at 1000 °C}$
What is the concentration of CO in an equilibrium mixture with [H2O] = 0.500 M at 1000 °C?
Q13.4.14
Sodium sulfate 10−hydrate, $\ce{Na2SO4 \cdot 10H2O}$, dehydrates according to the equation
$\ce{Na2SO4⋅10H2O}(s)⇌\ce{Na2SO4}(s)+\ce{10H2O}(g) \hspace{20px}$
with $K_p=4.08×10^{−25}$ at 25°C. What is the pressure of water vapor at equilibrium with a mixture of $\ce{Na2SO4·10H2O}$ and $\ce{NaSO4}$?
S13.4.14
$P_{\ce{H2O}}=3.64×10^{−3}\:\ce{atm}$
Q13.4.15
Calcium chloride 6−hydrate, CaCl2·6H2O, dehydrates according to the equation
$\ce{CaCl2⋅6H2O}(s)⇌\ce{CaCl2}(s)+\ce{6H2O}(g) \hspace{20px} K_P=5.09×10^{−44}\textrm{ at 25 °C}$
What is the pressure of water vapor at equilibrium with a mixture of CaCl2·6H2O and CaCl2?
Q13.4.16
A student solved the following problem and found the equilibrium concentrations to be [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:
$\ce{2SO2}(g)+\ce{O2}(g)⇌\ce{2SO3}(g) \hspace{20px} K_c=4.32$
What are the equilibrium concentrations of all species in a mixture that was prepared with [SO3] = 0.500 M, [SO2] = 0 M, and [O2] = 0.350 M?
S13.4.16
Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium.
Q13.4.16
A student solved the following problem and found [N2O4] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N2O4 in a mixture formed from a sample of NO2 with a concentration of 0.10 M?
$\ce{2NO2}(g)⇌\ce{N2O4}(g) \hspace{20px} K_c=160$
Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.
$\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_c=1.07×10^{−5}$ in chloroform
(b) Show that the change is small enough to be neglected.
S13.4.16
(a)
• [NO2] = 1.17 × 10−3 M
• [N2O4] = 0.128 M
(b) Percent error $=\dfrac{5.87×10^{−4}}{0.129}×100\%=0.455\%$. The change in concentration of N2O4 is far less than the 5% maximum allowed.
Q13.4.17
Assume that the change in concentration of COCl2 is small enough to be neglected in the following problem.
1. Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl2 with an initial concentration of 0.3166 M. $\ce{COCl2}(g)⇌\ce{CO}(g)+\ce{Cl2}(g) \hspace{20px} K_c=2.2×10^{−10}$
2. Show that the change is small enough to be neglected.
Q13.4.18
Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.
$\ce{2H2S}(g)⇌\ce{2H2}(g)+\ce{S2}(g) \hspace{20px} K_P=2.2×10^{−6}$
(b) Show that the change is small enough to be neglected.
S13.4.18
(a)
• [H2S] = 0.810 atm
• [H2] = 0.014 atm
• S2] = 0.0072 atm
(b) The 2x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is $\dfrac{0.014}{0.824}×100\%=1.7\%$, which is less than allowed by the “5% test.” The error is, indeed, negligible.
Q13.4.19
What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl2O] = 1.00 M comes to equilibrium at 25 °C?
$\ce{H2O}(g)+\ce{Cl2O}(g)⇌\ce{2HOCl}(g) \hspace{20px} K_c=0.0900$
Q13.4.20
What are the concentrations of PCl5, PCl3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCl5 with [PCl5] = 2.00 M?
$\ce{PCl5}(g)⇌\ce{PCl3}(g)+\ce{Cl2}(g) \hspace{20px} K_c=0.0211$
S13.4.20
[PCl3] = 1.80 M; [PC3] = 0.195 M; [PCl3] = 0.195 M.
Q13.4.21
Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl2 produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:
$\ce{2NOCl}(g)⇌\ce{2NO}(g)+\ce{Cl2}(g) \hspace{20px} K_P=4.0×10^{−4}$
Q13.4.22
Calculate the equilibrium concentrations of NO, O2, and NO2 in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.10 M O2. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)
$\ce{2NO}(g)+\ce{O2}(g)⇌\ce{2NO2}(g) \hspace{20px} K_c=2.3×10^5\textrm{ at 250 °C}$
S13.4.22
• [NO2] = 0.19 M
• [NO] = 0.0070 M
• [O2] = 0.0035 M
Q13.4.23
Calculate the equilibrium concentrations that result when 0.25 M O2 and 1.0 M HCl react and come to equilibrium.
$\ce{4HCl}(g)+\ce{O2}(g)⇌\ce{2Cl2}(g)+\ce{2H2O}(g) \hspace{20px} K_c=3.1×10^{13}$
Q13.4.24
One of the important reactions in the formation of smog is represented by the equation
$\ce{O3}(g)+\ce{NO}(g)⇌\ce{NO2}(g)+\ce{O2}(g) \hspace{20px} K_P=6.0×10^{34}$
What is the pressure of O3 remaining after a mixture of O3 with a pressure of 1.2 × 10−8 atm and NO with a pressure of 1.2 × 10−8 atm comes to equilibrium? (Hint: KP is large; assume the reaction goes to completion then comes back to equilibrium.)
S13.4.24
$P_{\ce{O3}}=4.9×10^{−26}\:\ce{atm}$
Q13.4.24
Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl2. (Hint: KP is small; assume the reverse reaction goes to completion then comes back to equilibrium.)
$\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g) \hspace{20px} K_P=2.5×10^3$
Q13.4.25
Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C.
$\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}$
507 g
Q13.4.26
Butane exists as two isomers, n−butane and isobutane.
KP = 2.5 at 25 °C
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
Q13.4.27
What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.050 for the decomposition reaction of CaCO3 at that temperature?
$\ce{CaCO3}(s)⇌\ce{CaO}(s)+\ce{CO2}(g)$
330 g
Q13.4.28
The equilibrium constant (Kc) for this reaction is 1.60 at 990 °C:
$\ce{H2}(g)+\ce{CO2}(g)⇌\ce{H2O}(g)+\ce{CO}(g)$
Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H2, 2.00 mol of CO2, 0.750 mol of H2O, and 1.00 mol of CO to a 5.00-L container at 990 °C.
Q13.4.29
At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N2O4 and NO2 are $P_{\ce{N2O4}}=0.70\:\ce{atm}$ and $P_{\ce{NO2}}=0.30\:\ce{atm}$.
1. Predict how the pressures of NO2 and N2O4 will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?
2. Calculate the partial pressures of NO2 and N2O4 when they are at equilibrium at 9.0 atm and 25 °C.
S13.4.29
(a) Both gases must increase in pressure.
(b) $P_{\ce{N2O4}}=\textrm{8.0 atm and }P_{\ce{NO2}}=1.0\:\ce{atm}$
Q13.4.30
In a 3.0-L vessel, the following equilibrium partial pressures are measured: N2, 190 torr; H2, 317 torr; NH3, 1.00 × 103 torr.
$\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)$
1. How will the partial pressures of H2, N2, and NH3 change if H2 is removed from the system? Will they increase, decrease, or remain the same?
2. Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
Q13.4.31
The equilibrium constant (Kc) for this reaction is 5.0 at a given temperature.
$\ce{CO}(g)+\ce{H2O}(g) <=> \ce{CO2}(g)+\ce{H2}(g)\)] 1. On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture? 2. Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established. S13.4.31 (a) 0.33 mol. (b) [CO]2 = 0.50 M Added H2 forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H2 is added. Q13.4.32a Antimony pentachloride decomposes according to this equation: $\ce{SbCl5}(g)⇌\ce{SbCl3}(g)+\ce{Cl2}(g)$ An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature? Q13.4.32b Consider the reaction between H2 and O2 at 1000 K \[\ce{2H2}(g)+\ce{O2}(g)⇌\ce{2H2O}(g) \hspace{20px} K_P=\dfrac{(P_{\ce{H2O}})^2}{(P_{\ce{O2}})(P_{\ce{H2}})^3}=1.33×10^{20}$
If 0.500 atm of H2 and 0.500 atm of O2 are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?
S13.4.32b
$P_{\ce{H2}}=8.64×10^{−11}\:\ce{atm}$
$P_{\ce{O2}}=0.250\:\ce{atm}$
$P_{\ce{H2O}}=0.500\:\ce{atm}$
Q13.4.33
An equilibrium is established according to the following equation
$\ce{Hg2^2+}(aq)+\ce{NO3−}(aq)+\ce{3H+}(aq)⇌\ce{2Hg^2+}(aq)+\ce{HNO2}(aq)+\ce{H2O}(l) \hspace{20px} K_c=4.6$
What will happen in a solution that is 0.20 M each in $\ce{Hg2^2+}$, $\ce{NO3−}$, H+, Hg2+, and HNO2?
1. $\ce{Hg2^2+}$ will be oxidized and $\ce{NO3−}$ reduced.
2. $\ce{Hg2^2+}$ will be reduced and $\ce{NO3−}$ oxidized.
3. Hg2+ will be oxidized and HNO2 reduced.
4. Hg2+ will be reduced and HNO2 oxidized.
5. There will be no change because all reactants and products have an activity of 1.
Q13.4.34
Consider the equilibrium
$\ce{4NO2}(g)+\ce{6H2O}(g)⇌\ce{4NH3}(g)+\ce{7O2}(g)$
1. What is the expression for the equilibrium constant (Kc) of the reaction?
2. How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
3. If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO2?
4. If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?
S13.4.34
(a) $K_c=\ce{\dfrac{[NH3]^4[O2]^7}{[NO2]^4[H2O]^6}}$. (b) [NH3] must increase for Qc to reach Kc. (c) That decrease in pressure would decrease [NO2]. (d) $P_{\ce{O2}}=49\:\ce{torr}$
Q13.4.35
The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as
$\ce{HbO2}(aq)+\ce{H3O+}(aq)+\ce{CO2}(g)⇌\ce{CO2−Hb−H+}+\ce{O2}(g)+\ce{H2O}(l)$
1. (a) Write the equilibrium constant expression for this reaction.
2. (b) Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxyhemoglobin in the blood passing through the muscle.
Q13.4.36
The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.
$\ce{C12H22O11}(aq)+\ce{H2O}(l)⟶\ce{C6H12O6}(aq)+\ce{C6H12O6}(aq)$
Rate = k[C12H22O11]
In neutral solution, k = 2.1 × 10−11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 105 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.
S13.4.36
[fructose] = 0.15 M
Q13.4.37
The density of trifluoroacetic acid vapor was determined at 118.1 °C and 468.5 torr, and found to be 2.784 g/L. Calculate Kc for the association of the acid.
Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25 °C, a value of KP = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g), NO2(g), and NO(g).
S13.4.37
$P_{\ce{N2O3}}=\textrm{1.90 atm and }P_{\ce{NO}}=P_{\ce{NO2}}=\textrm{1.90 atm}$
Q13.4.38
A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N2, 1.00 M; H2, 0.50 M; and NH3, 0.25 M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M?
Q13.4.39
A 0.010 M solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010 M solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.
(a) Which acid has the larger equilibrium constant for ionization
HA $[\ce{HA}(aq)⇌\ce{A-}(aq)+\ce{H+}(aq)]$ or HB $[\ce{HB}(aq)⇌\ce{H+}(aq)+\ce{B-}(aq)]$?
(b) What are the equilibrium constants for the ionization of these acids?
(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A or B), and the hydrogen ion (H+). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)
S13.4.39
(a) HB ionizes to a greater degree and has the larger Kc.
(b) Kc(HA) = 5 × 10−4
Kc(HB) = 3 × 10−3 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/13%3A_Fundamental_Equilibrium_Concepts/13.E%3A_Fundamental_Equilibrium_Concepts_%28Exercises%29.txt |
This chapter will illustrate the chemistry of acid-base reactions and equilibria, and provide you with tools for quantifying the concentrations of acids and bases in solutions.
14: Acid-Base Equilibria
Learning Objectives
• Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
• Write equations for acid and base ionization reactions
• Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
• Describe the acid-base behavior of amphiprotic substances
Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
In an earlier chapter on chemical reactions, we defined acids and bases as Arrhenius did: We identified an acid as a compound that dissolves in water to yield hydronium ions (H3O+) and a base as a compound that dissolves in water to yield hydroxide ions ($\ce{OH-}$). This definition is not wrong; it is simply limited.
Later, we extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, $\ce{H^+}$. A proton is what remains when a normal hydrogen atom, $\ce{^1_1H}$, loses an electron. A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). In a subsequent chapter of this text we will introduce the most general model of acid-base behavior introduced by the American chemist G. N. Lewis.
Acids may be compounds such as HCl or H2SO4, organic acids like acetic acid ($\ce{CH_3COOH}$) or ascorbic acid (vitamin C), or H2O. Anions (such as $\ce{HSO_4^-}$, $\ce{H_2PO_4^-}$, $\ce{HS^-}$, and $\ce{HCO_3^-}$) and cations (such as $\ce{H_3O^+}$, $\ce{NH_4^+}$, and $\ce{[Al(H_2O)_6]^{3+}}$) may also act as acids. Bases fall into the same three categories. Bases may be neutral molecules (such as $\ce{H_2O}$, $\ce{NH_3}$, and $\ce{CH_3NH_2}$), anions (such as $\ce{OH^-}$, $\ce{HS^-}$, $\ce{HCO_3^-}$, $\ce{CO_3^{2−}}$, $\ce{F^-}$, and $\ce{PO_4^{3−}}$), or cations (such as $\ce{[Al(H_2O)_5OH]^{2+}}$). The most familiar bases are ionic compounds such as $\ce{NaOH}$ and $\ce{Ca(OH)_2}$, which contain the hydroxide ion, $\ce{OH^-}$. The hydroxide ion in these compounds accepts a proton from acids to form water:
$\ce{H^+ + OH^- \rightarrow H_2O} \label{14.11}$
We call the product that remains after an acid donates a proton the conjugate base of the acid. This species is a base because it can accept a proton (to re-form the acid):
$\text{acid} \rightleftharpoons \text{proton} + \text{conjugate base}\label{14.12a}$
$\ce{HF \rightleftharpoons H^+ + F^-} \label{14.12b}$
$\ce{H_2SO_4 \rightleftharpoons H^+ + HSO_4^{−}}\label{14.12c}$
$\ce{H_2O \rightleftharpoons H^+ + OH^-}\label{14.12d}$
$\ce{HSO_4^- \rightleftharpoons H^+ + SO_4^{2−}}\label{14.12e}$
$\ce{NH_4^+ \rightleftharpoons H^+ + NH_3} \label{14.12f}$
We call the product that results when a base accepts a proton the base’s conjugate acid. This species is an acid because it can give up a proton (and thus re-form the base):
$\text{base} + \text{proton} \rightleftharpoons \text{conjugate acid} \label{14.13a}$
$\ce{OH^- +H^+ \rightleftharpoons H2O}\label{14.13b}$
$\ce{H_2O + H^+ \rightleftharpoons H3O+}\label{14.13c}$
$\ce{NH_3 +H^+ \rightleftharpoons NH4+}\label{14.13d}$
$\ce{S^{2-} +H^+ \rightleftharpoons HS-}\label{14.13e}$
$\ce{CO_3^{2-} +H^+ \rightleftharpoons HCO3-}\label{14.13f}$
$\ce{F^- +H^+ \rightleftharpoons HF} \label{14.13g}$
In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$:
The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
When we add a base to water, a base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding pyridine to water yields hydroxide ions and pyridinium ions:
Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic treated in a later section of this chapter. In the preceding paragraphs we saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:
This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization. Pure water undergoes autoionization to a very slight extent. Only about two out of every $10^9$ molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water (Kw):
$\ce{H_2O}_{(l)}+\ce{H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)}+\ce{OH^-}_{(aq)}\;\;\; K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{14.14}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of $1.0 \times 10^{−14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for $K_\ce{w}$ is approximately $5.1 \times 10^{−13}$, roughly 50-times larger than the value at 25 °C.
Example $1$: Ion Concentrations in Pure Water
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, $\ce{[H_3O^+]} = \ce{[OH^- ]}$. At 25 °C:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]}=\ce{[H_3O^+]^2}=\ce{[OH^- ]^2}=1.0 \times 10^{−14} \nonumber$
So:
$\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber$
The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal $1.0 \times 10^{−7}\; M$.
Exercise $1$
The ion product of water at 80 °C is $2.4 \times 10^{−13}$. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?
Answer
$\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7}\; M$
It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium. Example 14.12 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.
Example $2$: The Inverse Proportionality of $\ce{[H_3O^+]}$ and $\ce{[OH^- ]}$
The Inverse Proportionality of [H3O+] and [OH-] A solution of carbon dioxide in water has a hydronium ion concentration of $2.0 \times 10^{−6}\; M$. What is the concentration of hydroxide ion at 25 °C?
Solution
We know the value of the ion-product constant for water at 25 °C:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber$
$K_\ce{w}=\ce{[H3O+][OH^- ]}=1.0 \times 10^{−14} \nonumber$
Thus, we can calculate the missing equilibrium concentration.
Rearrangement of the Kw expression yields that $[\ce{OH^- }]$ is directly proportional to the inverse of [H3O+]:
$[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber$
The hydroxide ion concentration in water is reduced to $5.0 \times 10^{−9}\: M$ as the hydrogen ion concentration increases to $2.0 \times 10^{−6}\; M$. This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the $\ce{[OH^- ]}$ is reduced relative to that in pure water.
A check of these concentrations confirms that our arithmetic is correct:
\begin{align*} K_\ce{w} &=\ce{[H_3O^+][OH^- ]} \[4pt] &=(2.0 \times 10^{−6})(5.0 \times 10^{−9})\[4pt] &=1.0 \times 10^{−14} \end{align*} \nonumber
Exercise $2$
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?
Answer
$\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber$
Amphiprotic Species
Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. Another term used to describe such species is amphoteric, which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here:
$\ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{CO^{2-}}_{3(aq)} + \ce{H_3O^+}_{(aq)} \label{14.15a}$
$\ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{H_2CO}_{3(aq)} + \ce{OH^-}_{(aq)} \label{14.15b}$
Example $3$: The Acid-Base Behavior of an Amphoteric Substance
Write separate equations representing the reaction of $\ce{HSO3-}$
1. as an acid with $\ce{OH^-}$
2. as a base with HI
Solution
1. $\ce{HSO3-}(aq)+ \ce{OH^-}(aq)\rightleftharpoons \ce{SO3^2-}(aq)+ \ce{H_2O}_{(l)}$
2. $\ce{HSO3-}(aq)+\ce{HI}(aq)\rightleftharpoons \ce{H2SO3}(aq)+\ce{I-}(aq)$
Exercise $3$
Write separate equations representing the reaction of $\ce{H2PO4-}$
1. as a base with HBr
2. as an acid with $\ce{OH^-}$
Answer a
$\ce{H2PO4-}(aq)+\ce{HBr}(aq)\rightleftharpoons \ce{H3PO4}(aq)+\ce{Br-}(aq)$
Answer b
$\ce{H2PO4-}(aq)+\ce{OH^-} (aq)\rightleftharpoons \ce{HPO4^2-}(aq)+ \ce{H_2O}_{(l)}$
Summary
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3O+, and the hydroxide ion, $\ce{OH^-}$ when it undergoes autoionization:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber$
The ion product of water, Kw is the equilibrium constant for the autoionization reaction:
$K_\ce{w}=\mathrm{[H_3O^+][OH^- ]=1.0 \times 10^{−14} \; at\; 25°C} \nonumber$
Key Equations
• $K_{\ce w} = \ce{[H3O+][OH^- ]} = 1.0 \times 10^{−14}\textrm{ (at 25 °C)} \nonumber$
Glossary
acid ionization
reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
amphiprotic
species that may either gain or lose a proton in a reaction
amphoteric
species that can act as either an acid or a base
autoionization
reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionization
reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
Brønsted-Lowry acid
proton donor
Brønsted-Lowry base
proton acceptor
conjugate acid
substance formed when a base gains a proton
conjugate base
substance formed when an acid loses a proton
ion-product constant for water (Kw)
equilibrium constant for the autoionization of water | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.1%3A_Brnsted-Lowry_Acids_and_Bases.txt |
Learning Objectives
• Explain the characterization of aqueous solutions as acidic, basic, or neutral
• Express hydronium and hydroxide ion concentrations on the pH and pOH scales
• Perform calculations relating pH and pOH
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ($K_w$). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm:
$\mathrm{pX=−\log X} \label{1}$
The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:
$\mathrm{pH=-\log[H_3O^+]}\label{$2$}$
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
$\mathrm{[H_3O^+]=10^{−pH}}\label{$3$}$
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
$\mathrm{pOH=-\log [OH^−]}\label{$4$}$
or
$\mathrm{[OH^-]=10^{−pOH}} \label{$5$}$
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_w$ expression:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{$6$}$
$-\log K_\ce{w}=\mathrm{-\log([H_3O^+][OH^−])=-\log[H_3O^+] + -\log[OH^-]}\label{$7$}$
$\mathrm{p\mathit{K}_w=pH + pOH} \label{$8$}$
At 25 °C, the value of $K_w$ is $1.0 \times 10^{−14}$, and so:
$\mathrm{14.00=pH + pOH} \label{$9$}$
The hydronium ion molarity in pure water (or any neutral solution) is $1.0 \times 10^{-7}\; M$ at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
$\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{10}$
$\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{11}$
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \times 10^{-7}\; M$ and hydroxide ion molarities less than $1.0 \times 10^{-7}\; M$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \times 10^{-7}\; M$ and hydroxide ion molarities greater than $1.0 \times 10^{-7}\; M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
When $pH=7$ Solutions are not Neutral
Since the autoionization constant $K_w$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the hydronium molarity of pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:
\begin{align*} pH &=-\log[\ce{H_3O^+}] \[4pt] &= -\log(4.9\times 10^{−7}) \[4pt] &=6.31 \label{12} \end{align*}
\begin{align*} pOH &=-\log[\ce{OH^-}]\[4pt] & =-\log(4.9\times 10^{−7}) \[4pt] &=6.31 \label{13}\end{align*}
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table $1$).
Table $1$: Summary of Relations for Acidic, Basic and Neutral Solutions
Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] pH < 7
neutral [H3O+] = [OH] pH = 7
basic [H3O+] < [OH] pH > 7
Figure $1$ shows the relationships between [H3O+], [OH], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.
Example $1$: Calculation of pH from $\ce{[H_3O^+]}$
What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of $1.2 \times 10^{−3}\; M$?
Solution
\begin{align*} pH &=-\log [H_3O^+] \[4pt] &= -\log(1.2 \times 10^{−3}) \[4pt] &=−(−2.92) \[4pt]&=2.92 \end{align*} \nonumber
Exercise $1$
Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water:
$\ce{CO2(aq) + H2O (l) \rightleftharpoons H2CO3(aq)} \nonumber$
Air-saturated water has a hydronium ion concentration caused by the dissolved $\ce{CO_2}$ of $2.0 \times 10^{−6}\; M$, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.
Answer
5.70
Example $2$: Calculation of Hydronium Ion Concentration from pH
Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline).
Solution
$\mathrm{pH=-\log[H_3O^+]=7.3} \nonumber$
$\mathrm{\log[H_3O^+]=−7.3} \nonumber$
$\mathrm{[H_3O^+]=10^{−7.3}} \nonumber$
or
$[\ce{H_3O^+}]=\textrm{antilog of} −7.3 \nonumber$
$[\ce{H_3O^+}]=5\times 10^{−8}\;M \nonumber$
(On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10−7.3.)
Exercise $2$
Calculate the hydronium ion concentration of a solution with a pH of −1.07.
Answer
12 M
Environmental Science
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid:
$\ce{H2O (l) + CO2(g) ⟶ H2CO3(aq)} \label{14}$
$\ce{H2CO3(aq) \rightleftharpoons H^+(aq) + HCO3^- (aq)} \label{15}$
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
$\ce{H2O (l) + SO3(g) ⟶ H2SO4(aq)} \label{16}$
$\ce{H2SO4(aq) ⟶ H^+(aq) + HSO4^- (aq)} \label{17}$
Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $2$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
Example $3$: Calculation of pOH
What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH?
Solution
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH] = 0.0125 M:
$\mathrm{pOH=-\log[OH^− ]=-\log 0.0125} \nonumber$
$=−(−1.903)=1.903 \nonumber$
The pH can be found from the $\ce{pOH}$:
$\mathrm{pH+pOH=14.00} \nonumber$
$\mathrm{pH=14.00−pOH=14.00−1.903=12.10} \nonumber$
Exercise $3$
The hydronium ion concentration of vinegar is approximately $4 \times 10^{−3}\; M$. What are the corresponding values of pOH and pH?
Answer
pOH = 11.6,
pH = 14.00 - pOH = 2.4
The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure $3$).
The pH of a solution may also be visually estimated using colored indicators (Figure $3$).
Summary
The concentration of hydronium ion in a solution of an acid in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of H3O+ in a solution can be expressed as the pH of the solution; $\ce{pH} = -\log \ce{H3O+}$. The concentration of OH can be expressed as the pOH of the solution: $\ce{pOH} = -\log[\ce{OH-}]$. In pure water, pH = 7.00 and pOH = 7.00
Key Equations
• $\ce{pH}=-\log[\ce{H3O+}]$
• $\ce{pOH} = -\log[\ce{OH-}]$
• [H3O+] = 10−pH
• [OH] = 10−pOH
• pH + pOH = pKw = 14.00 at 25 °C
Glossary
acidic
describes a solution in which [H3O+] > [OH]
basic
describes a solution in which [H3O+] < [OH]
neutral
describes a solution in which [H3O+] = [OH]
pH
logarithmic measure of the concentration of hydronium ions in a solution
pOH
logarithmic measure of the concentration of hydroxide ions in a solution | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.2%3A_pH_and_pOH.txt |
Learning Objectives
• Assess the relative strengths of acids and bases according to their ionization constants
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems
We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber$
Water is the base that reacts with the acid $\ce{HA}$, $\ce{A^{−}}$ is the conjugate base of the acid $\ce{HA}$, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of $\ce{H3O+}$ and $\ce{A^{−}}$ when the acid ionizes in water; Figure $1$ lists several strong acids. A weak acid gives small amounts of $\ce{H3O+}$ and $\ce{A^{−}}$.
Figure $1$: Some of the common strong acids and bases are listed here.
Six Strong Acids Six Strong Bases
$\ce{HClO4}$ perchloric acid $\ce{LiOH}$ lithium hydroxide
$\ce{HCl}$ hydrochloric acid $\ce{NaOH}$ sodium hydroxide
$\ce{HBr}$ hydrobromic acid $\ce{KOH}$ potassium hydroxide
$\ce{HI}$ hydroiodic acid $\ce{Ca(OH)2}$ calcium hydroxide
$\ce{HNO3}$ nitric acid $\ce{Sr(OH)2}$ strontium hydroxide
$\ce{H2SO4}$ sulfuric acid $\ce{Ba(OH)2}$ barium hydroxide
The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid $\ce{HA}$:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber$
we write the equation for the ionization constant as:
$K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}} \nonumber$
where the concentrations are those at equilibrium. As noted in the section on equilibrium constants, although water is a reactant in the reaction, it is the solvent as well, so its activity has a value of 1, which does not change the value of $K_a$.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
The larger the $K_a$ of an acid, the larger the concentration of $\ce{H3O+}$ and $\ce{A^{−}}$ relative to the concentration of the nonionized acid, $\ce{HA}$. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase.
The following data on acid-ionization constants indicate the order of acid strength: $\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}$
\begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \[4pt] \ce{HSO4-}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned} \nonumber
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:
$\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon}$
Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
Example $1$: Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Solution
The percent ionization for an acid is:
$\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \nonumber$
The chemical equation for the dissociation of the nitrous acid is:
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \nonumber$
Since $10^{−pH} = \ce{[H3O+]}$, we find that $10^{−2.09} = 8.1 \times 10^{−3}\, M$, so that percent ionization (Equation \ref{PercentIon}) is:
$\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \nonumber$
Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.
Exercise $1$
Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89.
Answer
1.3% ionized
We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:
$\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq) \nonumber$
Water is the acid that reacts with the base, $\ce{HB^{+}}$ is the conjugate acid of the base $\ce{B}$, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH and HB+ when it reacts with water; Figure $1$ lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.
As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, $\ce{B}$:
$\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq), \nonumber$
we write the equation for the ionization constant as:
$K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}} \nonumber$
where the concentrations are those at equilibrium. Again, we do not see water in the equation because water is the solvent and has an activity of 1. The chemical reactions and ionization constants of the three bases shown are:
\begin{aligned} \ce{NO2-}(aq)+\ce{H2O}(l) &⇌\ce{HNO2}(aq)+\ce{OH-}(aq) \quad &K_\ce{b}=2.17×10^{−11} \[4pt] \ce{CH3CO2-}(aq)+\ce{H2O}(l) &⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) &K_\ce{b}=5.6×10^{−10} \[4pt] \ce{NH3}(aq)+\ce{H2O}(l) &⇌\ce{NH4+}(aq)+\ce{OH-}(aq) &K_\ce{b}=1.8×10^{−5} \end{aligned} \nonumber
A table of ionization constants of weak bases appears in Table E2. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution.
Consider the ionization reactions for a conjugate acid-base pair, $\ce{HA − A^{−}}$:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber$
with $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$.
$\ce{A-}(aq)+\ce{H2O}(l)⇌\ce{OH-}(aq)+\ce{HA}(aq) \nonumber$
with $K_\ce{b}=\ce{\dfrac{[HA][OH]}{[A- ]}}$.
Adding these two chemical equations yields the equation for the autoionization for water:
\begin{align*} \cancel{\ce{HA}(aq)}+\ce{H2O}(l)+\cancel{\ce{A-}(aq)}+\ce{H2O}(l) &⇌ \ce{H3O+}(aq)+\cancel{\ce{A-}(aq)}+\ce{OH-}(aq)+\cancel{\ce{HA}(aq)} \[4pt] \ce{2H2O}(l) &⇌\ce{H3O+}(aq)+\ce{OH-}(aq) \end{align*} \nonumber
As shown in the previous chapter on equilibrium, the $K$ expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ $K$ expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:
$K_\ce{a}×K_\ce{b}=\ce{\dfrac{[H3O+][A- ]}{[HA]}×\dfrac{[HA][OH- ]}{[A- ]}}=\ce{[H3O+][OH- ]}=K_\ce{w} \nonumber$
For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion ($\ce{CH3COO-}$), is 5.6 × 10−10. The product of these two constants is indeed equal to $K_w$:
$K_\ce{a}×K_\ce{b}=(1.8×10^{−5})×(5.6×10^{−10})=1.0×10^{−14}=K_\ce{w} \nonumber$
The extent to which an acid, $\ce{HA}$, donates protons to water molecules depends on the strength of the conjugate base, $\ce{A^{−}}$, of the acid. If $\ce{A^{−}}$ is a strong base, any protons that are donated to water molecules are recaptured by $\ce{A^{−}}$. Thus there is relatively little $\ce{A^{−}}$ and $\ce{H3O+}$ in solution, and the acid, $\ce{HA}$, is weak. If $\ce{A^{−}}$ is a weak base, water binds the protons more strongly, and the solution contains primarily $\ce{A^{−}}$ and $\ce{H3O^{+}}$—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure $2$).
Figure $3$ lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.
The first six acids in Figure $3$ are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.
Those acids that lie between the hydronium ion and water in Figure $3$ form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure $3$ exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.
The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure $3$. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.
Example $2$: The Product Ka × Kb = Kw
Use the $K_b$ for the nitrite ion, $\ce{NO2-}$, to calculate the $K_a$ for its conjugate acid.
Solution
Kb for $\ce{NO2-}$ is given in this section as 2.17 × 10−11. The conjugate acid of $\ce{NO2-}$ is HNO2; Ka for HNO2 can be calculated using the relationship:
$K_\ce{a}×K_\ce{b}=1.0×10^{−14}=K_\ce{w} \nonumber$
Solving for Ka, we get:
\begin{align*} K_\ce{a} &=\dfrac{K_\ce{w}}{K_\ce{b}} \[4pt] &=\dfrac{1.0×10^{−14}}{2.17×10^{−11}} \[4pt] &=4.6×10^{−4} \end{align*} \nonumber
This answer can be verified by finding the Ka for HNO2 in Table E1
Exercise $2$
We can determine the relative acid strengths of $\ce{NH4+}$ and $\ce{HCN}$ by comparing their ionization constants. The ionization constant of $\ce{HCN}$ is given in Table E1 as 4.9 × 10−10. The ionization constant of $\ce{NH4+}$ is not listed, but the ionization constant of its conjugate base, $\ce{NH3}$, is listed as 1.8 × 10−5. Determine the ionization constant of $\ce{NH4+}$, and decide which is the stronger acid, $\ce{HCN}$ or $\ce{NH4+}$.
Answer
$\ce{NH4+}$ is the slightly stronger acid (Ka for $\ce{NH4+}$ = 5.6 × 10−10).
The Ionization of Weak Acids and Weak Bases
Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).
Acetic acid ($\ce{CH3CO2H}$) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure $4$). The remaining weak acid is present in the nonionized form.
For acetic acid, at equilibrium:
$K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber$
Table $1$: Ionization Constants of Some Weak Acids
Ionization Reaction Ka at 25 °C
$\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}$ 1.2 × 10−2
$\ce{HF + H2O ⇌ H3O+ + F-}$ 3.5 × 10−4
$\ce{HNO2 + H2O ⇌ H3O+ + NO2-}$ 4.6 × 10−4
$\ce{HNCO + H2O ⇌ H3O+ + NCO-}$ 2 × 10−4
$\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}$ 1.8 × 10−4
$\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}$ 1.8 × 10−5
$\ce{HCIO + H2O ⇌ H3O+ + CIO-}$ 2.9 × 10−8
$\ce{HBrO + H2O ⇌ H3O+ + BrO-}$ 2.8 × 10−9
$\ce{HCN + H2O ⇌ H3O+ + CN-}$ 4.9 × 10−10
Table $1$ gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1.
At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).
For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation:
$\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber$
This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.
We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.4.5). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, $K_b$, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:
$K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}} \nonumber$
The ionization constants of several weak bases are given in Table $2$ and Table E2.
Table $2$: Ionization Constants of Some Weak Bases
Ionization Reaction Kb at 25 °C
$\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}$ 5.9 × 10−4
$\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}$ 4.4 × 10−4
$\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}$ 6.3 × 10−5
$\ce{NH3 + H2O ⇌ NH4+ + OH-}$ 1.8 × 10−5
$\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}$ 4.3 × 10−10
Example $3$: Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and $\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M$. What is the value of $K_a$ for acetic acid?
Solution
We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \[4pt] &=1.77×10^{−5} \end{align*} \nonumber
Exercise $3$
What is the equilibrium constant for the ionization of the $\ce{HSO4-}$ ion, the weak acid used in some household cleansers:
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber$
In one mixture of NaHSO4 and Na2SO4 at equilibrium, $\ce{[H3O+]}$ = 0.027 M; $\ce{[HSO4- ]}=0.29\:M$; and $\ce{[SO4^2- ]}=0.13\:M$.
Answer
$K_a$ for $\ce{HSO_4^-}= 1.2 ×\times 10^{−2}$
Example $4$: Determination of Kb from Equilibrium Concentrations
Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, $\ce{[C8H10N4O2H+]}$ = 5.0 × 10−3 M, and [OH] = 2.5 × 10−3 M?
Solution
At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber$
so
$K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber$
Exercise $4$
What is the equilibrium constant for the ionization of the $\ce{HPO4^2-}$ ion, a weak base:
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber$
In a solution containing a mixture of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ at equilibrium with:
• $[\ce{OH^{−}}] = 1.3 × 10^{−6} M$
• $\ce{[H2PO4^{-}]=0.042\:M}$ and
• $\ce{[HPO4^{2-}]=0.341\:M}$.
Answer
Kb for $\ce{HPO4^2-}=1.6×10^{−7}$
Example $5$: Determination of Ka or Kb from pH
The pH of a 0.0516-M solution of nitrous acid, $\ce{HNO2}$, is 2.34. What is its $K_a$?
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber$
Solution
We determine an equilibrium constant starting with the initial concentrations of HNO2, $\ce{H3O+}$, and $\ce{NO2-}$ as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)
We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:
We can summarize the various concentrations and changes as shown here. Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction.
To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate $\ce{[H3O+]}$, the equilibrium concentration of $\ce{H3O+}$, from the pH:
$\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber$
The change in concentration of $\ce{H3O+}$, $x_{\ce{[H3O+]}}$, is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration, $\mathrm{[H_3O^+]_i}$. The initial concentration of $\ce{H3O+}$ is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).
The change in concentration of $\ce{NO2-}$ is equal to the change in concentration of $\ce{[H3O+]}$. For each 1 mol of $\ce{H3O+}$ that forms, 1 mol of $\ce{NO2-}$ forms. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration.
Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:
Finally, we calculate the value of the equilibrium constant using the data in the table:
$K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber$
Exercise $5$
The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.
Answer
$K_b = 1.8 × 10^{−5}$
Example $6$: Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings.
What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?
$\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber$
Solution
1. Determine x and equilibrium concentrations. The equilibrium expression is:
$\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber$
Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction so we do not need to consider it when setting up the ICE table.
The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):
2. Solve for $x$ and the equilibrium concentrations. At equilibrium:
\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*} \nonumber
Now solve for $x$. Because the initial concentration of acid is reasonably large and $K_a$ is very small, we assume that $x << 0.534$, which permits us to simplify the denominator term as $(0.534 − x) = 0.534$. This gives:
$K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber$
Solve for $x$ as follows:
\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \[4pt] &=9.6×10^{−5} \[4pt] x &=\sqrt{9.6×10^{−5}} \[4pt] &=9.8×10^{−3} \end{align*} \nonumber
To check the assumption that $x$ is small compared to 0.534, we calculate:
\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*} \nonumber
$x$ is less than 5% of the initial concentration; the assumption is valid.
We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:
\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \[4pt] &=9.8×10^{−3}\:M \end{align*} \nonumber
The pH of the solution can be found by taking the negative log of the $\ce{[H3O+]}$, so:
$pH = −\log(9.8×10^{−3})=2.01 \nonumber$
Exercise $6$: acetic acid
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber$
Hint
Determine $\ce{[CH3CO2- ]}$ at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or $\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100$.
Answer
percent ionization = 1.3%
The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.
Example $7$: Equilibrium Concentrations in a Solution of a Weak Base
Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:
$\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \nonumber$
Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example $6$. The reactants and products will be different and the numbers will be different, but the logic will be the same:
1. Determine x and equilibrium concentrations. The table shows the changes and concentrations:
2. Solve for $x$ and the equilibrium concentrations. At equilibrium:
$K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber$
If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives:
$x=4.0×10^{−3} \nonumber$
This change is less than 5% of the initial concentration (0.25), so the assumption is justified.
Recall that, for this computation, $x$ is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):
\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \[4pt] &=4.0×10^{−3}\:M \end{align*} \nonumber
Then calculate pOH as follows:
$\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber$
Using the relation introduced in the previous section of this chapter:
$\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber$
permits the computation of pH:
$\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber$
Check the work. A check of our arithmetic shows that $K_b = 6.3 \times 10^{−5}$.
Exercise $7$
1. Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 6.3 × 10−5.
2. Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or $\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%$
Answer a
$7.56 × 10^{−4}\, M$, 2.33%
Answer b
2.33%
Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.
Example $8$: Equilibrium Concentrations in a Solution of a Weak Acid
Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the $\ce{HSO4-}$ ion, a weak acid. What is the pH of a 0.50-M solution of $\ce{HSO4-}$?
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber$
Solution
We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of $\ce{HSO4-}$ so that we can use $\ce{[H3O+]}$ to determine the pH. As in the previous examples, we can approach the solution by the following steps:
1. Determine $x$ and equilibrium concentrations. This table shows the changes and concentrations:
2. Solve for $x$ and the concentrations.
As we begin solving for $x$, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of $x$.
At equilibrium:
$K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber$
If we assume that x is small and approximate (0.50 − x) as 0.50, we find:
$x=7.7×10^{−2} \nonumber$
When we check the assumption, we confirm:
$\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber$
which for this system is
$\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber$
The value of $x$ is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find $x$.
The equation:
$K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber$
gives
$6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber$
or
$x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber$
This equation can be solved using the quadratic formula. For an equation of the form
$ax^{2+} + bx + c=0, \nonumber$
$x$ is given by the quadratic equation:
$x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber$
In this problem, $a = 1$, $b = 1.2 × 10^{−3}$, and $c = −6.0 × 10^{−3}$.
Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:
$x=7.2×10^{−2} \nonumber$
Now determine the hydronium ion concentration and the pH:
\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \[4pt] &=7.2×10^{−2}\:M \end{align*} \nonumber
The pH of this solution is:
$\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber$
Exercise $8$
1. Show that the quadratic formula gives $x = 7.2 × 10^{−2}$.
2. Calculate the pH in a 0.010-M solution of caffeine, a weak base:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber$
Hint
It will be necessary to convert [OH] to $\ce{[H3O+]}$ or pOH to pH toward the end of the calculation.
Answer
pH 11.16
The Relative Strengths of Strong Acids and Bases
Strong acids, such as $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order $\ce{HCl < HBr < HI}$, and so $\ce{HI}$ is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water.
Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O2−, and the amide ion, $\ce{NH2-}$, are such strong bases that they react completely with water:
$\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq) \nonumber$
$\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq) \nonumber$
Thus, O2− and $\ce{NH2-}$ appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is $\ce{HF < HCl < HBr < HI}$. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\ce{CH4 < NH3 < H2O < HF}$; across the third row, it is $\ce{SiH4 < PH3 < H2S < HCl}$ (see Figure $6$).
Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula OnE(OH)m, and include sulfuric acid, $\ce{O2S(OH)2}$, sulfurous acid, $\ce{OS(OH)2}$, nitric acid, $\ce{O2NOH}$, perchloric acid, $\ce{O3ClOH}$, aluminum hydroxide, $\ce{Al(OH)3}$, calcium hydroxide, $\ce{Ca(OH)2}$, and potassium hydroxide, $\ce{KOH}$:
If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.
If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.
Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure $7$).
Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate $\ce{Al(H2O)3(OH)3}$, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, $\ce{Al(H2O)3(OH)3}$, is converted into the soluble ion, $\ce{[Al(H2O)2(OH)4]-}$, by reaction with hydroxide ion:
$[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber$
In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The $\ce{Al(H2O)3(OH)3}$ compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion $\ce{[Al(H2O)6]^3+}$ by reaction with hydronium ion:
$\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber$
In this case, protons are transferred from hydronium ions in solution to $\ce{Al(H2O)3(OH)3}$, and the compound functions as a base.
Summary
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].
Key Equations
• $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$
• $K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}$
• $K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})$
• $\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100$
Glossary
acid ionization constant (Ka)
equilibrium constant for the ionization of a weak acid
base ionization constant (Kb)
equilibrium constant for the ionization of a weak base
leveling effect of water
any acid stronger than $\ce{H3O+}$, or any base stronger than OH will react with water to form $\ce{H3O+}$, or OH, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong
oxyacid
compound containing a nonmetal and one or more hydroxyl groups
percent ionization
ratio of the concentration of the ionized acid to the initial acid concentration, times 100 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.3%3A_Relative_Strengths_of_Acids_and_Bases.txt |
Learning Objectives
• Predict whether a salt solution will be acidic, basic, or neutral
• Calculate the concentrations of the various species in a salt solution
• Describe the process that causes solutions of certain metal ions to be acidic
As we have seen in the section on chemical reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word “neutralization” seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall now see.
Acid-Base Neutralization
A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:
1. A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the conjugate partners produced are of negligible strength: $\ce{HCl}(aq)+\ce{NaOH}(aq)⇌\ce{NaCl}(aq)+\ce{H2O}(l) \nonumber$
2. A strong acid and a weak base yield a weakly acidic solution, not because of the strong acid involved, but because of the conjugate acid of the weak base.
3. A weak acid and a strong base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of the weak acid and the conjugate acid of the strong base. The conjugate acid of the strong base is a weaker acid than water and has no effect on the acidity of the resulting solution. However, the conjugate base of the weak acid is a weak base and ionizes slightly in water. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.
4. A weak acid plus a weak base can yield either an acidic, basic, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acid and the weak base will have the same strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral. To predict whether a particular combination will be acidic, basic or neutral, tabulated K values of the conjugates must be compared.
Stomach Antacids
Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO3. The reaction,
$CaCO_3(s)+2HCl(aq)⇌CaCl_2(aq)+H_2O(l)+CO_2(g) \nonumber$
not only neutralizes stomach acid, it also produces CO2(g), which may result in a satisfying belch.
Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)2. It works according to the reaction:
$Mg(OH)_2(s)⇌Mg^{2+}(aq)+2OH^-(aq) \nonumber$
The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that :
$H_3O^+ + OH^- ⇌ 2H_2O(l) \nonumber$
This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, Al(OH)3, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.
Culinary Aspects of Chemistry
Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO3 is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.
Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure $1$). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.
Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.
Salts of Weak Bases and Strong Acids
When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH4Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:
$\ce{NH3}(aq)+\ce{HCl}(aq)⟶\ce{NH4Cl}(aq) \nonumber$
A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:
$\ce{NH4+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NH3}(aq) \nonumber$
The equilibrium equation for this reaction is simply the ionization constant. Ka, for the acid $\ce{NH4+}$:
$\ce{\dfrac{[H3O+][NH3]}{[NH4+]}}=K_\ce{a} \nonumber$
We will not find a value of Ka for the ammonium ion in Table E1. However, it is not difficult to determine Ka for $\ce{NH4+}$ from the value of the ionization constant of water, Kw, and Kb, the ionization constant of its conjugate base, NH3, using the following relationship:
$K_\ce{w}=K_\ce{a}×K_\ce{b} \nonumber$
This relation holds for any base and its conjugate acid or for any acid and its conjugate base.
Example $1$: pH of a Solution of a Salt of a Weak Base and a Strong Acid
Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, $\ce{[C6H5NH3+]Cl}$, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?
$\ce{C6H5NH3+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H5NH2}(aq) \nonumber$
Solution
The new step in this example is to determine Ka for the $\ce{C6H5NH3+}$ ion. The $\ce{C6H5NH3+}$ ion is the conjugate acid of a weak base. The value of Ka for this acid is not listed in Table E1, but we can determine it from the value of Kb for aniline, C6H5NH2, which is given as 4.6 × 10−10 :
$\mathrm{\mathit{K}_a(for\:C_6H_5NH_3^+)×\mathit{K}_b(for\:C_6H_5NH_2)=\mathit{K}_w=1.0×10^{−14}} \nonumber$
$\mathrm{\mathit{K}_a(for\:C_6H_5NH_3^+)=\dfrac{\mathit{K}_w}{\mathit{K}_b(for\:C_6H_5NH_2)}=\dfrac{1.0×10^{−14}}{4.6×10^{−10}}=2.3×10^{−5}} \nonumber$
Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H3O+, and the pH:
With these steps we find [H3O+] = 2.3 × 10−3 M and pH = 2.64
Exercise $1$
1. Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of $\ce{C6H5NH3+}$ is 2.3 × 10−3 and the pH is 2.64.
2. What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions $\ce{NH4+}$ and $\ce{NO3-}$. Use the data in Table E1 to determine Kb for the ammonium ion. Which is the stronger acid $\ce{C6H5NH3+}$ or $\ce{NH4+}$?
Answer a
$K_a\ce{(for\:NH4+)}=5.6×10^{−10}$, [H3O+] = 7.5 × 10−6 M
Answer b
$\ce{C6H5NH3+}$ is the stronger acid (a) (b) .
Salts of Weak Acids and Strong Bases
When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH3CO2, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:
$\ce{CH3CO2H}(aq)+\ce{NaOH}(aq)⟶\ce{NaCH3CO2}(aq)+\ce{H2O}(aq) \nonumber$
A solution of this salt contains sodium ions and acetate ions. The sodium ion has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:
$\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) \nonumber$
The equilibrium equation for this reaction is the ionization constant, Kb, for the base $\ce{CH3CO2-}$. The value of Kb can be calculated from the value of the ionization constant of water, Kw, and Ka, the ionization constant of the conjugate acid of the anion using the equation:
$_\ce{w}=K_\ce{a}×K_\ce{b} \nonumber$
For the acetate ion and its conjugate acid we have:
$\mathrm{\mathit{K}_b(for\:\ce{CH_3CO_2^-})=\dfrac{\mathit{K}_w}{\mathit{K}_a(for\:CH_3CO_2H)}=\dfrac{1.0×10^{−14}}{1.8×10^{−5}}=5.6×10^{−10}} \nonumber$
Some handbooks do not report values of Kb. They only report ionization constants for acids. If we want to determine a Kb value using one of these handbooks, we must look up the value of Ka for the conjugate acid and convert it to a Kb value.
Example $2$: Equilibrium of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with $\ce{[CH3CO2- ]}=0.050\:M$ and [OH] = 2.5 × 10−6 M at equilibrium. The reaction is:
$\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) \nonumber$
Solution
We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.
The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine Kb as follows:
$\mathrm{\mathit{K}_b(for\:\ce{CH_3CO_2^-})=\dfrac{\mathit{K}_w}{\mathit{K}_a(for\:CH_3CO_2H)}=\dfrac{1.0×10^{−14}}{1.8×10^{−5}}=5.6×10^{−10}} \nonumber$
Now find the missing concentration:
$K_\ce{b}=\ce{\dfrac{[CH3CO2H][OH- ]}{[CH3CO2- ]}}=5.6×10^{−10} \nonumber$
$=\dfrac{[\ce{CH3CO2H}](2.5×10^{−6})}{(0.050)}=5.6×10^{−10} \nonumber$
Solving this equation we get [CH3CO2H] = 1.1 × 10−5 M.
Exercise $2$
What is the pH of a 0.083-M solution of CN? Use 4.9 × 10−10 as Ka for HCN. Hint: We will probably need to convert pOH to pH or find [H3O+] using [OH] in the final stages of this problem.
Answer
11.16
Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base
In a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the Ka of the weak acid and the Kb of the weak base. If Ka > Kb, the solution is acidic, and if Kb > Ka, the solution is basic.
Example $3$: Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. KBr
2. NaHCO3
3. NH4Cl
4. Na2HPO4
5. NH4F
Solution
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:
1. The K+ cation and the Br anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.
2. The Na+ cation is a spectator, and will not affect the pH of the solution; the $\ce{HCO3-}$ anion is amphiprotic.The Ka of $\ce{HCO3-}$ is 4.7 × 10−11, so the Kb of its conjugate base is $\dfrac{1.0×10^{−14}}{4.7×10^{−11}}=2.1×10^{−4}$. Since Kb >> Ka, the solution is basic.
3. The $\ce{NH4+}$ ion is acidic and the Cl ion is a spectator. The solution will be acidic.
4. The Na+ ion is a spectator and will not affect the pH of the solution, while the $\ce{HPO4^{2-}}$ ion is amphiprotic. The Ka of $\ce{HPO_4^{2-}}$ is 4.2 × 10−13 and its Kb is $\dfrac{1.0×10^{−14}}{4.2×10^{−13}}=2.4×10^{−2}$. Because Kb >> Ka, the solution is basic.
5. The $\ce{NH4+}$ ion is listed as being acidic, and the F ion is listed as a base, so we must directly compare the Ka and the Kb of the two ions. Ka of $\ce{NH4+}$ is 5.6 × 10−10, which seems very small, yet the Kb of F is 1.4 × 10−11, so the solution is acidic, since Ka > Kb.
Exercise $3$
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. K2CO3
2. CaCl2
3. KH2PO4
4. (NH4)2CO3
5. AlBr3
Answer a
basic
Answer b
neutral
Answer c
acidic
Answer d
basic
Answer e
acidic
The Ionization of Hydrated Metal Ions
If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, $\ce{Al(H2O)6^3+}$, dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this $\ce{Al(H2O)6^3+}$ cluster as well:
$\ce{Al(NO3)3}(s)+\ce{6H2O}(l)⟶\ce{Al(H2O)6^3+}(aq)+\ce{3NO3-}(aq) \nonumber$
We frequently see the formula of this ion simply as “Al3+(aq)”, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, H3O+ to H+. However, in this case, the hydrated aluminum ion is a weak acid (Figure $2$) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:
$\ce{Al(H2O)6^3+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Al(H2O)5(OH)^2+}(aq) \hspace{20px} K_\ce{a}=1.4×10^{−5} \nonumber$
As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:
$\ce{Al(H2O)6^3+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Al(H2O)5(OH)^2+}(aq) \nonumber$
$\ce{Al(H2O)5(OH)^2+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Al(H2O)4(OH)2+}(aq) \nonumber$
$\ce{Al(H2O)4(OH)2+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq) \nonumber$
Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the left side of the equilibrium expressions and as bases when they appear on the right side.
However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:
$\ce{Fe(H2O)6^3+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Fe(H2O)5(OH)^2+}(aq) \hspace{20px} K_\ce{a}=2.74 \nonumber$
$\ce{Cu(H2O)6^2+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Cu(H2O)5(OH)+}(aq) \hspace{20px} K_\ce{a}=~6.3 \nonumber$
$\ce{Zn(H2O)4^2+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Zn(H2O)3(OH)+}(aq) \hspace{20px} K_\ce{a}=9.6 \nonumber$
Example $4$: Hydrolysis of [Al(H2O)6]3+
Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion $\ce{[Al(H2O)6]^3+}$ in solution.
Solution
In spite of the unusual appearance of the acid, this is a typical acid ionization problem.
1. Determine the direction of change. The equation for the reaction and Ka are:
$\ce{Al(H2O)6^3+}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{Al(H2O)5(OH)^2+}(aq) \hspace{20px} K_\ce{a}=1.4×10^{−5}$
The reaction shifts to the right to reach equilibrium.
2. Determine x and equilibrium concentrations. Use the table
:
Solve for x and the equilibrium concentrations. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:
1. $K_\ce{a}=\ce{\dfrac{[H3O+][Al(H2O)5(OH)^2+]}{[Al(H2O)6^3+]}} \nonumber$
$=\dfrac{(x)(x)}{0.10−x}=1.4 \times 10^{−5}$
Solving this equation gives:
$x=1.2×10^{−3}\:M \nonumber$
From this we find:
$\ce{[H3O+]}=0+x=1.2×10^{−3}\:M \nonumber$
$\mathrm{pH=−log[H_3O^+]=2.92(an\: acidic\: solution)} \nonumber$
Check the work. The arithmetic checks; when 1.2 × 10−3 M is substituted for x, the result = Ka.
Exercise $4$
What is $\ce{[Al(H2O)5(OH)^2+]}$ in a 0.15-M solution of Al(NO3)3 that contains enough of the strong acid HNO3 to bring [H3O+] to 0.10 M?
Answer
2.1 × 10−5 M
The constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.
Summary
The characteristic properties of aqueous solutions of Brønsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Brønsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic. Solutions that contain salts or hydrated metal ions have a pH that is determined by the extent of the hydrolysis of the ions in the solution. The pH of the solutions may be calculated using familiar equilibrium techniques, or it may be qualitatively determined to be acidic, basic, or neutral depending on the relative Ka and Kb of the ions involved. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.4%3A_Hydrolysis_of_Salt_Solutions.txt |
Learning Objectives
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as $\ce{HCl}$, $\ce{HNO3}$, and $\ce{HCN}$ that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:
$\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq) \nonumber$
$\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq) \nonumber$
$\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq) \nonumber$
Even though it contains four hydrogen atoms, acetic acid, $\ce{CH3CO2H}$, is also monoprotic because only the hydrogen atom from the carboxyl group ($\ce{-COOH}$) reacts with bases:
Similarly, monoprotic bases are bases that will accept a single proton.
Diprotic Acids
Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:
• The first ionization is
$\ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq) \nonumber$
with $K_{\ce a1} > 10^2;\: {complete\: dissociation}$.
• The second ionization is
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq) \nonumber$
with $K_{\ce a2}=1.2×10^{−2}$.
This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, $\ce{H2CO3}$, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.
• First Ionization
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \nonumber$
with
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=4.3×10^{−7} \nonumber$
The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.
• Second Ionization
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
with
$K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11} \nonumber$
$K_{\ce{H2CO3}}$ is larger than $K_{\ce{HCO3-}}$ by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the $\ce{HCO3-}$ formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and $\ce{HCO3-}$ are practically equal in a pure aqueous solution of H2CO3.
If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H3O+ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.
Example $1$: Ionization of a Diprotic Acid
When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are $\ce{[H3O+]}$, $\ce{[HCO3- ]}$, and $\ce{[CO3^2- ]}$ in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1}$
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2}$
Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than $\ce{HCO3-}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem:
1. Using the customary four steps, we determine the concentration of H3O+ and $\ce{HCO3-}$ produced by ionization of H2CO3.
2. Then we determine the concentration of $\ce{CO3^2-}$ in a solution with the concentration of H3O+ and $\ce{HCO3-}$ determined in (1).
To summarize:
1. First Ionization: Determine the concentrations of $\ce{H3O+}$ and $\ce{HCO3-}$.
Since \ref{step1} is has a much bigger $K_{a1}=4.3×10^{−7}$ than $K_{a2}=4.7×10^{−11}$ for \ref{step2}, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem).
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber$
As for the ionization of any other weak acid:
An abbreviated table of changes and concentrations shows:
Abbreviated table of changes and concentrations
ICE Table $\ce{H2CO3}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{HCO3-}(aq)$
Initial (M) $0.033 \:M$ - $0$ $0$
Change (M) $- x$ - $+x$ $+x$
Equilibrium (M) $0.033 \:M - x$ - $x$ $x$
Substituting the equilibrium concentrations into the equilibrium constant gives us:
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7} \nonumber$
Solving the preceding equation making our standard assumptions gives:
$x=1.2×10^{−4} \nonumber$
Thus:
$\ce{[H2CO3]}=0.033\:M \nonumber$
$\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber$
2. Second Ionization: Determine the concentration of $CO_3^{2-}$ in a solution at equilibrium.
Since the \ref{step1} is has a much bigger $K_a$ than \ref{step2}, we can the equilibrium conditions calculated from first part of example as the initial conditions for an ICER Table for the \ref{step2}:
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
ICER Table for the \ref{step2}:
ICE Table $\ce{HCO3-}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{CO3^2-}(aq)$
Initial (M) $1.2×10^{−4}\:M$ - $1.2×10^{−4}\:M$ $0$
Change (M) $- y$ - $+y$ $+y$
Equilibrium (M) $1.2×10^{−4}\:M - y$ - $1.2×10^{−4}\:M + y$ $y$
\begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}} \[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*} \nonumber
To avoid solving a quadratic equation, we can assume $y \ll 1.2×10^{−4}\:M$ so
$K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber$
Rearranging to solve for $y$
$y \approx \dfrac{ (4.7×10^{−11})(1.2×10^{−4}\:M )}{ 1.2×10^{−4}\:M} \nonumber$
$[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber$
To summarize:
In part 1 of this example, we found that the $\ce{H2CO3}$ in a 0.033-M solution ionizes slightly and at equilibrium $[\ce{H2CO3}] = 0.033\, M$, $[\ce{H3O^{+}}] = 1.2 × 10^{−4}$, and $\ce{[HCO3- ]}=1.2×10^{−4}\:M$. In part 2, we determined that $\ce{[CO3^2- ]}=5.6×10^{−11}\:M$.
Exercise $2$: Hydrogen Sulfide
The concentration of $H_2S$ in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate $\ce{[H3O+]}$, $\ce{[HS^{−}]}$, and $\ce{[S^{2−}]}$ in the solution:
$\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=8.9×10^{−8} \nonumber$
$\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19} \nonumber$
Answer
$[\ce{H2S}] = 0.1 M$, $\ce{[H3O+]} = [HS^{−}] = 0.0001\, M$, $[S^{2−}] = 1 × 10^{−19}\, M$
We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).
Triprotic Acids
A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:
• The first ionization is
$\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \nonumber$
with $K_{\ce a1}=7.5×10^{−3}$.
• The second ionization is
$\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq) \nonumber$
with $K_{\ce a2}=6.2×10^{−8}$.
• The third ionization is
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \nonumber$
with $K_{\ce a3}=4.2×10^{−13}$.
As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106. This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, the calculations can be broken down into a series of parts similar to those for diprotic acids.
Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:
$\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber$
and
$\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq) \nonumber$
Summary
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.
Glossary
diprotic acid
acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps
diprotic base
base capable of accepting two protons. The protons are accepted in two steps
monoprotic acid
acid containing one ionizable hydrogen atom per molecule
stepwise ionization
process in which an acid is ionized by losing protons sequentially
triprotic acid
acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.5%3A_Polyprotic_Acids.txt |
Learning Objectives
• Describe the composition and function of acid–base buffers
• Calculate the pH of a buffer before and after the addition of added acid or base
A mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure $1$). A solution of acetic acid ($\ce{CH3COOH}$ and sodium acetate $\ce{CH3COONa}$) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia ($\ce{NH3(aq)}$) and ammonium chloride ($\ce{NH4Cl(aq)}$).
How Buffers Work
A mixture of acetic acid and sodium acetate is acidic because the Ka of acetic acid is greater than the Kb of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. Then more of the acetic acid reacts with water, restoring the hydronium ion concentration almost to its original value:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
The pH changes very little. If we add an acid such as hydrochloric acid, most of the hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid molecules:
$\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)⟶\ce{CH3CO2H}(aq)+\ce{H2O}(l) \nonumber$
Thus, there is very little increase in the concentration of the hydronium ion, and the pH remains practically unchanged (Figure $2$).
A mixture of ammonia and ammonium chloride is basic because the Kb for ammonia is greater than the Ka for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value:
$\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l) \nonumber$
If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value:
$\ce{H3O+}(aq)+\ce{NH3}(aq)⟶\ce{NH4+}(aq)+\ce{H2O}(l) \nonumber$
The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid.
Example $1$: pH Changes in Buffered and Unbuffered Solutions
Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds.
1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
2. Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.
Solution
1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):
1. Determine the direction of change. The equilibrium in a mixture of H3O+, $\ce{CH3CO2-}$, and CH3CO2H is:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
The equilibrium constant for CH3CO2H is not given, so we look it up in Table E1: Ka = 1.8 × 10−5. With [CH3CO2H] = $\ce{[CH3CO2- ]}$ = 0.10 M and [H3O+] = ~0 M, the reaction shifts to the right to form H3O+.
2. Determine x and equilibrium concentrations. A table of changes and concentrations follows:
• Solve for x and the equilibrium concentrations. We find:
$x=1.8×10^{−5}\:M \nonumber$
• and
$\ce{[H3O+]}=0+x=1.8×10^{−5}\:M \nonumber$
Thus:
$\mathrm{pH=−log[H_3O^+]=−log(1.8×10^{−5})} \nonumber$
$=4.74 \nonumber$
4. Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = Ka.
(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.
First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:
1. Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains:
$\mathrm{0.0010\cancel{L}×\left(\dfrac{0.10\:mol\: NaOH}{1\cancel{L}}\right)=1.0×10^{−4}\:mol\: NaOH} \nonumber$
2. Determine the moles of CH2CO2H. Before reaction, 0.100 L of the buffer solution contains:
$\mathrm{0.100\cancel{L}×\left(\dfrac{0.100\:mol\:CH_3CO_2H}{1\cancel{L}}\right)=1.00×10^{−2}\:mol\:CH_3CO_2H} \nonumber$
3. Solve for the amount of NaCH3CO2 produced. The 1.0 × 10−4 mol of NaOH neutralizes 1.0 × 10−4 mol of CH3CO2H, leaving:
$\mathrm{(1.0×10^{−2})−(0.01×10^{−2})=0.99×10^{−2}\:mol\:CH_3CO_2H} \nonumber$
and producing 1.0 × 10−4 mol of NaCH3CO2. This makes a total of:
[\mathrm{(1.0×10^{−2})+(0.01×10^{−2})=1.01×10^{−2}\:mol\:NaCH_3CO_2} \nonumber \]
4. Find the molarity of the products. After reaction, CH3CO2H and NaCH3CO2 are contained in 101 mL of the intermediate solution, so:
$\ce{[CH3CO2H]}=\mathrm{\dfrac{9.9×10^{−3}\:mol}{0.101\:L}}=0.098\:M \nonumber$
$\ce{[NaCH3CO2]}=\mathrm{\dfrac{1.01×10^{−2}\:mol}{0.101\:L}}=0.100\:M \nonumber$
Now we calculate the pH after the intermediate solution, which is 0.098 M in CH3CO2H and 0.100 M in NaCH3CO2, comes to equilibrium. The calculation is very similar to that in part (a) of this example:
This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution.
(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 × 10−5-M solution of HCl). The volume of the final solution is 101 mL.
Solution
This 1.8 × 10−5-M solution of HCl has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:
$\mathrm{0.100\:L×\left(\dfrac{1.8×10^{−5}\:mol\: HCl}{1\:L}\right)=1.8×10^{−6}\:mol\: HCl}$
As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 × 10−4 mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:
$(1.0×10^{−4})−(1.8×10^{−6})=9.8×10^{−5}\:M$
The concentration of NaOH is:
$\dfrac{9.8×10^{−5}\:M\:\ce{NaOH}}{0.101\:\ce{L}}=9.7×10^{−4}\:M$
The pOH of this solution is:
$\mathrm{pOH=−log[OH^- ]=−log(9.7×10^{−4})=3.01}$
The pH is:
$\mathrm{pH=14.00−pOH=10.99}$
The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).
Exercise $1$
Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 × 10−5 M HCl solution from 4.74 to 3.00.
Answer
Initial pH of 1.8 × 10−5 M HCl; pH = −log[H3O+] = −log[1.8 × 10−5] = 4.74
Moles of H3O+ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L × 0.0010 L = 1.0 × 10−4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:
$\mathrm{pH=−log[H_3O^+]=−log\left(\dfrac{total\: moles\:H_3O^+}{total\: volume}\right)=−log\left(\dfrac{1.0×10^{−4}\:mol+1.8×10^{−6}\:mol}{101\:mL\left(\dfrac{1\:L}{1000\:mL}\right)}\right)=3.00} \nonumber$
Buffer Capacity
Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure $3$). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.
The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.
Selection of Suitable Buffer Mixtures
There are two useful rules of thumb for selecting buffer mixtures:
1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure $4$ shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.
1. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.
Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H2CO3, and the bicarbonate ion, $\ce{HCO3-}$. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:
$\ce{H3O+}(aq)+\ce{HCO3-}(aq)⟶\ce{H2CO3}(aq)+\ce{H2O}(l) \nonumber$
When an excess of the hydroxide ion is present, it is removed by the reaction:
$\ce{OH-}(aq)+\ce{H2CO3}(aq)⟶\ce{HCO3-}(aq)+\ce{H2O}(l) \nonumber$
The pH of human blood thus remains very near 7.35, that is, slightly basic. Variations are usually less than 0.1 of a pH unit. A change of 0.4 of a pH unit is likely to be fatal.
The Henderson-Hasselbalch Approximation
The ionization-constant expression for a solution of a weak acid can be written as:
$K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}} \nonumber$
Rearranging to solve for [H3O+], we get:
$\ce{[H3O+]}=K_\ce{a}×\ce{\dfrac{[HA]}{[A- ]}} \nonumber$
Taking the negative logarithm of both sides of this equation, we arrive at:
$\mathrm{−log[H_3O^+]=−log\mathit{K}_a − log\dfrac{[HA]}{[A^- ]}} \nonumber$
which can be written as
$\mathrm{pH=p\mathit{K}_a+log\dfrac{[A^- ]}{[HA]}} \nonumber$
where pKa is the negative of the common logarithm of the ionization constant of the weak acid (pKa = −log Ka). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch approximation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.
Lawrence Joseph Henderson and Karl Albert Hasselbalch
Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.
In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.
Medicine: The Buffer System in Blood
The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:
$\ce{CO2}(g)+\ce{2H2O}(l)⇌\ce{H2CO3}(aq)⇌\ce{HCO3-}(aq)+\ce{H3O+}(aq) \nonumber$
The concentration of carbonic acid, H2CO3 is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, $\ce{HCO3-}$, is around 0.024 M. Using the Henderson-Hasselbalch equation and the pKa of carbonic acid at body temperature, we can calculate the pH of blood:
$\mathrm{pH=p\mathit{K}_a+\log\dfrac{[base]}{[acid]}=6.1+\log\dfrac{0.024}{0.0012}=7.4} \nonumber$
The fact that the H2CO3 concentration is significantly lower than that of the $\ce{HCO3-}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.
Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the $\ce{HCO3-}$ ion, producing H2CO3. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO2 from the blood through the lungs driving the equilibrium reaction such that [H3O+] is lowered. If the blood is too alkaline, a lower breath rate increases CO2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H+] and restoring an appropriate pH.
Summary
A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).
Key Equations
• pKa = −log Ka
• pKb = −log Kb
• $\mathrm{pH=p\mathit{K}_a+\log\dfrac{[A^- ]}{[HA]}}$
Glossary
buffer capacity
amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)
buffer
mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added
Henderson-Hasselbalch equation
equation used to calculate the pH of buffer solutions | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.6%3A_Buffers.txt |
Learning Objectives
• Interpret titration curves for strong and weak acid-base systems
• Compute sample pH at important stages of a titration
• Explain the function of acid-base indicators
As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the changes in the concentrations of the acidic and basic species present in a solution during the process of a titration.
Titration Curves
Previously, when we studied acid-base reactions in solution, we focused only on the point at which the acid and base were stoichiometrically equivalent. No consideration was given to the pH of the solution before, during, or after the neutralization.
Example $1$: Calculating pH for Titration Solutions: Strong Acid/Strong Base
A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is shown in Figure $1$. Calculate the pH at these volumes of added base solution:
1. 0.00 mL
2. 12.50 mL
3. 25.00 mL
4. 37.50 mL
Solution
Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of H3O+ is $\ce{[H3O+]_0}=0.100\:M$. When the base solution is added, it also dissociates completely, providing OH ions. The H3O+ and OH ions neutralize each other, so only those of the two that were in excess remain, and their concentration determines the pH. Thus, the solution is initially acidic (pH < 7), but eventually all the hydronium ions present from the original acid are neutralized, and the solution becomes neutral. As more base is added, the solution turns basic.
The total initial amount of the hydronium ions is:
$\mathrm{n(H^+)_0=[H_3O^+]_0×0.02500\: L=0.002500\: mol} \nonumber$
Once X mL of the 0.100-M base solution is added, the number of moles of the OH ions introduced is:
$\mathrm{n(OH^-)_0=0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber$
The total volume becomes:
$V=\mathrm{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber$
The number of moles of H3O+ becomes:
$\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0=0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber$
The concentration of H3O+ is:
$\mathrm{[H_3O^+]=\dfrac{n(H^+)}{V}=\dfrac{0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber$
$\mathrm{=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×X\: mL}{25.00\: mL+X\: mL}} \nonumber$
with the definition of $\mathrm{pH}$:
$\mathrm{pH=−\log([H_3O^+])} \label{phdef}$
The preceding calculations work if $\mathrm{n(H^+)_0-n(OH^-)_0>0}$ and so n(H+) > 0. When $\mathrm{n(H^+)_0=n(OH^-)_0}$, the H3O+ ions from the acid and the OH ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH particles to neutralize them. Therefore, in this case:
$\ce{[H3O+]}=\ce{[OH- ]},\:\ce{[H3O+]}=K_\ce{w}=1.0\times 10^{-14};\:\ce{[H3O+]}=1.0\times 10^{-7} \nonumber$
$\mathrm{pH=-log(1.0\times 10^{-7})=7.00} \nonumber$
Finally, when $\mathrm{n(OH^-)_0>n(H^+)_0}$, there are not enough H3O+ ions to neutralize all the OH ions, and instead of $\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0}$, we calculate: $\mathrm{n(OH^-)=n(OH^-)_0-n(H^+)_0}$
In this case:
$\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)-0.002500\: mol}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber$
$\mathrm{=\dfrac{0.100\:\mathit{M}×X\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+X\: mL}} \nonumber$
then using the definition of $pOH$ and its relationship to $pH$ in room temperature aqueous solutios (Equation \ref{phdef}):
\begin{align} pH &=14-pOH \nonumber \&=14+\log([OH^-]) \nonumber\end{align} \nonumber
Let us now consider the four specific cases presented in this problem:
(a) X = 0 mL
$\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL}=0.1\:\mathit{M}} \nonumber$
then using the definition of $pH$ (Equation \ref{phdef}):
\begin{align} pH &= −\log(0.100) \nonumber \ &= 1.000 \nonumber\end{align} \nonumber
(b) X = 12.50 mL
$\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×12.50\: mL}{25.00\: mL+12.50\: mL}=0.0333\:\mathit{M}} \nonumber$
then using the definition of $pH$ (Equation \ref{phdef}):
\begin{align} pH &= −\log(0.0333) \nonumber \ &= 1.477 \nonumber\end{align} \nonumber
(c) X = 25.00 mL
Since the volumes and concentrations of the acid and base solutions are the same:
$\mathrm{n(H^+)_0=n(OH^-)_0} \nonumber$
and
$pH = 7.000 \nonumber$
as described earlier.
(d) X = 37.50 mL
In this case:
$\mathrm{n(OH^-)_0>n(H^+)_0} \nonumber$
$\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×35.70\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+37.50\: mL}=0.0200\:\mathit{M}} \nonumber$
then using the definition of $pH$ (Equation \ref{phdef}):
\begin{align}[pH = 14 − pOH \nonumber\ &= 14 + \log([OH^{−}]) \nonumber \ &= 14 + \log(0.0200) \nonumber \ &= 12.30 \nonumber \end{align} \nonumber
Exercise $1$
Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO3(aq) and 0.200 M NaOH (titrant) at the listed volumes of added base:
1. 0.00 mL,
2. 15.0 mL,
3. 25.0 mL, and
4. 40.0 mL.
Answer a
0.00: 1.000
Answer b
15.0: 1.5111
Answer c
25.0: 7e. Do not delete this text first.
Answer d
40.0: 12.523
In Example $1$, we calculated pH at four points during a titration. Table $1$ shows a detailed sequence of changes in the pH of a strong acid and a weak acid in a titration with NaOH.
Table $1$: pH Values in the Titrations of a Strong Acid with a Strong Base and of a Weak Acid with a Strong Base
Volume of 0.100 M NaOH Added (mL) Moles of NaOH Added pH Values 0.100 M HCl1 pH Values 0.100 M $CH_3CO_2H$2
0.0 0.0 1.00 2.87
5.0 0.00050 1.18 4.14
10.0 0.00100 1.37 4.57
15.0 0.00150 1.60 4.92
20.0 0.00200 1.95 5.35
22.0 0.00220 2.20 5.61
24.0 0.00240 2.69 6.13
24.5 0.00245 3.00 6.44
24.9 0.00249 3.70 7.14
25.0 0.00250 7.00 8.72
25.1 0.00251 10.30 10.30
25.5 0.00255 11.00 11.00
26.0 0.00260 11.29 11.29
28.0 0.00280 11.75 11.75
30.0 0.00300 11.96 11.96
35.0 0.00350 12.22 12.22
40.0 0.00400 12.36 12.36
45.0 0.00450 12.46 12.46
50.0 0.00500 12.52 12.52
1. Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH.
2. Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH.
The simplest acid-base reactions are those of a strong acid with a strong base. Table $1$ shows data for the titration of a 25.0-mL sample of 0.100 M hydrochloric acid with 0.100 M sodium hydroxide. The values of the pH measured after successive additions of small amounts of NaOH are listed in the first column of this table, and are graphed in Figure $1$, in a form that is called a titration curve. The pH increases slowly at first, increases rapidly in the middle portion of the curve, and then increases slowly again. The point of inflection (located at the midpoint of the vertical part of the curve) is the equivalence point for the titration. It indicates when equivalent quantities of acid and base are present. For the titration of a strong acid with a strong base, the equivalence point occurs at a pH of 7.00 and the points on the titration curve can be calculated using solution stoichiometry (Table $1$ and Figure $1$).
The titration of a weak acid with a strong base (or of a weak base with a strong acid) is somewhat more complicated than that just discussed, but it follows the same general principles. Let us consider the titration of 25.0 mL of 0.100 M acetic acid (a weak acid) with 0.100 M sodium hydroxide and compare the titration curve with that of the strong acid. Table $1$ gives the pH values during the titration, Figure $\PageIndex{1b}$ shows the titration curve.
Although the initial volume and molarity of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionized. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH:
$\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(l)+\ce{OH-}(aq) \nonumber$
After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases.
Example $2$: Titration of a Weak Acid with a Strong Base
The titration curve shown in Figure $\PageIndex{1b}$ is for the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The reaction can be represented as:
$\ce{CH3CO2H + OH- ⟶ CH3CO2- + H2O} \nonumber$
1. What is the initial pH before any amount of the NaOH solution has been added? Ka = 1.8 × 10−5 for CH3CO2H.
2. Find the pH after 25.00 mL of the NaOH solution have been added.
3. Find the pH after 12.50 mL of the NaOH solution has been added.
4. Find the pH after 37.50 mL of the NaOH solution has been added.
Solution
(a) Assuming that the dissociated amount is small compared to 0.100 M, we find that:
$K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}≈\ce{\dfrac{[H3O+]^2}{[CH3CO2H]_0}} \nonumber$
and
$\ce{[H3O+]}=\sqrt{K_\ce{a}×\ce{[CH3CO2H]}}=\sqrt{1.8\times 10^{-5}×0.100}=1.3\times 10^{-3} \nonumber$
$\mathrm{pH=-\log(1.3\times 10^{-3})=2.87} \nonumber$
(b) After 25.00 mL of NaOH are added, the number of moles of NaOH and CH3CO2H are equal because the amounts of the solutions and their concentrations are the same. All of the CH3CO2H has been converted to $\ce{CH3CO2-}$. The concentration of the $\ce{CH3CO2-}$ ion is:
$\mathrm{\dfrac{0.00250\: mol}{0.0500\: L}=0.0500\: \ce{MCH3CO2-}} \nonumber$
The equilibrium that must be focused on now is the basicity equilibrium for $\ce{CH3CO2-}$:
$\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) \nonumber$
so we must determine Kb for the base by using the ion product constant for water:
$K_\ce{b}=\ce{\dfrac{[CH3CO2H][OH- ]}{[CH3CO2- ]}} \nonumber$
$K_\ce{a}=\ce{\dfrac{[CH3CO2- ][H+]}{[CH3CO2H]}},\textrm{ so }\ce{\dfrac{[CH3CO2H]}{[CH3CO2- ]}}=\dfrac{\ce{[H+]}}{K_\ce{a}}. \nonumber$
Since Kw = [H+][OH]:
\begin{align} K_\ce{b} &=\dfrac{\ce{[H+][OH- ]}}{K_\ce{a}} \ &=\dfrac{K_\ce{w}}{K_\ce{a}} \ &=\dfrac{1.0\times 10^{-14}}{1.8\times 10^{-5}} \ &=5.6\times 10^{-10} \end{align} \nonumber
Let us denote the concentration of each of the products of this reaction, CH3CO2H and OH, as x. Using the assumption that x is small compared to 0.0500 M, $K_\ce{b}=\dfrac{x^2}{0.0500\:M}$, and then:
$x=\ce{[OH- ]}=5.3\times 10^{−6} \nonumber$
$\ce{pOH}=-\log(5.3\times 10^{-6})=5.28 \nonumber$
$\ce{pH}=14.00−5.28=8.72 \nonumber$
Note that the pH at the equivalence point of this titration is significantly greater than 7.
(c) In (a), 25.00 mL of the NaOH solution was added, and so practically all the CH3CO2H was converted into $\ce{CH3CO2-}$. In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH3CO2H is converted into $\ce{CH3CO2-}$. The total initial number of moles of CH3CO2H is 0.02500L × 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH3CO2H and $\ce{CH3CO2-}$ are both approximately equal to $\mathrm{\dfrac{0.00250\: mol}{2}=0.00125\: mol}$, and their concentrations are the same.
Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation:
$\ce{pH}=\ce p K_\ce{a}+\log\ce{\dfrac{[Base]}{[Acid]}}=-\log(\mathit{K}_\ce{a})+\log\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]}}=-\log(1.8\times 10^{-5})+\log(1)$
(as the concentrations of $\ce{CH3CO2-}$ and CH3CO2H are the same)
Thus:
$\ce{pH}=−\log(1.8\times 10^{−5})=4.74 \nonumber$
(the pH = the pKa at the halfway point in a titration of a weak acid)
(d) After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L × 0.100 M = 0.003750 mol NaOH. Since this is past the equivalence point, the excess hydroxide ions will make the solution basic, and we can again use stoichiometric calculations to determine the pH:
$\mathrm{[OH^-]=\dfrac{(0.003750\: mol−0.00250\: mol)}{0.06250\: L}}=2.00\times 10^{−2}\:M$
So:
$\mathrm{pOH=−\log(2.00\times 10^{−2})=1.70,\: and\: pH=14.00−1.70=12.30}$
Note that this result is the same as for the strong acid-strong base titration example provided, since the amount of the strong base added moves the solution past the equivalence point.
Exercise $2$
Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH(aq) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base:
1. 0.00 mL,
2. 15.0 mL,
3. 25.0 mL, and
4. 30.0 mL.
Answer a
0.00 mL: 2.37
Answer b
15.0 mL: 3.92
Answer c
25.00 mL: 8.29
Answer d
30.0 mL: 12.097
Acid-Base Indicators
Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × 10−9 M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × 10−9 M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acid-base indicators are either weak organic acids or weak organic bases.
The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use $\ce{HIn}$ as a simple representation for the complex methyl orange molecule:
$\underbrace{\ce{HIn (aq)}}_{\ce{red}}+\ce{H2O (l)}⇌\ce{H3O^{+} (aq)}+\underbrace{\ce{In^{-} (aq)}}_{\ce{yellow}} \nonumber$
$K_\ce{a}=\ce{\dfrac{[H3O+][In- ]}{[HIn]}}=4.0\times 10^{−4} \nonumber$
The anion of methyl orange, $\ce{In^{-}}$, is yellow, and the nonionized form, $\ce{HIn}$, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.
An indicator’s color is the visible result of the ratio of the concentrations of the two species In and $\ce{HIn}$. If most of the indicator (typically about 60−90% or more) is present as $\ce{In^{-}}$, then we see the color of the $\ce{In^{-}}$ ion, which would be yellow for methyl orange. If most is present as $\ce{HIn}$, then we see the color of the $\ce{HIn}$ molecule: red for methyl orange. For methyl orange, we can rearrange the equation for Ka and write:
$\mathrm{\dfrac{[In^-]}{[HIn]}=\dfrac{[substance\: with\: yellow\: color]}{[substance\: with\: red\: color]}=\dfrac{\mathit{K}_a}{[H_3O^+]}} \label{ABeq2}$
Equation \ref{ABeq2} shows us how the ratio of $\ce{\dfrac{[In- ]}{[HIn]}}$ varies with the concentration of hydronium ion. The above expression describing the indicator equilibrium can be rearranged:
\begin{align} \dfrac{[H_3O^+]}{\mathit{K}_a} &=\dfrac{[HIn]}{[In^- ]} \[8pt] \log\left(\dfrac{[H_3O^+]}{\mathit{K}_a}\right) &= \log\left(\dfrac{[HIn]}{[In^- ]}\right) \[8pt] \log([H_3O^+])-\log(\mathit{K}_a) &=-\log\left(\dfrac{[In^-]}{[HIn]}\right) \[8pt] -pH+p\mathit{K}_a & =-\log\left(\dfrac{[In^-]}{[HIn]}\right) \[8pt] pH &=p\mathit{K}_a+\log\left(\dfrac{[In^-]}{[HIn]}\right) \end {align} \nonumber
or in general terms
$pH=p\mathit{K}_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{HHeq}$
Equation \ref{HHeq} is the same as the Henderson-Hasselbalch equation, which can be used to describe the equilibrium of indicators.
When [H3O+] has the same numerical value as Ka, the ratio of [In] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In), and the solution appears orange in color. When the hydronium ion concentration increases to 8 × 10−4 M (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 × 10−5 M (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the color-change interval of methyl orange; the pronounced color change takes place between these pH values.
There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure $2$ presents several indicators, their colors, and their color-change intervals.
Titration curves help us pick an indicator that will provide a sharp color change at the equivalence point. The best selection would be an indicator that has a color change interval that brackets the pH at the equivalence point of the titration.
The color change intervals of three indicators are shown in Figure $3$. The equivalence points of both the titration of the strong acid and of the weak acid are located in the color-change interval of phenolphthalein. We can use it for titrations of either strong acid with strong base or weak acid with strong base.
Litmus is a suitable indicator for the HCl titration because its color change brackets the equivalence point. However, we should not use litmus for the CH3CO2H titration because the pH is within the color-change interval of litmus when only about 12 mL of NaOH has been added, and it does not leave the range until 25 mL has been added. The color change would be very gradual, taking place during the addition of 13 mL of NaOH, making litmus useless as an indicator of the equivalence point.
We could use methyl orange for the HCl titration, but it would not give very accurate results: (1) It completes its color change slightly before the equivalence point is reached (but very close to it, so this is not too serious); (2) it changes color, as Figure $2$ demonstrates, during the addition of nearly 0.5 mL of NaOH, which is not so sharp a color change as that of litmus or phenolphthalein; and (3) it goes from yellow to orange to red, making detection of a precise endpoint much more challenging than the colorless to pink change of phenolphthalein. Figure $2$ shows us that methyl orange would be completely useless as an indicator for the CH3CO2H titration. Its color change begins after about 1 mL of NaOH has been added and ends when about 8 mL has been added. The color change is completed long before the equivalence point (which occurs when 25.0 mL of NaOH has been added) is reached and hence provides no indication of the equivalence point.
We base our choice of indicator on a calculated pH, the pH at the equivalence point. At the equivalence point, equimolar amounts of acid and base have been mixed, and the calculation becomes that of the pH of a solution of the salt resulting from the titration.
Summary
A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator.
Glossary
acid-base indicator
organic acid or base whose color changes depending on the pH of the solution it is in
color-change interval
range in pH over which the color change of an indicator takes place
titration curve
plot of the pH of a solution of acid or base versus the volume of base or acid added during a titration | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.7%3A_Acid-Base_Titrations.txt |
14.1: Brønsted-Lowry Acids and Bases
Q14.1.1
Write equations that show NH3 as both a conjugate acid and a conjugate base.
S14.1.1
One example for NH3 as a conjugate acid: $\ce{NH2- + H+ ⟶ NH3}$; as a conjugate base: $\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l)$
Q14.1.2
Write equations that show $\ce{H2PO4-}$ acting both as an acid and as a base.
Q14.1.3
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
1. $\ce{H3O+}$
2. HCl
3. NH3
4. CH3CO2H
5. $\ce{NH4+}$
6. $\ce{HSO4-}$
S14.1.3
1. $\ce{H3O+}(aq)⟶\ce{H+}(aq)+ \ce{H_2O}_{(l)}$ ;
2. $\ce{HCl}(l)⟶\ce{H+}(aq)+\ce{Cl-}(aq)$;
3. $\ce{NH3}(aq)⟶\ce{H+}(aq)+\ce{NH2-}(aq)$;
4. $\ce{CH3CO2H}(aq)⟶\ce{H+}(aq)+\ce{CH3CO2-}(aq)$;
5. $\ce{NH4+}(aq)⟶\ce{H+}(aq)+\ce{NH3}(aq)$;
6. $\ce{HSO4-}(aq)⟶\ce{H+}(aq)+\ce{SO4^2-}(aq)$
Q14.1.4
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
1. HNO3
2. $\ce{PH4+}$
3. H2S
4. CH3CH2COOH
5. $\ce{H2PO4-}$
6. HS
Q14.1.5
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
1. H2O
2. $\ce{OH-}$
3. NH3
4. CN
5. S2−
6. $\ce{H2PO4-}$
S14.1.5
1. $\ce{H_2O}_{(l)} + \ce{H^+} (aq)⟶\ce{H3O+}(aq)$
2. $\ce{OH-} (aq) + \ce{H^+} (aq)⟶ \ce{H_2O}_{(l)}$
3. $\ce{NH3}(aq) + \ce{H^+} (aq)⟶\ce{NH4+}(aq)$;
4. $\ce{CN-}(aq) + \ce{H^+} (aq)⟶\ce{HCN}(aq)$
5. $\ce{S^2-}(aq) + \ce{H^+} (aq)⟶\ce{HS-}(aq)$
6. $\ce{H2PO4-}(aq) + \ce{H^+} (aq)⟶\ce{H3PO4}(aq)$
Q14.1.6
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
1. HS
2. $\ce{PO4^3-}$
3. $\ce{NH2-}$
4. C2H5OH
5. O2−
6. $\ce{H2PO4-}$
Q14.1.7
What is the conjugate acid of each of the following? What is the conjugate base of each?
1. $\ce{OH-}$
2. H2O
3. $\ce{HCO3-}$
4. NH3
5. $\ce{HSO4-}$
6. H2O2
7. HS
8. $\ce{H5N2+}$
S14.1.7
H2O, O2−; H3O+, $\ce{OH^-}$ ; H2CO3, $\ce{CO3^2-}$; $\ce{NH4+}$, $\ce{NH2-}$; H2SO4, $\ce{SO4^2-}$; $\ce{H3O2+}$, $\ce{HO2-}$; H2S; S2−; $\ce{H6N2^2+}$, H4N2
Q14.1.8
What is the conjugate acid of each of the following? What is the conjugate base of each?
1. H2S
2. $\ce{H2PO4-}$
3. PH3
4. HS
5. $\ce{HSO3-}$
6. $\ce{H3O2+}$
7. H4N2
8. CH3OH
Q14.1.9
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
1. $\ce{HNO3 + H2O ⟶ H3O+ + NO3-}$
2. $\ce{CN- + H2O ⟶ HCN + OH-}$
3. $\ce{H2SO4 + Cl- ⟶ HCl + HSO4-}$
4. $\ce{HSO4- + OH- ⟶ SO4^2- + H2O}$
5. $\ce{O^2- + H2O ⟶ 2OH-}$
6. $\ce{[Cu(H2O)3(OH)]+ + [Al(H2O)6]^3+ ⟶ [Cu(H2O)4]^2+ + [Al(H2O)5(OH)]^2+}$
7. $\ce{H2S + NH2- ⟶ HS- + NH3}$
S14.1.9
The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. HNO3(BA), H2O(BB), H3O+(CA), $\ce{NO3- (CB)}$; CN(BB), H2O(BA), HCN(CA), $\ce{OH^-}$ (CB); H2SO4(BA), Cl(BB), HCl(CA), $\ce{HSO4- (CB)}$; $\ce{HSO4- (BA)}$, OH-(BB), $\ce{SO4^2- (CB)}$, H2O(CA); O2−(BB), H2O(BA) $\ce{OH^-}$ (CB and CA); [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB); H2S(BA), $\ce{NH2- (BB)}$, HS(CB), NH3(CA)
Q14.1.10
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
1. $\ce{NO2- + H2O ⟶ HNO2 + OH-}$
2. $\ce{HBr + H2O ⟶ H3O+ + Br-}$
3. $\ce{HS- + H2O ⟶ H2S + OH-}$
4. $\ce{H2PO4- + OH- ⟶HPO4^2- + H2O}$
5. $\ce{H2PO4- + HCl ⟶ H3PO4 + Cl-}$
6. $\ce{[Fe(H2O)5(OH)]^2+ + [Al(H2O)6]^3+ ⟶ [Fe(H2O)6]^3+ + [Al(H2O)5(OH)]^2+}$
7. $\ce{CH3OH + H- ⟶ CH3O- + H2}$
Q14.1.11
What are amphiprotic species? Illustrate with suitable equations.
S14.1.11
Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2O.
• As an acid: $\ce{H2O}(aq) + \ce{NH3}(aq) \rightleftharpoons \ce{NH4+}(aq) + \ce{OH-}(aq)$.
• As a base: $\ce{H2O}(aq) + \ce{HCl}(aq) \rightleftharpoons \ce{H3O+}(aq) + \ce{Cl-}(aq)$
Q14.1.12
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:
1. H2O
2. $\ce{H2PO4-}$
3. S2−
4. $\ce{CO3^2-}$
5. $\ce{HSO4-}$
Q14.1.13
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
1. NH3
2. $\ce{HPO4-}$
3. Br
4. $\ce{NH4+}$
5. $\ce{ASO4^3-}$
S14.113
amphiprotic: $\ce{NH3 + H3O+ ⟶ NH4OH + H2O}$, $\ce{NH3 + OCH3- ⟶ NH2- + CH3OH}$; $\ce{HPO4^2- + OH- ⟶ PO4^3- + H2O}$, $\ce{HPO4^2- + HClO4 ⟶ H2PO4- + ClO4-}$; not amphiprotic: Br; $\ce{NH4+}$; $\ce{AsO4^3-}$
Q14.1.14
Is the self ionization of water endothermic or exothermic? The ionization constant for water (Kw) is $2.9 \times 10^{-14}$ at 40 °C and $9.6 \times 10^{-14}$ at 60 °C.
14.2: pH and pOH
Q14.2.1
Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 × 10−7 M. Kw is 2.9 × 10−14 at 40 °C.
S14.2.1
In a neutral solution [H3O+] = [OH]. At 40 °C,
[H3O+] = [OH] = (2.910−14)1/2 = 1.7 × 10−7.
Q14.2.2
The ionization constant for water (Kw) is 2.9 × 10−14 at 40 °C. Calculate [H3O+], [OH], pH, and pOH for pure water at 40 °C.
Q14.2.3
The ionization constant for water (Kw) is 9.614 × 10−14 at 60 °C. Calculate [H3O+], [OH], pH, and pOH for pure water at 60 °C.
S14.2.3
x = 3.101 × 10−7 M = [H3O+] = [OH]
pH = -log 3.101 × 10−7 = −(−6.5085) = 6.5085
pOH = pH = 6.5085
Q14.2.4
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
1. 0.200 M HCl
2. 0.0143 M NaOH
3. 3.0 M HNO3
4. 0.0031 M Ca(OH)2
Q14.2.5
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
1. 0.000259 M HClO4
2. 0.21 M NaOH
3. 0.000071 M Ba(OH)2
4. 2.5 M KOH
S14.2.5
pH = 3.587; pOH = 10.413; pH = 0.68; pOH = 13.32; pOH = 3.85; pH = 10.15; pH = −0.40; pOH = 14.4
Q14.2.6
What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
Q14.2.6
What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
S14.2.6
[H3O+] = 3.0 × 10−7 M; [OH] = 3.3 × 10−8 M
Q14.2.7
Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See below Figure for useful information.
Q14.2.8
Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure for useful information.
S14.2.9
[H3O+] = 1 × 10−2 M; [OH] = 1 × 10−12 M
Q14.2.9
The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10−6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?
Q14.2.10
The hydroxide ion concentration in household ammonia is 3.2 × 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?
S14.2.10
[OH] = 3.1 × 10−12 M
14.3: Relative Strengths of Acids and Bases
Q14.3.1
Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
Q14.3.2
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH, which causes the solution to be basic.
Q14.3.3
Use this list of important industrial compounds (and Figure) to answer the following questions regarding: CaO, Ca(OH)2, CH3CO2H, CO2, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.
1. Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
2. List those compounds in that can behave as Brønsted-Lowry acids with strengths lying between those of H3O+ and H2O.
3. List those compounds in that can behave as Brønsted-Lowry bases with strengths lying between those of H2O and OH.
Q14.3.4
The odor of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
S14.3.4
[H2O] > [CH3CO2H] > $\ce{[H3O+]}$ ≈ $\ce{[CH3CO2- ]}$ > [OH]
Q14.3.5
Household ammonia is a solution of the weak base NH3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.
Q14.3.4
Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.
S14.3.4
The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.
Q14.3.7
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.
Q14.3.8
Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
S14.3.8
$\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}$
Q14.3.9
Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.
Q14.3.10
What is the ionization constant at 25 °C for the weak acid $\ce{CH3NH3+}$, the conjugate acid of the weak base CH3NH2, Kb = 4.4 × 10−4.
S14.3.10
$K_\ce{a}=2.3×10^{−11}$
Q14.3.11
What is the ionization constant at 25 °C for the weak acid $\ce{(CH3)2NH2+}$, the conjugate acid of the weak base (CH3)2NH, Kb = 7.4 × 10−4?
Q14.3.12
Which base, CH3NH2 or (CH3)2NH, is the strongest base? Which conjugate acid, $\ce{(CH3)2NH2+}$ or (CH3)2NH, is the strongest acid?
S14.3.12
The strongest base or strongest acid is the one with the larger Kb or Ka, respectively. In these two examples, they are (CH3)2NH and $\ce{CH3NH3+}$.
Q14.3.3
Which is the stronger acid, $\ce{NH4+}$ or HBrO?
Q14.3.14
Which is the stronger base, (CH3)3N or $\ce{H2BO3-}$?
triethylamine.
Q14.3.15
Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
1. H2O or HF
2. B(OH)3 or Al(OH)3
3. $\ce{HSO3-}$ or $\ce{HSO4-}$
4. NH3 or H2S
5. H2O or H2Te
Q14.3.16
Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
1. $\ce{HSO4-}$ or $\ce{HSeO4-}$
2. NH3 or H2O
3. PH3 or HI
4. NH3 or PH3
5. H2S or HBr
S14.3.16
1. $\ce{HSO4-}$; higher electronegativity of the central ion. H2O;
2. NH3 is a base and water is neutral, or decide on the basis of Ka values. HI;
3. PH3 is weaker than HCl; HCl is weaker than HI. Thus, PH3 is weaker than HI.
4. PH3; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group.
5. HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.
Q14.3.17
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
1. acidity: HCl, HBr, HI
2. basicity: H2O, OH, H, Cl
3. basicity: Mg(OH)2, Si(OH)4, ClO3(OH) (Hint: Formula could also be written as HClO4).
4. acidity: HF, H2O, NH3, CH4
Q14.3.18
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
1. acidity: NaHSO3, NaHSeO3, NaHSO4
2. basicity: $\ce{BrO2-}$, $\ce{ClO2-}$, $\ce{IO2-}$
3. acidity: HOCl, HOBr, HOI
4. acidity: HOCl, HOClO, HOClO2, HOClO3
5. basicity: $\ce{NH2-}$, HS, HTe, $\ce{PH2-}$
6. basicity: BrO, $\ce{BrO2-}$, $\ce{BrO3-}$, $\ce{BrO4-}$
S14.3.18
1. NaHSeO3 < NaHSO3 < NaHSO4; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner.
2. $\ce{ClO2- < BrO2- < IO2-}$; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
3. HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
4. HOCl < HOClO < HOClO2 < HOClO3; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).
5. $\ce{HTe- < HS- << PH2- < NH2-}$; $\ce{PH2-}$ and $\ce{NH2-}$ are anions of weak bases, so they act as strong bases toward H+. $\ce{HTe-}$ and HS are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion.
6. $\ce{BrO4- < BrO3- < BrO2- < BrO-}$; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
Q14.3.19
Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN, is the stronger base? See Table.
Q14.3.20
The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxyl group (−CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).
$\ce{[H2O] > [C6H4OH(CO2H)] > [H+]0 > [C6H4OH(CO2)- ] ≫ [C6H4O(CO2H)- ] > [OH- ]}$
What do we represent when we write:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)?$
Q14.3.21
Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?
S14.3.21
Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.
Q14.3.22
Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.
Q14.3.23
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?
S14.3.23
1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration.
2. Assume we can neglect the contribution of water to the equilibrium concentration of H3O+.
Q14.3.24
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base?
Q14.3.25
Which of the following will increase the percent of NH3 that is converted to the ammonium ion in water (Hint: Use LeChâtelier’s principle.)?
1. addition of NaOH
2. addition of HCl
3. addition of NH4Cl
S14.3.25
The addition of HCl
Q14.3.26
Which of the following will increase the percent of HF that is converted to the fluoride ion in water?
1. addition of NaOH
2. addition of HCl
3. addition of NaF
Q14.3.27
What is the effect on the concentrations of $\ce{NO2-}$, HNO2, and OH when the following are added to a solution of KNO2 in water:
1. HCl
2. HNO2
3. NaOH
4. NaCl
5. KNO
The equation for the equilibrium is:
$\ce{NO2-}(aq)+\ce{H2O}(l)⇌\ce{HNO2}(aq)+\ce{OH-}(aq)$
S14.3.27
1. Adding HCl will add H3O+ ions, which will then react with the OH ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentration of $\ce{NO2-}$ ions.
2. Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of $\ce{NO2-}$ ions and decreasing the concentration of OH ions.
3. Adding NaOH adds OH ions, which shifts the equilibrium to the left, increasing the concentration of $\ce{NO2-}$ ions and decreasing the concentrations of HNO2.
4. Adding NaCl has no effect on the concentrations of the ions.
5. Adding KNO2 adds $\ce{NO2-}$ ions and shifts the equilibrium to the right, increasing the HNO2 and OH ion concentrations.
Q14.3.28
What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?
1. HCl
2. KF
3. NaCl
4. KOH
5. HF
The equation for the equilibrium is:
$\ce{HF}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{F-}(aq)$
Q14.3.29
Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?
S14.3.29
This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H3O+] produced by the stronger acid.
Q14.3.30
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
CH3CO2H: $\ce{[H3O+]}$ = 1.34 × 10−3 M;
$\ce{[CH3CO2- ]}$ = 1.34 × 10−3 M;
[CH3CO2H] = 9.866 × 10−2 M;
ClO: [OH] = 4.0 × 10−4 M;
[HClO] = 2.38 × 10−5 M;
[ClO] = 0.273 M;
HCO2H: [HCO2H] = 0.524 M;
$\ce{[H3O+]}$ = 9.8 × 10−3 M; $\ce{[HCO2- ]}$ = 9.8 × 10−3 M;
$\ce{C6H5NH3+ : [C6H5NH3+]}$ = 0.233 M;
[C6H5NH2] = 2.3 × 10−3 M;
$\ce{[H3O+]}$ = 2.3 × 10−3 M
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
NH3: [OH] = 3.1 × 10−3 M;
$\ce{[NH4+]}$ = 3.1 × 10−3 M;
[NH3] = 0.533 M;
HNO2: $\ce{[H3O+]}$ = 0.011 M;
$\ce{[NO2- ]}$ = 0.0438 M;
[HNO2] = 1.07 M;
(CH3)3N: [(CH3)3N] = 0.25 M;
[(CH3)3NH+] = 4.3 × 10−3 M;
[OH] = 4.3 × 10−3 M;
$\ce{NH4+ : [NH4+]}$ = 0.100 M;
[NH3] = 7.5 × 10−6 M;
[H3O+] = 7.5 × 10−6 M
1. $K_\ce{b}=1.8×10^{−5};$
2. $K_\ce{a}=4.5×10^{−4};$
3. $K_\ce{b}=7.4×10^{−5};$
4. $K_\ce{a}=5.6×10^{−10}$
Q14.3.31
Determine Kb for the nitrite ion, $\ce{NO2-}$. In a 0.10-M solution this base is 0.0015% ionized.
Q14.3.32
Determine Ka for hydrogen sulfate ion, $\ce{HSO4-}$. In a 0.10-M solution the acid is 29% ionized.
S14.3.32
$K_\ce{a}=1.2×10^{−2}$
Q14.3.33
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
1. F
2. $\ce{NH4+}$
3. $\ce{AsO4^3-}$
4. $\ce{(CH3)2NH2+}$
5. $\ce{NO2-}$
6. $\ce{HC2O4-}$ (as a base)
Q14.3.52
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
1. HTe (as a base)
2. $\ce{(CH3)3NH+}$
3. $\ce{HAsO4^3-}$ (as a base)
4. $\ce{HO2-}$ (as a base)
5. $\ce{C6H5NH3+}$
6. $\ce{HSO3-}$ (as a base)
S14.3.52
1. $K_\ce{b}=4.3×10^{−12};$
2. $K_\ce{a}=1.4×10^{−10};$
3. $K_\ce{b}=1×10^{−7};$
4. $K_\ce{b}=4.2×10^{−3};$
5. $K_\ce{b}=4.2×10^{−3};$
6. $K_\ce{b}=8.3×10^{−13}$
Q14.3.53
For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?
1. 3 × 10−8 M HNO3
2. 0.10 g HCl in 1.0 L of solution
3. 0.00080 g NaOH in 0.50 L of solution
4. 1 × 10−7 M Ca(OH)2
5. 0.0245 M KNO3
Q14.3.54
Even though both NH3 and C6H5NH2 are weak bases, NH3 is a much stronger acid than C6H5NH2. Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH3 and 0.10 M in C6H5NH2?
1. $\ce{[OH- ]}=\ce{[NH4+]}$
2. $\ce{[NH4+]}=\ce{[C6H5NH3+]}$
3. $\ce{[OH- ]}=\ce{[C6H5NH3+]}$
4. [NH3] = [C6H5NH2]
5. both a and b are correct
is the correct statement.
Q14.3.55
Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO2H and 0.10 M in HClO.
Q14.3.56
Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO2 and 0.120 M in HBrO.
S14.3.56
[H3O+] = 7.5 × 10−3 M
[HNO2] = 0.126 [OH] = 1.3 × 10−12 M [BrO] = 3.2 × 10−8 M [HBrO] = 0.120 M
Q14.3.57
Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH3NH2 and 0.10 M in C5H5N (Kb = 1.7 × 10−9).
Q14.3.58
Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH3 and 0.100 M in C6H5NH2.
S14.3.58
[OH] = $\ce{[NO4+]}$ = 0.0014 M
[NH3] = 0.144 M [H3O+] = 6.9 × 10−12 M $\ce{[C6H5NH3+]}$ = 3.9 × 10−8 M [C6H5NH2] = 0.100 M
Q14.3.59
Using the Ka values in Appendix H, place $\ce{Al(H2O)6^3+}$ in the correct location in Figure.
Q14.3.60
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I.
1. 0.0092 M HClO, a weak acid
2. 0.0784 M C6H5NH2, a weak base
3. 0.0810 M HCN, a weak acid
4. 0.11 M (CH3)3N, a weak base
5. 0.120 M $\ce{Fe(H2O)6^2+}$ a weak acid, Ka = 1.6 × 10−7
S14.3.60
$\ce{\dfrac{[H3O+][ClO- ]}{[HClO]}}=\dfrac{(x)(x)}{(0.0092−x)}≈\dfrac{(x)(x)}{0.0092}=3.5×10^{−8}$
Solving for x gives 1.79 × 10−5 M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [ClO] = 1.8 × 10−5 M[HClO] = 0.00092 M [OH] = 5.6 × 10−10 M;
$\ce{\dfrac{[C6H5NH3+][OH- ]}{[C6H5NH2]}}=\dfrac{(x)(x)}{(0.0784−x)}≈\dfrac{(x)(x)}{0.0784}=4.6×10^{−10}$
Solving for x gives 6.01 × 10−6 M.
This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: $\ce{[CH3CO2- ]}$ = [OH] = 6.0 × 10−6 M [C6H5NH2] = 0.00784 [H3O+] = 1.7× 10−9 M; $\ce{\dfrac{[H3O+][CN- ]}{[HCN]}}=\dfrac{(x)(x)}{(0.0810−x)}≈\dfrac{(x)(x)}{0.0810}=4×10^{−10}$ Solving for x gives 5.69 × 10−6 M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H3O+] = [CN] = 5.7 × 10−6 M [HCN] = 0.0810 M [OH] = 1.8 × 10−9 M; $\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{(0.11−x)}≈\dfrac{(x)(x)}{0.11}=7.4×10^{−5}$ Solving for x gives 2.85 × 10−3 M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [(CH3)3NH+] = [OH] = 2.9 × 10−3 M [(CH3)3N] = 0.11 M [H3O+] = 3.5 × 10−12 M; $\ce{\dfrac{[Fe(H2O)5(OH)+][H3O+]}{[Fe(H2O)6^2+]}}=\dfrac{(x)(x)}{(0.120−x)}≈\dfrac{(x)(x)}{0.120}=1.6×10^{−7}$ Solving for x gives 1.39 × 10−4 M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [Fe(H2O)5(OH)+] = [H3O+] = 1.4 × 10−4 M $\ce{[Fe(H2O)6^2+]}$ = 0.120 M [OH] = 7.2 × 10−11 M
Q14.3.61
Propionic acid, C2H5CO2H (Ka = 1.34 × 10−5), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698-M solution of C2H5CO2H?
Q14.3.62
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm3, what is the pH?
pH = 2.41
Q14.3.63
The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after strenuous exercise, is 1.36 × 10−4. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?
Q14.3.64
Nicotine, C10H14N2, is a base that will accept two protons (K1 = 7 × 10−7, K2 = 1.4 × 10−11). What is the concentration of each species present in a 0.050-M solution of nicotine?
S14.3.64
[C10H14N2] = 0.049 M
[C10H14N2H+] = 1.9 × 10−4 M $\ce{[C10H14N2H2^2+]}$ = 1.4 × 10−11 M [OH] = 1.9 × 10−4 M [H3O+] = 5.3 × 10−11 M
Q14.3.65
The pH of a 0.20-M solution of HF is 1.92. Determine Ka for HF from these data.
Q14.3.66
The pH of a 0.15-M solution of $\ce{HSO4-}$ is 1.43. Determine Ka for $\ce{HSO4-}$ from these data.
S14.3.66
$K_\ce{a}=1.2×10^{−2}$
Q14.3.67
The pH of a 0.10-M solution of caffeine is 11.16. Determine Kb for caffeine from these data:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq)$
Q14.3.68
The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.
S14.3.68
$K_\ce{b}=1.77×10^{−5}$
14.4: Hydrolysis of Salt Solutions
Q14.4.1
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. Al(NO3)3
2. RbI
3. KHCO2
4. CH3NH3Br
Q14.4.2
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. FeCl3
2. K2CO3
3. NH4Br
4. KClO4
S14.4.2
acidic; basic; acidic; neutral
Q14.4.3
Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 × 10−6. Is a solution of novocaine acidic or basic? What are [H3O+], [OH], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.
14.5: Polyprotic Acids
Q15.5.1
Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid:
• $\ce{[H3O+]}$,
• $[OH^−]$
• $[H_2CO_3]$
• $\ce{[HCO3- ]}$
• $\ce{[CO3^2- ]}$
No calculations are needed to answer this question.
S15.5.1
[H3O+] and $\ce{[HCO3- ]}$ are equal, H3O+ and $\ce{HCO3-}$ are practically equal
Q15.5.2
Calculate the concentration of each species present in a 0.050-M solution of H2S.
Q15.5.3
Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.
S15.5.3
$\ce{C6H4(CO2H)2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H4(CO2H)(CO2)-}(aq) \hspace{20px} K_\ce{a}=1.1×10^{−3}$
$\ce{C6H4(CO2H)(CO2)}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H4(CO2)2^2-}(aq) \hspace{20px} K_\ce{a}=3.9×10^{−6}$
[C6H4(CO2H)2] 7.2 × 10−3 M, [C6H4(CO2H)(CO2)] = [H3O+] 2.8 × 10−3 M, $\ce{[C6H4(CO2)2^2- ]}$3.9 × 10−6 M, [OH] 3.6 × 10−12 M
Q15.5.4
Salicylic acid, HOC6H4CO2H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
1. Both functional groups of salicylic acid ionize in water, with Ka = 1.0 × 10−3 for the—CO2H group and 4.2 × 10−13 for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).
2. Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH3CO2C6H4CO2H. The −CO2H functional group is still present, but its acidity is reduced, Ka = 3.0 × 10−4. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
3. Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid:
$\ce{CH3CO2C6H4CO2H}(aq)+\ce{H2O}(l)⟶\ce{HOC6H4CO2H}(aq)+\ce{CH3CO2H}(aq)$
1. Which of the acids salicylic acid or acetic acid produces more hydronium ions in solution such a solution?
2. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of 0.50 g of aspirin dissolved in enough water to give 75 mL of solution?
Q15.5.5
The ion HTe is an amphiprotic species; it can act as either an acid or a base.
1. What is Ka for the acid reaction of HTe with H2O?
2. What is Kb for the reaction in which HTe functions as a base in water?
3. Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe] in a 0.10 M solution of H2Te.
S15.5.5
1. $K_{\ce a2}=1×10^{−5};$
2. $K_\ce{b}=4.3×10^{−12};$
3. $\ce{\dfrac{[Te^2- ][H3O+]}{[HTe- ]}}=\dfrac{(x)(0.0141+x)}{(0.0141−x)}≈\dfrac{(x)(0.0141)}{0.0141}=1×10^{−5}$. Solving for x gives 1 × 10−5 M. Therefore, compared with 0.014 M, this value is negligible (0.071%).
14.6: Buffers
Q14.6.1
Explain why a buffer can be prepared from a mixture of NH4Cl and NaOH but not from NH3 and NaOH.
Q14.6.2
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H3PO4 and a salt of its conjugate base NaH2PO4.
S14.6.2
Excess H3O+ is removed primarily by the reaction:
$\ce{H3O+}(aq)+\ce{H2PO4-}(aq)⟶\ce{H3PO4}(aq)+\ce{H2O}(l)$ Excess base is removed by the reaction: $\ce{OH-}(aq)+\ce{H3PO4}(aq)⟶\ce{H2PO4-}(aq)+\ce{H2O}(l)$
Q14.6.3
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH3 and a salt of its conjugate acid NH4Cl.
Q14.6.4
What is [H3O+] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2?
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5}$
S14.6.4
[H3O+] = 1.5 × 10−4 M
Q14.6.5
What is [H3O+] in a solution of 0.075 M HNO2 and 0.030 M NaNO2?
S14.6.6
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \hspace{20px} K_\ce{a}=4.5×10^{−5}$
Q14.6.6
What is [OH] in a solution of 0.125 M CH3NH2 and 0.130 M CH3NH3Cl?
S14.6.6
$\ce{CH3NH2}(aq)+\ce{H2O}(l)⇌\ce{CH3NH3+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=4.4×10^{−4}$
[OH] = 4.2 × 10−4 M
Q14.6.7
What is [OH] in a solution of 1.25 M NH3 and 0.78 M NH4NO3?
S14.6.7
$\ce{NH3}(aq)+\ce{H2O}(l)⇌\ce{NH4+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=1.8×10^{−5}$
Q14.6.8
What concentration of NH4NO3 is required to make [OH] = 1.0 × 10−5 in a 0.200-M solution of NH3?
S14.6.8
[NH4NO3] = 0.36 M
Q14.6.9A
What concentration of NaF is required to make [H3O+] = 2.3 × 10−4 in a 0.300-M solution of HF?
Q14.6.9B
What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
1. HCl
2. KCH3CO2
3. NaCl
4. KOH
5. CH3CO2H
S14.6.10
1. The added HCl will increase the concentration of H3O+ slightly, which will react with $\ce{CH3CO2-}$ and produce CH3CO2H in the process. Thus, $\ce{[CH3CO2- ]}$ decreases and [CH3CO2H] increases.
2. The added KCH3CO2 will increase the concentration of $\ce{[CH3CO2- ]}$ which will react with H3O+ and produce CH3CO2 H in the process. Thus, [H3O+] decreases slightly and [CH3CO2H] increases.
3. The added NaCl will have no effect on the concentration of the ions.
4. The added KOH will produce OH ions, which will react with the H3O+, thus reducing [H3O+]. Some additional CH3CO2H will dissociate, producing $\ce{[CH3CO2- ]}$ ions in the process. Thus, [CH3CO2H] decreases slightly and $\ce{[CH3CO2- ]}$ increases.
5. The added CH3CO2H will increase its concentration, causing more of it to dissociate and producing more $\ce{[CH3CO2- ]}$ and H3O+ in the process. Thus, [H3O+] increases slightly and $\ce{[CH3CO2- ]}$ increases.
Q14.6.11
What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
1. KI
2. NH3
3. HI
4. NaOH
5. NH4Cl
What will be the pH of a buffer solution prepared from 0.20 mol NH3, 0.40 mol NH4NO3, and just enough water to give 1.00 L of solution?
pH = 8.95
Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH2PO4, and enough water to make 0.500 L of solution.
How much solid NaCH3CO2•3H2O must be added to 0.300 L of a 0.50-M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
37 g (0.27 mol)
What mass of NH4Cl must be added to 0.750 L of a 0.100-M solution of NH3 to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)
Q14.6.1
A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 × 10−5 as Ka for acetic acid.
1. What is the pH of the solution?
2. Is the solution acidic or basic?
Q14.6.1
What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
1. pH = 5.222;
2. The solution is acidic. (c) pH = 5.221
Q14.6.1
A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L.
1. What is the pH of this buffer solution?
2. Is the solution acidic or basic?
3. What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?
Which acid in [link] is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.
To prepare the best buffer for a weak acid HA and its salt, the ratio $\dfrac{\ce{[H3O+]}}{K_\ce{a}}$ should be as close to 1 as possible for effective buffer action. The [H3O+] concentration in a buffer of pH 3.1 is [H3O+] = 10−3.1 = 7.94 × 10−4 M
We can now solve for Ka of the best acid as follows:
$\dfrac{\ce{[H3O+]}}{K_\ce{a}}=1$
$K_\ce{a}=\dfrac{\ce{[H3O+]}}{1}=7.94×10^{−4}$
In [link], the acid with the closest Ka to 7.94 × 10−4 is HF, with a Ka of 7.2 × 10−4.
Which acid in [link] is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.
Which base in [link] is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.
For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio $\dfrac{\ce{[OH- ]}}{K_\ce{b}}$ that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH] is [OH] = 10−pOH = 10−3.35 = 4.467 × 10−4 M.
We can now solve for Kb of the best base as follows: $\dfrac{\ce{[OH- ]}}{K_\ce{b}}=1$ Kb = [OH] = 4.47 × 10−4 In [link], the base with the closest Kb to 4.47 × 10−4 is CH3NH2, with a Kb = 4.4 × 10−4.
Which base in [link] is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.
Q14.6.4
Saccharin, C7H4NSO3H, is a weak acid (Ka = 2.1 × 10−2). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 × 10−3 g of sodium saccharide, Na(C7H4NSO3), what are the final concentrations of saccharine and sodium saccharide in the solution?
S14.6.4
The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:
9.75 × 10−6 mol. This ionizes initially to form saccharin ions, A, with: [A] = 3.9 × 10−5 M
Q14.6.5
What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C5H9NO4, a diprotic acid; K1 = 8.5 × 10−5, K2 = 3.39 × 10−10) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?
14.7: Acid-Base Titrations
Q14.7.1
Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
S14.7.1
At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.
Q14.7.2
Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
Q14.7.3
Why can we ignore the contribution of water to the concentrations of H3O+ in the solutions of following acids:
• 0.0092 M HClO, a weak acid
• 0.0810 M HCN, a weak acid
• 0.120 M $\ce{Fe(H2O)6^2+}$ a weak acid, Ka = 1.6 × 10−7
but not the contribution of water to the concentration of OH?
S14.7.3
In an acid solution, the only source of OH ions is water. We use Kw to calculate the concentration. If the contribution from water was neglected, the concentration of OH would be zero.
Q14.7.4
We can ignore the contribution of water to the concentration of OH in a solution of the following bases:
0.0784 M C6H5NH2, a weak base
0.11 M (CH3)3N, a weak base
but not the contribution of water to the concentration of H3O+?
Q14.7.5
Draw a curve for a series of solutions of HF. Plot [H3O+]total on the vertical axis and the total concentration of HF (the sum of the concentrations of both the ionized and nonionized HF molecules) on the horizontal axis. Let the total concentration of HF vary from 1 × 10−10 M to 1 × 10−2 M.
Q14.7.6
Draw a curve similar to that shown in Figure for a series of solutions of NH3. Plot [OH] on the vertical axis and the total concentration of NH3 (both ionized and nonionized NH3 molecules) on the horizontal axis. Let the total concentration of NH3 vary from 1 × 10−10 M to 1 × 10−2 M.
Q14.7.7
Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.
1. no KOH added
2. 20 mL of KOH solution added
3. 39 mL of KOH solution added
4. 40 mL of KOH solution added
5. 41 mL of KOH solution added
1. pH = 2.50;
2. pH = 4.01;
3. pH = 5.60;
4. pH = 8.35;
5. pH = 11.08
Q14.7.8
The indicator dinitrophenol is an acid with a Ka of 1.1 × 10−4. In a 1.0 × 10−4-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow). | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.E%3A_Acid-Base_Equilibria_%28Exercises%29.txt |
We previously learned about aqueous solutions and their importance, as well as about solubility rules. While this gives us a picture of solubility, that picture is not complete if we look at the rules alone. Solubility equilibrium, which we will explore in this chapter, is a more complex topic that allows us to determine the extent to which a slightly soluble ionic solid will dissolve, and the conditions under which precipitation.
15: Equilibria of Other Reaction Classes
Learning Objectives
• Write chemical equations and equilibrium expressions representing solubility equilibria
• Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations
The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home’s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions.
In some cases, we want to prevent dissolution from occurring. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca5(PO4)3(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the decay. On the other hand, sometimes we want a substance to dissolve. We want the calcium carbonate in a chewable antacid to dissolve because the $\ce{CO3^2-}$ ions produced in this process help soothe an upset stomach.
In this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Chatelier’s principle. We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a solution.
The Solubility Product Constant
Silver chloride is what’s known as a sparingly soluble ionic solid (Figure $1$). Recall from the solubility rules in an earlier chapter that halides of Ag+ are not normally soluble. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag+ and Cl ions in equilibrium with undissolved silver chloride:
$\ce{AgCl}(s)\mathrm{\xrightleftharpoons[\:precipitation\:]{\:dissolution\:}}\ce{Ag+}(aq)+\ce{Cl-}(aq) \nonumber$
This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but at the same time, Ag+ and Cl ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.
The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the solubility product (Ksp) of the solid. Recall from the chapter on solutions and colloids that we use an ion’s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:
$\ce{AgCl}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{Cl-}(aq) \nonumber$
with
$K_\ce{sp}=[\ce{Ag+}(aq)][\ce{Cl-}(aq)] \nonumber$
When looking at dissolution reactions such as this, the solid is listed as a reactant, whereas the ions are listed as products. The solubility product constant, as with every equilibrium constant expression, is written as the product of the concentrations of each of the ions, raised to the power of their stoichiometric coefficients. Here, the solubility product constant is equal to Ag+ and Cl when a solution of silver chloride is in equilibrium with undissolved AgCl. There is no denominator representing the reactants in this equilibrium expression since the reactant is a pure solid; therefore [AgCl] does not appear in the expression for Ksp.
Table $1$: Common Solubility Products by Decreasing Equilibrium Constants
Substance Ksp at 25 °C
CuCl 1.2 × 10–6
CuBr 6.27 × 10–9
AgI 1.5 × 10–16
PbS 7 × 10–29
Al(OH)3 2 × 10–32
Fe(OH)3 4 × 10–38
Some common solubility products are listed in Table $1$ according to their Ksp values, whereas a more extensive compilation of products appears in Table E3. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small Ksp represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.
Example $1$: Writing Equations and Solubility Products
Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:
1. AgI, silver iodide, a solid with antiseptic properties
2. CaCO3, calcium carbonate, the active ingredient in many over-the-counter chewable antacids
3. Mg(OH)2, magnesium hydroxide, the active ingredient in Milk of Magnesia
4. Mg(NH4)PO4, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium
5. Ca5(PO4)3OH, the mineral apatite, a source of phosphate for fertilizers
(Hint: When determining how to break (d) and (e) up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)
Solution
1. $\ce{AgI}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{I-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ag+][I- ]}$
2. $\ce{CaCO3}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ca^2+][CO3^2- ]}$
3. $\ce{Mg(OH)2}(s) \rightleftharpoons \ce{Mg^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Mg^2+][OH- ]^2}$
4. $\ce{Mg(NH4)PO4}(s) \rightleftharpoons \ce{Mg^2+}(aq)+\ce{NH4+}(aq)+\ce{PO4^3-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Mg^2+][NH4+][PO4^3- ]}$
5. $\ce{Ca5(PO4)3OH}(s) \rightleftharpoons \ce{5Ca^2+}(aq)+\ce{3PO4^3-}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ca^2+]^5[PO4^3- ]^3[OH- ]}$
Exercise $1$
Write the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:
1. BaSO4
2. Ag2SO4
3. Al(OH)3
4. Pb(OH)Cl
Answer a
$\ce{BaSO4}(s) \rightleftharpoons \ce{Ba^2+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ba^2+][SO4^2- ]}$
Answer b
$\ce{Ag2SO4}(s) \rightleftharpoons \ce{2Ag+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ag+]^2[SO4^2- ]}$
Answer c
$\ce{Al(OH)3}(s) \rightleftharpoons \ce{Al^2+}(aq)+\ce{3OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Al^3+][OH- ]^3}$
Answer d
$\ce{Pb(OH)Cl}(s) \rightleftharpoons \ce{Pb^2+}(aq)+\ce{OH-}(aq)+\ce{Cl-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Pb^2+][OH- ][Cl- ]}$
Now we will extend the discussion of Ksp and show how the solubility product constant is determined from the solubility of its ions, as well as how Ksp can be used to determine the molar solubility of a substance.
Ksp and Solubility
Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:
$\ce{M}_p\ce{X}_q(s) \rightleftharpoons p\mathrm{M^{m+}}(aq)+q\mathrm{X^{n−}}(aq) \nonumber$
In this case, we calculate the solubility product by taking the solid’s solubility expressed in units of moles per liter (mol/L), known as its molar solubility.
Example $2$: Calculation of Ksp from Equilibrium Concentrations
We began the chapter with an informal discussion of how the mineral fluorite is formed. Fluorite, CaF2, is a slightly soluble solid that dissolves according to the equation:
$\ce{CaF2}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{2F-}(aq) \nonumber$
The concentration of Ca2+ in a saturated solution of CaF2 is 2.1 × 10–4 M; therefore, that of F is 4.2 × 10–4 M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?
Solution
First, write out the Ksp expression, then substitute in concentrations and solve for Ksp:
$\ce{CaF2}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{2F-}(aq) \nonumber$
A saturated solution is a solution at equilibrium with the solid. Thus:
\begin{align*} K_\ce{sp} &=\ce{[Ca^2+][F^{-}]^2} \[4pt] &=(2.1×10^{−4})(4.2×10^{−4})^2 \[4pt] &=3.7×10^{−11} \end{align*} \nonumber
As with other equilibrium constants, we do not include units with Ksp.
Exercise $2$
In a saturated solution that is in contact with solid Mg(OH)2, the concentration of Mg2+ is 3.7 × 10–5 M. What is the solubility product for Mg(OH)2?
$\ce{Mg(OH)2}(s) \rightleftharpoons \ce{Mg^2+}(aq)+\ce{2OH-}(aq) \nonumber$
Answer
2.0 × 10–13
Example $3$: Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide, CuBr, is 6.3 × 10–9. Calculate the molar solubility of copper bromide.
Solution
The solubility product constant of copper(I) bromide is $6.3 \times 10^{–9}$.
The reaction is:
$\ce{CuBr}(s) \rightleftharpoons \ce{Cu+}(aq)+\ce{Br-}(aq) \nonumber$
First, write out the solubility product equilibrium constant expression:
$K_\ce{sp}=\ce{[Cu+][Br- ]} \nonumber$
Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
\begin{align*} K_{sp} &=\ce{[Cu+][Br- ]} \[4pt] 6.3×10^{−9} &=(x)(x)=x^2 \[4pt] x &=\sqrt{(6.3×10^{−9})}=7.9×10^{−5} \end{align*} \nonumber
Therefore, the molar solubility of CuBr is 7.9 × 10–5 M.
Exercise $3$
The Ksp of AgI is 1.5 × 10–16. Calculate the molar solubility of silver iodide.
Answer
1.2 × 10–8 M
Example $4$: Determination of Molar Solubility from Ksp, Part II
Determination of Molar Solubility from Ksp, Part II The Ksp of calcium hydroxide, Ca(OH)2, is 8.0 × 10–6. Calculate the molar solubility of calcium hydroxide.
Solution
The solubility product constant of calcium hydroxide is 1.3 × 10–6.
The reaction is:
$\ce{Ca(OH)2}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{2OH-}(aq) \nonumber$
First, write out the solubility product equilibrium constant expression:
$K_\ce{sp}=\ce{[Ca^2+][OH- ]^2} \nonumber$
Create an ICE table, leaving the Ca(OH)2 column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
\begin{align*} K_\ce{sp} &=\ce{[Ca^2+][OH- ]^2} \[4pt] 1.3×10^{−6} &=(x)(2x)^2=(x)(4x^2)=4x^3 \[4pt] x &=\sqrt[3]{\dfrac{1.3×10^{-6}}{4}}=6.9×10^{-3} \end{align*} \nonumber
Therefore, the molar solubility of Ca(OH)2 is 6.9 × 10–3 M.
Exercise $4$
The Ksp of PbI2 is 1.4 × 10–8. Calculate the molar solubility of lead(II) iodide.
Answer
1.5 × 10–3 M
Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. Example $5$ shows how to perform those unit conversions before determining the solubility product equilibrium.
Example $5$: Determination of Ksp from Gram Solubility
Many of the pigments used by artists in oil-based paints (Figure $2$) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is 4.6 × 10–6 g/L. Determine the solubility product equilibrium constant for PbCrO4.
Solution
We are given the solubility of PbCrO4 in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb2+ and $\ce{CrO4^2-}$, then Ksp:
1. Use the molar mass of PbCrO4 $\mathrm{\left(\dfrac{323.2\:g}{1\:mol}\right)}$ to convert the solubility of PbCrO4 in grams per liter into moles per liter:
$\mathrm{[PbCrO_4]=\dfrac{4.6×10^{−6}\:g\: PbCrO_4}{1\:L}×\dfrac{1\:mol\: PbCrO_4}{323.2\:g\: PbCrO_4}}$
$\mathrm{=\dfrac{1.4×10^{−8}\:mol\: PbCrO_4}{1\:L}}$
$=1.4×10^{−8}\:M$
• The chemical equation for the dissolution indicates that 1 mol of PbCrO4 gives 1 mol of Pb2+(aq) and 1 mol of $\ce{CrO_4^{2-}}(aq)$:
$\ce{PbCrO4}(s) \rightleftharpoons \ce{Pb^2+}(aq)+\ce{CrO4^2-}(aq)$
Thus, both [Pb2+] and $\ce{[CrO4^2- ]}$ are equal to the molar solubility of PbCrO4:
$\ce{[Pb^2+]}=\ce{[CrO4^2- ]}=1.4×10^{−8}\:M$
• Solve. Ksp = [Pb2+] $\ce{[CrO4^2- ]}$ = (1.4 × 10–8)(1.4 × 10–8) = 2.0 × 10–16
Exercise $5$
The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.46 grams per liter at 20 °C. What is its solubility product?
Answer
2.08 × 10–4
Example $6$: Calculating the Solubility of Hg2Cl2
Calomel, Hg2Cl2, is a compound composed of the diatomic ion of mercury(I), $\ce{Hg2^2+}$, and chloride ions, Cl. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:
$\ce{Hg2Cl2}(s) \rightleftharpoons \ce{Hg2^2+}(aq)+\ce{2Cl-}(aq) \hspace{20px} K_\ce{sp}=1.1×10^{−18} \nonumber$
Calculate the molar solubility of Hg2Cl2.
Solution
The molar solubility of Hg2Cl2 is equal to the concentration of $\ce{Hg2^2+}$ ions because for each 1 mol of Hg2Cl2 that dissolves, 1 mol of $\ce{Hg2^2+}$ forms:
1. Determine the direction of change. Before any Hg2Cl2 dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium.
2. Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:
Note that the change in the concentration of Cl (2x) is twice as large as the change in the concentration of $\ce{Hg2^2+}$ (x) because 2 mol of Cl forms for each 1 mol of $\ce{Hg2^2+}$ that forms. Hg2Cl2 is a pure solid, so it does not appear in the calculation.
1. Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for Ksp and calculate the value of x:
\begin{align*} K_\ce{sp} &=\ce{[Hg2^2+][Cl- ]^2} \[4pt] 1.1×10^{−18} &=(x)(2x)^2 \[4pt] 4x^3 &=1.1×10^{−18} \[4pt] x &=\sqrt[3]{\left(\dfrac{1.1×10^{-18}}{4}\right)}=6.5×10^{-7}\:M \end{align*} \nonumber
So the concentrations are
\begin{align*} \ce{[Hg2^2+]} &=6.5×10^{−7}\:M=6.5×10^{−7}\:M \[4pt] \ce{[Cl- ]} &=2x=2(6.5×10^{−7})=1.3×10^{−6}\:M \end{align*} \nonumber
The molar solubility of Hg2Cl2 is equal to $\ce{[Hg2^2+]}$, or 6.5 × 10–7 M.
Check the work.
At equilibrium, Q = Ksp:
$Q=\ce{[Hg2^2+][Cl- ]^2}=(6.5×10^{−7})(1.3×10^{−6})^2=1.1×10^{−18} \nonumber$
The calculations check.
Exercise $6$
Determine the molar solubility of MgF2 from its solubility product: Ksp = 6.4 × 10–9.
Answer
1.2 × 10–3 M
Tabulated Ksp values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: Q equals Ksp at equilibrium; if Q is less than Ksp, the solid will dissolve until Q equals Ksp; if Q is greater than Ksp, precipitation will occur at a given temperature until Q equals Ksp.
Using Barium Sulfate for Medical Imaging
Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the Ksp of barium sulfate is 1.1 × 10–10, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure $3$).
Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate’s movement as it passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.
Predicting Precipitation
The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:
$\ce{CaCO3}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) \nonumber$
We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient $\ce{(Q=[Ca^2+][CO3^2- ])}$ is equal to the solubility product (Ksp = 4.8 × 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains $\ce{CO3^2-}$ ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and $\ce{CO3^2-}$ ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form.
We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as Example $7$ shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)
Example $7$: Precipitation of Mg(OH)2
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH ion:
$\ce{Mg(OH)2}(s) \rightleftharpoons \ce{Mg^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=8.9×10^{−12} \nonumber$
The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH] of 0.0010 M?
Solution
This problem asks whether the reaction:
$\ce{Mg(OH)2}(s) \rightleftharpoons \ce{Mg^2+}(aq)+\ce{2OH-}(aq) \nonumber$
shifts to the left and forms solid Mg(OH)2 when [Mg2+] = 0.0537 M and [OH] = 0.0010 M. The reaction shifts to the left if Q is greater than Ksp. Calculation of the reaction quotient under these conditions is shown here:
$\mathrm{Q=[Mg^{2+}][OH^-]^2=(0.0537)(0.0010)^2=5.4×10^{−8}} \nonumber$
Because Q is greater than Ksp (Q = 5.4 × 10–8 is larger than Ksp = 8.9 × 10–12), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp.
Exercise $7$
Use the solubility products in Table E3 to determine whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and $\ce{[HPO4^2- ]}$ = 0.001 M.
Answer
No precipitation of CaHPO4; Q = 1 × 10–7, which is less than Ksp
Example $8$: Precipitation of AgCl upon Mixing Solutions
Does silver chloride precipitate when equal volumes of a 2.0 × 10–4-M solution of AgNO3 and a 2.0 × 10–4-M solution of NaCl are mixed?
(Note: The solution also contains Na+ and $\ce{NO3-}$ ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
$\ce{AgCl}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{Cl-}(aq)$
The solubility product is 1.8 × 10–10 (Table E3).
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl] are both equal to:
$\dfrac{1}{2}(2.0×10^{−4})\:M=1.0×10^{−4}\:M$
The reaction quotient, Q, is momentarily greater than Ksp for AgCl, so a supersaturated solution is formed:
$Q=\ce{[Ag+][Cl- ]}=(1.0×10^{−4})(1.0×10^{−4})=1.0×10^{−8}>K_\ce{sp}$
Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to Ksp.
Exercise $8$
Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of $\ce{ClO4-}$? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)
Answer
No, Q = 4.0 × 10–3, which is less than Ksp = 1.07 × 10–2
In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.
Example $9$: Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, $\ce{C2O4^2-}$, for this purpose (Figure $4$). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4•H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 × 10–3 M. What concentration of $\ce{C2O4^2-}$ ion must be established before CaC2O4•H2O begins to precipitate?
Solution
The equilibrium expression is:
$\ce{CaC2O4}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{C2O4^2-}(aq) \nonumber$
For this reaction (Table E3):
$K_\ce{sp}=\ce{[Ca^2+][C2O4^2- ]}=1.96×10^{−8} \nonumber$
CaC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.
Solid CaC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Ca2+], we can solve for the concentration of $\ce{C2O4^2-}$ that is necessary to produce the first trace of solid:
$Q=K_\ce{sp}=\ce{[Ca^2+][C2O4^2- ]}=1.96×10^{−8}$
$(2.2×10^{−3})\ce{[C2O4^2- ]}=1.96×10^{−8}$
$\ce{[C2O4^2- ]}=\dfrac{1.96×10^{−8}}{2.2×10^{−3}}=8.9×10^{−6}$
A concentration of $\ce{[C2O4^2- ]}$ = 8.9 × 10–6 M is necessary to initiate the precipitation of CaC2O4 under these conditions.
Exercise $9$
If a solution contains 0.0020 mol of $\ce{CrO4^2-}$ per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.
Answer
4.5 × 10–9 M
It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example $8$—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.
Example $10$: Concentrations Following Precipitation
Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 × 10–6 M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)2, what pH is required to keep [Mn2+] equal to 1.8 × 10–6 M?
Solution
The dissolution of Mn(OH)2 is described by the equation:
$\ce{Mn(OH)2}(s) \rightleftharpoons \ce{Mn^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=2×10^{−13} \nonumber$
We need to calculate the concentration of OH when the concentration of Mn2+ is 1.8 × 10–6 M. From that, we calculate the pH. At equilibrium:
$K_\ce{sp}=\ce{[Mn^2+][OH- ]^2} \nonumber$
or
$(1.8×10^{−6})\ce{[OH- ]^2}=2×10^{−13} \nonumber$
so
$\ce{[OH- ]}=3.3×10^{−4}\:M \nonumber$
Now we calculate the pH from the pOH:
$\mathrm{pOH=−\log[OH^-]=−\log(3.3×10^{−4})=3.48}$
$\mathrm{pH=14.00−pOH=14.00−3.48=10.52}$
If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.52, the manganese ion will be reduced to a concentration of 1.8 × 10–6 M; at that concentration or less, the ion will not stain clothing.
Exercise $10$
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 × 10–2 M. Calculate the pH at which [Mg2+] is diminished to 1.0 × 10–5 M by the addition of Ca(OH)2.
Answer
10.97
Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 × 10–10), AgBr (Ksp = 5.0 × 10–13), and AgI (Ksp = 1.5 × 10–16) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl, Br, and I; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl, Br, and I to a solution of Ag+.
When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller Ksp) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the Ksp values of the two compounds differ by two orders of magnitude or more (e.g., 10–2 vs. 10–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.
The Role of Precipitation in Wastewater Treatment
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure $5$). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions $\ce{(PO4^2- )}$ are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.
One common way to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)2. The lime is converted into calcium carbonate, a strong base, in the water. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3(OH), which then precipitates out of the solution:
$\ce{5Ca^2+ + 3PO4^3- + OH- \rightleftharpoons Ca10(PO4)6⋅(OH)2}(s) \nonumber$
The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.
Selective precipitation can also be used in qualitative analysis. In this method, reagents are added to an unknown chemical mixture in order to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the addition of the reagent can be used to determine whether the ion is present in the solution.
Example $11$: Precipitation of Silver Halides
A solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?
Solution
The two equilibria involved are:
$\ce{AgCl}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{Cl-}(aq) \hspace{20px} K_\ce{sp}=1.6×10^{−10}$
$\ce{AgI}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{I-}(aq) \hspace{20px} K_\ce{sp}=1.5×10^{−16}$
If the solution contained about equal concentrations of Cl and I, then the silver salt with the smallest Ksp (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgI begins to precipitate. The salt that forms at the lower [Ag+] precipitates first.
For AgI: AgI precipitates when Q equals Ksp for AgI (1.5 × 10–16). When [I] = 0.0010 M:
$Q=\ce{[Ag+][I- ]}=\ce{[Ag+]}(0.0010)=1.5×10^{−16}$
$\ce{[Ag+]}=\dfrac{1.5×10^{−16}}{0.0010}=1.5×10^{−13}$
AgI begins to precipitate when [Ag+] is 1.5 × 10–13 M.
For AgCl: AgCl precipitates when Q equals Ksp for AgCl (1.6 × 10–10). When [Cl] = 0.10 M:
$Q_\ce{sp}=\ce{[Ag+][Cl- ]}=\ce{[Ag+]}(0.10)=1.6×10^{−10}$
$\ce{[Ag+]}=\dfrac{1.6×10^{−10}}{0.10}=1.6×10^{−9}\:M$
AgCl begins to precipitate when [Ag+] is 1.6 × 10–9 M.
AgI begins to precipitate at a lower [Ag+] than AgCl, so AgI begins to precipitate first.
Exercise $11$
If silver nitrate solution is added to a solution which is 0.050 M in both Cl and Br ions, at what [Ag+] would precipitation begin, and what would be the formula of the precipitate?
Answer
[Ag+] = 1.0 × 10–11 M; AgBr precipitates first
Common Ion Effect
As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH3CO2, is added. We can explain this effect using Le Chatelier’s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of $\ce{H3O+}$ to compensate for the increased acetate ion concentration. This increases the concentration of CH3CO2H:
$\ce{CH3CO2H + H2O \rightleftharpoons H3O+ + CH3CO2-}\)] Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the common ion effect. The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved: \[\ce{AgI}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{I-}(aq) \nonumber$
If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Chatelier’s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.
Example $12$: Common Ion Effect
Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 × 10–28.
Solution
The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr2 concentration as a contributor of cadmium ions:
$\ce{CdS}(s) \rightleftharpoons \ce{Cd^2+}(aq)+\ce{S^2-}(aq) \nonumber$
$K_\ce{sp}=\ce{[Cd^2+][S^2- ]}=1.0×10^{−28}$
$(0.010+x)(x)=1.0×10^{−28}$ $x^2+0.010x−1.0×10^{−28}=0$
We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the Ksp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our Ksp expression, we would now get:
$K_\ce{sp}=\ce{[Cd^2+][S^2- ]}=1.0×10^{−28}$
$(0.010)(x)=1.0×10^{−28}$
$x=1.0×10^{−26}$
Therefore, the molar solubility of CdS in this solution is 1.0 × 10–26 M.
Exercise $12$
Calculate the molar solubility of aluminum hydroxide, Al(OH)3, in a 0.015-M solution of aluminum nitrate, Al(NO3)3. The Ksp of Al(OH)3 is 2 × 10–32.
Answer
4 × 10–11
Summary
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Ksp, of the solid. When we have a heterogeneous equilibrium involving the slightly soluble solid MpXq and its ions Mm+ and Xn–:
$\ce{M}_p\ce{X}_q(s) \rightleftharpoons p\mathrm{M^{m+}}(aq)+q\mathrm{X^{n−}}(aq) \nonumber$
We write the solubility product expression as:
$K_\ce{sp}=\mathrm{[M^{m+}]}^p\mathrm{[X^{n−}]}^q \nonumber$
The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its Ksp, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions. A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product. A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Chatelier’s principle applies and more precipitate comes out of solution so that the molar solubility is reduced.
Glossary
common ion effect
effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base
molar solubility
solubility of a compound expressed in units of moles per liter (mol/L)
selective precipitation
process in which ions are separated using differences in their solubility with a given precipitating reagent
solubility product (Ksp)
equilibrium constant for the dissolution of a slightly soluble electrolyte | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/15%3A_Equilibria_of_Other_Reaction_Classes/15.1%3A_Precipitation_and_Dissolution.txt |
Learning Objectives
• Explain the Lewis model of acid-base chemistry
• Write equations for the formation of adducts and complex ions
• Perform equilibrium calculations involving formation constants
In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.
A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.
Definition: Lewis Acids and Bases
• A Lewis acid is any species (molecule or ion) that can accept a pair of electrons, and a Lewis base is any species (molecule or ion) that can donate a pair of electrons.
A Lewis acid-base reaction occurs when a base donates a pair of electrons to an acid. A Lewis acid-base adduct, a compound that contains a coordinate covalent bond between the Lewis acid and the Lewis base, is formed. The following equations illustrate the general application of the Lewis concept.
The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell. Being short of the preferred octet, BF3 is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:
In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:
Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:
Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:
The last displacement reaction shows how the reaction of a Brønsted-Lowry acid with a base fits into the Lewis concept. A Brønsted-Lowry acid such as HCl is an acid-base adduct according to the Lewis concept, and proton transfer occurs because a more stable acid-base adduct is formed. Thus, although the definitions of acids and bases in the two theories are quite different, the theories overlap considerably.
Many slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the complex ion $\ce{Ag(NH3)2+}$. The Lewis structure of the $\ce{Ag(NH3)2+}$ ion is:
The equations for the dissolution of AgCl in a solution of NH3 are:
$\ce{AgCl}(s)⟶\ce{Ag+}(aq)+\ce{Cl-}(aq) \nonumber$
$\ce{Ag+}(aq)+\ce{2NH3}(aq)⟶\ce{Ag(NH3)2+}(aq) \nonumber$
$\textrm{Net: }\ce{AgCl}(s)+\ce{2NH3}(aq)⟶\ce{Ag(NH3)2+}(aq)+\ce{Cl-}(aq) \nonumber$
Aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion $\ce{Al(OH)4-}$. The Lewis structure of the $\ce{Al(OH)4-}$ ion is:
The equations for the dissolution are:
$\ce{Al(OH)3}(s)⟶\ce{Al^3+}(aq)+\ce{3OH-}(aq) \nonumber$
$\ce{Al^3+}(aq)+\ce{4OH-}(aq)⟶\ce{Al(OH)4-}(aq) \nonumber$
$\textrm{Net: }\ce{Al(OH)3}(s)+\ce{OH-}(aq)⟶\ce{Al(OH)4-}(aq) \nonumber$
Mercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S2– ion:
$\ce{HgS}(s)⟶\ce{Hg^2+}(aq)+\ce{S^2-}(aq) \nonumber$
$\ce{Hg^2+}(aq)+\ce{2S^2-}(aq)⟶\ce{HgS2^2-}(aq) \nonumber$
$\textrm{Net: }\ce{HgS}(s)+\ce{S^2-}(aq)⟶\ce{HgS2^2-}(aq) \nonumber$
A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called ligands. These ligands can be neutral molecules like H2O or NH3, or ions such as CN or OH. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands aggregate themselves around the central atom, creating a new ion with a charge equal to the sum of the charges and, most often, a transitional metal ion. This more complex arrangement is why the resulting ion is called a complex ion. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters.
The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a formation constant (Kf) (sometimes called a stability constant). For example, the complex ion $\ce{Cu(CN)2-}$ is shown here:
It forms by the reaction:
$\ce{Cu+}(aq)+\ce{2CN-}(aq)⇌\ce{Cu(CN)2-}(aq) \nonumber$
At equilibrium:
$K_\ce{f}=Q=\ce{\dfrac{[Cu(CN)2- ]}{[Cu+][CN- ]^2}} \nonumber$
The inverse of the formation constant is the dissociation constant (Kd), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. Table E4 and Table $1$ are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of Ksp values, the stoichiometry of the compound must be considered.
Table $1$: Common Complex Ions by Decreasing Formation Constants
Substance Kf at 25 °C
$\ce{[Cd(CN)4]^2-}$ 3 × 1018
$\ce{Ag(NH3)2+}$ 1.7 × 107
$\ce{[AlF6]^3-}$ 7 × 1019
As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag+ ([Ag+] = 1.3 × 10–5 M):
$\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq) \nonumber$
However, if NH3 is present in the water, the complex ion, $\ce{Ag(NH3)2+}$, can form according to the equation:
$\ce{Ag+}(aq)+\ce{2NH3}(aq)⇌\ce{Ag(NH3)2+}(aq) \nonumber$
with
$K_\ce{f}=\ce{\dfrac{[Ag(NH3)2+]}{[Ag+][NH3]^2}}=1.7×10^7 \nonumber$
The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH3 to form $\ce{Ag(NH3)2+}$. As a consequence, the concentration of silver ions, [Ag+], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag+][Cl], falls below the solubility product of AgCl:
$Q=\ce{[Ag+][Cl- ]}<K_\ce{sp} \nonumber$
More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.
Example $1$: Dissociation of a Complex Ion
Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to $\ce{Ag(NH3)2+}$.
Solution
We use the familiar path to solve this problem:
1. Determine the direction of change. The complex ion $\ce{Ag(NH3)2+}$ is in equilibrium with its components, as represented by the equation: $\ce{Ag+}(aq)+\ce{2NH3}(aq)⇌\ce{Ag(NH3)2+}(aq) \nonumber$
We write the equilibrium as a formation reaction because Table E4 lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant (K_f = 1.7 \times 10^7\), and $Q=\dfrac{0.10}{0 \times 0} = \infty$ (it is infinitely large), so the reaction shifts to the left to reach equilibrium. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/15%3A_Equilibria_of_Other_Reaction_Classes/15.2%3A_Lewis_Acids_and_Bases.txt |
Learning Objectives
• Describe examples of systems involving two (or more) simultaneous chemical equilibria
• Calculate reactant and product concentrations for multiple equilibrium systems
• Compare dissolution and weak electrolyte formation
There are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction).
The ocean is a unique example of a system with multiple equilibria, or multiple states of solubility equilibria working simultaneously. Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H2CO3). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions $\ce{(HCO3- )}$, which can further ionize into more hydrogen ions and carbonate ions $\ce{(CO3^2- )}$:
$\ce{CO2}(g)⇌\ce{CO2}(aq) \nonumber$
$\ce{CO2}(aq)+\ce{H2O}⇌\ce{H2CO3}(aq) \nonumber$
$\ce{H2CO3}(aq)⇌\ce{H+}(aq)+\ce{HCO3-}(aq) \nonumber$
$\ce{HCO3-}(aq)⇌\ce{H+}(aq)+\ce{CO3^2-}(aq) \nonumber$
The excess H+ ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure $1$). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world’s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.
Learn more about ocean acidification and how it affects other marine creatures.
Slightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO3, FeS, and Ca3(PO4)2 in HCl because their basic anions react to form weak acids (H2CO3, H2S, and $\ce{H2PO4-}$). The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Chatelier’s principle.
Of particular relevance to us is the dissolution of hydroxylapatite, Ca5(PO4)3OH, in acid. Apatites are a class of calcium phosphate minerals (Figure $2$); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca5(PO4)3OH and dissolved Ca2+, $\ce{PO4^3-}$, and OH ions:
$\ce{Ca5(PO4)3OH}(s)⟶\ce{5Ca^2+}(aq)+\ce{3PO4^3-}(aq)+\ce{OH-}(aq) \nonumber$
When exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid:
$\ce{PO4^3-}(aq)+\ce{H3O+}⇌\ce{H2PO4^2-}+\ce{H2O} \nonumber$
$\ce{PO4^2-}(aq)+\ce{H3O+}⇌\ce{H2PO4-}+\ce{H2O} \nonumber$
$\ce{H2PO4- + H3O+ ⇌ H3PO4 + H2O} \nonumber$
Hydroxide ion reacts to form water:
$\ce{OH-}(aq)+\ce{H3O+}⇌\ce{2H2O} \nonumber$
These reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Chatelier’s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF2. They function by replacing the OH ion in hydroxylapatite with F ion, producing fluorapatite, Ca5(PO4)3F:
$\ce{NaF + Ca5(PO4)3OH ⇌ Ca5(PO4)3F + Na+ + OH-} \nonumber$
The resulting Ca5(PO4)3F is slightly less soluble than Ca5(PO4)3OH, and F is a weaker base than OH. Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.
Role of Fluoride in Preventing Tooth Decay
As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca5(PO4)3F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure $3$).
Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.
When acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid:
$\ce{H3O+}(aq)+\ce{CO3^2-}(aq)⟶\ce{HCO3-}(aq)+\ce{H2O}(l) \nonumber$
Calcium hydrogen carbonate, Ca(HCO3)2, is soluble, so limestone and marble objects slowly dissolve in acid rain.
If calcium carbonate is added to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation:
$\ce{2H3O+}(aq)+\ce{CO3^2-}(aq)⟶\ce{H2CO3}(aq)+\ce{2H2O}(l) \nonumber$
(Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved:
$\ce{H2CO3}(aq)⟶\ce{CO2}(g)+\ce{H2O}(l) \nonumber$
These reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved.
Example $1$: Prevention of Precipitation of Mg(OH)2
Calculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH)2 in a solution with [Mg2+] = 0.10 M and [NH3] = 0.10 M.
Solution
Two equilibria are involved in this system:
• Reaction (1): $\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq); \hspace{20px} K_\ce{sp}=8.9×10^{−12}$
• Reaction (2): $\ce{NH3}(aq)+\ce{H2O}(l)⇌\ce{NH4+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{sp}=1.8×10^{−5}$
To prevent the formation of solid Mg(OH)2, we must adjust the concentration of OH so that the reaction quotient for Equation (1), Q = [Mg2+][OH]2, is less than Ksp for Mg(OH)2. (To simplify the calculation, we determine the concentration of OH when Q = Ksp.) [OH] can be reduced by the addition of $\ce{NH4+}$, which shifts Reaction (2) to the left and reduces [OH].
1. We determine the [OH] at which Q = Ksp when [Mg2+] = 0.10 M: $Q=\ce{[Mg^2+][OH- ]^2}=(0.10)\ce{[OH- ]^2}=8.9×10^{−12} \nonumber$
$\ce{[OH- ]}=9.4×10^{−6}\:M \nonumber$
Solid Mg(OH)2 will not form in this solution when [OH] is less than 9.4 × 10–6 M.
• We calculate the $\mathit{[NH_4^+]}$ needed to decrease [OH] to 9.4 × 10–6 M when [NH3] = 0.10. $K_\ce{b}=\ce{\dfrac{[NH4+][OH- ]}{[NH3]}}=\dfrac{\ce{[NH4+]}(9.4×10^{−6})}{0.10}=1.8×10^{−5} \nonumber$ $\ce{[NH4+]}=0.19\:M \nonumber$
When $\ce{[NH4+]}$ equals 0.19 M, [OH] will be 9.4 × 10–6 M. Any $\ce{[NH4+]}$ greater than 0.19 M will reduce [OH] below 9.4 × 10–6 M and prevent the formation of Mg(OH)2.
Exercise $1$
Consider the two equilibria:
$\ce{ZnS}(s)⇌\ce{Zn^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=1×10^{−27} \nonumber$
$\ce{2H2O}(l)+\ce{H2S}(aq)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26} \nonumber$
and calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 M in Zn2+ and saturated with H2S (0.10 M H2S).
Answer
$\ce{[H3O+]}>0.2\:M \nonumber$
([S2–] is less than 2 × 10–26 M and precipitation of ZnS does not occur.)
Therefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions.
Example $2$: Multiple Equilibria
Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion $\ce{Ag(S2O3)2^3-}$ (Kf = 4.7 × 1013). The reaction with silver bromide is:
What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $\ce{Ag(S2O3)2^3-}$?
Solution
Two equilibria are involved when AgBr dissolves in a solution containing the $\ce{S2O3^2-}$ ion:
• Reaction (1): $\ce{AgBr}(s)⇌\ce{Ag+}(aq)+\ce{Br-}(aq) \hspace{20px} K_\ce{sp}=5.0×10^{−13}$
• Reaction (2): $\ce{Ag+}(aq)+\ce{S2O3^2-}(aq)⇌\ce{Ag(S2O3)2^3-}(aq) \hspace{20px} K_\ce{f}=4.7×10^{13}$
In order for 1.00 g of AgBr to dissolve, the [Ag+] in the solution that results must be low enough for Q for Reaction (1) to be smaller than Ksp for this reaction. We reduce [Ag+] by adding $\ce{S2O3^2-}$ and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na2S2O3 is needed to provide the necessary $\ce{S2O3^2-}$.
1. We calculate the [Br] produced by the complete dissolution of 1.00 g of AgBr (5.33 × 10–3 mol AgBr) in 1.00 L of solution: $\ce{[Br- ]}=5.33×10^{−3}\:M \nonumber$
• We use [Br] and Ksp to determine the maximum possible concentration of Ag+ that can be present without causing reprecipitation of AgBr: $\ce{[Ag+]}=9.4×10^{−11}\:M \nonumber$
• We determine the $\mathit{[S_2O_3^{2-}]}$ required to make [Ag+] = 9.4 × 10–11 M after the remaining Ag+ ion has reacted with $\mathit{S_2O_3^{2-}}$ according to the equation: $\ce{Ag+ + 2S2O3^2- ⇌ Ag(S2O3)2^3-} \hspace{20px} K_\ce{f}=4.7×10^{13} \nonumber$
Because 5.33 × 10–3 mol of AgBr dissolves:
$(5.33×10^{−3})−(9.4×10^{−11})=5.33×10^{−3}\ce{\:mol\:Ag(S2O3)2^3-}\ \nonumber ] Thus, at equilibrium: $\ce{[Ag(S2O3)2^3- ]}$ = 5.33 × 10–3 M, [Ag+] = 9.4× 10–11 M, and Q = Kf = 4.7 × 1013: \[K_\ce{f}=\ce{\dfrac{[Ag(S2O3)2^3- ]}{[Ag+][S2O3^2- ]^2}}=4.7×10^{13} \nonumber$ $\ce{[S2O3^2- ]}=1.1×10^{−3}\:M \nonumber$
When $\ce{[S2O3^2- ]}$ is 1.1 × 10–3 M, [Ag+] is 9.4 × 10–11 M and all AgBr remains dissolved.
• We determine the total number of moles of $\mathit{S_2O_3^{2-}}$ that must be added to the solution. This equals the amount that reacts with Ag+ to form $\ce{Ag(S2O3)2^3-}$ plus the amount of free $\ce{S2O3^2-}$ in solution at equilibrium. To form 5.33 × 10–3 mol of $\ce{Ag(S2O3)2^3-}$ requires 2 × (5.33 × 10–3) mol of $\ce{S2O3^2-}$. In addition, 1.1 × 10–3 mol of unreacted $\ce{S2O3^2-}$ is present (Step 3). Thus, the total amount of $\ce{S2O3^2-}$ that must be added is: $\mathrm{2×(5.33×10^{−3}mol\:S_2O_3^{2-}) + 1.1×10^{−3}\:mol\:S_2O_3^{2-} = 1.18×10^{−2}\: mol\:S_2O_3^{2-}} \nonumber$
• We determine the mass of Na2S2O3 required to give 1.18 × 10–2 mol $\mathit{S_2O_3^{2-}}$ using the molar mass of Na2S2O3: $\mathrm{1.18×10^{−2}\:mol\:S_2O_3^{2-}×\dfrac{158.1\:g\:Na_2S_2O_3}{1\:mol\:Na_2S_2O_3}=1.9\:g\:Na_2S_2O_3} \nonumber$
Thus, 1.00 L of a solution prepared from 1.9 g Na2S2O3 dissolves 1.0 g of AgBr.
Exercise $2$
AgCl(s), silver chloride, is well known to have a very low solubility: $\ce{Ag}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)$, Ksp = 1.6 × 10–10. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed: $\ce{Ag+}(aq)+\ce{2NH3}(aq)⇌\ce{Ag(NH3)2+}(aq)$, Kf = 1.7 × 107. What mass of NH3 is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of $\ce{Ag(NH3)2+}$?
Answer
1.00 L of a solution prepared with 4.81 g NH3 dissolves 2.0 g of AgCl.
Dissolution versus Weak Electrolyte Formation
We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Chatelier’s principle. For example, one way to control the concentration of manganese(II) ion, Mn2+, in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:
$\ce{Mn(OH)2}(s) ⇌ \ce{Mn^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Mn^2+][OH- ]^2} \nonumber$
This could be important to a laundry because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn2+ ion while increasing the amount of solid Mn(OH)2 in the equilibrium mixture, as predicted by Le Chatelier’s principle.
Example $3$: Solubility Equilibrium of a Slightly Soluble Solid
What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH when each of the following are added to a mixture of solid Mg(OH)2 in water at equilibrium?
1. MgCl2
2. KOH
3. an acid
4. NaNO3
5. Mg(OH)2
Solution
The equilibrium among solid Mg(OH)2 and a solution of Mg2+ and OH is:
$\ce{Mg(OH)2}(s) ⇌ \ce{Mg^2+}(aq)+\ce{2OH-}(aq) \nonumber$
(a) The reaction shifts to the left to relieve the stress produced by the additional Mg2+ ion, in accordance with Le Chatelier’s principle. In quantitative terms, the added Mg2+ causes the reaction quotient to be larger than the solubility product (Q > Ksp), and Mg(OH)2 forms until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is less and [Mg2+] is greater than in the solution of Mg(OH)2 in pure water. More solid Mg(OH)2 is present.
(b) The reaction shifts to the left to relieve the stress of the additional OH ion. Mg(OH)2 forms until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is greater and [Mg2+] is less than in the solution of Mg(OH)2 in pure water. More solid Mg(OH)2 is present.
(c) The concentration of OH is reduced as the OH reacts with the acid. The reaction shifts to the right to relieve the stress of less OH ion. In quantitative terms, the decrease in the OH concentration causes the reaction quotient to be smaller than the solubility product (Q < Ksp), and additional Mg(OH)2 dissolves until the reaction quotient again equals Ksp. At the new equilibrium, [OH] is less and [Mg2+] is greater than in the solution of Mg(OH)2 in pure water. More Mg(OH)2 is dissolved.
(d) NaNO3 contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg2+ and OH. (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.)
(e) The addition of solid Mg(OH)2 has no effect on the solubility of Mg(OH)2 or on the concentration of Mg2+ and OH. The concentration of Mg(OH)2 does not appear in the equation for the reaction quotient:
$Q=\ce{[Mg^2+][OH- ]^2} \nonumber$
Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.
Exercise $3$
What is the effect on the amount of solid NiCO3 that dissolves and the concentrations of Ni2+ and $\ce{CO3^2-}$ when each of the following are added to a mixture of the slightly soluble solid NiCO3 and water at equilibrium?
1. Ni(NO3)2
2. KClO4
3. NiCO3
4. K2CO3
5. HNO3 (reacts with carbonate giving $\ce{HCO3-}$ or H2O and CO2)
Answer
(a) mass of NiCO3(s) increases, [Ni2+] increases, $\ce{[CO3^2- ]}$ decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO3; (d) mass of NiCO3(s) increases, [Ni2+] decreases, $\ce{[CO3^2- ]}$ increases; (e) mass of NiCO3(s) decreases, [Ni2+] increases, $\ce{[CO3^2- ]}$ decreases
Summary
Several systems we encounter consist of multiple equilibria, systems where two or more equilibria processes are occurring simultaneously. Some common examples include acid rain, fluoridation, and dissolution of carbon dioxide in sea water. When looking at these systems, we need to consider each equilibrium separately and then combine the individual equilibrium constants into one solubility product or reaction quotient expression using the tools from the first equilibrium chapter. Le Chatelier’s principle also must be considered, as each reaction in a multiple equilibria system will shift toward reactants or products based on what is added to the initial reaction and how it affects each subsequent equilibrium reaction.
Glossary
multiple equilibrium
system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/15%3A_Equilibria_of_Other_Reaction_Classes/15.3%3A_Coupled_Equilibria.txt |
15.1: Precipitation and Dissolution
Q15.1.1
Complete the changes in concentrations for each of the following reactions:
\begin{alignat}{3} &\ce{AgI}(s)⟶&&\ce{Ag+}(aq)\,+\,&&\ce{I-}(aq)\ & &&x &&\underline{\hspace{45px}} \end{alignat}
\begin{alignat}{3} &\ce{CaCO3}(s)⟶&&\ce{Ca^2+}(aq)\,+\,&&\ce{CO3^2-}(aq)\ & &&\underline{\hspace{45px}} &&x \end{alignat}
\begin{alignat}{3} &\ce{Mg(OH)2}(s)⟶&&\ce{Mg^2+}(aq)\,+\,&&\ce{2OH-}(aq)\ & &&x &&\underline{\hspace{45px}} \end{alignat}
\begin{alignat}{3} &\ce{Mg3(PO4)2}(s)⟶&&\ce{3Mg^2+}(aq)\,+\,&&\ce{2PO4^3-}(aq)\ & && &&x\underline{\hspace{45px}} \end{alignat}
\begin{alignat}{3} &\ce{Ca5(PO4)3OH}(s)⟶&&\ce{5Ca^2+}(aq)\,+\,&&\ce{3PO4^3-}(aq)\,+\,&&\ce{OH-}(aq)\ & &&\underline{\hspace{45px}} &&\underline{\hspace{45px}} &&x \end{alignat}
S15.1.1
\begin{alignat}{3} &\ce{AgI}(s)⟶&&\ce{Ag+}(aq)\,+\,&&\ce{I-}(aq)\ & &&x &&\underline{x} \end{alignat}
\begin{alignat}{3} &\ce{CaCO3}(s)⟶&&\ce{Ca^2+}(aq)\,+\,&&\ce{CO3^2-}(aq)\ & &&\underline{x} &&x \end{alignat}
\begin{alignat}{3} &\ce{Mg(OH)2}(s)⟶&&\ce{Mg^2+}(aq)\,+\,&&\ce{2OH-}(aq)\ & &&x &&\underline{2x} \end{alignat}
\begin{alignat}{3} &\ce{Mg3(PO4)2}(s)⟶&&\ce{3Mg^2+}(aq)\,+\,&&\ce{2PO4^3-}(aq)\ & &&\underline{3x} &&2x \end{alignat}
\begin{alignat}{3} &\ce{Ca5(PO4)3OH}(s)⟶&&\ce{5Ca^2+}(aq)\,+\,&&\ce{3PO4^3-}(aq)\,+\,&&\ce{OH-}(aq)\ & &&\underline{5x} &&\underline{3x} &&x \end{alignat}
Q15.1.2
Complete the changes in concentrations for each of the following reactions:
\begin{alignat}{3} &\ce{BaSO4}(s)⟶&&\ce{Ba^2+}(aq)\,+\,&&\ce{SO4^2-}(aq)\ & &&x &&\underline{\hspace{45px}} \end{alignat}
\begin{alignat}{3} &\ce{Ag2SO4}(s)⟶&&\ce{2Ag+}(aq)\,+\,&&\ce{SO4^2-}(aq)\ & &&\underline{\hspace{45px}} &&x \end{alignat}
\begin{alignat}{3} &\ce{Al(OH)3}(s)⟶&&\ce{Al^3+}(aq)\,+\,&&\ce{3OH-}(aq)\ & &&x &&\underline{\hspace{45px}} \end{alignat}
\begin{alignat}{3} &\ce{Pb(OH)Cl}(s)⟶&&\ce{Pb^2+}(aq)\,+\,&&\ce{OH-}(aq)\,+\,&&\ce{Cl-}(aq)\ & &&\underline{\hspace{45px}} &&x &&\underline{\hspace{45px}} \end{alignat}
\begin{alignat}{3} &\ce{Ca3(AsO4)2}(s)⟶&&\ce{3Ca^2+}(aq)\,+\,&&\ce{2AsO4^3-}(aq)\ & &&3x &&\underline{\hspace{45px}} \end{alignat}
Q15.1.3
How do the concentrations of Ag+ and $\ce{CrO4^2-}$ in a saturated solution above 1.0 g of solid Ag2CrO4 change when 100 g of solid Ag2CrO4 is added to the system? Explain.
S15.1.3
There is no change. A solid has an activity of 1 whether there is a little or a lot.
Q15.1.4
How do the concentrations of Pb2+ and S2– change when K2S is added to a saturated solution of PbS?
Q15.1.5
What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?
S15.1.5
The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.
Q15.1.6
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO3, CuI, PbCO3, PbCl2, Tl2S, KClO4?
Q15.1.7
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO4, CaF2, Hg2I2, MnCO3, ZnS, PbS?
S15.1.7
CaF2, MnCO3, and ZnS
Q15.1.8
Write the ionic equation for dissolution and the solubility product (Ksp) expression for each of the following slightly soluble ionic compounds:
1. PbCl2
2. Ag2S
3. Sr3(PO4)2
4. SrSO4
Q15.1.9
Write the ionic equation for the dissolution and the Ksp expression for each of the following slightly soluble ionic compounds:
1. LaF3
2. CaCO3
3. Ag2SO4
4. Pb(OH)2
Q15.1.10
1. $\ce{LaF3}(s)⇌\ce{La^3+}(aq)+\ce{3F-}(aq) \hspace{20px} K_\ce{sp}=\ce{[La^3+][F- ]^3};$
2. $\ce{CaCO3}(s)⇌\ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ca^2+][CO3^2- ]};$
3. $\ce{Ag2SO4}(s)⇌\ce{2Ag+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Ag+]^2[SO4^2- ]};$
4. $\ce{Pb(OH)2}(s)⇌\ce{Pb^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Pb^2+][OH- ]^2}$
Q15.1.11
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
1. BaSiF6, 0.026 g/100 mL (contains $\ce{SiF6^2-}$ ions)
2. Ce(IO3)4, 1.5 × 10–2 g/100 mL
3. Gd2(SO4)3, 3.98 g/100 mL
4. (NH4)2PtBr6, 0.59 g/100 mL (contains $\ce{PtBr6^2-}$ ions)
Q15.1.12
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
1. BaSeO4, 0.0118 g/100 mL
2. Ba(BrO3)2•H2O, 0.30 g/100 mL
3. NH4MgAsO4•6H2O, 0.038 g/100 mL
4. La2(MoO4)3, 0.00179 g/100 mL
S15.1.12
(a)1.77 × 10–7; 1.6 × 10–6; 2.2 × 10–9; 7.91 × 10–22
Q15.1.13
Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF2, Hg2Cl2, PbI2, or Sn(OH)2.
Q15.1.14
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
1. KHC4H4O6
2. PbI2
3. Ag4[Fe(CN)6], a salt containing the $\ce{Fe(CN)4-}$ ion
4. Hg2I2
S15.1.15
2 × 10–2 M; 1.3 × 10–3 M; 2.27 × 10–9 M; 2.2 × 10–10 M
Q15.1.16
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
1. Ag2SO4
2. PbBr2
3. AgI
4. CaC2O4•H2O
Q15.1.X
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
1. AgCl(s) in 0.025 M NaCl
2. CaF2(s) in 0.00133 M KF
3. Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4
4. Zn(OH)2(s) in a solution buffered at a pH of 11.45
S15.1.X
7.2 × 10−9 M = [Ag+], [Cl] = 0.025 M
Check: $\dfrac{7.2×10^{−9}\:M}{0.025\:M}×100\%=2.9×10^{−5}\%$, an insignificant change;
2.2 × 10−5 M = [Ca2+], [F] = 0.0013 M
Check: $\dfrac{2.25×10^{−5}\:M}{0.00133\:M}×100\%=1.69\%$. This value is less than 5% and can be ignored.
0.2238 M = $\ce{[SO4^2- ]}$; [Ag+] = 2.30 × 10–9 M
Check: $\dfrac{1.15×10^{−9}}{0.2238}×100\%=5.14×10^{−7}$; the condition is satisfied.
[OH] = 2.8 × 10–3 M; 5.7 × 10−12 M = [Zn2+]
Check: $\dfrac{5.7×10^{−12}}{2.8×10^{−3}}×100\%=2.0×10^{−7}\%$; x is less than 5% of [OH] and is, therefore, negligible.
Q15.1.X
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
1. TlCl(s) in 1.250 M HCl
2. PbI2(s) in 0.0355 M CaI2
3. Ag2CrO4(s) in 0.225 L of a solution containing 0.856 g of K2CrO4
4. Cd(OH)2(s) in a solution buffered at a pH of 10.995
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.
1. TlCl(s) in 0.025 M TlNO3
2. BaF2(s) in 0.0313 M KF
3. MgC2O4 in 2.250 L of a solution containing 8.156 g of Mg(NO3)2
4. Ca(OH)2(s) in an unbuffered solution initially with a pH of 12.700
S15.1.X
• [Cl] = 7.6 × 10−3 M
Check: $\dfrac{7.6×10^{−3}}{0.025}×100\%=30\%$
This value is too large to drop x. Therefore solve by using the quadratic equation:
• [Ti+] = 3.1 × 10–2 M
• [Cl] = 6.1 × 10–3
• [Ba2+] = 1.7 × 10–3 M
Check: $\dfrac{1.7×10^{−3}}{0.0313}×100\%=5.5\%$
This value is too large to drop x, and the entire equation must be solved.
• [Ba2+] = 1.6 × 10–3 M
• [F] = 0.0329 M;
• Mg(NO3)2 = 0.02444 M
$\ce{[C2O4^2- ]}=3.5×10^{−3}$
Check: $\dfrac{3.5×10^{−3}}{0.02444}×100\%=14\%$
This value is greater than 5%, so the quadratic equation must be used:
• $\ce{[C2O4^2- ]}=3.5×10^{−3}\:M$
• [Mg2+] = 0.0275 M
• [OH] = 0.0501 M
• [Ca2+] = 3.15 × 10–3
Check: $\dfrac{3.15×10^{−3}}{0.050}×100\%=6.28\%$
This value is greater than 5%, so a more exact method, such as successive approximations, must be used.
• [Ca2+] = 2.8 × 10–3 M
• [OH] = 0.053 × 10–2 M
Q15.1.X
Explain why the changes in concentrations of the common ions in Exercise can be neglected.
Q15.1.X
Explain why the changes in concentrations of the common ions in Exercise cannot be neglected.
S15.1.X
The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.
Q15.1.X
Calculate the solubility of aluminum hydroxide, Al(OH)3, in a solution buffered at pH 11.00.
Q15.1.X
Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.
S15.1.X
CaSO4∙2H2O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L.
Q15.1.X
Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract. This use of BaSO4 is possible because of its low solubility. Calculate the molar solubility of BaSO4 and the mass of barium present in 1.00 L of water saturated with BaSO4.
Q15.1.X
Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 × 10–3 M) of $\ce{SO4^2-}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO4 (“gyp” water) as a result or passing through soil containing gypsum, CaSO4•2H2O, meet these standards? What is $\ce{SO4^2-}$ in such water?
S15.1.X
4.9 × 10–3 M = $\ce{[SO4^2- ]}$ = [Ca2+]; Since this concentration is higher than 2.60 × 10–3 M, “gyp” water does not meet the standards.
Q15.1.X
Perform the following calculations:
1. Calculate [Ag+] in a saturated aqueous solution of AgBr.
2. What will [Ag+] be when enough KBr has been added to make [Br] = 0.050 M?
3. What will [Br] be when enough AgNO3 has been added to make [Ag+] = 0.020 M?
The solubility product of CaSO4•2H2O is 2.4 × 10–5. What mass of this salt will dissolve in 1.0 L of 0.010 M $\ce{SO4^2-}$?
S15.1.X
Mass (CaSO4•2H2O) = 0.34 g/L
Q15.1.X
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products).
1. TlCl
2. BaF2
3. Ag2CrO4
4. CaC2O4•H2O
5. the mineral anglesite, PbSO4
Q15.1.X
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products).
1. AgI
2. Ag2SO4
3. Mn(OH)2
4. Sr(OH)2•8H2O
5. the mineral brucite, Mg(OH)2
S15.1.X
[Ag+] = [I] = 1.2 × 10–8 M; [Ag+] = 2.86 × 10–2 M, $\ce{[SO4^2- ]}$ = 1.43 × 10–2 M; [Mn2+] = 2.2 × 10–5 M, [OH] = 4.5 × 10–5 M; [Sr2+] = 4.3 × 10–2 M, [OH] = 8.6 × 10–2 M; [Mg2+] = 1.6 × 10–4 M, [OH] = 3.1 × 10–4 M.
Q15.1.X
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:
1. AgBr: [Ag+] = 5.7 × 10–7 M, [Br] = 5.7 × 10–7 M
2. CaCO3: [Ca2+] = 5.3 × 10–3 M, $\ce{[CO3^2- ]}$ = 9.0 × 10–7 M
3. PbF2: [Pb2+] = 2.1 × 10–3 M, [F] = 4.2 × 10–3 M
4. Ag2CrO4: [Ag+] = 5.3 × 10–5 M, 3.2 × 10–3 M
5. InF3: [In3+] = 2.3 × 10–3 M, [F] = 7.0 × 10–3 M
Q15.1.X
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:
1. TlCl: [Tl+] = 1.21 × 10–2 M, [Cl] = 1.2 × 10–2 M
2. Ce(IO3)4: [Ce4+] = 1.8 × 10–4 M, $\ce{[IO3- ]}$ = 2.6 × 10–13 M
3. Gd2(SO4)3: [Gd3+] = 0.132 M, $\ce{[SO4^2- ]}$ = 0.198 M
4. Ag2SO4: [Ag+] = 2.40 × 10–2 M, $\ce{[SO4^2- ]}$ = 2.05 × 10–2 M
5. BaSO4: [Ba2+] = 0.500 M, $\ce{[SO4^2- ]}$ = 2.16 × 10–10 M
S15.1.X
2.0 × 10–4; 5.1 × 10–17; 1.35 × 10–4; 1.18 × 10–5; 1.08 × 10–10
Q15.1.X
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for Ksp values.)
1. KClO4: [K+] = 0.01 M, $\ce{[ClO4- ]}$ = 0.01 M
2. K2PtCl6: [K+] = 0.01 M, $\ce{[PtCl6^2- ]}$ = 0.01 M
3. PbI2: [Pb2+] = 0.003 M, [I] = 1.3 × 10–3 M
4. Ag2S: [Ag+] = 1 × 10–10 M, [S2–] = 1 × 10–13 M
Q15.1.X
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for Ksp values.)
1. CaCO3: [Ca2+] = 0.003 M, $\ce{[CO3^2- ]}$ = 0.003 M
2. Co(OH)2: [Co2+] = 0.01 M, [OH] = 1 × 10–7 M
3. CaHPO4: [Ca2+] = 0.01 M, $\ce{[HPO4^2- ]}$ = 2 × 10–6 M
4. Pb3(PO4)2: [Pb2+] = 0.01 M, $\ce{[PO4^3- ]}$ = 1 × 10–13 M
S15.1.X
1. CaCO3 does precipitate.
2. The compound does not precipitate.
3. The compound does not precipitate.
4. The compound precipitates.
Q15.1.X
Calculate the concentration of Tl+ when TlCl just begins to precipitate from a solution that is 0.0250 M in Cl.
Q15.1.X
Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.
1.42 × 10−9 M
Q15.1.X
Calculate the concentration of Sr2+ when SrF2 starts to precipitate from a solution that is 0.0025 M in F.
Q15.1.X
Calculate the concentration of $\ce{PO4^3-}$ when Ag3PO4 starts to precipitate from a solution that is 0.0125 M in Ag+.
9.2 × 10−13 M
Q15.1.X
Calculate the concentration of F required to begin precipitation of CaF2 in a solution that is 0.010 M in Ca2+.
Q15.1.X
Calculate the concentration of Ag+ required to begin precipitation of Ag2CO3 in a solution that is 2.50 × 10–6 M in $\ce{CO3^2-}$.
S15.1.X
[Ag+] = 1.8 × 10–3 M
Q15.1.X
What [Ag+] is required to reduce $\ce{[CO3^2- ]}$ to 8.2 × 10–4 M by precipitation of Ag2CO3?
Q15.1.X
What [F] is required to reduce [Ca2+] to 1.0 × 10–4 M by precipitation of CaF2?
6.2 × 10–4
Q15.1.X
A volume of 0.800 L of a 2 × 10–4-M Ba(NO3)2 solution is added to 0.200 L of 5 × 10–4 M Li2SO4. Does BaSO4 precipitate? Explain your answer.
Q15.1.X
Perform these calculations for nickel(II) carbonate.
1. With what volume of water must a precipitate containing NiCO3 be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO3 (Ksp = 1.36 × 10–7).
2. If the NiCO3 were a contaminant in a sample of CoCO3 (Ksp = 1.0 × 10–12), what mass of CoCO3 would have been lost? Keep in mind that both NiCO3 and CoCO3 dissolve in the same solution.
S15.1.X
2.28 L; 7.3 × 10–7 g
Q15.1.X
Iron concentrations greater than 5.4 × 10–6 M in water used for laundry purposes can cause staining. What [OH] is required to reduce [Fe2+] to this level by precipitation of Fe(OH)2?
Q15.1.X
A solution is 0.010 M in both Cu2+ and Cd2+. What percentage of Cd2+ remains in the solution when 99.9% of the Cu2+ has been precipitated as CuS by adding sulfide?
S15.1.X
100% of it is dissolved
Q15.1.X
A solution is 0.15 M in both Pb2+ and Ag+. If Cl is added to this solution, what is [Ag+] when PbCl2 begins to precipitate?
Q15.1.X
What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the Ksp values given in Appendix J.)
1. $\ce{Hg2^2+}$ and Cu2+
2. $\ce{SO4^2-}$ and Cl
3. Hg2+ and Co2+
4. Zn2+ and Sr2+
5. Ba2+ and Mg2+
6. $\ce{CO3^2-}$ and OH
S15.1.X
1. $\ce{Hg2^2+}$ and Cu2+: Add $\ce{SO4^2-}$.
2. $\ce{SO4^2-}$ and Cl: Add Ba2+.
3. Hg2+ and Co2+: Add S2–.
4. Zn2+ an Sr2+: Add OH until [OH] = 0.050 M.
5. Ba2+ and Mg2+: Add $\ce{SO4^2-}$.
6. $\ce{CO3^2-}$ and OH: Add Ba2+.
Q15.1.X
A solution contains 1.0 × 10–5 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
Q15.1.X
A solution contains 1.0 × 10–2 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?
S15.1.X
AgI will precipitate first.
Q15.1.X
The calcium ions in human blood serum are necessary for coagulation. Potassium oxalate, K2C2O4, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC2O4•H2O. It is necessary to remove all but 1.0% of the Ca2+ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca2+ per 100 mL of serum, what mass of K2C2O4 is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the Ksp value for CaC2O4 in serum is the same as in water.)
Q15.1.X
About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca3(PO4)2. The normal mid range calcium content excreted in the urine is 0.10 g of Ca2+ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?
4 × 10−9 M
Q15.1.X
The pH of normal urine is 6.30, and the total phosphate concentration ($\ce{[PO4^3- ]}$ + $\ce{[HPO4^2- ]}$ + $\ce{[H2PO4- ]}$ + [H3PO4]) is 0.020 M. What is the minimum concentration of Ca2+ necessary to induce kidney stone formation? (See Exercise for additional information.)
Q15.1.X
Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:
$\ce{Mg^2+}(aq)+\ce{Ca(OH)2}(aq)⟶\ce{Mg(OH)2}(s)+\ce{Ca^2+}(aq)$
$\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)⟶\ce{MgCl2}(s)+\ce{2H2O}(l)$
$\ce{MgCl2}(l)\xrightarrow{\ce{electrolysis}}\ce{Mg}(s)+\ce{Cl2}(g)$
Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium as Mg2+(aq) by mass. What mass, in kilograms, of Ca(OH)2 is required to precipitate 99.9% of the magnesium in 1.00 × 103 L of sea water?
3.99 kg
Q15.1.X
Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb2+ and Fe2+ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H2S ([H2S] = 0.10 M). What concentrations of Pb2+ and Fe2+ remain in the solution? For a saturated solution of H2S we can use the equilibrium:
$\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}$
(Hint: The $\ce{[H3O+]}$ changes as metal sulfides precipitate.)
Q15.1.X
Perform the following calculations involving concentrations of iodate ions:
1. The iodate ion concentration of a saturated solution of La(IO3)3 was found to be 3.1 × 10–3 mol/L. Find the Ksp.
2. Find the concentration of iodate ions in a saturated solution of Cu(IO3)2 (Ksp = 7.4 × 10–8).
S15.1.X
3.1 × 10–11; [Cu2+] = 2.6 × 10–3; $\ce{[IO3- ]}$ = 5.3 × 10–3
Q15.1.X
Calculate the molar solubility of AgBr in 0.035 M NaBr (Ksp = 5 × 10–13).
Q15.1.X
How many grams of Pb(OH)2 will dissolve in 500 mL of a 0.050-M PbCl2 solution (Ksp = 1.2 × 10–15)?
S15.1.X
1.8 × 10–5 g Pb(OH)2
Q15.1.X
Use the simulation from the earlier Link to Learning to complete the following exercise:. Using 0.01 g CaF2, give the Ksp values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.
Q15.1.X
How many grams of Milk of Magnesia, Mg(OH)2 (s) (58.3 g/mol), would be soluble in 200 mL of water. Ksp = 7.1 × 10–12. Include the ionic reaction and the expression for Ksp in your answer. (Kw = 1 × 10–14 = [H3O+][OH])
S15.1.X
$\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}+\ce{2OH-}$
$K_\ce{sp}=\ce{[Mg^2+][OH- ]^2}$
$1.14 × 10−3 g Mg(OH)2$
Q15.1.X
Two hypothetical salts, LM2 and LQ, have the same molar solubility in H2O. If Ksp for LM2 is 3.20 × 10–5, what is the Ksp value for LQ?
Q15.1.X
Which of the following carbonates will form first? Which of the following will form last? Explain.
1. $\ce{MgCO3} \hspace{20px} K_\ce{sp}=3.5×10^{−8}$
2. $\ce{CaCO3} \hspace{20px} K_\ce{sp}=4.2×10^{−7}$
3. $\ce{SrCO3} \hspace{20px} K_\ce{sp}=3.9×10^{−9}$
4. $\ce{BaCO3} \hspace{20px} K_\ce{sp}=4.4×10^{−5}$
5. $\ce{MnCO3} \hspace{20px} K_\ce{sp}=5.1×10^{−9}$
S15.1.X
SrCO3 will form first, since it has the smallest Ksp value it is the least soluble. BaCO3 will be the last to precipitate, it has the largest Ksp value.
Q15.1.X
How many grams of Zn(CN)2(s) (117.44 g/mol) would be soluble in 100 mL of H2O? Include the balanced reaction and the expression for Ksp in your answer. The Ksp value for Zn(CN)2(s) is 3.0 × 10–16.
15.2: Lewis Acids and Bases
Q15.2.X
Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
S15.2.X
when the amount of solid is so small that a saturated solution is not produced
Q15.2.X
Explain why the addition of NH3 or HNO3 to a saturated solution of Ag2CO3 in contact with solid Ag2CO3 increases the solubility of the solid.
Q15.2.X
Calculate the cadmium ion concentration, [Cd2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3)2 with 1.150 L of 0.100 NH3(aq).
2.35 × 10–4 M
Q15.2.X
Explain why addition of NH3 or HNO3 to a saturated solution of Cu(OH)2 in contact with solid Cu(OH)2 increases the solubility of the solid.
S15.2.X
Sometimes equilibria for complex ions are described in terms of dissociation constants, Kd. For the complex ion $\ce{AlF6^3-}$ the dissociation reaction is:
5 × 1023
Q15.2.X
Using the value of the formation constant for the complex ion $\ce{Co(NH3)6^2+}$, calculate the dissociation constant.
Q15.2.X
Using the dissociation constant, Kd = 7.8 × 10–18, calculate the equilibrium concentrations of Cd2+ and CN in a 0.250-M solution of $\ce{Cd(CN)4^2-}$.
S15.2.X
[Cd2+] = 9.5 × 10–5 M; [CN] = 3.8 × 10–4 M
Q15.2.X
Using the dissociation constant, Kd = 3.4 × 10–15, calculate the equilibrium concentrations of Zn2+ and OH in a 0.0465-M solution of $\ce{Zn(OH)4^2-}$.
Q15.2.X
Using the dissociation constant, Kd = 2.2 × 10–34, calculate the equilibrium concentrations of Co3+ and NH3 in a 0.500-M solution of $\ce{Co(NH3)6^3+}$.
S15.2.X
[Co3+] = 3.0 × 10–6 M; [NH3] = 1.8 × 10–5 M
Q15.2.X
Using the dissociation constant, Kd = 1 × 10–44, calculate the equilibrium concentrations of Fe3+ and CN in a 0.333 M solution of $\ce{Fe(CN)6^3-}$.
Q15.2.X
Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 × 10–2 mol of silver cyanide, AgCN.
1.3 g
Q15.2.X
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 × 10–3 mol of silver bromide.
Q15.2.X
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na2S2O3•5H2O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $\ce{Ag(S2O3)2^3-}$ (Kf = 4.7 × 1013)?
0.80 g
Q15.2.X
We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H+) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
Q15.2.X
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
1. $\ce{CO2 + OH- ⟶ HCO3-}$
2. $\ce{B(OH)3 + OH- ⟶ B(OH)4-}$
3. $\ce{I- + I2 ⟶ I3-}$
4. $\ce{AlCl3 + Cl- ⟶ AlCl4-}$ (use Al-Cl single bonds)
5. $\ce{O^2- + SO3 ⟶ SO4^2-}$
(a)
;
(b)
;
(c)
;
(d)
;
(e)
Q15.2.X
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
1. $\ce{CS2 + SH- ⟶ HCS3-}$
2. $\ce{BF3 + F- ⟶ BF4-}$
3. $\ce{I- + SnI2 ⟶ SnI3-}$
4. $\ce{Al(OH)3 + OH- ⟶ Al(OH)4-}$
5. $\ce{F- + SO3 ⟶ SFO3-}$
Q15.2.X
Using Lewis structures, write balanced equations for the following reactions:
1. $\ce{HCl}(g)+\ce{PH3}(g)⟶$
2. $\ce{H3O+ + CH3- ⟶}$
3. $\ce{CaO + SO3 ⟶}$
4. $\ce{NH4+ + C2H5O- ⟶}$
S15.2.X
(a)
;
$\ce{H3O+ + CH3- ⟶ CH4 + H2O}$
;
$\ce{CaO + SO3 ⟶ CaSO4}$
;
$\ce{NH4+ + C2H5O- ⟶ C2H5OH + NH3}$
Q15.2.X
Calculate $\ce{[HgCl4^2- ]}$ in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl2 solution.
Q15.2.X
In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN goes to completion, producing the $\ce{Ag(CN)2-}$ complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of $\ce{Ag(CN)2-}$. How many grams of NaCN were in the original sample?
0.0281 g
Q15.2.X
What are the concentrations of Ag+, CN, and $\ce{Ag(CN)2-}$ in a saturated solution of AgCN?
Q15.2.X
In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO3 acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F ion (for example, BF3 or SbF5). Write balanced chemical equations for the reaction of pure HNO3 with pure HF and of pure HF with BF3.
S15.2.X
$\ce{HNO3}(l)+\ce{HF}(l)⟶\ce{H2NO3+}+\ce{F-}$; $\ce{HF}(l)+\ce{BF3}(g)⟶\ce{H+}+\ce{BF4}$
Q15.2.X
The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH2, and a carboxylic acid group, –CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3.
Q15.2.X
Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.
Q15.2.X
Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH2 and $\ce{−CO2-}$ groups.)
Q15.2.X
Boric acid, H3BO3, is not a Brønsted-Lowry acid but a Lewis acid.
1. Write an equation for its reaction with water.
2. Predict the shape of the anion thus formed.
3. What is the hybridization on the boron consistent with the shape you have predicted?
S15.2.X
$\ce{H3BO3 + H2O ⟶ H4BO4- + H+}$; The electronic and molecular shapes are the same—both tetrahedral. The tetrahedral structure is consistent with sp3 hybridization.
15.3: Multiple Equilibria
Q15.3.1
A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?
Q15.3.2
Calculate the equilibrium concentration of Ni2+ in a 1.0-M solution [Ni(NH3)6](NO3)2.
0.014 M
Q15.3.3
Calculate the equilibrium concentration of Zn2+ in a 0.30-M solution of $\ce{Zn(CN)4^2-}$.
Q15.3.4
Calculate the equilibrium concentration of Cu2+ in a solution initially with 0.050 M Cu2+ and 1.00 M NH3.
1.0 × 10–13 M
Q15.3.5
Calculate the equilibrium concentration of Zn2+ in a solution initially with 0.150 M Zn2+ and 2.50 M CN.
Q15.3.6
Calculate the Fe3+ equilibrium concentration when 0.0888 mole of K3[Fe(CN)6] is added to a solution with 0.0.00010 M CN.
9 × 10−22 M
Q15.3.7
Calculate the Co2+ equilibrium concentration when 0.100 mole of [Co(NH3)6](NO3)2 is added to a solution with 0.025 M NH3. Assume the volume is 1.00 L.
Q15.3.8
The equilibrium constant for the reaction $\ce{Hg^2+}(aq)+\ce{2Cl-}(aq)⇌\ce{HgCl2}(aq)$ is 1.6 × 1013. Is HgCl2 a strong electrolyte or a weak electrolyte? What are the concentrations of Hg2+ and Cl in a 0.015-M solution of HgCl2?
S15.3.2
6.2 × 10–6 M = [Hg2+]; 1.2 × 10–5 M = [Cl]; The substance is a weak electrolyte because very little of the initial 0.015 M HgCl2 dissolved.
Q15.3.9
Calculate the molar solubility of Sn(OH)2 in a buffer solution containing equal concentrations of NH3 and $\ce{NH4+}$.
Q15.3.X
Calculate the molar solubility of Al(OH)3 in a buffer solution with 0.100 M NH3 and 0.400 M $\ce{NH4+}$.
S15.3.2
[OH] = 4.5 × 10−5; [Al3+] = 2.1 × 10–20 (molar solubility)
Q15.3.10
What is the molar solubility of CaF2 in a 0.100-M solution of HF? Ka for HF = 7.2 × 10–4.
Q15.3.X
What is the molar solubility of BaSO4 in a 0.250-M solution of NaHSO4? Ka for $\ce{HSO4-}$ = 1.2 × 10–2.
S15.3.2
• $\ce{[SO4^2- ]}=0.049\:M$
• [Ba2+] = 2.2 × 10–9 (molar solubility)
Q15.3.X
What is the molar solubility of Tl(OH)3 in a 0.10-M solution of NH3?
Q15.3.X
What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?
S15.3.2
• [OH] = 7.6 × 10−3 M
• [Pb2+] = 4.8 × 10–12 (molar solubility)
Q15.3.X
A solution of 0.075 M CoBr2 is saturated with H2S ([H2S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?
$\ce{CoS}(s)⇌\ce{Co^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.5×10^{−27}$
$\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}$
A 0.125-M solution of Mn(NO3)2 is saturated with H2S ([H2S] = 0.10 M). At what pH does MnS begin to precipitate?
$\ce{MnS}(s)⇌\ce{Mn^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.3×10^{−22}$
$\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}$
3.27
Q15.3.X
Calculate the molar solubility of BaF2 in a buffer solution containing 0.20 M HF and 0.20 M NaF.
Q15.3.X
Calculate the molar solubility of CdCO3 in a buffer solution containing 0.115 M Na2CO3 and 0.120 M NaHCO3
S15.3.2
• $\ce{[CO3^2- ]}=0.115\:M$
• [Cd2+] = 3 × 10−12 M
Q15.3.X
To a 0.10-M solution of Pb(NO3)2 is added enough HF(g) to make [HF] = 0.10 M.
1. Does PbF2 precipitate from this solution? Show the calculations that support your conclusion.
2. What is the minimum pH at which PbF2 precipitates?
Calculate the concentration of Cd2+ resulting from the dissolution of CdCO3 in a solution that is 0.010 M in H2CO3.
1 × 10−5 M
Q15.3.X
Both AgCl and AgI dissolve in NH3.
1. What mass of AgI dissolves in 1.0 L of 1.0 M NH3?
2. What mass of AgCl dissolves in 1.0 L of 1.0 M NH3?
Calculate the volume of 1.50 M CH3CO2H required to dissolve a precipitate composed of 350 mg each of CaCO3, SrCO3, and BaCO3.
S15.3.2
0.0102 L (10.2 mL)
Q15.3.X
Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)2?
Q15.3.X
What mass of NaCN must be added to 1 L of 0.010 M Mg(NO3)2 in order to produce the first trace of Mg(OH)2?
5 × 10−3 g
Q15.3.X
Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)
Q15.3.X
The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
Solve the following problem:
$\ce{MgF2}(s)⇌\ce{Mg^2+}(aq)+\ce{2F-}(aq)$
In a saturated solution of MgF2 at 18 °C, the concentration of Mg2+ is 1.21 × 10–3 M. The equilibrium is represented by the preceding equation.
1. Write the expression for the solubility-product constant, Ksp, and calculate its value at 18 °C.
2. Calculate the equilibrium concentration of Mg2+ in 1.000 L of saturated MgF2 solution at 18 °C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.
3. Predict whether a precipitate of MgF2 will form when 100.0 mL of a 3.00 × 10–3-M solution of Mg(NO3)2 is mixed with 200.0 mL of a 2.00 × 10–3-M solution of NaF at 18 °C. Show the calculations to support your prediction.
4. At 27 °C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 × 10–3 M. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.
S15.3.2
Ksp = [Mg2+][F]2 = (1.21 × 10–3)(2 × 1.21 × 10–3)2 = 7.09 × 10–9; 7.09 × 10–7 M
Determine the concentration of Mg2+ and F that will be present in the final volume. Compare the value of the ion product [Mg2+][F]2 with Ksp. If this value is larger than Ksp, precipitation will occur. 0.1000 L × 3.00 × 10–3 M Mg(NO3)2 = 0.3000 L × M Mg(NO3)2 M Mg(NO3)2 = 1.00 × 10–3 M 0.2000 L × 2.00 × 10–3 M NaF = 0.3000 L × M NaF M NaF = 1.33 × 10–3 M ion product = (1.00 × 10–3)(1.33 × 10–3)2 = 1.77 × 10–9 This value is smaller than Ksp, so no precipitation will occur. MgF2 is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Chatelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.
Q15.3.X
Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: CuCl, CaCO3, MnS, PbBr2, CaF2? Explain your answer.
Q15.3.X
Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: AgBr, BaF2, Ca3(PO4)3, ZnS, PbI2? Explain your answer.
BaF2, Ca3(PO4)2, ZnS; each is a salt of a weak acid, and the $\ce{[H3O+]}$ from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations
Q15.3.X
What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH when each of the following are added to a mixture of solid Mg(OH)2 and water at equilibrium?
1. MgCl2
2. KOH
3. HClO4
4. NaNO3
5. Mg(OH)2
Q15.3.X
What is the effect on the amount of CaHPO4 that dissolves and the concentrations of Ca2+ and $\ce{HPO4-}$ when each of the following are added to a mixture of solid CaHPO4 and water at equilibrium?
1. CaCl2
2. HCl
3. KClO4
4. NaOH
5. CaHPO4
S15.3.X
Effect on amount of solid CaHPO4, [Ca2+], [OH]: increase, increase, decrease; decrease, increase, decrease; no effect, no effect, no effect; decrease, increase, decrease; increase, no effect, no effect
Q15.3.X
Identify all chemical species present in an aqueous solution of Ca3(PO4)2 and list these species in decreasing order of their concentrations. (Hint: Remember that the $\ce{PO4^3-}$ ion is a weak base.)
Q15.3.X
A volume of 50 mL of 1.8 M NH3 is mixed with an equal volume of a solution containing 0.95 g of MgCl2. What mass of NH4Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)2?
7.1 g | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/15%3A_Equilibria_of_Other_Reaction_Classes/15.E%3A_Equilibria_of_Other_Reaction_Classes_%28Exercises%29.txt |
Electrochemistry deals with chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. The reactions involve electron transfer, and so they are oxidation-reduction (or redox) reactions. Many metals may be purified or electroplated using electrochemical methods. Devices such as automobiles, smartphones, electronic tablets, watches, pacemakers, and many others use batteries for power. Batteries use chemical reactions that produce electricity spontaneously and that can be converted into useful work. All electrochemical systems involve the transfer of electrons in a reacting system. In many systems, the reactions occur in a region known as the cell, where the transfer of electrons occurs at electrodes.
16: Electrochemistry
Learning Objectives
• Define electrochemistry and a number of important associated terms
• Split oxidation-reduction reactions into their oxidation half-reactions and reduction half-reactions
• Produce balanced oxidation-reduction equations for reactions in acidic or basic solution
• Identify oxidizing agents and reducing agents
Electricity refers to a number of phenomena associated with the presence and flow of electric charge. Electricity includes such diverse things as lightning, static electricity, the current generated by a battery as it discharges, and many other influences on our daily lives. The flow or movement of charge is an electric current (Figure $1$). Electrons or ions may carry the charge. The elementary unit of charge is the charge of a proton, which is equal in magnitude to the charge of an electron. The SI unit of charge is the coulomb (C) and the charge of a proton is 1.602 × 10−19 C. The presence of an electric charge generates an electric field. Electric current is the rate of flow of charge.
The SI unit for electrical current is the SI base unit called the ampere (A), which is a flow rate of 1 coulomb of charge per second (1 A = 1 C/s). An electric current flows in a path, called an electric circuit. In most chemical systems, it is necessary to maintain a closed path for current to flow. The flow of charge is generated by an electrical potential difference, or potential, between two points in the circuit. Electrical potential is the ability of the electric field to do work on the charge. The SI unit of electrical potential is the volt (V). When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule (J) of energy. Table $1$ summarizes some of this information about electricity.
Table $1$: Common Electrical Terms
Quantity Definition Measure or Unit
Electric charge Charge on a proton 1.602 × 10−19 C
Electric current The movement of charge ampere = A = 1 C/s
Electric potential The force trying to move the charge volt = V = J/C
Electric field The force acting upon other charges in the vicinity
Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced chemical equation were relatively simple. In this section, we will concentrate on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important partly for balancing more complicated reactions and partly because many aspects of electrochemistry are easier to discuss in terms of half-reactions. There are alternate methods of balancing these reactions; however, there are no good alternatives to half-reactions for discussing what is occurring in many systems. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier.
Electrochemical reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. It helps to see this in an actual problem. Consider the following unbalanced oxidation-reduction reaction in acidic solution:
$\ce{MnO4-}(aq)+\ce{Fe^2+}(aq)⟶\ce{Mn^2+}(aq)+\ce{Fe^3+}(aq) \nonumber$
We can start by collecting the species we have so far into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction. Each of these half-reactions contain the same element in two different oxidation states. The Fe2+ has lost an electron to become Fe3+; therefore, the iron underwent oxidation. The reduction is not as obvious; however, the manganese gained five electrons to change from Mn7+ to Mn2+.
\begin{align*} &\textrm{oxidation (unbalanced): }\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)\ &\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)⟶\ce{Mn^2+}(aq) \end{align*} \nonumber
In acidic solution, there are hydrogen ions present, which are often useful in balancing half-reactions. It may be necessary to use the hydrogen ions directly or as a reactant that may react with oxygen to generate water. Hydrogen ions are very important in acidic solutions where the reactants or products contain hydrogen and/or oxygen. In this example, the oxidation half-reaction involves neither hydrogen nor oxygen, so hydrogen ions are not necessary to the balancing. However, the reduction half-reaction does involve oxygen. It is necessary to use hydrogen ions to convert this oxygen to water.
$\textrm{charge not balanced: }\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
The situation is different in basic solution because the hydrogen ion concentration is lower and the hydroxide ion concentration is higher. After finishing this example, we will examine how basic solutions differ from acidic solutions. A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.
The iron atoms in the oxidation half-reaction are balanced (mass balance); however, the charge is unbalanced, since the charges on the ions are not equal. It is necessary to use electrons to balance the charge. The way to balance the charge is by adding electrons to one side of the equation. Adding a single electron on the right side gives a balanced oxidation half-reaction:
$\textrm{oxidation (balanced): }\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-} \nonumber$
You should check the half-reaction for the number of each atom type and the total charge on each side of the equation. The charges include the actual charges of the ions times the number of ions and the charge on an electron times the number of electrons.
\begin{align*} &\textrm{Fe: } \mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{Charge: } \mathrm{Does\:[1×(+2)]=[1×(+3)+1×(−1)]?\:Yes.} \end{align*} \nonumber
If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent.
Now return to the reduction half-reaction equation:
$\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) × (1) + (8) × (+1)], or +7, while the total charge on the right side is [(1) × (+2) + (4) × (0)], or +2. The difference between +7 and +2 is five; therefore, it is necessary to add five electrons to the left side to achieve charge balance.
$\textrm{Reduction (balanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
You should check this half-reaction for each atom type and for the charge, as well:
\begin{align*} &\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(8×1) =(4×2)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×4) =(4×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?\:Yes.} \end{align*} \nonumber
Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when $\ce{MnO4-}$ was reduced to Mn2+. In all reduction half-reactions, electrons appear as reactants (on the left side). As discussed in the earlier chapter, the species that was reduced, $\ce{MnO4-}$ in this case, is also called the oxidizing agent. We now have two balanced half-reactions.
\begin{align*} &\textrm{oxidation: }\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-}\ &\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \end{align*} \nonumber
It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives
\begin{align*} &\textrm{oxidation: }5×(\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-})\ &\underline{\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)}\ &\textrm{overall: }\ce{5Fe^2+}(aq)+\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{5Fe^3+}(aq)+\ce{Mn^2+}(aq)+\ce{4H2O}(l) \end{align*} \nonumber
The electrons do not appear in the final answer because the oxidation electrons are the same electrons as the reduction electrons and they “cancel.” Carefully check each side of the overall equation to verify everything was combined correctly:
\begin{align*} &\textrm{Fe: }\mathrm{Does\:(5×1)=(5×1)?\:Yes.}\ &\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(8×1)=(4×2)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×4)=(4×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[5×(+2)+1×(−1)+8×(+1)]=[5×(+3)+1×(+2)]?\:Yes.} \end{align*} \nonumber
Everything checks, so this is the overall equation in acidic solution. If something does not check, the most common error occurs during the multiplication of the individual half-reactions.
Now suppose we wanted the solution to be basic. Recall that basic solutions have excess hydroxide ions. Some of these hydroxide ions will react with hydrogen ions to produce water. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. However, it is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions and many metal ions will precipitate as the metal hydroxide. We just produced the following reaction, which we want to change to a basic reaction:
$\ce{5Fe^2+}(aq)+\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{5Fe^3+}(aq)+\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
However, under basic conditions, $\ce{MnO4-}$ normally reduces to MnO2 and iron will be present as either Fe(OH)2 or Fe(OH)3. For these reasons, under basic conditions, this reaction will be
$\ce{3Fe(OH)2}(s)+\ce{MnO4-}(aq)+\ce{2H2O}(l)⟶\ce{3Fe(OH)3}(s)+\ce{MnO2}(s)+\ce{OH-}(aq) \nonumber$
(Under very basic conditions $\ce{MnO4-}$ will reduce to $\ce{MnO4^2-}$, instead of MnO2.)
It is still possible to balance any oxidation-reduction reaction as an acidic reaction and then, when necessary, convert the equation to a basic reaction. This will work if the acidic and basic reactants and products are the same or if the basic reactants and products are used before the conversion from acidic or basic. There are very few examples in which the acidic and basic reactions will involve the same reactants and products. However, balancing a basic reaction as acidic and then converting to basic will work. To convert to a basic reaction, it is necessary to add the same number of hydroxide ions to each side of the equation so that all the hydrogen ions (H+) are removed and mass balance is maintained. Hydrogen ion combines with hydroxide ion (OH) to produce water.
Let us now try a basic equation. We will start with the following basic reaction:
$\ce{Cl-}(aq)+\ce{MnO4-}(aq)⟶\ce{ClO3-}(aq)+\ce{MnO2}(s) \nonumber$
Balancing this as acid gives
$\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{2H+}(aq)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{H2O}(l) \nonumber$
In this case, it is necessary to add two hydroxide ions to each side of the equation to convert the two hydrogen ions on the left into water:
$\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{(2H+ + 2OH- )}(aq)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{H2O}(l)+\ce{2OH-}(aq) \nonumber$
$\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{(2H2O)}(l)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{H2O}(l)+\ce{2OH-}(aq) \nonumber$
Note that both sides of the equation show water. Simplifying should be done when necessary, and gives the desired equation. In this case, it is necessary to remove one H2O from each side of the reaction arrows.
$\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{H2O}(l)⟶\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{2OH-}(aq) \nonumber$
Again, check each side of the overall equation to make sure there are no errors:
\begin{align*} &\textrm{Cl: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{Mn: }\mathrm{Does\:(2×1)=(2×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(1×2)=(2×1)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(2×4+1×1)=(3×1+2×2+2×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[1×(−1)+2×(−1)]=[1×(−1)+2×(−1)]?\:Yes.} \end{align*} \nonumber
Everything checks, so this is the overall equation in basic solution.
Example $1$: Balancing Acidic Oxidation-Reduction Reactions
Balance the following reaction equation in acidic solution:
$\ce{MnO4-}(aq)+\ce{Cr^3+}(aq)⟶\ce{Mn^2+}(aq)+\ce{Cr2O7^2-}(aq) \nonumber$
Solution
This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction.
\begin{align*} &\textrm{oxidation (unbalanced): }\ce{Cr^3+}(aq)⟶\ce{Cr2O7^2-}(aq)\ &\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)⟶\ce{Mn^2+}(aq) \end{align*} \nonumber
Starting with the oxidation half-reaction, we can balance the chromium
$\textrm{oxidation (unbalanced): }\ce{2Cr^3+}(aq)⟶\ce{Cr2O7^2-}(aq) \nonumber$
In acidic solution, we can use or generate hydrogen ions (H+). Adding seven water molecules to the left side provides the necessary oxygen; the “left over” hydrogen appears as 14 H+ on the right:
$\textrm{oxidation (unbalanced): }\ce{2Cr^3+}(aq)+\ce{7H2O}(l)⟶\ce{Cr2O7^2-}(aq)+\ce{14H+}(aq) \nonumber$
The left side of the equation has a total charge of [2 × (+3) = +6], and the right side a total charge of [−2 + 14 × (+1) = +12]. The difference is six; adding six electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):
$\textrm{oxidation (balanced): }\ce{2Cr^3+}(aq)+\ce{7H2O}(l)⟶\ce{Cr2O7^2-}(aq)+\ce{14H+}(aq)+\ce{6e-} \nonumber$
Checking the half-reaction:
\begin{align*} &\textrm{Cr: }\mathrm{Does\:(2×1)=(1×2)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(7×2)=(14×1)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(7×1) =(1×7)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[2×(+3)]=[1×(−2)+14×(+1)+6×(−1)]?\:Yes.} \end{align*} \nonumber
Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H+ to convert the oxygen into four water molecules:
$\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
Then add five electrons to the left side to balance the charge:
$\textrm{reduction (balanced): }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
Make sure to check the half-reaction:
\begin{align*} &\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(8×1)=(4×2)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×4)=(4×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[1×(−1)+8×(+1)+5×(−1)]=[1×(+2)]?\:Yes.} \end{align*} \nonumber
Collecting what we have so far:
\begin{align*} &\textrm{oxidation: }\ce{2Cr^3+}(aq)+\ce{7H2O}(l)⟶\ce{Cr2O7^2-}(aq)+\ce{14H+}(aq)+\ce{6e-}\ &\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l) \end{align*} \nonumber
The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:
$\ce{10Cr^3+}(aq)+\ce{35H2O}(l)+\ce{6MnO4-}(aq)+\ce{48H+}(aq)⟶\ce{5Cr2O7^2-}(aq)+\ce{70H+}(aq)+\ce{6Mn^2+}(aq)+\ce{24H2O}(l) \nonumber$
$\ce{10Cr^3+}(aq)+\ce{11H2O}(l)+\ce{6MnO4-}(aq)⟶\ce{5Cr2O7^2-}(aq)+\ce{22H+}(aq)+\ce{6Mn^2+}(aq) \nonumber$
Checking each side of the equation:
\begin{align*} &\textrm{Mn: }\mathrm{Does\:(6×1)=(6×1)?\:Yes.}\ &\textrm{Cr: }\mathrm{Does\:(10×1)=(5×2)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(11×2)=(22×1)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(11×1+6×4)=(5×7)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[10×(+3)+6×(−1)]=[5×(−2)+22×(+1)+6×(+2)]?\:Yes.} \end{align*} \nonumber
This is the balanced equation in acidic solution.
Exercise $1$
Balance the following equation in acidic solution:
$\ce{Hg2^2+ + Ag ⟶ Hg + Ag+} \nonumber$
Answer
$\ce{Hg2^2+}(aq)+\ce{2Ag}(s)⟶\ce{2Hg}(l)+\ce{2Ag+}(aq) \nonumber$
Example $2$: Balancing Basic Oxidation-Reduction Reactions
Balance the following reaction equation in basic solution:
$\ce{MnO4-}(aq)+\ce{Cr(OH)3}(s)⟶\ce{MnO2}(s)+\ce{CrO4^2-}(aq) \nonumber$
Solution
This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction
\begin{align*} &\textrm{oxidation (unbalanced): }\ce{Cr(OH)3}(s)⟶\ce{CrO4^2-}(aq)\ &\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)⟶\ce{MnO2}(s) \end{align*} \nonumber
Starting with the oxidation half-reaction, we can balance the chromium
$\textrm{oxidation (unbalanced): }\ce{Cr(OH)3}(s)⟶\ce{CrO4^2-}(aq) \nonumber$
In acidic solution, we can use or generate hydrogen ions (H+). Adding one water molecule to the left side provides the necessary oxygen; the “left over” hydrogen appears as five H+ on the right side:
$\textrm{oxidation (unbalanced): }\ce{Cr(OH)3}(s)+\ce{H2O}(l)⟶\ce{CrO4^2-}(aq)+\ce{5H+}(aq) \nonumber$
The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 × (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):
$\textrm{oxidation (balanced): }\ce{Cr(OH)3}(s)+\ce{H2O}(l)⟶\ce{CrO4^2-}(aq)+\ce{5H+}(aq)+\ce{3e-} \nonumber$
Checking the half-reaction:
\begin{align*} &\textrm{Cr: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(1×3+1×2)=(5×1)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×3+1×1)=(4×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does \:[0=[1×(−2)+5×(+1)+3×(−1)]?\:Yes.} \end{align*} \nonumber
Now work on the reduction. It is necessary to convert the four O atoms in the MnO4 minus the two O atoms in MnO2 into two water molecules. To do this, add four H+ to convert the oxygen into two water molecules:
$\textrm{reduction (unbalanced): }\ce{MnO4-}(aq)+\ce{4H+}(aq)⟶\ce{MnO2}(s)+\ce{2H2O}(l) \nonumber$
Then add three electrons to the left side to balance the charge:
$\textrm{reduction (balanced): }\ce{MnO4-}(aq)+\ce{4H+}(aq)+\ce{3e-}⟶\ce{MnO2}(s)+\ce{2H2O}(l) \nonumber$
Make sure to check the half-reaction:
\begin{align*} &\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(4×1)=(2×2)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×4)=(1×2+2×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[1×(−1)+4×(+1)+3×(−1)]=[0]?\:Yes.} \end{align*} \nonumber
Collecting what we have so far:
\begin{align*} &\textrm{oxidation: }\ce{Cr(OH)3}(s)+\ce{H2O}(l)⟶\ce{CrO4^2-}(aq)+\ce{5H+}(aq)+\ce{3e-}\ &\textrm{reduction: }\ce{MnO4-}(aq)+\ce{4H+}(aq)+\ce{3e-}⟶\ce{MnO2}(s)+\ce{2H2O}(l) \end{align*} \nonumber
In this case, both half reactions involve the same number of electrons; therefore, simply add the two half-reactions together.
$\ce{MnO4-}(aq)+\ce{4H+}(aq)+\ce{Cr(OH)3}(s)+\ce{H2O}(l)⟶\ce{CrO4^2-}(aq)+\ce{MnO2}(s)+\ce{2H2O}(l)+\ce{5H+}(aq) \nonumber$
$\ce{MnO4-}(aq)+\ce{Cr(OH)3}(s)⟶\ce{CrO4^2-}(aq)+\ce{MnO2}(s)+\ce{H2O}(l)+\ce{H+}(aq) \nonumber$
Checking each side of the equation:
\begin{align*} &\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{Cr: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(1×3)=(2×1+1×1)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×4+1×3)=(1×4+1×2+1×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[1×(−1)]=[1×(−2)+1×(+1)]?\:Yes.} \end{align*} \nonumber
This is the balanced equation in acidic solution. For a basic solution, add one hydroxide ion to each side and simplify:
$\ce{OH-}(aq)+\ce{MnO4-}(aq)+\ce{Cr(OH)3}(s)⟶\ce{CrO4^2-}(aq)+\ce{MnO2}(s)+\ce{H2O}(l)+\ce{(H+ + OH- )}(aq) \nonumber$
$\ce{OH-}(aq)+\ce{MnO4-}(aq)+\ce{Cr(OH)3}(s)⟶\ce{CrO4^2-}(aq)+\ce{MnO2}(s)+\ce{2H2O}(l) \nonumber$
Checking each side of the equation:
\begin{align*} &\textrm{Mn: }\mathrm{Does\:(1×1)=(1×1)?\: Yes.}\ &\textrm{Cr: }\mathrm{Does\:(1×1)=(1×1)?\:Yes.}\ &\textrm{H: }\mathrm{Does\:(1×1+1×3)=(2×2)?\:Yes.}\ &\textrm{O: }\mathrm{Does\:(1×1+1×4+1×3)=(1×4+1×2+2×1)?\:Yes.}\ &\textrm{Charge: }\mathrm{Does\:[1×(−1)+1×(−1)]=[1×(−2)]?\:Yes.} \end{align*} \nonumber
This is the balanced equation in basic solution.
Exercise $2$
Balance the following in the type of solution indicated.
1. $\ce{H2 + Cu^2+ ⟶ Cu \:\:\:(acidic\: solution)}$
2. $\ce{H2 + Cu(OH)2 ⟶ Cu\:\:\:(basic\: solution)}$
3. $\ce{Fe + Ag+ ⟶ Fe^2+ + Ag}$
4. Identify the oxidizing agents in reactions (a), (b), and (c).
5. Identify the reducing agents in reactions (a), (b), and (c).
Answer a
$\ce{H2}(g)+\ce{Cu^2+}(aq)⟶\ce{2H+}(aq)+\ce{Cu}(s)$
Answer b
$\ce{H2}(g)+\ce{Cu(OH)2}(s)⟶\ce{2H2O}(l)+\ce{Cu}(s)$
Answer c
$\ce{Fe}(s)+\ce{2Ag+}(aq)⟶\ce{Fe^2+}(aq)+\ce{2Ag}(s)$
Answer d
oxidizing agent = species reduced: Cu2+, Cu(OH)2, Ag+
Answer e
reducing agent = species oxidized: H2, H2, Fe.
Summary
An electric current consists of moving charge. The charge may be in the form of electrons or ions. Current flows through an unbroken or closed circular path called a circuit. The current flows through a conducting medium as a result of a difference in electrical potential between two points in a circuit. Electrical potential has the units of energy per charge. In SI units, charge is measured in coulombs (C), current in amperes $\mathrm{\left(A=\dfrac{C}{s}\right)}$, and electrical potential in volts $\mathrm{\left(V=\dfrac{J}{C}\right)}$.
Oxidation is the loss of electrons, and the species that is oxidized is also called the reducing agent. Reduction is the gain of electrons, and the species that is reduced is also called the oxidizing agent. Oxidation-reduction reactions can be balanced using the half-reaction method. In this method, the oxidation-reduction reaction is split into an oxidation half-reaction and a reduction half-reaction. The oxidation half-reaction and reduction half-reaction are then balanced separately. Each of the half-reactions must have the same number of each type of atom on both sides of the equation and show the same total charge on each side of the equation. Charge is balanced in oxidation half-reactions by adding electrons as products; in reduction half-reactions, charge is balanced by adding electrons as reactants. The total number of electrons gained by reduction must exactly equal the number of electrons lost by oxidation when combining the two half-reactions to give the overall balanced equation. Balancing oxidation-reduction reaction equations in aqueous solutions frequently requires that oxygen or hydrogen be added or removed from a reactant. In acidic solution, hydrogen is added by adding hydrogen ion (H+) and removed by producing hydrogen ion; oxygen is removed by adding hydrogen ion and producing water, and added by adding water and producing hydrogen ion. A balanced equation in basic solution can be obtained by first balancing the equation in acidic solution, and then adding hydroxide ion to each side of the balanced equation in such numbers that all the hydrogen ions are converted to water.
Glossary
circuit
path taken by a current as it flows because of an electrical potential difference
current
flow of electrical charge; the SI unit of charge is the coulomb (C) and current is measured in amperes $\mathrm{\left(1\: A=1\:\dfrac{C}{s}\right)}$
electrical potential
energy per charge; in electrochemical systems, it depends on the way the charges are distributed within the system; the SI unit of electrical potential is the volt $\mathrm{\left(1\: V=1\:\dfrac{J}{C}\right)}$
half-reaction method
method that produces a balanced overall oxidation-reduction reaction by splitting the reaction into an oxidation “half” and reduction “half,” balancing the two half-reactions, and then combining the oxidation half-reaction and reduction half-reaction in such a way that the number of electrons generated by the oxidation is exactly canceled by the number of electrons required by the reduction
oxidation half-reaction
the “half” of an oxidation-reduction reaction involving oxidation; the half-reaction in which electrons appear as products; balanced when each atom type, as well as the charge, is balanced
reduction half-reaction
the “half” of an oxidation-reduction reaction involving reduction; the half-reaction in which electrons appear as reactants; balanced when each atom type, as well as the charge, is balanced | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.1%3A_Balancing_Oxidation-Reduction_Reactions.txt |
Learning Objectives
• Use cell notation to describe galvanic cells
• Describe the basic components of galvanic cells
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.
Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate (Figure $1$). As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. The reaction may be split into its two half-reactions. Half-reactions separate the oxidation from the reduction, so each can be considered individually.
\begin{align} &\textrm{oxidation: }\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2e-}\ &\underline{\textrm{reduction: }2×(\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s))\hspace{40px}\ce{or}\hspace{40px}\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)}\ &\textrm{overall: }\ce{2Ag+}(aq)+\ce{Cu}(s)⟶\ce{2Ag}(s)+\ce{Cu^2+}(aq) \end{align} \nonumber
The equation for the reduction half-reaction had to be doubled so the number electrons “gained” in the reduction half-reaction equaled the number of electrons “lost” in the oxidation half-reaction.
Galvanic or voltaic cells involve spontaneous electrochemical reactions in which the half-reactions are separated (Figure $2$) so that current can flow through an external wire. The beaker on the left side of the figure is called a half-cell, and contains a 1 M solution of copper(II) nitrate [Cu(NO3)2] with a piece of copper metal partially submerged in the solution. The copper metal is an electrode. The copper is undergoing oxidation; therefore, the copper electrode is the anode. The anode is connected to a voltmeter with a wire and the other terminal of the voltmeter is connected to a silver electrode by a wire. The silver is undergoing reduction; therefore, the silver electrode is the cathode. The half-cell on the right side of the figure consists of the silver electrode in a 1 M solution of silver nitrate (AgNO3). At this point, no current flows—that is, no significant movement of electrons through the wire occurs because the circuit is open. The circuit is closed using a salt bridge, which transmits the current with moving ions. The salt bridge consists of a concentrated, nonreactive, electrolyte solution such as the sodium nitrate (NaNO3) solution used in this example. As electrons flow from left to right through the electrode and wire, nitrate ions (anions) pass through the porous plug on the left into the copper(II) nitrate solution. This keeps the beaker on the left electrically neutral by neutralizing the charge on the copper(II) ions that are produced in the solution as the copper metal is oxidized. At the same time, the nitrate ions are moving to the left, sodium ions (cations) move to the right, through the porous plug, and into the silver nitrate solution on the right. These added cations “replace” the silver ions that are removed from the solution as they were reduced to silver metal, keeping the beaker on the right electrically neutral. Without the salt bridge, the compartments would not remain electrically neutral and no significant current would flow. However, if the two compartments are in direct contact, a salt bridge is not necessary. The instant the circuit is completed, the voltmeter reads +0.46 V, this is called the cell potential. The cell potential is created when the two dissimilar metals are connected, and is a measure of the energy per unit charge available from the oxidation-reduction reaction. The volt is the derived SI unit for electrical potential
$\mathrm{volt=\mathit{V}=\dfrac{J}{C}} \nonumber$
In this equation, A is the current in amperes and C the charge in coulombs. Note that volts must be multiplied by the charge in coulombs (C) to obtain the energy in joules (J).
When the electrochemical cell is constructed in this fashion, a positive cell potential indicates a spontaneous reaction and that the electrons are flowing from the left to the right. There is a lot going on in Figure $2$, so it is useful to summarize things for this system:
• Electrons flow from the anode to the cathode: left to right in the standard galvanic cell in the figure.
• The electrode in the left half-cell is the anode because oxidation occurs here. The name refers to the flow of anions in the salt bridge toward it.
• The electrode in the right half-cell is the cathode because reduction occurs here. The name refers to the flow of cations in the salt bridge toward it.
• Oxidation occurs at the anode (the left half-cell in the figure).
• Reduction occurs at the cathode (the right half-cell in the figure).
• The cell potential, +0.46 V, in this case, results from the inherent differences in the nature of the materials used to make the two half-cells.
• The salt bridge must be present to close (complete) the circuit and both an oxidation and reduction must occur for current to flow.
There are many possible galvanic cells, so a shorthand notation is usually used to describe them. The cell notation (sometimes called a cell diagram) provides information about the various species involved in the reaction. This notation also works for other types of cells. A vertical line, │, denotes a phase boundary and a double line, ‖, the salt bridge. Information about the anode is written to the left, followed by the anode solution, then the salt bridge (when present), then the cathode solution, and, finally, information about the cathode to the right. The cell notation for the galvanic cell in Figure $2$ is then
$\ce{Cu}(s)│\ce{Cu^2+}(aq,\: 1\:M)║\ce{Ag+}(aq,\: 1\:M)│\ce{Ag}(s) \nonumber$
Note that spectator ions are not included and that the simplest form of each half-reaction was used. When known, the initial concentrations of the various ions are usually included.
One of the simplest cells is the Daniell cell. It is possible to construct this battery by placing a copper electrode at the bottom of a jar and covering the metal with a copper sulfate solution. A zinc sulfate solution is floated on top of the copper sulfate solution; then a zinc electrode is placed in the zinc sulfate solution. Connecting the copper electrode to the zinc electrode allows an electric current to flow. This is an example of a cell without a salt bridge, and ions may flow across the interface between the two solutions.
Some oxidation-reduction reactions involve species that are poor conductors of electricity, and so an electrode is used that does not participate in the reactions. Frequently, the electrode is platinum, gold, or graphite, all of which are inert to many chemical reactions. One such system is shown in Figure $3$. Magnesium undergoes oxidation at the anode on the left in the figure and hydrogen ions undergo reduction at the cathode on the right. The reaction may be summarized as
\begin{align} &\textrm{oxidation: }\ce{Mg}(s)⟶\ce{Mg^2+}(aq)+\ce{2e-}\ &\textrm{reduction: }\ce{2H+}(aq)+\ce{2e-}⟶\ce{H2}(g)\ &\overline{\textrm{overall: }\ce{Mg}(s)+\ce{2H+}(aq)⟶\ce{Mg^2+}(aq)+\ce{H2}(g)} \end{align} \nonumber
The cell used an inert platinum wire for the cathode, so the cell notation is
$\ce{Mg}(s)│\ce{Mg^2+}(aq)║\ce{H+}(aq)│\ce{H2}(g)│\ce{Pt}(s) \nonumber$
The magnesium electrode is an active electrode because it participates in the oxidation-reduction reaction. Inert electrodes, like the platinum electrode in Figure $3$, do not participate in the oxidation-reduction reaction and are present so that current can flow through the cell. Platinum or gold generally make good inert electrodes because they are chemically unreactive.
Example $1$: Using Cell Notation
Consider a galvanic cell consisting of
$\ce{2Cr}(s)+\ce{3Cu^2+}(aq)⟶\ce{2Cr^3+}(aq)+\ce{3Cu}(s) \nonumber$
Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Cr is oxidized when three electrons are lost to form Cr3+, and Cu2+ is reduced as it gains two electrons to form Cu. Balancing the charge gives
\begin{align} &\textrm{oxidation: }\ce{2Cr}(s)⟶\ce{2Cr^3+}(aq)+\ce{6e-}\ &\textrm{reduction: }\ce{3Cu^2+}(aq)+\ce{6e-}⟶\ce{3Cu}(s)\ &\overline{\textrm{overall: }\ce{2Cr}(s)+\ce{3Cu^2+}(aq)⟶\ce{2Cr^3+}(aq)+\ce{3Cu}(s)} \end{align} \nonumber
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so:
$\ce{Cr}(s)│\ce{Cr^3+}(aq)║\ce{Cu^2+}(aq)│\ce{Cu}(s). \nonumber$
Oxidation occurs at the anode and reduction at the cathode.
Example $2$: Using Cell Notation
Consider a galvanic cell consisting of
$\ce{5Fe^2+}(aq)+\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{5Fe^3+}(aq)+\ce{Mn^2+}(aq)+\ce{4H2O}(l) \nonumber$
Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Fe2+ undergoes oxidation when one electron is lost to form Fe3+, and MnO4 is reduced as it gains five electrons to form Mn2+. Balancing the charge gives
\begin{align} &\textrm{oxidation: }5(\ce{Fe^2+}(aq)⟶\ce{Fe^3+}(aq)+\ce{e-})\ &\underline{\textrm{reduction: }\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)}\ &\textrm{overall: }\ce{5Fe^2+}(aq)+\ce{MnO4-}(aq)+\ce{8H+}(aq)⟶\ce{5Fe^3+}(aq)+\ce{Mn^2+}(aq)+\ce{4H2O}(l) \end{align} \nonumber
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode. No concentrations were specified so:
$\ce{Pt}(s)│\ce{Fe^2+}(aq),\: \ce{Fe^3+}(aq)║\ce{MnO4-}(aq),\: \ce{H+}(aq),\: \ce{Mn^2+}(aq)│\ce{Pt}(s). \nonumber$
Oxidation occurs at the anode and reduction at the cathode.
Exercise $1$
Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.
Answer
From the information given in the problem:
\begin{align} &\textrm{anode (oxidation): }\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{2e-}\ &\textrm{cathode (reduction): }\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\ &\overline{\textrm{overall: }\ce{Zn}(s)+\ce{Cu^2+}(aq)⟶\ce{Zn^2+}(aq)+\ce{Cu}(s)} \end{align} \nonumber
Using cell notation:
$\ce{Zn}(s)│\ce{Zn^2+}(aq)║\ce{Cu^2+}(aq)│\ce{Cu}(s) \nonumber. \nonumber$
Summary
Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire. One half-cell, normally depicted on the left side in a figure, contains the anode. Oxidation occurs at the anode. The anode is connected to the cathode in the other half-cell, often shown on the right side in a figure. Reduction occurs at the cathode. Adding a salt bridge completes the circuit allowing current to flow. Anions in the salt bridge flow toward the anode and cations in the salt bridge flow toward the cathode. The movement of these ions completes the circuit and keeps each half-cell electrically neutral. Electrochemical cells can be described using cell notation. In this notation, information about the reaction at the anode appears on the left and information about the reaction at the cathode on the right. The salt bridge is represented by a double line, ‖. The solid, liquid, or aqueous phases within a half-cell are separated by a single line, │. The phase and concentration of the various species is included after the species name. Electrodes that participate in the oxidation-reduction reaction are called active electrodes. Electrodes that do not participate in the oxidation-reduction reaction but are there to allow current to flow are inert electrodes. Inert electrodes are often made from platinum or gold, which are unchanged by many chemical reactions.
Glossary
active electrode
electrode that participates in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidation-reduction reaction
anode
electrode in an electrochemical cell at which oxidation occurs; information about the anode is recorded on the left side of the salt bridge in cell notation
cathode
electrode in an electrochemical cell at which reduction occurs; information about the cathode is recorded on the right side of the salt bridge in cell notation
cell notation
shorthand way to represent the reactions in an electrochemical cell
cell potential
difference in electrical potential that arises when dissimilar metals are connected; the driving force for the flow of charge (current) in oxidation-reduction reactions
galvanic cell
electrochemical cell that involves a spontaneous oxidation-reduction reaction; electrochemical cells with positive cell potentials; also called a voltaic cell
inert electrode
electrode that allows current to flow, but that does not otherwise participate in the oxidation-reduction reaction in an electrochemical cell; the mass of an inert electrode does not change during the oxidation-reduction reaction; inert electrodes are often made of platinum or gold because these metals are chemically unreactive.
voltaic cell
another name for a galvanic cell | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.2%3A_Galvanic_Cells.txt |
Learning Objectives
• Determine standard cell potentials for oxidation-reduction reactions
• Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices
The cell potential results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in Figure 17.4.1 and is called the standard hydrogen electrode (SHE). The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is
$\ce{2H+}(aq,\: 1\:M)+\ce{2e-}⇌\ce{H2}(g,\:1\: \ce{atm}) \hspace{20px} E°=\mathrm{0\: V} \nonumber$
E° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). The voltage is defined as zero for all temperatures.
A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (Figure $2$). In cell notation, the reaction is
$\ce{Pt}(s)│\ce{H2}(g,\:1\: \ce{atm})│\ce{H+}(aq,\:1\:M)║\ce{Cu^2+}(aq,\:1\:M)│\ce{Cu}(s) \nonumber$
Electrons flow from the anode to the cathode. The reactions, which are reversible, are
\begin{align*} &\textrm{Anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq) + \ce{2e-}\ &\textrm{Cathode (reduction): }\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\ &\overline{\textrm{Overall: }\ce{Cu^2+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{Cu}(s)} \end{align*} \nonumber
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction.
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode} \nonumber$
$\mathrm{+0.34\: V}=E^\circ_{\ce{Cu^2+/Cu}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Cu^2+/Cu}}−0=E^\circ_{\ce{Cu^2+/Cu}} \nonumber$
Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure $2$, where
$\ce{Pt}(s)│\ce{H2}(g,\:1\: \ce{atm})│\ce{H+}(aq,\: 1\:M)║\ce{Ag+}(aq,\: 1\:M)│\ce{Ag}(s) \nonumber$
Electrons flow from left to right, and the reactions are
\begin{align*} &\textrm{anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2e-}\ &\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\ &\overline{\textrm{overall: }\ce{2Ag+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2Ag}(s)} \end{align*} \nonumber
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode} \nonumber$
$\mathrm{+0.80\: V}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Ag+/Ag}}−0=E^\circ_{\ce{Ag+/Ag}} \nonumber$
It is important to note that the potential is not doubled for the cathode reaction.
The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential, $E^\circ_\ce{cell}$, for any cell. For example, for the following cell:
$\ce{Cu}(s)│\ce{Cu^2+}(aq,\:1\:M)║\ce{Ag+}(aq,\:1\:M)│\ce{Ag}(s) \nonumber$
\begin{align*} &\textrm{anode (oxidation): }\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2e-}\ &\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\ &\overline{\textrm{overall: }\ce{Cu}(s)+\ce{2Ag+}(aq)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)} \end{align*} \nonumber
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{Cu^2+/Cu}}=\mathrm{0.80\: V−0.34\: V=0.46\: V} \nonumber$
Again, note that when calculating $E^\circ_\ce{cell}$, standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in Table $1$. A more complete list is provided in Tables P1 or P2.
Table $1$: Selected Standard Reduction Potentials at 25 °C
Half-Reaction E° (V)
$\ce{F2}(g)+\ce{2e-}⟶\ce{2F-}(aq)$ +2.866
$\ce{PbO2}(s)+\ce{SO4^2-}(aq)+\ce{4H+}(aq)+\ce{2e-}⟶\ce{PbSO4}(s)+\ce{2H2O}(l)$ +1.69
$\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)$ +1.507
$\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s)$ +1.498
$\ce{Cl2}(g)+\ce{2e-}⟶\ce{2Cl-}(aq)$ +1.35827
$\ce{O2}(g)+\ce{4H+}(aq)+\ce{4e-}⟶\ce{2H2O}(l)$ +1.229
$\ce{Pt^2+}(aq)+\ce{2e-}⟶\ce{Pt}(s)$ +1.20
$\ce{Br2}(aq)+\ce{2e-}⟶\ce{2Br-}(aq)$ +1.0873
$\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s)$ +0.7996
$\ce{Hg2^2+}(aq)+\ce{2e-}⟶\ce{2Hg}(l)$ +0.7973
$\ce{Fe^3+}(aq)+\ce{e-}⟶\ce{Fe^2+}(aq)$ +0.771
$\ce{MnO4-}(aq)+\ce{2H2O}(l)+\ce{3e-}⟶\ce{MnO2}(s)+\ce{4OH-}(aq)$ +0.558
$\ce{I2}(s)+\ce{2e-}⟶\ce{2I-}(aq)$ +0.5355
$\ce{NiO2}(s)+\ce{2H2O}(l)+\ce{2e-}⟶\ce{Ni(OH)2}(s)+\ce{2OH-}(aq)$ +0.49
$\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)$ +0.34
$\ce{Hg2Cl2}(s)+\ce{2e-}⟶\ce{2Hg}(l)+\ce{2Cl-}(aq)$ +0.26808
$\ce{AgCl}(s)+\ce{e-}⟶\ce{Ag}(s)+\ce{Cl-}(aq)$ +0.22233
$\ce{Sn^4+}(aq)+\ce{2e-}⟶\ce{Sn^2+}(aq)$ +0.151
$\ce{2H+}(aq)+\ce{2e-}⟶\ce{H2}(g)$ 0.00
$\ce{Pb^2+}(aq)+\ce{2e-}⟶\ce{Pb}(s)$ −0.1262
$\ce{Sn^2+}(aq)+\ce{2e-}⟶\ce{Sn}(s)$ −0.1375
$\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s)$ −0.257
$\ce{Co^2+}(aq)+\ce{2e-}⟶\ce{Co}(s)$ −0.28
$\ce{PbSO4}(s)+\ce{2e-}⟶\ce{Pb}(s)+\ce{SO4^2-}(aq)$ −0.3505
$\ce{Cd^2+}(aq)+\ce{2e-}⟶\ce{Cd}(s)$ −0.4030
$\ce{Fe^2+}(aq)+\ce{2e-}⟶\ce{Fe}(s)$ −0.447
$\ce{Cr^3+}(aq)+\ce{3e-}⟶\ce{Cr}(s)$ −0.744
$\ce{Mn^2+}(aq)+\ce{2e-}⟶\ce{Mn}(s)$ −1.185
$\ce{Zn(OH)2}(s)+\ce{2e-}⟶\ce{Zn}(s)+\ce{2OH-}(aq)$ −1.245
$\ce{Zn^2+}(aq)+\ce{2e-}⟶\ce{Zn}(s)$ −0.7618
$\ce{Al^3+}(aq)+\ce{3e-}⟶\ce{Al}(s)$ −1.662
$\ce{Mg^2+}(aq)+\ce{2e-}⟶\ce{Mg}(s)$ −2.372
$\ce{Na+}(aq)+\ce{e-}⟶\ce{Na}(s)$ −2.71
$\ce{Ca^2+}(aq)+\ce{2e-}⟶\ce{Ca}(s)$ −2.868
$\ce{Ba^2+}(aq)+\ce{2e-}⟶\ce{Ba}(s)$ −2.912
$\ce{K+}(aq)+\ce{e-}⟶\ce{K}(s)$ −2.931
$\ce{Li+}(aq)+\ce{e-}⟶\ce{Li}(s)$ −3.04
Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions.
Example $1$: Cell Potentials from Standard Reduction Potentials
What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.
Solution
Using Table $1$, the reactions involved in the galvanic cell, both written as reductions, are
$\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V} \nonumber$
$\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s) \hspace{20px} E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V} \nonumber$
Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:
\begin{align*} &\textrm{Anode (oxidation): }\ce{Ni}(s)⟶\ce{Ni^2+}(aq)+\ce{2e-} \hspace{20px} E^\circ_\ce{anode}=E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V}\ &\textrm{Cathode (reduction): }\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_\ce{cathode}=E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V} \end{align*} \nonumber
The least common factor is six, so the overall reaction is
$\ce{3Ni}(s)+\ce{2Au^3+}(aq)⟶\ce{3Ni^2+}(aq)+\ce{2Au}(s)$
The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=\mathrm{1.498\: V−(−0.257\: V)=1.755\: V} \nonumber$
From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au3+ is reduced, so it is the oxidizing agent.
Exercise $1$
A galvanic cell consists of a Mg electrode in 1 M Mg(NO3)2 solution and a Ag electrode in 1 M AgNO3 solution. Calculate the standard cell potential at 25 °C.
Answer
$\ce{Mg}(s)+\ce{2Ag+}(aq)⟶\ce{Mg^2+}(aq)+\ce{2Ag}(s) \hspace{20px} E^\circ_\ce{cell}=\mathrm{0.7996\: V−(−2.372\: V)=3.172\: V} \nonumber$
Summary
Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.
Key Equations
• $E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}$
Glossary
standard cell potential $(E^\circ_\ce{cell})$
the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K; can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half-reaction occurring at the cathode
standard hydrogen electrode (SHE)
the electrode consists of hydrogen gas bubbling through hydrochloric acid over an inert platinum electrode whose reduction at standard conditions is assigned a value of 0 V; the reference point for standard reduction potentials
standard reduction potential (E°)
the value of the reduction under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K; tabulated values used to calculate standard cell potentials | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.3%3A_Standard_Reduction_Potentials.txt |
Learning Objectives
• Relate cell potentials to free energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
We will now extend electrochemistry by determining the relationship between $E^\circ_\ce{cell}$ and the thermodynamics quantities such as ΔG° (Gibbs free energy) and K (the equilibrium constant). In galvanic cells, chemical energy is converted into electrical energy, which can do work. The electrical work is the product of the charge transferred multiplied by the potential difference (voltage):
$\mathrm{electrical\: work=volts \times (charge\: in\: coulombs)=J} \nonumber$
The charge on 1 mole of electrons is given by Faraday’s constant (F)
\begin{align*} F &=\dfrac{6.022 \times 10^{23}\:e^-}{mol} \times \dfrac{1.602 \times 10^{−19}\:C}{e^-} \[4pt] &=9.648 \times 10^4\:\dfrac{C}{mol} \[4pt] &=9.648 \times 10^4\:\dfrac{J}{V⋅mol} \end{align*} \nonumber
Therefore
$\mathrm{total\: charge=(number\: of\: moles\: of\: e^-)} \times F=nF \nonumber$
In this equation, $n$ is the number of moles of electrons for the balanced oxidation-reduction reaction. The measured cell potential is the maximum potential the cell can produce and is related to the electrical work (wele) by
$E_\ce{cell}=\dfrac{−w_\ce{ele}}{nF}\hspace{40px}\ce{or}\hspace{40px}w_\ce{ele}=−nFE_\ce{cell} \nonumber$
The negative sign for the work indicates that the electrical work is done by the system (the galvanic cell) on the surroundings. In an earlier chapter, the free energy was defined as the energy that was available to do work. In particular, the change in free energy was defined in terms of the maximum work ($w_{max}$), which, for electrochemical systems, is $w_{ele}$.
\begin{align*} ΔG&=w_\ce{max}=w_\ce{ele} \[4pt] &=−nFE_\ce{cell} \end{align*} \nonumber
We can verify the signs are correct when we realize that $n$ and $F$ are positive constants and that galvanic cells, which have positive cell potentials, involve spontaneous reactions. Thus, spontaneous reactions, which have $ΔG < 0$, must have $E_{cell} > 0$. If all the reactants and products are in their standard states, this becomes
$ΔG°=−nFE^\circ_\ce{cell} \nonumber$
This provides a way to relate standard cell potentials to equilibrium constants, since
$ΔG°=−RT\ln K \nonumber$
$−nFE^\circ_\ce{cell}=−RT\ln K \nonumber$
or
$E^\circ_\ce{cell}=\dfrac{RT}{nF}\ln K \nonumber$
Most of the time, the electrochemical reactions are run at standard temperature (298.15 K). Collecting terms at this temperature yields
\begin{align*} E^\circ_\ce{cell}&=\dfrac{RT}{nF}\:\ln K \[4pt] &=\dfrac{\left(8.314\:\dfrac{\ce{J}}{\textrm{K⋅mol}}\right)(298.15\:K)}{n \times 96,485\: \textrm{C/V⋅mol}}\:\ln K \[4pt] &=\dfrac{\mathrm{0.0257\: V}}{n}\:\ln K \end{align*} \nonumber
where $n$ is the number of moles of electrons. The logarithm in equations involving cell potentials is often expressed using base 10 logarithms (i.e., $\log_{10}$ or just $\log$), which changes the constant by a factor of 2.303:
$E^\circ_\ce{cell}=\dfrac{\mathrm{0.0592\: V}}{n}\:\log K \nonumber$
Thus, if ΔG°, K, or $E^\circ_\ce{cell}$ is known or can be calculated, the other two quantities can be readily determined. The relationships are shown graphically in Figure $1$.
Given any one of the quantities, the other two can be calculated.
Example $1$: Equilibrium Constants, Potentials, & Free Energy Changes
What is the standard Gibbs free energy change and equilibrium constant for the following reaction at 25 °C?
$\ce{2Ag+}(aq)+\ce{Fe}(s)⇌\ce{2Ag}(s)+\ce{Fe^2+}(aq) \nonumber$
Solution
The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Table P1.
\begin{align*} &\textrm{anode (oxidation): }\ce{Fe}(s)⟶\ce{Fe^2+}(aq)+\ce{2e-} \hspace{40px} E^\circ_{\ce{Fe^2+/Fe}}=\mathrm{−0.447\: V} \nonumber\ &\textrm{cathode (reduction): }2 \times (\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s)) \hspace{40px} E^\circ_{\ce{Ag+/Ag}}=\mathrm{0.7996\: V} \nonumber\ &E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{Fe^2+/Fe}}=\mathrm{+1.247\: V} \nonumber \end{align*} \nonumber
Remember that the cell potential for the cathode is not multiplied by two when determining the standard cell potential. With n = 2, the equilibrium constant is then
$E^\circ_\ce{cell}=\dfrac{\mathrm{0.0592\: V}}{n}\:\log K \nonumber$
\begin{align*} K&=10^{n \times E^\circ_\ce{cell}/\mathrm{0.0592\: V}} \[4pt] &=10^{2 \times \mathrm{1.247\: V/0.0592\: V}} \[4pt] &=10^{42.128} \[4pt] &=1.3 \times 10^{42}\end{align*} \nonumber
The standard free energy is then
$ΔG°=−nFE^\circ_\ce{cell} \nonumber$
$ΔG°=\mathrm{−2 \times 96,485\:\dfrac{J}{\textrm{V⋅mol}} \times 1.247\: V=−240.6\:\dfrac{kJ}{mol}} \nonumber$
Check your answer: A positive standard cell potential means a spontaneous reaction, so the standard free energy change should be negative, and an equilibrium constant should be >1.
Exercise $1$
What is the standard Gibbs free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?
$\ce{Sn}(s)+\ce{2Cu^2+}(aq)⇌\ce{Sn^2+}(aq)+\ce{2Cu+}(aq) \nonumber$
Answer
Spontaneous; n = 2; $E^\circ_\ce{cell}=\mathrm{+0.291\: V}$; $ΔG°=\mathrm{−56.2\:\dfrac{kJ}{mol}}$; $K = 6.8 \times 10^9$.
Now that the connection has been made between the free energy and cell potentials, nonstandard concentrations follow. Recall that
$ΔG=ΔG°+RT\ln Q \nonumber$
where $Q$ is the reaction quotient (see the chapter on equilibrium fundamentals). Converting to cell potentials:
$−nFE_\ce{cell}=−nFE^\circ_\ce{cell}+RT\ln Q \label{nernst1A}$
or
$E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{RT}{nF}\:\ln Q \label{nernst1B}$
Equation \ref{nernst1B} is the generalized Nernst equation that is applicable at any temperature. However, is can be simplified for reactions occuring at 25 °C (298.15 K) by rewriting it as
$E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0257\: V}}{n}\:\ln Q \label{nernst2A}$
or
$E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0592\: V}}{n}\log_{10} Q \label{nernst2B}$
If the temperature is not 298.15 K, it is necessary to recalculate the potential with Equation\ref{nernst1B}. With the Nernst equation, it is possible to calculate the cell potential at nonstandard conditions. This adjustment is necessary because potentials determined under different conditions will have different values.
Example $2$: Cell Potentials at Nonstandard Conditions
Consider the following reaction at room temperature:
$\ce{Co}(s)+\ce{Fe^2+}(aq,\:1.94\:M)⟶\ce{Co^2+}(aq,\: 0.15\:M)+\ce{Fe}(s) \nonumber$
Is the process spontaneous?
Solution
There are two ways to solve the problem. If the thermodynamic information in Table T1 were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in Table P1. Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from Table P1 and the problem,
\begin{align*} &\textrm{Anode (oxidation): }\ce{Co}(s)⟶\ce{Co^2+}(aq)+\ce{2e-} \hspace{40px} E^\circ_{\ce{Co^2+/Co}}=\mathrm{−0.28\: V}\ &\textrm{Cathode (reduction): }\ce{Fe^2+}(aq)+\ce{2e-}⟶\ce{Fe}(s) \hspace{40px} E^\circ_{\ce{Fe^2+/Fe}}=\mathrm{−0.447\: V}\ &E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=\mathrm{−0.447\: V−(−0.28\: V)=−0.17\: V} \end{align*} \nonumber
The process is not spontaneous under standard conditions. Using the Nernst equation and the concentrations stated in the problem and $n = 2$,
$Q=\ce{\dfrac{[Co^2+]}{[Fe^2+]}}=\dfrac{0.15\:M}{1.94\:M}=0.077 \nonumber$
Now we can insert these into the Nernst Equation at room temperature (Equation \ref{nernst2B})
\begin{align*} E_\ce{cell} &=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0592\: V}}{n}\:\log Q \[4pt] &=\mathrm{−0.17\: V−\dfrac{0.0592\: V}{2}\:\log 0.077} \[4pt] &=\mathrm{−0.17\: V+0.033\: V=−0.14\: V} \end{align*} \nonumber
The process is (still) nonspontaneous.
Exercise $2$
What is the cell potential for the following reaction at room temperature?
$\ce{Al}(s)│\ce{Al^3+}(aq,\:0.15\:M)║\ce{Cu^2+}(aq,\:0.025\:M)│\ce{Cu}(s) \nonumber$
What are the values of n and Q for the overall reaction? Is the reaction spontaneous under these conditions?
Answer
n = 6; Q = 1440; Ecell = +1.97 V, spontaneous.
Finally, we will take a brief look at a special type of cell called a concentration cell. In a concentration cell, the electrodes are the same material and the half-cells differ only in concentration. Since one or both compartments is not standard, the cell potentials will be unequal; therefore, there will be a potential difference, which can be determined with the aid of the Nernst equation.
Example $3$: Concentration Cells
What is the cell potential of the concentration cell described by
$\ce{Zn}(s)│\ce{Zn^2+}(aq,\: 0.10\:M)║\ce{Zn^2+}(aq,\: 0.50\:M)│\ce{Zn}(s) \nonumber$
Solution
From the information given:
\begin{align*} &\textrm{Anode: }\ce{Zn}(s)⟶\ce{Zn^2+}(aq,\: 0.10\:M)+\ce{2e-} \hspace{40px} E^\circ_\ce{anode}=\mathrm{−0.7618\: V}\ &\textrm{Cathode: }\ce{Zn^2+}(aq,\: 0.50\:M)+\ce{2e-}⟶\ce{Zn}(s) \hspace{40px} E^\circ_\ce{cathode}=\mathrm{−0.7618\: V}\ &\overline{\textrm{Overall: }\ce{Zn^2+}(aq,\: 0.50\:M)⟶\ce{Zn^2+}(aq,\: 0.10\:M) \hspace{40px} E^\circ_\ce{cell}=\mathrm{0.000\: V}} \end{align*} \nonumber
The standard cell potential is zero because the anode and cathode involve the same reaction; only the concentration of Zn2+ changes. Substituting into the Nernst equation,
$E_\ce{cell}=\mathrm{0.000\: V−\dfrac{0.0592\: V}{2}\:\log\dfrac{0.10}{0.50}=+0.021\: V} \nonumber$
and the process is spontaneous at these conditions.
Check your answer: In a concentration cell, the standard cell potential will always be zero. To get a positive cell potential (spontaneous process) the reaction quotient Q must be <1. Q < 1 in this case, so the process is spontaneous.
Exercise $3$
What value of Q for the previous concentration cell would result in a voltage of 0.10 V? If the concentration of zinc ion at the cathode was 0.50 M, what was the concentration at the anode?
Answer
Q = 0.00042; [Zn2+]cat = 2.1 \times 10−4 M.
Summary
Electrical work (wele) is the negative of the product of the total charge (Q) and the cell potential (Ecell). The total charge can be calculated as the number of moles of electrons (n) times the Faraday constant (F = 96,485 C/mol e). Electrical work is the maximum work that the system can produce and so is equal to the change in free energy. Thus, anything that can be done with or to a free energy change can also be done to or with a cell potential. The Nernst equation relates the cell potential at nonstandard conditions to the logarithm of the reaction quotient. Concentration cells exploit this relationship and produce a positive cell potential using half-cells that differ only in the concentration of their solutes.
Key Equations
• $E^\circ_\ce{cell}=\dfrac{RT}{nF}\:\ln K$
• $E^\circ_\ce{cell}=\dfrac{\mathrm{0.0257\: V}}{n}\:\ln K=\dfrac{\mathrm{0.0592\: V}}{n}\:\log K \hspace{40px} \mathrm{(at\: 298.15\:\mathit{K})}$
• $E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{RT}{nF}\:\ln Q \hspace{40px} \textrm{(Nernst equation)}$
• $E_\ce{cell}=E^\circ_\ce{cell}−\dfrac{\mathit{0.0257\: V}}{n}\:\ln Q=E^\circ_\ce{cell}−\dfrac{\mathrm{0.0592\: V}}{n}\:\log Q \hspace{40px} \mathrm{(at\: 298.15\:\mathit{K})}$
• ΔG = −nFEcell
• $ΔG^∘=−nFE^\circ_\ce{cell}$
• $w_\ce{ele}=w_\ce{max}=−nFE_\ce{cell}$
Glossary
concentration cell
galvanic cell in which the two half-cells are the same except for the concentration of the solutes; spontaneous when the overall reaction is the dilution of the solute
electrical work (wele)
negative of total charge times the cell potential; equal to wmax for the system, and so equals the free energy change (ΔG)
Faraday’s constant (F)
charge on 1 mol of electrons; F = 96,485 C/mol e
Nernst equation
equation that relates the logarithm of the reaction quotient (Q) to nonstandard cell potentials; can be used to relate equilibrium constants to standard cell potentials | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.4%3A_The_Nernst_Equation.txt |
Learning Objectives
• Classify batteries as primary or secondary
• List some of the characteristics and limitations of batteries
• Provide a general description of a fuel cell
A battery is an electrochemical cell or series of cells that produces an electric current. In principle, any galvanic cell could be used as a battery. An ideal battery would never run down, produce an unchanging voltage, and be capable of withstanding environmental extremes of heat and humidity. Real batteries strike a balance between ideal characteristics and practical limitations. For example, the mass of a car battery is about 18 kg or about 1% of the mass of an average car or light-duty truck. This type of battery would supply nearly unlimited energy if used in a smartphone, but would be rejected for this application because of its mass. Thus, no single battery is “best” and batteries are selected for a particular application, keeping things like the mass of the battery, its cost, reliability, and current capacity in mind. There are two basic types of batteries: primary and secondary. A few batteries of each type are described next.
Visit this site to learn more about batteries.
Primary Batteries
Primary batteries are single-use batteries because they cannot be recharged. A common primary battery is the dry cell (Figure $1$). The dry cell is a zinc-carbon battery. The zinc can serves as both a container and the negative electrode. The positive electrode is a rod made of carbon that is surrounded by a paste of manganese(IV) oxide, zinc chloride, ammonium chloride, carbon powder, and a small amount of water. The reaction at the anode can be represented as the ordinary oxidation of zinc:
$\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{2e-} \hspace{20px} E^\circ_{\ce{Zn^2+/Zn}}=\mathrm{−0.7618\: V} \nonumber$
The reaction at the cathode is more complicated, in part because more than one reaction occurs. The series of reactions that occurs at the cathode is approximately
$\ce{2MnO2}(s)+\ce{2NH4Cl}(aq)+\ce{2e-}⟶\ce{Mn2O3}(s)+\ce{2NH3}(aq)+\ce{H2O}(l)+\ce{2Cl-} \nonumber$
The overall reaction for the zinc–carbon battery can be represented as
$\ce{2MnO2}(s) + \ce{2NH4Cl}(aq) + \ce{Zn}(s) ⟶ \ce{Zn^2+}(aq) + \ce{Mn2O3}(s) + \ce{2NH3}(aq) + \ce{H2O}(l) + \ce{2Cl-} \nonumber$
with an overall cell potential which is initially about 1.5 V, but decreases as the battery is used. It is important to remember that the voltage delivered by a battery is the same regardless of the size of a battery. For this reason, D, C, A, AA, and AAA batteries all have the same voltage rating. However, larger batteries can deliver more moles of electrons. As the zinc container oxidizes, its contents eventually leak out, so this type of battery should not be left in any electrical device for extended periods.
Alkaline batteries (Figure $2$) were developed in the 1950s partly to address some of the performance issues with zinc–carbon dry cells. They are manufactured to be exact replacements for zinc-carbon dry cells. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are
\begin{align*} &\textrm{anode: }\ce{Zn}(s)+\ce{2OH-}(aq)⟶\ce{ZnO}(s)+\ce{H2O}(l)+\ce{2e-} \hspace{40px} E^\circ_\ce{anode}=\mathrm{−1.28\: V}\ &\underline{\textrm{cathode: }\ce{2MnO2}(s)+\ce{H2O}(l)+\ce{2e-}⟶\ce{Mn2O3}(s)+\ce{2OH-}(aq) \hspace{40px} E^\circ_\ce{cathode}=\mathrm{+0.15\: V}}\ &\textrm{overall: }\ce{Zn}(s)+\ce{2MnO2}(s)⟶\ce{ZnO}(s)+\ce{Mn2O3}(s) \hspace{40px} E^\circ_\ce{cell}=\mathrm{+1.43\: V} \end{align*} \nonumber
An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so these should also be removed from devices for long-term storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.
Secondary Batteries
Secondary batteries are rechargeable. These are the types of batteries found in devices such as smartphones, electronic tablets, and automobiles.
Nickel-cadmium, or NiCd, batteries (Figure $3$) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a “jelly-roll” design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are
\begin{align*} &\textrm{anode: }\ce{Cd}(s)+\ce{2OH-}(aq)⟶\ce{Cd(OH)2}(s)+\ce{2e-}\ &\underline{\textrm{cathode: }\ce{NiO2}(s)+\ce{2H2O}(l)+\ce{2e-}⟶\ce{Ni(OH)2}(s)+\ce{2OH-}(aq)}\ &\textrm{overall: }\ce{Cd}(s)+\ce{NiO2}(s)+\ce{2H2O}(l)⟶\ce{Cd(OH)2}(s)+\ce{Ni(OH)2}(s) \end{align*} \nonumber
The voltage is about 1.2 V to 1.25 V as the battery discharges. When properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be opened or put into the regular trash.
Lithium ion batteries (Figure $4$) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are
\begin{align*} &\textrm{anode: }\ce{LiCoO2}⇌\ce{Li}_{1-x}\ce{CoO2}+x\ce{Li+}+x\ce{e-}\ &\textrm{cathode: }x\ce{Li+}+x\ce{e-}+x\ce{C6}⇌x\ce{LiC6}\ &\overline{\textrm{overall: }\ce{LiCoO2}+x\ce{C6}⇌\ce{Li}_{1-x}\ce{CoO2}+x\ce{LiC6}} \end{align*} \nonumber
With the coefficients representing moles, x is no more than about 0.5 moles. The battery voltage is about 3.7 V. Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.
The lead acid battery (Figure $5$) is the type of secondary battery used in your automobile. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are
\begin{align*} &\textrm{anode: }\ce{Pb}(s)+\ce{HSO4-}(aq)⟶\ce{PbSO4}(s)+\ce{H+}(aq)+\ce{2e-}\ &\underline{\textrm{cathode: } \ce{PbO2}(s)+\ce{HSO4-}(aq)+\ce{3H+}(aq)+\ce{2e-}⟶\ce{PbSO4}(s)+\ce{2H2O}(l)}\ &\textrm{overall: }\ce{Pb}(s)+\ce{PbO2}(s)+\ce{2H2SO4}(aq)⟶\ce{2PbSO4}(s)+\ce{2H2O}(l) \end{align*} \nonumber
Each cell produces 2 V, so six cells are connected in series to produce a 12-V car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.
Fuel Cells
A fuel cell is a device that converts chemical energy into electrical energy. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines (Figure $6$).
In a hydrogen fuel cell, the reactions are
\begin{align*} &\textrm{anode: }\ce{2H2 + 2O^2- ⟶ 2H2O + 4e-}\ &\underline{\textrm{cathode: }\ce{O2 + 4e- ⟶ 2O^2-}\hspace{55px}}\ &\textrm{overall: }\ce{2H2 + O2 ⟶ 2H2O} \end{align*} \nonumber
The voltage is about 0.9 V. The efficiency of fuel cells is typically about 40% to 60%, which is higher than the typical internal combustion engine (25% to 35%) and, in the case of the hydrogen fuel cell, produces only water as exhaust. Currently, fuel cells are rather expensive and contain features that cause them to fail after a relatively short time.
Summary
Batteries are galvanic cells, or a series of cells, that produce an electric current. When cells are combined into batteries, the potential of the battery is an integer multiple of the potential of a single cell. There are two basic types of batteries: primary and secondary. Primary batteries are “single use” and cannot be recharged. Dry cells and (most) alkaline batteries are examples of primary batteries. The second type is rechargeable and is called a secondary battery. Examples of secondary batteries include nickel-cadmium (NiCd), lead acid, and lithium ion batteries. Fuel cells are similar to batteries in that they generate an electrical current, but require continuous addition of fuel and oxidizer. The hydrogen fuel cell uses hydrogen and oxygen from the air to produce water, and is generally more efficient than internal combustion engines.
Summary
alkaline battery
primary battery that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an exact replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell
battery
galvanic cell or series of cells that produces a current; in theory, any galvanic cell
dry cell
primary battery, also called a zinc-carbon battery; can be used in any orientation because it uses a paste as the electrolyte; tends to leak electrolyte when stored
fuel cell
devices that produce an electrical current as long as fuel and oxidizer are continuously added; more efficient than internal combustion engines
lead acid battery
secondary battery that consists of multiple cells; the lead acid battery found in automobiles has six cells and a voltage of 12 V
lithium ion battery
very popular secondary battery; uses lithium ions to conduct current and is light, rechargeable, and produces a nearly constant potential as it discharges
nickel-cadmium battery
(NiCd battery) secondary battery that uses cadmium, which is a toxic heavy metal; heavier than lithium ion batteries, but with similar performance characteristics
primary battery
single-use nonrechargeable battery
secondary battery
battery that can be recharged | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.5%3A_Batteries_and_Fuel_Cells.txt |
Learning Objectives
• Define corrosion
• List some of the methods used to prevent or slow corrosion
Corrosion is usually defined as the degradation of metals due to an electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion in the United States is significant, with estimates in excess of half a trillion dollars a year.
Changing Colors
The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color. When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide ($\ce{Cu_2O}$), which is red, and then to copper(II) oxide, which is black.
$\ce{2Cu(s)} +\ce{1/2O2(g)} \rightarrow \underset{\text{red}}{\ce{Cu2O(s)}} \nonumber$
$\ce{Cu2O(s)} +\ce{1/2O2(g)} \rightarrow \underset{\text{black}}{\ce{2CuO(s)}} \nonumber$
Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, sulfur trioxide, carbon dioxide, and water all reacted with the $\ce{CuO}$.
$\ce{2CuO(s)}+\ce{CO2(g)} + \ce{H2O(l)} \rightarrow \underset{\text{green}}{\ce{Cu_2CO3(OH)2(s)}} \nonumber$
$\ce{3CuO(s)}+\ce{2CO2(g)}+\ce{H2O(l)} \rightarrow \underset{\text{blue}}{\ce{Cu_2(CO_3)_2(OH)2(s)}} \nonumber$
$\ce{4CuO(s)}+\ce{SO3(g)}+\ce{3H2O(l)} \rightarrow \underset{\text{green}}{\ce{Cu_4SO_4(OH)6(s)}} \nonumber$
These three compounds are responsible for the characteristic blue-green patina seen today. Fortunately, formation of the patina created a protective layer on the surface, preventing further corrosion of the copper skin. The formation of the protective layer is a form of passivation, which is discussed further in a later chapter.
Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. The main steps in the rusting of iron appear to involve the following. Once exposed to the atmosphere, iron rapidly oxidizes.
$\textrm{anode: }\ce{Fe}_{(s)} \rightarrow \ce{Fe^{2+}}_{(aq)}+\ce{2e^-}\;\;\; E^\circ_{\ce{Fe^{2+}/Fe}}=\mathrm{−0.44\: V} \nonumber$
The electrons reduce oxygen in the air in acidic solutions.
$\textrm{cathode: }\ce{O}_{2(g)}+\ce{4H^+}_{(aq)}+\ce{4e^-} \rightarrow \ce{2H_2O}_{(l)}\;\; E^\circ_{\ce{O_2/O_2}}=\mathrm{+1.23\; V} \nonumber$
$\textrm{overall: }\ce{2Fe}_{(s)}+\ce{O}_{2(g)}+\ce{4H^+}_{(aq)} \rightarrow \ce{2Fe^{2+}}_{(aq)}+\ce{2H_2O}_{(l)} \;\;\;E^\circ_\ce{cell}=\mathrm{+1.67\; V} \nonumber$
What we call rust is hydrated iron(III) oxide, which forms when iron(II) ions react further with oxygen.
$\ce{4Fe^{2+}}_{(aq)}+\ce{O}_{2(g)}+(4+2x)\ce{H_2O}_{(l)} \rightarrow \ce{2Fe_2O_3} \cdot x\ce{H_2O}_{(s)}+\ce{8H^+}_{(aq)} \nonumber$
The number of water molecules is variable, so it is represented by x. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.
One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion.
Other strategies include alloying the iron with other metals. For example, stainless steel is mostly iron with a bit of chromium. The chromium tends to collect near the surface, where it forms an oxide layer that protects the iron.
Zinc-plated or galvanized iron uses a different strategy. Zinc is more easily oxidized than iron because zinc has a lower reduction potential. Since zinc has a lower reduction potential, it is a more active metal. Thus, even if the zinc coating is scratched, the zinc will still oxidize before the iron. This suggests that this approach should work with other active metals.
Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium. This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode, and so does not oxidize (corrode). When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended.
Summary
Corrosion is the degradation of a metal caused by an electrochemical process. Large sums of money are spent each year repairing the effects of, or preventing, corrosion. Some metals, such as aluminum and copper, produce a protective layer when they corrode in air. The thin layer that forms on the surface of the metal prevents oxygen from coming into contact with more of the metal atoms and thus “protects” the remaining metal from further corrosion. Iron corrodes (forms rust) when exposed to water and oxygen. The rust that forms on iron metal flakes off, exposing fresh metal, which also corrodes. One way to prevent, or slow, corrosion is by coating the metal. Coating prevents water and oxygen from contacting the metal. Paint or other coatings will slow corrosion, but they are not effective once scratched. Zinc-plated or galvanized iron exploits the fact that zinc is more likely to oxidize than iron. As long as the coating remains, even if scratched, the zinc will oxidize before the iron. Another method for protecting metals is cathodic protection. In this method, an easily oxidized and inexpensive metal, often zinc or magnesium (the sacrificial anode), is electrically connected to the metal that must be protected. The more active metal is the sacrificial anode, and is the anode in a galvanic cell. The “protected” metal is the cathode, and remains unoxidized. One advantage of cathodic protection is that the sacrificial anode can be monitored and replaced if needed.
Glossary
cathodic protection
method of protecting metal by using a sacrificial anode and effectively making the metal that needs protecting the cathode, thus preventing its oxidation
corrosion
degradation of metal through an electrochemical process
galvanized iron
method for protecting iron by covering it with zinc, which will oxidize before the iron; zinc-plated iron
sacrificial anode
more active, inexpensive metal used as the anode in cathodic protection; frequently made from magnesium or zinc | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.6%3A_Corrosion.txt |
Learning Objectives
• Describe electrolytic cells and their relationship to galvanic cells
• Perform various calculations related to electrolysis
In galvanic cells, chemical energy is converted into electrical energy. The opposite is true for electrolytic cells. In electrolytic cells, electrical energy causes nonspontaneous reactions to occur in a process known as electrolysis. The charging electric barttery shows one such process. Electrical energy is converted into the chemical energy in the battery as it is charged. Once charged, the battery can be used to power the automobile. The same principles are involved in electrolytic cells as in galvanic cells. We will look at three electrolytic cells and the quantitative aspects of electrolysis.
The Electrolysis of Molten Sodium Chloride
In molten sodium chloride, the ions are free to migrate to the electrodes of an electrolytic cell. A simplified diagram of the cell commercially used to produce sodium metal and chlorine gas is shown in Figure $1$. Sodium is a strong reducing agent and chlorine is used to purify water, and is used in antiseptics and in paper production. The reactions are
\begin{align} &\textrm{anode: }\ce{2Cl-}(l) ⟶\ce{Cl2}(g)+\ce{2e-} \hspace{20px}E^\circ_{\ce{Cl2/Cl-}}=\mathrm{+1.3\: V}\ &\textrm{cathode: }\ce{Na+}(l)+\ce{e-} ⟶\ce{Na}(l) \hspace{20px}E^\circ_{\ce{Na+/Na}}=\mathrm{−2.7\: V}\ & \overline{\textrm{overall: } \ce{2Na+}(l)+\ce{2Cl-}(l) ⟶\ce{2Na}(l)+\ce{Cl2}(g) \hspace{20px}E^\circ_\ce{cell}=\mathrm{−4.0\: V}} \end{align} \nonumber
The power supply (battery) must supply a minimum of 4 V, but, in practice, the applied voltages are typically higher because of inefficiencies in the process itself.
The Electrolysis of Water
It is possible to split water into hydrogen and oxygen gas by electrolysis. Acids are typically added to increase the concentration of hydrogen ion in solution (Figure $2$). The reactions are
\begin{align} &\textrm{anode: }\ce{2H2O}(l)⟶\ce{O2}(g)+\ce{4H+}(aq)+\ce{4e-} \hspace{20px} E^\circ_\ce{anode}=\mathrm{+1.229\: V}\ &\textrm{cathode: }\ce{2H+}(aq)+\ce{2e-}⟶\ce{H2}(g) \hspace{20px} E^\circ_\ce{cathode}=\mathrm{0\: V}\ &\overline{\textrm{overall: }\ce{2H2O}(l)⟶\ce{2H2}(g)+\ce{O2}(g) \hspace{20px} E^\circ_\ce{cell}=\mathrm{−1.229\: V}} \end{align} \nonumber
Note that the sulfuric acid is not consumed and that the volume of hydrogen gas produced is twice the volume of oxygen gas produced. The minimum applied voltage is 1.229 V.
The Electrolysis of Aqueous Sodium Chloride
The electrolysis of aqueous sodium chloride is the more common example of electrolysis because more than one species can be oxidized and reduced. Considering the anode first, the possible reactions are
\begin{align} &\textrm{(i) }\ce{2Cl-}(aq)⟶\ce{Cl2}(g)+\ce{2e-} \hspace{20px} E^\circ_\ce{anode}=\mathrm{+1.35827\: V}\ &\textrm{(ii) }\ce{2H2O}(l)⟶\ce{O2}(g)+\ce{4H+}(aq)+\ce{4e-} \hspace{20px} E^\circ_\ce{anode}=\mathrm{+1.229\: V} \end{align} \nonumber
These values suggest that water should be oxidized at the anode because a smaller potential would be needed—using reaction (ii) for the oxidation would give a less-negative cell potential. When the experiment is run, it turns out chlorine, not oxygen, is produced at the anode. The unexpected process is so common in electrochemistry that it has been given the name overpotential. The overpotential is the difference between the theoretical cell voltage and the actual voltage that is necessary to cause electrolysis. It turns out that the overpotential for oxygen is rather high and effectively makes the reduction potential more positive. As a result, under normal conditions, chlorine gas is what actually forms at the anode.
Now consider the cathode. Three reductions could occur:
\begin{align} &\textrm{(iii) }\ce{2H+}(aq)+\ce{2e-}⟶\ce{H2}(g) \hspace{20px} E^\circ_\ce{cathode}=\mathrm{0\: V}\ &\textrm{(iv) }\ce{2H2O}(l)+\ce{2e-}⟶\ce{H2}(g)+\ce{2OH-}(aq) \hspace{20px} E^\circ_\ce{cathode}=\mathrm{−0.8277\:V}\ &\textrm{(v) }\ce{Na+}(aq)+\ce{e-}⟶\ce{Na}(s) \hspace{20px} E^\circ_\ce{cathode}=\mathrm{−2.71\: V} \end{align} \nonumber
Reaction (v) is ruled out because it has such a negative reduction potential. Under standard state conditions, reaction (iii) would be preferred to reaction (iv). However, the pH of a sodium chloride solution is 7, so the concentration of hydrogen ions is only 1× 10−7 M. At such low concentrations, reaction (iii) is unlikely and reaction (iv) occurs. The overall reaction is then
$\textrm{overall: }\ce{2H2O}(l)+\ce{2Cl-}(aq)⟶\ce{H2}(g)+\ce{Cl2}(g)+\ce{2OH-}(aq) \hspace{20px} E^\circ_\ce{cell}=\mathrm{−2.186\: V} \nonumber$
As the reaction proceeds, hydroxide ions replace chloride ions in solution. Thus, sodium hydroxide can be obtained by evaporating the water after the electrolysis is complete. Sodium hydroxide is valuable in its own right and is used for things like oven cleaner, drain opener, and in the production of paper, fabrics, and soap.
Electroplating
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. We can get an idea of how this works by investigating how silver-plated tableware is produced (Figure $3$).
In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. As the potential is increased, current flows. Silver metal is lost at the anode as it goes into solution.
$\textrm{anode: }\ce{Ag}(s)⟶\ce{Ag+}(aq)+\ce{e-} \nonumber$
The mass of the cathode increases as silver ions from the solution are deposited onto the spoon
$\textrm{cathode: }\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s) \nonumber$
The net result is the transfer of silver metal from the anode to the cathode. The quality of the object is usually determined by the thickness of the deposited silver and the rate of deposition.
Quantitative Aspects of Electrolysis
The amount of current that is allowed to flow in an electrolytic cell is related to the number of moles of electrons. The number of moles of electrons can be related to the reactants and products using stoichiometry. Recall that the SI unit for current (I) is the ampere (A), which is the equivalent of 1 coulomb per second (1 A = $\mathrm{1\: \dfrac{C}{s}}$). The total charge (Q, in coulombs) is given by
$Q=I×t=n×F \nonumber$
where
• t is the time in seconds,
• n the number of moles of electrons, and
• F is the Faraday constant.
Moles of electrons can be used in stoichiometry problems. The time required to deposit a specified amount of metal might also be requested, as in the second of the following examples.
Example $1$: Converting Current to Moles of Electrons
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?
Solution
Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time
$n=\dfrac{Q}{F}=\mathrm{\dfrac{\dfrac{10.23\: C}{s}×1\: hr×\dfrac{60\: min}{hr}×\dfrac{60\:s}{min}}{96,485\: C/mol\: e^-}=\dfrac{36,830\: C}{96,485\: C/mol\:e^-}=0.3817\: mol\: e^-} \nonumber$
From the problem, the solution contains AgNO3, so the reaction at the cathode involves 1 mole of electrons for each mole of silver
$\textrm{cathode: }\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s) \nonumber$
The atomic mass of silver is 107.9 g/mol, so
$\mathrm{mass\: Ag=0.3817\: mol\: e^-×\dfrac{1\: mol\: Ag}{1\: mol\: e^-}×\dfrac{107.9\: g\: Ag}{1\: mol\: Ag}=41.19\: g\: Ag} \nonumber$
Check your answer: From the stoichiometry, 1 mole of electrons would produce 1 mole of silver. Less than one-half a mole of electrons was involved and less than one-half a mole of silver was produced.
Exercise $1$
Aluminum metal can be made from aluminum ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 2.50 × 103 A passed through the solution for 15.0 minutes? Assume the yield is 100%.
Answer
$\ce{Al^3+}(aq)+\ce{3e-}⟶\ce{Al}(s)$; 7.77 mol Al = 210.0 g Al.
Example $2$: Time Required for Deposition
In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m2 from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm3.
Solution This problem brings in a number of topics covered earlier. An outline of what needs to be done is:
• If the total charge can be determined, the time required is just the charge divided by the current
• The total charge can be obtained from the amount of Cr needed and the stoichiometry
• The amount of Cr can be obtained using the density and the volume Cr required
• The volume Cr required is the thickness times the area
Solving in steps, and taking care with the units, the volume of Cr required is
$\mathrm{volume=\left(0.010\: mm×\dfrac{1\: cm}{10\: mm}\right)×\left(3.3\:m^2×\left(\dfrac{10,000\:cm^2}{1\:m^2}\right)\right)=33\: cm^3} \nonumber$
Cubic centimeters were used because they match the volume unit used for the density. The amount of Cr is then
$\mathrm{mass=volume×density=33\cancel{cm^3}×\dfrac{7.19\: g}{\cancel{cm^3}}=237\: g\: Cr} \nonumber$
$\mathrm{mol\: Cr=237\: g\: Cr×\dfrac{1\: mol\: Cr}{52.00\: g\: Cr}=4.56\: mol\: Cr} \nonumber$
Since the solution contains chromium(III) ions, 3 moles of electrons are required per mole of Cr. The total charge is then
$Q=\mathrm{4.56\: mol\: Cr×\dfrac{3\:mol\: e^-}{1\: mol\: Cr}×\dfrac{96485\: C}{mol\: e^-}=1.32×10^6\:C} \nonumber$
The time required is then
$t=\dfrac{Q}{I}=\mathrm{\dfrac{1.32×10^6\:C}{33.46\: C/s}=3.95×10^4\:s=11.0\: hr} \nonumber$
Check your answer: In a long problem like this, a single check is probably not enough. Each of the steps gives a reasonable number, so things are probably correct. Pay careful attention to unit conversions and the stoichiometry.
Exercise $2$
What mass of zinc is required to galvanize the top of a 3.00 m × 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO3)2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm3.
Answer
231 g Zn required 23,000 minutes.
Summary
Using electricity to force a nonspontaneous process to occur is electrolysis. Electrolytic cells are electrochemical cells with negative cell potentials (meaning a positive Gibbs free energy), and so are nonspontaneous. Electrolysis can occur in electrolytic cells by introducing a power supply, which supplies the energy to force the electrons to flow in the nonspontaneous direction. Electrolysis is done in solutions, which contain enough ions so current can flow. If the solution contains only one material, like the electrolysis of molten sodium chloride, it is a simple matter to determine what is oxidized and what is reduced. In more complicated systems, like the electrolysis of aqueous sodium chloride, more than one species can be oxidized or reduced and the standard reduction potentials are used to determine the most likely oxidation (the half-reaction with the largest [most positive] standard reduction potential) and reduction (the half-reaction with the smallest [least positive] standard reduction potential). Sometimes unexpected half-reactions occur because of overpotential. Overpotential is the difference between the theoretical half-reaction reduction potential and the actual voltage required. When present, the applied potential must be increased, making it possible for a different reaction to occur in the electrolytic cell. The total charge, Q, that passes through an electrolytic cell can be expressed as the current (I) multiplied by time (Q = It) or as the moles of electrons (n) multiplied by Faraday’s constant (Q = nF). These relationships can be used to determine things like the amount of material used or generated during electrolysis, how long the reaction must proceed, or what value of the current is required.
Summary
electrolysis
process using electrical energy to cause a nonspontaneous process to occur
electrolytic cell
electrochemical cell in which electrolysis is used; electrochemical cell with negative cell potentials
electroplating
depositing a thin layer of one metal on top of a conducting surface
overpotential
difference between the theoretical potential and actual potential in an electrolytic cell; the “extra” voltage required to make some nonspontaneous electrochemical reaction to occur
16.E: Electrochemistry (Exercises)
Migrated to ADAPT | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/16%3A_Electrochemistry/16.7%3A_Electrolysis.txt |
Chemical kinetics is the study of rates of chemical processes and includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction.
17: Kinetics
The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun’s rays is critical to the lizard’s survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. In the absence of warmth, the lizard is an easy meal for predators.
From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. When planning to run a chemical reaction, we should ask at least two questions. The first is: “Will the reaction produce the desired products in useful quantities?” The second question is: “How rapidly will the reaction occur?” A reaction that takes 50 years to produce a product is about as useful as one that never gives a product at all. A third question is often asked when investigating reactions in greater detail: “What specific molecular-level processes take place as the reaction occurs?” Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled.
The study of chemical kinetics concerns the second and third questions—that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. In this chapter, we will examine the factors that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to determine and describe the rate at which reactions occur.
17.1: Chemical Reaction Rates
Learning Objectives
• Define chemical reaction rate
• Derive rate expressions from the balanced equation for a given chemical reaction
• Calculate reaction rates from experimental data
A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.
The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.
For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. If we measure the concentration of hydrogen peroxide, H2O2, in an aqueous solution, we find that it changes slowly over time as the H2O2 decomposes, according to the equation:
$\ce{2H2O2}(aq)⟶\ce{2H2O}(l)+\ce{O2}(g) \nonumber$
The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:
\begin{align*} \ce{rate\: of\: decomposition\: of\: H_2O_2} &=\mathrm{−\dfrac{change\: in\: concentration\: of\: reactant}{time\: interval}}\[4pt] &=−\dfrac{[\ce{H2O2}]_{t_2}−[\ce{H2O2}]_{t_1}}{t_2−t_1}\[4pt] &=−\dfrac{Δ[\ce{H2O2}]}{Δt} \end{align*} \nonumber
This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus, $[\ce{H2O2}]_{t_1}$ represents the molar concentration of hydrogen peroxide at some time t1; likewise, $[\ce{H2O2}]_{t_2}$ represents the molar concentration of hydrogen peroxide at a later time t2; and Δ[H2O2] represents the change in molar concentration of hydrogen peroxide during the time interval Δt (that is, t2t1). Since the reactant concentration decreases as the reaction proceeds, Δ[H2O2] is a negative quantity; we place a negative sign in front of the expression because reaction rates are, by convention, positive quantities. Figure $1$ provides an example of data collected during the decomposition of H2O2.
To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:
$\dfrac{−Δ[\ce{H2O2}]}{Δt}=\mathrm{\dfrac{−(0.500\: mol/L−1.000\: mol/L)}{(6.00\: h−0.00\: h)}=0.0833\: mol\:L^{−1}\:h^{−1}} \nonumber$
Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:
$\dfrac{−Δ[\ce{H2O2}]}{Δt}=\mathrm{\dfrac{−(0.0625\:mol/L−0.125\:mol/L)}{(24.00\:h−18.00\:h)}=0.0104\:mol\:L^{−1}\:h^{−1}} \nonumber$
This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t0). A few moments later, the instantaneous rate at a specific moment—call it t1—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.
The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. If we plot the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H2O2 at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure $2$). We can use calculus to evaluating the slopes of such tangent lines, but the procedure for doing so is beyond the scope of this chapter.
Reaction Rates in Analysis: Test Strips for Urinalysis
Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure $2$). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.
The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:
$\ce{C6H12O6 + O2}\underset{\large\textrm{catalyst}}{\xrightarrow{\hspace{45px}}}\ce{C6H10O6 + H2O2} \label{eq1}$
$\ce{2H2O2 + 2I-}\underset{\large\textrm{catalyst}}{\xrightarrow{\hspace{45px}}}\ce{I2 + 2H2O + O2} \label{eq2}$
Equation $\ref{eq1}$ depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine (Equation $\ref{eq2}$), which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.
The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.
Relative Rates of Reaction
The rate of a reaction may be expressed in terms of the change in the amount of any reactant or product, and may be simply derived from the stoichiometry of the reaction. Consider the reaction represented by the following equation:
$\ce{2NH3}(g)⟶\ce{N2}(g)+\ce{3H2}(g) \nonumber$
The stoichiometric factors derived from this equation may be used to relate reaction rates in the same manner that they are used to related reactant and product amounts. The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:
$\mathrm{−\dfrac{Δmol\: NH_3}{Δ\mathit t}×\dfrac{1\: mol\: N_2}{2\: mol\: NH_3}=\dfrac{Δmol\:N_2}{Δ\mathit t}} \nonumber$
We can express this more simply without showing the stoichiometric factor’s units:
$−\dfrac{1}{2}\dfrac{\mathrm{Δmol\:NH_3}}{Δt}=\dfrac{\mathrm{Δmol\:N_2}}{Δt} \nonumber$
Note that a negative sign has been added to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). If the reactants and products are present in the same solution, the molar amounts may be replaced by concentrations:
$−\dfrac{1}{2}\dfrac{Δ[\ce{NH3}]}{Δt}=\dfrac{Δ[\ce{N2}]}{Δt} \nonumber$
Similarly, the rate of formation of H2 is three times the rate of formation of N2 because three moles of H2 form during the time required for the formation of one mole of N2:
$\dfrac{1}{3}\dfrac{Δ[\ce{H2}]}{Δt}=\dfrac{Δ[\ce{N2}]}{Δt} \nonumber$
Figure $3$ illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. We can see from the slopes of the tangents drawn at t = 500 seconds that the instantaneous rates of change in the concentrations of the reactants and products are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:
$\dfrac{2.91×10^{−6}\:M/\ce s}{9.71×10^{−6}\:M/\ce s}≈3 \nonumber$
Example $1$: Expressions for Relative Reaction Rates
The first step in the production of nitric acid is the combustion of ammonia:
$\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g) \nonumber$
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
Solution
Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:
$−\dfrac{1}{4}\dfrac{Δ[\ce{NH3}]}{Δt}=−\dfrac{1}{5}\dfrac{Δ[\ce{O2}]}{Δt}=\dfrac{1}{4}\dfrac{Δ[\ce{NO}]}{Δt}=\dfrac{1}{6}\dfrac{Δ[\ce{H2O}]}{Δt} \nonumber$
Exercise $1$
The rate of formation of Br2 is 6.0 × 10−6 mol/L/s in a reaction described by the following net ionic equation:
$\ce{5Br- + BrO3- + 6H+ ⟶ 3Br2 + 3H2O} \nonumber$
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
Answer
$−\dfrac{1}{5}\dfrac{Δ[\ce{Br-}]}{Δt}=−\dfrac{Δ[\ce{BrO3-}]}{Δt}=−\dfrac{1}{6}\dfrac{Δ[\ce{H+}]}{Δt}=\dfrac{1}{3}\dfrac{Δ[\ce{Br2}]}{Δt}=\dfrac{1}{3}\dfrac{Δ[\ce{H2O}]}{Δt} \nonumber$
Example $2$: Reaction Rate Expressions for Decomposition of H2O2
The graph in Figure $3$ shows the rate of the decomposition of H2O2 over time:
$\ce{2H2O2 ⟶ 2H2O + O2} \nonumber$
Based on these data, the instantaneous rate of decomposition of H2O2 at t = 11.1 h is determined to be 3.20 × 10−2 mol/L/h, that is:
$−\dfrac{Δ[\ce{H2O2}]}{Δt}=\mathrm{3.20×10^{−2}\:mol\: L^{−1}\:h^{−1}} \nonumber$
What is the instantaneous rate of production of H2O and O2?
Solution
Using the stoichiometry of the reaction, we may determine that:
$−\dfrac{1}{2}\dfrac{Δ[\ce{H2O2}]}{Δt}=\dfrac{1}{2}\dfrac{Δ[\ce{H2O}]}{Δt}=\dfrac{Δ[\ce{O2}]}{Δt} \nonumber$
Therefore:
$\mathrm{\dfrac{1}{2}×3.20×10^{−2}\:mol\:L^{−1}\:h^{−1}}=\dfrac{Δ[\ce{O2}]}{Δt} \nonumber$
and
$\dfrac{Δ[\ce{O2}]}{Δt}=\mathrm{1.60×10^{−2}\:mol\:L^{−1}\:h^{−1}} \nonumber$
Exercise $2$
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 × 10−6 mol/L/s, what is the rate of production of nitrogen and hydrogen?
Answer
1.05 × 10−6 mol/L/s, N2 and 3.15 × 10−6 mol/L/s, H2.
Summary
The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.
Glossary
average rate
rate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred
initial rate
instantaneous rate of a chemical reaction at t = 0 s (immediately after the reaction has begun)
instantaneous rate
rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time
rate of reaction
measure of the speed at which a chemical reaction takes place
rate expression
mathematical representation relating reaction rate to changes in amount, concentration, or pressure of reactant or product species per unit time | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.0%3A_Prelude_to_Kinetics.txt |
Learning Objectives
• Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates
The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. We can identify five factors that affect the rates of chemical reactions: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst.
The Chemical Nature of the Reacting Substances
The rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air overnight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive.
The State of Subdivision of the Reactants
Except for substances in the gaseous state or in solution, reactions occur at the boundary, or interface, between two phases. Hence, the rate of a reaction between two phases depends to a great extent on the surface contact between them. A finely divided solid has more surface area available for reaction than does one large piece of the same substance. Thus a liquid will react more rapidly with a finely divided solid than with a large piece of the same solid. For example, large pieces of iron react slowly with acids; finely divided iron reacts much more rapidly (Figure $1$). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.
Video $1$: The reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates.
Temperature of the Reactants
Chemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. We use a burner or a hot plate in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. In many cases, an increase in temperature of only 10 °C will approximately double the rate of a reaction in a homogeneous system.
Concentrations of the Reactants
The rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate ($\mathrm{CaCO_3}$) deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure $2$). As an acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:
$\ce{SO}_{2(g)}+\ce{H_2O}_{(g)}⟶\ce{H_2SO}_{3(aq)} \label{12.3.1}$
Calcium carbonate reacts with sulfurous acid as follows:
$\ce{CaCO}_{3(s)}+\ce{H_2SO}_{3(aq)}⟶\ce{CaSO}_{3(aq)}+\ce{CO}_{2(g)}+\ce{H_2O}_{(l)} \label{12.3.2}$
In a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about 20% oxygen.
Video $2$: Phosphorous burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen in is higher.
The Presence of a Catalyst
Hydrogen peroxide solutions foam when poured onto an open wound because substances in the exposed tissues act as catalysts, increasing the rate of hydrogen peroxide’s decomposition. However, in the absence of these catalysts (for example, in the bottle in the medicine cabinet) complete decomposition can take months. A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without itself being consumed by the reaction. Activation energy is the minimum amount of energy required for a chemical reaction to proceed in the forward direction. A catalyst increases the reaction rate by providing an alternative pathway or mechanism for the reaction to follow (Figure $3$). Catalysis will be discussed in greater detail later in this chapter as it relates to mechanisms of reactions.
Chemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the PhET Reactions & Rates interactive to explore how temperature, concentration, and the nature of the reactants affect reaction rates.
Summary
The rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway that causes the activation energy of the reaction to decrease.
Glossary
catalyst
substance that increases the rate of a reaction without itself being consumed by the reaction | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.2%3A_Factors_Affecting_Reaction_Rates.txt |
Learning Objectives
• Explain the form and function of a rate law
• Use rate laws to calculate reaction rates
• Use rate and concentration data to identify reaction orders and derive rate laws
As described in the previous module, the rate of a reaction is affected by the concentrations of reactants. Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In general, a rate law (or differential rate law, as it is sometimes called) takes this form:
$\ce{rate}=k[A]^m[B]^n[C]^p… \nonumber$
in which [A], [B], and [C] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents m, n, and p are usually positive integers (although it is possible for them to be fractions or negative numbers). The rate constant k and the exponents m, n, and p must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the concentration of A, B, or C, but it does vary with temperature and surface area.
The exponents in a rate law describe the effects of the reactant concentrations on the reaction rate and define the reaction order. Consider a reaction for which the rate law is:
$\ce{rate}=k[A]^m[B]^n \nonumber$
If the exponent m is 1, the reaction is first order with respect to A. If m is 2, the reaction is second order with respect to A. If n is 1, the reaction is first order in B. If n is 2, the reaction is second order in B. If m or n is zero, the reaction is zero order in A or B, respectively, and the rate of the reaction is not affected by the concentration of that reactant. The overall reaction order is the sum of the orders with respect to each reactant. If m = 1 and n = 1, the overall order of the reaction is second order (m + n = 1 + 1 = 2).
The rate law:
$\ce{rate}=k[\ce{H2O2}] \nonumber$
describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:
$\ce{rate}=k[\ce{C4H6}]^2 \nonumber$
describes a reaction that is second order in C4H6 and second order overall. The rate law:
$\ce{rate}=k[\ce{H+}][\ce{OH-}] \nonumber$
describes a reaction that is first order in H+, first order in OH, and second order overall.
Example $1$: Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber$
is second order in NO2 and zero order in CO at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
$\ce{rate}=k[\ce{NO2}]^m[\ce{CO}]^n \nonumber$
The reaction is second order in NO2; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:
$\ce{rate}=k[\ce{NO2}]^2[\ce{CO}]^0=k[\ce{NO2}]^2 \nonumber$
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
Exercise $\PageIndex{1A}$
The rate law for the reaction:
$\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g) \nonumber$
has been experimentally determined to be rate = k[NO]2[H2]. What are the orders with respect to each reactant, and what is the overall order of the reaction?
Answer
• order in NO = 2;
• order in H2 = 1;
• overall order = 3
Exercise $\PageIndex{1B}$
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel:
$\ce{CH3OH + CH3CH2OCOCH3 ⟶ CH3OCOCH3 + CH3CH2OH} \nonumber$
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be:
$\ce{rate}=k[\ce{CH3OH}] \nonumber$
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
Answer
• order in CH3OH = 1;
• order in CH3CH2OCOCH3 = 0;
• overall order = 1
It is sometimes helpful to use a more explicit algebraic method, often referred to as the method of initial rates, to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.
Example $2$: Determining a Rate Law from Initial Rates
Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure $1$). One such reaction is the combination of nitric oxide, NO, with ozone, O3:
$\ce{NO}(g)+\ce{O3}(g)⟶\ce{NO2}(g)+\ce{O2}(g) \nonumber$
This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.
Trial $[\ce{NO}]$ (mol/L) $[\ce{O3}]$ (mol/L) $\dfrac{Δ[\ce{NO2}]}{Δt}\:\mathrm{(mol\:L^{−1}\:s^{−1})}$
1 1.00 × 10−6 3.00 × 10−6 6.60 × 10−5
2 1.00 × 10−6 6.00 × 10−6 1.32 × 10−4
3 1.00 × 10−6 9.00 × 10−6 1.98 × 10−4
4 2.00 × 10−6 9.00 × 10−6 3.96 × 10−4
5 3.00 × 10−6 9.00 × 10−6 5.94 × 10−4
Determine the rate law and the rate constant for the reaction at 25 °C.
Solution
The rate law will have the form:
$\ce{rate}=k[\ce{NO}]^m[\ce{O3}]^n \nonumber$
We can determine the values of m, n, and k from the experimental data using the following three-part process:
1. Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.
2. Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus:
$\ce{rate}=k[\ce{NO}]^1[\ce{O3}]^1=k[\ce{NO}][\ce{O3}] \nonumber$
3. Determine the value of k from one set of concentrations and the corresponding rate.
\begin{align*} k&=\mathrm{\dfrac{rate}{[NO][O_3]}}\ &=\mathrm{\dfrac{6.60×10^{−5}\cancel{mol\: L^{−1}}\:s^{−1}}{(1.00×10^{−6}\cancel{mol\: L^{−1}})(3.00×10^{−6}\:mol\:L^{−1})}}\ &=\mathrm{2.20×10^7\:L\:mol^{−1}\:s^{−1}} \end{align*} \nonumber
The large value of k tells us that this is a fast reaction that could play an important role in ozone depletion if [NO] is large enough.
Exercise $2$
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
$\ce{CH3CHO}(g)⟶\ce{CH4}(g)+\ce{CO}(g) \nonumber$
Determine the rate law and the rate constant for the reaction from the following experimental data:
Trial $[\ce{CH3CHO}]$ (mol/L) $−\dfrac{Δ[\ce{CH3CHO}]}{Δt}\mathrm{(mol\:L^{−1}\:s^{−1})}$
1 1.75 × 10−3 2.06 × 10−11
2 3.50 × 10−3 8.24 × 10−11
3 7.00 × 10−3 3.30 × 10−10
Answer
$\ce{rate}=k[\ce{CH3CHO}]^2$ with k = 6.73 × 10−6 L/mol/s
Example $3$: Determining Rate Laws from Initial Rates
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:
$\ce{2NO}(g)+\ce{Cl2}(g)⟶\ce{2NOCl}(g) \nonumber$
Trial [NO] (mol/L) $[Cl_2]$ (mol/L) $−\dfrac{Δ[\ce{NO}]}{2Δt}\mathrm{(mol\:L^{−1}\:s^{−1})}$
1 0.10 0.10 0.00300
2 0.10 0.15 0.00450
3 0.15 0.10 0.00675
Solution
The rate law for this reaction will have the form:
$\ce{rate}=k[\ce{NO}]^m[\ce{Cl2}]^n \nonumber$
As in Example $2$, we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k. In this example, however, we will use a different approach to determine the values of m and n:
Determine the value of m from the data in which [NO] varies and [Cl2] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:
$\dfrac{\ce{rate}_x}{\ce{rate}_y}=\dfrac{k[\ce{NO}]^m_x[\ce{Cl2}]^n_x}{k[\ce{NO}]^m_y[\ce{Cl2}]^n_y} \nonumber$
Using the third trial and the first trial, in which [Cl2] does not vary, gives:
$\mathrm{\dfrac{rate\: 3}{rate\: 1}}=\dfrac{0.00675}{0.00300}=\dfrac{k(0.15)^m(0.10)^n}{k(0.10)^m(0.10)^n} \nonumber$
After canceling equivalent terms in the numerator and denominator, we are left with:
$\dfrac{0.00675}{0.00300}=\dfrac{(0.15)^m}{(0.10)^m} \nonumber$
which simplifies to:
$2.25=(1.5)^m \nonumber$
We can use natural logs to determine the value of the exponent m:
\begin{align*} \ln(2.25)&=m\ln(1.5) \dfrac{\ln(2.25)}{\ln(1.5)}&=m 2&=m \end{align*}
We can confirm the result easily, since:
$1.5^2=2.25$
• Determine the value of n from data in which [Cl2] varies and [NO] is constant. $\mathrm{\dfrac{rate\: 2}{rate\: 1}}=\dfrac{0.00450}{0.00300}=\dfrac{k(0.10)^m(0.15)^n}{k(0.10)^m(0.10)^n} \nonumber$
Cancelation gives:
$\dfrac{0.0045}{0.0030}=\dfrac{(0.15)^n}{(0.10)^n} \nonumber$
which simplifies to:
$1.5=(1.5)^n \nonumber$
Thus n must be 1, and the form of the rate law is:
$\ce{Rate}=k[\ce{NO}]^m[\ce{Cl2}]^n=k[\ce{NO}]^2[\ce{Cl2}] \nonumber$
• Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol−2 L2/s so that the rate is in terms of mol/L/s.
To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k:
\begin{align*} \mathrm{0.00300\:mol\:L^{−1}\:s^{−1}}&=k\mathrm{(0.10\:mol\:L^{−1})^2(0.10\:mol\:L^{−1})^1}\ k&=\mathrm{3.0\:mol^{−2}\:L^2\:s^{−1}} \end{align*}
Exercise $3$
Use the provided initial rate data to derive the rate law for the reaction whose equation is:
$\ce{OCl-}(aq)+\ce{I-}(aq)⟶\ce{OI-}(aq)+\ce{Cl-}(aq) \nonumber$
Trial [OCl] (mol/L) [I] (mol/L) Initial Rate (mol/L/s)
1 0.0040 0.0020 0.00184
2 0.0020 0.0040 0.00092
3 0.0020 0.0020 0.00046
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.
Answer
$\mathrm{\dfrac{rate\: 2}{rate\: 3}}=\dfrac{0.00092}{0.00046}=\dfrac{k(0.0020)^x(0.0040)^y}{k(0.0020)^x(0.0020)^y}$
2.00 = 2.00y
y = 1
$\mathrm{\dfrac{rate\: 1}{rate\: 2}}=\dfrac{0.00184}{0.00092}=\dfrac{k(0.0040)^x(0.0020)^y}{k(0.0020)^x(0.0040)^y}$
\begin{align*} 2.00&=\dfrac{2^x}{2^y}\ 2.00&=\dfrac{2^x}{2^1}\ 4.00&=2^x\ x&=2 \end{align*}
Substituting the concentration data from trial 1 and solving for k yields:
\begin{align*} \ce{rate}&=k[\ce{OCl-}]^2[\ce{I-}]^1\ 0.00184&=k(0.0040)^2(0.0020)^1\ k&=\mathrm{5.75×10^4\:mol^{−2}\:L^2\:s^{−1}} \end{align*}
Reaction Order and Rate Constant Units
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case. Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:
$\ce{NO2 + CO⟶NO + CO2}\hspace{20px}\ce{rate}=k[\ce{NO2}]^2\ \ce{CH3CHO⟶CH4 + CO}\hspace{20px}\ce{rate}=k[\ce{CH3CHO}]^2\ \ce{2N2O5⟶2NO2 + O2}\hspace{20px}\ce{rate}=k[\ce{N2O5}]\ \ce{2NO2 + F2⟶2NO2F}\hspace{20px}\ce{rate}=k[\ce{NO2}][\ce{F2}]\ \ce{2NO2Cl⟶2NO2 + Cl2}\hspace{20px}\ce{rate}=k[\ce{NO2Cl}]$
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
Reaction orders also play a role in determining the units for the rate constant k. In Example $2$, a second-order reaction, we found the units for k to be $\mathrm{L\:mol^{-1}\:s^{-1}}$, whereas in Example $3$, a third order reaction, we found the units for k to be mol−2 L2/s. More generally speaking, the units for the rate constant for a reaction of order $(m+n)$ are $\ce{mol}^{1−(m+n)}\ce L^{(m+n)−1}\ce s^{−1}$. Table $1$ summarizes the rate constant units for common reaction orders.
Table $1$: Rate Constants for Common Reaction Orders
Reaction Order Units of k
$(m+n)$ $\ce{mol}^{1−(m+n)}\ce L^{(m+n)−1}\ce s^{−1}$
zero mol/L/s
first s−1
second L/mol/s
third mol−2 L2 s−1
Note that the units in the table can also be expressed in terms of molarity (M) instead of mol/L. Also, units of time other than the second (such as minutes, hours, days) may be used, depending on the situation.
Summary
Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the amount of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.
Glossary
method of initial rates
use of a more explicit algebraic method to determine the orders in a rate law
overall reaction order
sum of the reaction orders for each substance represented in the rate law
rate constant (k)
proportionality constant in the relationship between reaction rate and concentrations of reactants
rate law
(also, rate equation) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants
reaction order
value of an exponent in a rate law, expressed as an ordinal number (for example, zero order for 0, first order for 1, second order for 2, and so on) | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.3%3A_Rate_Laws.txt |
Learning Objectives
• Explain the form and function of an integrated rate law
• Perform integrated rate law calculations for zero-, first-, and second-order reactions
• Define half-life and carry out related calculations
• Identify the order of a reaction from concentration/time data
The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.
Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.
First-Order Reactions
An equation relating the rate constant $k$ to the initial concentration $[A]_0$ and the concentration $[A]_t$ present after any given time $t$ can be derived for a first-order reaction and shown to be:
$\ln\left(\dfrac{[A]_t}{[A]_0}\right)=−kt \nonumber$
or alternatively
$\ln\left(\dfrac{[A]_0}{[A]_t}\right)=kt \nonumber$
or
$[A]=[A]_0e^{−kt} \nonumber$
Example $1$: The Integrated Rate Law for a First-Order Reaction
The rate constant for the first-order decomposition of cyclobutane, $\ce{C4H8}$ at 500 °C is 9.2 × 10−3 s−1:
$\ce{C4H8⟶2C2H4} \nonumber$
How long will it take for 80.0% of a sample of C4H8 to decompose?
Solution
We use the integrated form of the rate law to answer questions regarding time:
$\ln\left(\dfrac{[A]_0}{[A]}\right)=kt \nonumber$
There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]0, [A], and k, and need to find t.
The initial concentration of C4H8, [A]0, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:
\begin{align*} t&=\ln\dfrac{[x]}{[0.200x]}×\dfrac{1}{k}\[4pt] &=\mathrm{\ln\dfrac{0.100\:mol\: L^{−1}}{0.020\:mol\: L^{−1}}×\dfrac{1}{9.2×10^{−3}\:s^{−1}}}\[4pt] &=\mathrm{1.609×\dfrac{1}{9.2×10^{−3}\:s^{−1}}}\[4pt] &=\mathrm{1.7×10^2\:s} \end{align*} \nonumber
Exercise $1$
Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:
$\textrm{I-131 ⟶ Xe-131 + electron} \nonumber$
The decay is first-order with a rate constant of 0.138 d−1. All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?
Answer
16.7 days
We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:
\begin{align} \ln[A]&=(−k)(t)+\ln[A]_0 \label{in1st}\[4pt] y&=mx+b \end{align} \nonumber
A plot of $\ln[A]$ versus $t$ for a first-order reaction is a straight line with a slope of $−k$ and an intercept of $\ln[A]_0$. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in $A$.
Example $2$: Determination of Reaction Order by Graphing
Show that the data in this Figure can be represented by a first-order rate law by graphing ln[H2O2] versus time. Determine the rate constant for the rate of decomposition of H2O2 from this data.
Solution
The data from this Figure with the addition of values of ln[H2O2] are given in Figure $1$.
Solutions to Example 12.4.2
Trial Time (h) [H2O2] (M) ln[H2O2]
1 0 1.000 0.0
2 6.00 0.500 −0.693
3 12.00 0.250 −1.386
4 18.00 0.125 −2.079
5 24.00 0.0625 −2.772
The plot of ln[H2O2] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.
The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H2O2] versus time where:
$\ce{slope}=\dfrac{\textrm{change in }y}{\textrm{change in }x}=\dfrac{Δy}{Δx}=\dfrac{Δ\ln[\ce{H2O2}]}{Δt} \nonumber$
In order to determine the slope of the line, we need two values of ln[H2O2] at different values of t (one near each end of the line is preferable). For example, the value of ln[H2O2] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:
\begin{align*} \ce{slope}&=\mathrm{\dfrac{−1.386−(−0.693)}{12.00\: h−6.00\: h}}\[4pt] &=\mathrm{\dfrac{−0.693}{6.00\: h}}\[4pt] &=\mathrm{−1.155×10^{−2}\:h^{−1}}\[4pt] k&=\mathrm{−slope=−(−1.155×10^{−1}\:h^{−1})=1.155×10^{−1}\:h^{−1}} \end{align*} \nonumber
Exercise $2$
Graph the following data to determine whether the reaction $A⟶B+C$ is first order.
Data for the reaction $A⟶B+C$
Trial Time (s) [A]
1 4.0 0.220
2 8.0 0.144
3 12.0 0.110
4 16.0 0.088
5 20.0 0.074
Answer
The plot of ln[A] vs. t is not a straight line. The equation is not first order:
Second-Order Reactions
The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:
$\ce{Rate}=k[A]^2 \nonumber$
For these second-order reactions, the integrated rate law is:
$\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \label{int2nd}$
where the terms in the equation have their usual meanings as defined earlier.
Example $3$: The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows:
$\ce{2C4H6}(g)⟶\ce{C8H12(g)} \nonumber$
The reaction is second order with a rate constant equal to 5.76 × 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min?
Solution
We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:
$\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \nonumber$
We know three variables in this equation: [A]0 = 0.200 mol/L, k = 5.76 × 10−2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:
\begin{align*} \dfrac{1}{[A]}&=\mathrm{(5.76×10^{−2}\:L\: mol^{−1}\:min^{−1})(10\:min)+\dfrac{1}{0.200\:mol^{−1}}}\[4pt] \dfrac{1}{[A]}&=\mathrm{(5.76×10^{−1}\:L\: mol^{−1})+5.00\:L\: mol^{−1}}\[4pt] \dfrac{1}{[A]}&=\mathrm{5.58\:L\: mol^{−1}}\[4pt] [A]&=\mathrm{1.79×10^{−1}\:mol\: L^{−1}} \end{align*} \nonumber
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
Exercise $3$
If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?
Answer
0.0196 mol/L
The integrated rate law for our second-order reactions has the form of the equation of a straight line:
\begin{align*} \dfrac{1}{[A]}&=kt+\dfrac{1}{[A]_0}\[4pt] y&=mx+b \end{align*} \nonumber
A plot of $\dfrac{1}{[A]}$ versus t for a second-order reaction is a straight line with a slope of k and an intercept of $\dfrac{1}{[A]_0}$. If the plot is not a straight line, then the reaction is not second order.
Example $4$: Determination of Reaction Order by Graphing
Test the data given to show whether the dimerization of C4H6 is a first- or a second-order reaction.
Solution
Solutions to Example 12.4.4
Trial Time (s) [C4H6] (M)
1 0 1.00 × 10−2
2 1600 5.04 × 10−3
3 3200 3.37 × 10−3
4 4800 2.53 × 10−3
5 6200 2.08 × 10−3
In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C4H6] versus t and compare it with a plot of $\mathrm{\dfrac{1}{[C_4H_6]}}$ versus t. The values needed for these plots follow.
Solutions to Example 12.4.4
Time (s) $\dfrac{1}{[\ce{C4H6}]}\:(M^{−1})$ ln[C4H6]
0 100 −4.605
1600 198 −5.289
3200 296 −5.692
4800 395 −5.978
6200 481 −6.175
The plots are shown in Figure $2$. As you can see, the plot of ln[C4H6] versus t is not linear, therefore the reaction is not first order. The plot of $\dfrac{1}{[\ce{C4H6}]}$ versus t is linear, indicating that the reaction is second order.
Exercise $4$
Does the following data fit a second-order rate law?
Data for second-order rate law
Trial Time (s) [A] (M)
1 5 0.952
2 10 0.625
3 15 0.465
4 20 0.370
5 25 0.308
6 35 0.230
Answer
Yes. The plot of $\dfrac{1}{[A]}$ vs. t is linear:
Zero-Order Reactions
For zero-order reactions, the differential rate law is:
$\ce{Rate}=k[A]^0=k \nonumber$
A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants.
The integrated rate law for a zero-order reaction also has the form of the equation of a straight line:
\begin{align*} [A]&=−kt+[A]_0 \label{intzero}\[4pt] y&=mx+b \end{align*} \nonumber
A plot of $[A]$ versus $t$ for a zero-order reaction is a straight line with a slope of −k and an intercept of [A]0. Figure $3$ shows a plot of [NH3] versus t for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO2). The decomposition of NH3 on hot tungsten is zero order; the plot is a straight line. The decomposition of NH3 on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant:
$\ce{slope}=−k=\mathrm{1.3110^{−6}\:mol/L/s} \nonumber$
The Half-Life of a Reaction
The half-life of a reaction (t1/2) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H2O2 decreases from 1.000 M to 0.500 M. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of H2O2 decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.
First-Order Reactions
We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:
\begin{align*} \ln\dfrac{[A]_0}{[A]}&=kt\[4pt] t&=\ln\dfrac{[A]_0}{[A]}×\dfrac{1}{k} \end{align*} \nonumber
If we set the time t equal to the half-life, $t_{1/2}$, the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when $t=t_{1/2}$, $[A]=\dfrac{1}{2}[A]_0$.
Therefore:
\begin{align*} t_{1/2}&=\ln\dfrac{[A]_0}{\dfrac{1}{2}[A]_0}×\dfrac{1}{k}\[4pt] &=\ln 2×\dfrac{1}{k}=0.693×\dfrac{1}{k} \end{align*} \nonumber
Thus:
$t_{1/2}=\dfrac{0.693}{k} \nonumber$
We can see that the half-life of a first-order reaction is inversely proportional to the rate constant k. A fast reaction (shorter half-life) will have a larger k; a slow reaction (longer half-life) will have a smaller k.
Example $5$: Calculation of a First-order Rate Constant using Half-Life
Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in Figure $4$.
Solution The half-life for the decomposition of H2O2 is 2.16 × 104 s:
\begin{align*} t_{1/2}&=\dfrac{0.693}{k}\[4pt] k&=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{2.16×10^4\:\ce s}=3.21×10^{−5}\:\ce s^{−1} \end{align*} \nonumber
Exercise $1$
The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d−1. What is the half-life for this decay?
Answer
5.02 d.
Second-Order Reactions
We can derive the equation for calculating the half-life of a second order as follows:
$\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \nonumber$
or
$\dfrac{1}{[A]}−\dfrac{1}{[A]_0}=kt \nonumber$
If
$t=t_{1/2} \nonumber$
then
$[A]=\dfrac{1}{2}[A]_0 \nonumber$
and we can write:
\begin{align*} \dfrac{1}{\dfrac{1}{2}[A]_0}−\dfrac{1}{[A]_0}&=kt_{1/2}\4pt] 2[A]_0−\dfrac{1}{[A]_0}&=kt_{1/2}\[4pt] \dfrac{1}{[A]_0}&=kt_{1/2} \end{align*} Thus: \[t_{1/2}=\dfrac{1}{k[A]_0} \nonumber
For a second-order reaction, $t_{1/2}$ is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.
Zero-Order Reactions
We can derive an equation for calculating the half-life of a zero order reaction as follows:
$[A]=−kt+[A]_0 \nonumber$
When half of the initial amount of reactant has been consumed $t=t_{1/2}$ and $[A]=\dfrac{[A]_0}{2}$. Thus:
\begin{align*} \dfrac{[A]_0}{2}&=−kt_{1/2}+[A]_0\[4pt] kt_{1/2}&=\dfrac{[A]_0}{2} \end{align*} \nonumber
and
$t_{1/2}=\dfrac{[A]_0}{2k} \nonumber$
The half-life of a zero-order reaction increases as the initial concentration increases. Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table $1$.
Table $1$: Summary of Rate Laws for Zero-, First-, and Second-Order Reactions
Zero-Order First-Order Second-Order
rate law rate = k rate = k[A] rate = k[A]2
units of rate constant M s−1 s−1 M−1 s−1
integrated rate law [A] = −kt + [A]0 ln[A] = −kt + ln[A]0 $\dfrac{1}{[A]}=kt+\left(\dfrac{1}{[A]_0}\right)$
plot needed for linear fit of rate data [A] vs. t ln[A] vs. t $\dfrac{1}{[A]}$ vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = +slope
half-life $t_{1/2}=\dfrac{[A]_0}{2k}$ $t_{1/2}=\dfrac{0.693}{k}$ $t_{1/2}=\dfrac{1}{[A]_0k}$
Summary
Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.
The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.
Key Equations
• integrated rate law for zero-order reactions (Equation \ref{intzero}):
$[A]=−kt+[A]_0$ with $t_{1/2}=\dfrac{[A]_0}{2k}$
• integrated rate law for first-order reactions (Equation \ref{in1st}):
$\ln[A]=−kt+ \ln[A]_0$ with $t_{1/2}=\dfrac{0.693}{k}$
• integrated rate law for second-order reactions (Equation \ref{int2nd}):
$\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0}$ with $t_{1/2}=\dfrac{1}{[A]_0k}$
Glossary
half-life of a reaction (tl/2)
time required for half of a given amount of reactant to be consumed
integrated rate law
equation that relates the concentration of a reactant to elapsed time of reaction | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.4%3A_Integrated_Rate_Laws.txt |
Learning Objectives
• Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
• Define the concepts of activation energy and transition state
• Use the Arrhenius equation in calculations relating rate constants to temperature
We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates. Collision theory is based on the following postulates:
Postulates of Collision theory
1. The rate of a reaction is proportional to the rate of reactant collisions: $\mathrm{reaction\: rate ∝ \dfrac{\#\,collisions}{time}} \nonumber$
2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
3. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).
We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen:
$\ce{2CO}(g)+\ce{O2}(g)⟶\ce{2CO2}(g) \nonumber$
Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient quantity, the reaction is spontaneous at high temperature and pressure.
The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:
$\ce{CO}(g)+\ce{O2}(g)⟶\ce{CO2}(g)+\ce{O}(g) \nonumber$
Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure $1$. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms $\ce{(O=C=O)}$. This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.
If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. Every reaction requires a certain amount of activation energy for it to proceed in the forward direction, yielding an appropriate activated complex along the way. As Figure $2$ demonstrates, even a collision with the correct orientation can fail to form the reaction product. In the study of reaction mechanisms, each of these three arrangements of atoms is called a proposed activated complex or transition state.
In most circumstances, it is impossible to isolate or identify a transition state or activated complex. In the reaction between carbon monoxide and oxygen to form carbon dioxide, activated complexes have only been observed spectroscopically in systems that utilize a heterogeneous catalyst. The gas-phase reaction occurs too rapidly to isolate any such chemical compound.
Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.
Activation Energy and the Arrhenius Equation
The minimum energy necessary to form a product during a collision between reactants is called the activation energy ($E_a$). The kinetic energy of reactant molecules plays an important role in a reaction because the energy necessary to form a product is provided by a collision of a reactant molecule with another reactant molecule. (In single-reactant reactions, activation energy may be provided by a collision of the reactant molecule with the wall of the reaction vessel or with molecules of an inert contaminant.) If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly: Only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, the fraction of molecules possessing the necessary kinetic energy will be large; most collisions between molecules will result in reaction, and the reaction will occur rapidly.
Figure $3$ shows the energy relationships for the general reaction of a molecule of $A$ with a molecule of $B$ to form molecules of $C$ and $D$:
$A+B⟶C+D \nonumber$
The figure shows that the energy of the transition state is higher than that of the reactants $A$ and $B$ by an amount equal to $E_a$, the activation energy. Thus, the sum of the kinetic energies of $A$ and $B$ must be equal to or greater than Ea to reach the transition state. After the transition state has been reached, and as $C$ and $D$ begin to form, the system loses energy until its total energy is lower than that of the initial mixture. This lost energy is transferred to other molecules, giving them enough energy to reach the transition state. The forward reaction (that between molecules $A$ and $B$) therefore tends to take place readily once the reaction has started. In Figure $3$, $ΔH$ represents the difference in enthalpy between the reactants ($A$ and $B$) and the products ($C$ and $D$). The sum of $E_a$ and $ΔH$ represents the activation energy for the reverse reaction:
$C+D⟶A+B \nonumber$
We can use the Arrhenius equation to relate the activation energy and the rate constant, k, of a given reaction:
$k=Ae^{−E_a/RT} \label{Arrhenius}$
In this equation,
• $R$ is the ideal gas constant, which has a value 8.314 J/mol/K,
• $T$ is temperature on the Kelvin scale,
• $E_a$ is the activation energy in joules per mole,
• $e$ is the constant 2.7183, and
• $A$ is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.
Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, $e^{−E_a/RT}$, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.
At one extreme, the system does not contain enough energy for collisions to overcome the activation barrier. In such cases, no reaction occurs. At the other extreme, the system has so much energy that every collision with the correct orientation can overcome the activation barrier, causing the reaction to proceed. In such cases, the reaction is nearly instantaneous.
The Arrhenius equation (Equation \ref{Arrhenius}) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of $E_a$ results in a smaller value for $e^{−E_a/RT}$, reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller $E_a$ has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of $e^{−E_a/RT}$, a larger rate constant, and a faster rate for the reaction.
An increase in temperature has the same effect as a decrease in activation energy. A larger fraction of molecules has the necessary energy to react (Figure $4$), as indicated by an increase in the value of $e^{−E_a/RT}$. The rate constant is also directly proportional to the frequency factor, $A$. Hence a change in conditions or reactants that increases the number of collisions with a favorable orientation for reaction results in an increase in $A$ and, consequently, an increase in $k$.
A convenient approach to determining $E_a$ for a reaction involves the measurement of $k$ at different temperatures and using of an alternate version of the Arrhenius equation that takes the form of linear equation:
\begin{align*} \ln k&=\left(\dfrac{−E_a}{R}\right)\left(\dfrac{1}{T}\right)+\ln A\ y&=mx+b \end{align*} \nonumber
Thus, a plot of $\ln k$ versus $\dfrac{1}{T}$ gives a straight line with the slope $\dfrac{-E_\ce{a}}{R}$, from which Ea may be determined. The intercept gives the value of $\ln A$. This is sometimes call an Arrhenius Plot.
Example $1$
Determination of Ea The variation of the rate constant with temperature for the decomposition of HI(g) to H2(g) and I2(g) is given here. What is the activation energy for the reaction?
$\ce{2HI}(g)⟶\ce{H2}(g)+\ce{I2}(g) \nonumber$
variation of the rate constant with temperature for the decomposition of HI(g) to H2(g) and I2(g)
T (K) k (L/mol/s)
555 3.52 × 10−7
575 1.22 × 10−6
645 8.59 × 10−5
700 1.16 × 10−3
781 3.95 × 10−2
Solution
Values of $\dfrac{1}{T}$ and ln k are:
Solutions to Example 12.5.1
$\mathrm{\dfrac{1}{T}\:(K^{−1})}$ ln k
1.80 × 10−3 −14.860
1.74 × 10−3 −13.617
1.55 × 10−3 −9.362
1.43 × 10−3 −6.759
1.28 × 10−3 −3.231
Figure $5$ is a graph of ln k versus $\dfrac{1}{T}$. To determine the slope of the line, we need two values of ln k, which are determined from the line at two values of $\dfrac{1}{T}$ (one near each end of the line is preferable). For example, the value of ln k determined from the line when $\dfrac{1}{T}=1.25×10^{−3}$ is −2.593; the value when $\dfrac{1}{T}=1.78×10^{−3}$ is −14.447.
The slope of this line is given by the following expression:
\begin{align*} \ce{Slope}&=\dfrac{Δ(\ln k)}{Δ\left(\dfrac{1}{T}\right)}\ &=\mathrm{\dfrac{(−14.447)−(−2.593)}{(1.78×10^{−3}\:K^{−1})−(1.25×10^{−3}\:K^{−1})}}\ &=\mathrm{\dfrac{−11.854}{0.53×10^{−3}\:K^{−1}}=2.2×10^4\:K}\ &=−\dfrac{E_\ce{a}}{R} \end{align*} \nonumber
Thus:
\begin{align*} E_\ce{a} &=\mathrm{−slope×\mathit R=−(−2.2×10^4\:K×8.314\: J\: mol^{−1}\:K^{−1})} \[4pt] &=\mathrm{1.8×10^5\:J\: mol^{−1}} \end{align*} \nonumber
In many situations, it is possible to obtain a reasonable estimate of the activation energy without going through the entire process of constructing the Arrhenius plot. The Arrhenius equation:
$\ln k=\left(\dfrac{−E_\ce{a}}{R}\right)\left(\dfrac{1}{T}\right)+\ln A \nonumber$
can be rearranged as shown to give:
$\dfrac{Δ(\ln k)}{Δ\left(\dfrac{1}{T}\right)}=−\dfrac{E_\ce{a}}{R} \nonumber$
or
$\ln\dfrac{k_1}{k_2}=\dfrac{E_\ce{a}}{R}\left(\dfrac{1}{T_2}−\dfrac{1}{T_1}\right) \nonumber$
This equation can be rearranged to give a one-step calculation to obtain an estimate for the activation energy:
$E_\ce{a}=−R\left( \dfrac{\ln k_2−\ln k_1}{\left(\dfrac{1}{T_2}\right)−\left(\dfrac{1}{T_1}\right)}\right ) \nonumber$
Using the experimental data presented here, we can simply select two data entries. For this example, we select the first entry and the last entry:
First and Last Entry
T (K) k (L/mol/s) $\dfrac{1}{T}\:(K^{-1})$ ln k
555 3.52 × 10−7 1.80 × 10−3 −14.860
781 3.95 × 10−2 1.28 × 10−3 −3.231
After calculating $\dfrac{1}{T}$ and ln k, we can substitute into the equation:
$E_\ce{a}=\mathrm{−8.314\:J\:mol^{−1}\:K^{−1}\left(\dfrac{−3.231−(−14.860)}{1.28×10^{−3}\:K^{−1}−1.80×10^{−3}\:K^{−1}}\right)} \nonumber$
and the result is Ea = 185,900 J/mol.
This method is very effective, especially when a limited number of temperature-dependent rate constants are available for the reaction of interest.
Exercise $1$
The rate constant for the rate of decomposition of N2O5 to NO and O2 in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:
$\ce{2N2O5}(g)⟶\ce{4NO}(g)+\ce{3O2}(g) \nonumber$
Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.
Answer
113,000 J/mol
Summary
Chemical reactions require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant and its activation energy, temperature, and dependence on collision orientation.
Key Equations
• $k=Ae^{−E_a/RT}$
• $\ln k=\left(\dfrac{−E_\ce{a}}{R}\right)\left(\dfrac{1}{T}\right)+\ln A$
• $\ln\dfrac{k_1}{k_2}=\dfrac{E_\ce{a}}{R}\left(\dfrac{1}{T_2}−\dfrac{1}{T_1}\right)$
Glossary
activated complex
(also, transition state) unstable combination of reactant species representing the highest energy state of a reaction system
activation energy (Ea)
energy necessary in order for a reaction to take place
Arrhenius equation
mathematical relationship between the rate constant and the activation energy of a reaction
collision theory
model that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics
frequency factor (A)
proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.5%3A_Collision_Theory.txt |
Learning Objectives
• Distinguish net reactions from elementary reactions (steps)
• Identify the molecularity of elementary reactions
• Write a balanced chemical equation for a process given its reaction mechanism
• Derive the rate law consistent with a given reaction mechanism
A balanced equation for a chemical reaction indicates what is reacting and what is produced, but it reveals nothing about how the reaction actually takes place. The reaction mechanism (or reaction path) is the process, or pathway, by which a reaction occurs. A chemical reaction often occurs in steps, although it may not always be obvious to an observer. The decomposition of ozone, for example, appears to follow a mechanism with two steps:
$\ce{O3}(g)⟶\ce{O2}(g)+\ce{O}\ \ce{O}+\ce{O3}(g)⟶\ce{2O2}(g) \label{12.7.1}$
We call each step in a reaction mechanism an elementary reaction. Elementary reactions occur exactly as they are written and cannot be broken down into simpler steps. Elementary reactions add up to the overall reaction, which, for the decomposition, is:
$\ce{2O3}(g)⟶\ce{3O2}(g) \label{12.7.2}$
Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.
While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction does not involve the collision and reaction of two ozone molecules. Rather, it involves a molecule of ozone decomposing to an oxygen molecule and an intermediate oxygen atom; the oxygen atom then reacts with a second ozone molecule to give two oxygen molecules. These two elementary reactions occur exactly as they are shown in the reaction mechanism.
Unimolecular Elementary Reactions
The molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the rearrangement of a single reactant species to produce one or more molecules of product:
$A⟶\ce{products} \label{12.7.2b}$
The rate equation for a unimolecular reaction is:
$\ce{rate}=k[A] \label{12.7.3}$
A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:
$\ce{O3 ⟶ O2 + O} \label{12.7.4}$
illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism. However, some unimolecular reactions may have only a single reaction in the reaction mechanism. (In other words, an elementary reaction can also be an overall reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C4H8, to ethylene, C2H4, occurs via a unimolecular, single-step mechanism:
For these unimolecular reactions to occur, all that is required is the separation of parts of single reactant molecules into products.
Chemical bonds do not simply fall apart during chemical reactions. Energy is required to break chemical bonds. The activation energy for the decomposition of C4H8, for example, is 261 kJ per mole. This means that it requires 261 kilojoules to distort one mole of these molecules into activated complexes that decompose into products:
In a sample of C4H8, a few of the rapidly moving C4H8 molecules collide with other rapidly moving molecules and pick up additional energy. When the C4H8 molecules gain enough energy, they can transform into an activated complex, and the formation of ethylene molecules can occur. In effect, a particularly energetic collision knocks a C4H8 molecule into the geometry of the activated complex. However, only a small fraction of gas molecules travel at sufficiently high speeds with large enough kinetic energies to accomplish this. Hence, at any given moment, only a few molecules pick up enough energy from collisions to react.
The rate of decomposition of C4H8 is directly proportional to its concentration. Doubling the concentration of C4H8 in a sample gives twice as many molecules per liter. Although the fraction of molecules with enough energy to react remains the same, the total number of such molecules is twice as great. Consequently, there is twice as much C4H8 per liter, and the reaction rate is twice as fast:
$\ce{rate}=−\dfrac{Δ[\ce{C4H8}]}{Δt}=k[\ce{C4H8}] \label{12.7.5}$
A similar relationship applies to any unimolecular elementary reaction; the reaction rate is directly proportional to the concentration of the reactant, and the reaction exhibits first-order behavior. The proportionality constant is the rate constant for the particular unimolecular reaction.
Bimolecular Elementary Reactions
The collision and combination of two molecules or atoms to form an activated complex in an elementary reaction is called a bimolecular reaction. There are two types of bimolecular elementary reactions:
$A+B⟶\ce{products} \label{12.7.6}$
and
$2A⟶\ce{products} \label{12.7.7}$
For the first type, in which the two reactant molecules are different, the rate law is first-order in A and first order in B:
$\ce{rate}=k[A][B] \label{12.7.8}$
For the second type, in which two identical molecules collide and react, the rate law is second order in A:
$\ce{rate}=k[A][A]=k[A]^2 \label{12.7.9}$
Some chemical reactions have mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \label{12.7.10}$
(see Figure $1$)
Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is one example:
$\ce{O}(g)+\ce{O3}(g)⟶\ce{2O2}(g) \label{12.7.12}$
Termolecular Elementary Reactions
An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:
$\ce{2NO + O2 ⟶ 2NO2}\ \ce{rate}=k[\ce{NO}]^2[\ce{O2}] \label{12.7.13}$
Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:
$\ce{2NO + Cl2 ⟶ 2NOCl}\ \ce{rate}=k[\ce{NO}]^2[\ce{Cl2}] \label{12.7.14}$
Relating Reaction Mechanisms to Rate Laws
It's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure $2$.
As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, we must determine the overall rate law from experimental data and deduce the mechanism from the rate law (and sometimes from other data). The reaction of NO2 and CO provides an illustrative example:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{CO2}(g)+\ce{NO}(g) \nonumber$
For temperatures above 225 °C, the rate law has been found to be:
$\ce{rate}=k[\ce{NO2}][\ce{CO}] \nonumber$
The reaction is first order with respect to NO2 and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.
At temperatures below 225 °C, the reaction is described by a rate law that is second order with respect to NO2:
$\ce{rate}=k[\ce{NO2}]^2 \nonumber$
This is consistent with a mechanism that involves the following two elementary reactions, the first of which is slower and is therefore the rate-determining step:
$\ce{NO2}(g)+\ce{NO2}(g)⟶\ce{NO3}(g)+\ce{NO}(g)\:\ce{(slow)}\ \ce{NO3}(g)+\ce{CO}(g)⟶\ce{NO2}(g)+\ce{CO2}(g)\:\ce{(fast)} \nonumber$
The rate-determining step gives a rate law showing second-order dependence on the NO2 concentration, and the sum of the two equations gives the net overall reaction.
In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving an equilibrium reaction, the rate law for the overall reaction may be more difficult to derive.
An elementary reaction is at equilibrium when it proceeds in both the forward and reverse directions at equal rates. Consider the dimerization of NO to N2O2, with k1 used to represent the rate constant of the forward reaction and k-1 used to represent the rate constant of the reverse reaction:
$\ce{NO + NO ⇌ N2O2} \nonumber$
$\ce{rate_{forward}=rate_{reverse}} \nonumber$
$k_1[\ce{NO}]^2=k_{−1}[\ce{N2O2}] \nonumber$
If N2O2 was an intermediate in a mechanism, this expression could be rearranged to represent the concentration of N2O2 in the overall rate law expression using algebraic manipulation:
$\mathrm{\left(\dfrac{k_1[NO]^2}{k_{−1}}\right)=[N_2O_2]} \nonumber$
However, once again, intermediates cannot be listed as part of the overall rate law expression, though they can be included in an individual elementary reaction of a mechanism. Example $1$ will illustrate how to derive overall rate laws from mechanisms involving equilibrium steps preceding the rate-determining step.
Example $1$: Deriving the Overall Rate Law Expression for a Multistep Reaction
Mechanism Nitryl chloride (NO2Cl) decomposes to nitrogen dioxide (NO2) and chlorine gas (Cl2) according to the following mechanism:
1. $\ce{2NO2Cl}(g)⇌\ce{ClO2}(g)+\ce{N2O}(g)+\ce{ClO}(g)$ (fast, k1 represents the rate constant for the forward reaction and k−1 the rate constant for the reverse reaction)
2. $\ce{N2O}(g)+\ce{ClO2}(g)⇌\ce{NO2}(g)+\ce{NOCl}(g)$ (fast, k2 for the forward reaction, k−2 for the reverse reaction)
3. $\ce{NOCl + ClO ⟶ NO2 + Cl2}$ (slow, k3 the rate constant for the forward reaction)
Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression.
Solution
For the overall reaction, simply sum the three steps, cancel intermediates, and combine like formulas:
$\ce{2NO2Cl}(g)⟶\ce{2NO2}(g)+\ce{Cl2}(g) \nonumber$
Next, write the rate law expression for each elementary reaction. Remember that for elementary reactions that are part of a mechanism, the rate law expression can be derived directly from the stoichiometry:
\begin{align*} k_1\ce{[NO2Cl]2}&=k_{−1}\ce{[ClO2][N2O][ClO]}\ k_2\ce{[N2O][ClO2]}&=k_{−2}\ce{[NO2][NOCl]}\ \ce{Rate}&=k_3\ce{[NOCl][ClO]} \end{align*} \nonumber
The third step, which is the slow step, is the rate-determining step. Therefore, the overall rate law expression could be written as Rate = k3 [NOCl][ClO]. However, both NOCl and ClO are intermediates. Algebraic expressions must be used to represent [NOCl] and [ClO] such that no intermediates remain in the overall rate law expression.
• Using elementary reaction 1, $\ce{[ClO]}=\dfrac{k_1\ce{[NO2Cl]^2}}{k_{−1}\ce{[ClO2][N2O]}}$.
• Using elementary reaction 2, $\ce{[NOCl]}=\dfrac{k_2\ce{[N2O][ClO2]}}{k_{−2}\ce{[NO2]}}$.
Now substitute these algebraic expressions into the overall rate law expression and simplify:
\begin{align*} \ce{rate}&=k_3\left(\dfrac{k_2\ce{[N2O][ClO2]}}{k_{−2}\ce{[NO2]}}\right)\left(\dfrac{k_1\ce{[NO2Cl]^2}}{k_{−1}\ce{[ClO2][N2O]}}\right)\ \ce{rate}&=\dfrac{k_3k_2k_1\ce{[NO2Cl]^2}}{k_{−2}k_{−1}\ce{[NO2]}} \end{align*} \nonumber
Notice that this rate law shows an inverse dependence on the concentration of one of the product species, consistent with the presence of an equilibrium step in the reaction mechanism.
Exercise $1$
Atomic chlorine in the atmosphere reacts with ozone in the following pair of elementary reactions:
$\ce{Cl}+\ce{O3}(g)⟶\ce{ClO}(g)+\ce{O2}(g)\hspace{20px}(\textrm{rate constant }k_1)$
$\ce{ClO}(g)+\ce{O}⟶\ce{Cl}(g)+\ce{O2}(g)\hspace{20px}(\textrm{rate constant }k_2)$
Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression.
Answer
• overall reaction: $\ce{O3}(g)+\ce{O}⟶\ce{2O2}(g)$
• rate1 = k1[O3][Cl]; rate2 = k2[ClO][O]
• intermediate: ClO(g)
• overall rate = k2k1[O3][Cl][O]
Summary
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The overall rate of a reaction is determined by the rate of the slowest step, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.
Footnotes
1. This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
Glossary
bimolecular reaction
elementary reaction involving the collision and combination of two reactant species
elementary reaction
reaction that takes place precisely as depicted in its chemical equation
intermediate
molecule or ion produced in one step of a reaction mechanism and consumed in another
molecularity
number of reactant species (atoms, molecules or ions) involved in an elementary reaction
rate-determining step
(also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction
reaction mechanism
stepwise sequence of elementary reactions by which a chemical change takes place
termolecular reaction
elementary reaction involving the simultaneous collision and combination of three reactant species
unimolecular reaction
elementary reaction involving the rearrangement of a single reactant species to produce one or more molecules of product | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.6%3A_Reaction_Mechanisms.txt |
Learning Objectives
• Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams
• List examples of catalysis in natural and industrial processes
We have seen that the rate of many reactions can be accelerated by catalysts. A catalyst speeds up the rate of a reaction by lowering the activation energy; in addition, the catalyst is regenerated in the process. Several reactions that are thermodynamically favorable in the absence of a catalyst only occur at a reasonable rate when a catalyst is present. One such reaction is catalytic hydrogenation, the process by which hydrogen is added across an alkene C=C bond to afford the saturated alkane product. A comparison of the reaction coordinate diagrams (also known as energy diagrams) for catalyzed and uncatalyzed alkene hydrogenation is shown in Figure $1$.
Catalysts function by providing an alternate reaction mechanism that has a lower activation energy than would be found in the absence of the catalyst. In some cases, the catalyzed mechanism may include additional steps, as depicted in the reaction diagrams shown in Figure $2$ This lower activation energy results in an increase in rate as described by the Arrhenius equation. Note that a catalyst decreases the activation energy for both the forward and the reverse reactions and hence accelerates both the forward and the reverse reactions. Consequently, the presence of a catalyst will permit a system to reach equilibrium more quickly, but it has no effect on the position of the equilibrium as reflected in the value of its equilibrium constant (see the later chapter on chemical equilibrium).
Example $1$: Using Reaction Diagrams to Compare Catalyzed Reactions
The two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Identify which diagram suggests the presence of a catalyst, and determine the activation energy for the catalyzed reaction:
Solution
A catalyst does not affect the energy of reactant or product, so those aspects of the diagrams can be ignored; they are, as we would expect, identical in that respect. There is, however, a noticeable difference in the transition state, which is distinctly lower in diagram (b) than it is in (a). This indicates the use of a catalyst in diagram (b). The activation energy is the difference between the energy of the starting reagents and the transition state—a maximum on the reaction coordinate diagram. The reagents are at 6 kJ and the transition state is at 20 kJ, so the activation energy can be calculated as follows:
$E_\ce{a}=\mathrm{20\:kJ−6\:kJ=14\:kJ} \label{12.8.1}$
Exercise $1$
Determine which of the two diagrams here (both for the same reaction) involves a catalyst, and identify the activation energy for the catalyzed reaction:
Answer
Diagram (b) is a catalyzed reaction with an activation energy of about 70 kJ.
Homogeneous Catalysts
A homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product. As an important illustration of homogeneous catalysis, consider the earth’s ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction:
$\ce{3O2}(g)\xrightarrow{hv}\ce{2O3}(g) \label{12.8.2}$
Ozone is a relatively unstable molecule that decomposes to yield diatomic oxygen by the reverse of this equation. This decomposition reaction is consistent with the following mechanism:
$\ce{O3 ⟶ O2 + O\ O + O3 ⟶ 2O2} \label{12.8.3}$
The presence of nitric oxide, NO, influences the rate of decomposition of ozone. Nitric oxide acts as a catalyst in the following mechanism:
$\ce{NO}(g)+\ce{O3}(g)⟶\ce{NO2}(g)+\ce{O2}(g)\ \ce{O3}(g)⟶\ce{O2}(g)+\ce{O}(g)\ \ce{NO2}(g)+\ce{O}(g)⟶\ce{NO}(g)+\ce{O2}(g) \label{12.8.4}$
The overall chemical change for the catalyzed mechanism is the same as:
$\ce{2O3}(g)⟶\ce{3O2}(g) \label{12.8.5}$
The nitric oxide reacts and is regenerated in these reactions. It is not permanently used up; thus, it acts as a catalyst. The rate of decomposition of ozone is greater in the presence of nitric oxide because of the catalytic activity of NO. Certain compounds that contain chlorine also catalyze the decomposition of ozone.
Mario J. Molina
The 1995 Nobel Prize in Chemistry was shared by Paul J. Crutzen, Mario J. Molina (Figure $3$), and F. Sherwood Rowland “for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone." Molina, a Mexican citizen, carried out the majority of his work at the Massachusetts Institute of Technology (MIT).
In 1974, Molina and Rowland published a paper in the journal Nature (one of the major peer-reviewed publications in the field of science) detailing the threat of chlorofluorocarbon gases to the stability of the ozone layer in earth’s upper atmosphere. The ozone layer protects earth from solar radiation by absorbing ultraviolet light. As chemical reactions deplete the amount of ozone in the upper atmosphere, a measurable “hole” forms above Antarctica, and an increase in the amount of solar ultraviolet radiation— strongly linked to the prevalence of skin cancers—reaches earth’s surface. The work of Molina and Rowland was instrumental in the adoption of the Montreal Protocol, an international treaty signed in 1987 that successfully began phasing out production of chemicals linked to ozone destruction.
Molina and Rowland demonstrated that chlorine atoms from human-made chemicals can catalyze ozone destruction in a process similar to that by which NO accelerates the depletion of ozone. Chlorine atoms are generated when chlorocarbons or chlorofluorocarbons—once widely used as refrigerants and propellants—are photochemically decomposed by ultraviolet light or react with hydroxyl radicals. A sample mechanism is shown here using methyl chloride:
$\ce{CH3Cl + OH ⟶ Cl + other\: products} \nonumber$
Chlorine radicals break down ozone and are regenerated by the following catalytic cycle:
$\ce{Cl + O3 ⟶ ClO + O2}\ \ce{ClO + O ⟶ Cl + O2}\ \textrm{overall Reaction: }\ce{O3 + O ⟶ 2O2} \nonumber$
A single monatomic chlorine can break down thousands of ozone molecules. Luckily, the majority of atmospheric chlorine exists as the catalytically inactive forms Cl2 and ClONO2.
Glucose-6-Phosphate Dehydrogenase Deficiency
Enzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in Figure $4$, is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells (FIgure $5$).
A disruption in this pathway can lead to reduced glutathione in red blood cells; once all glutathione is consumed, enzymes and other proteins such as hemoglobin are susceptible to damage. For example, hemoglobin can be metabolized to bilirubin, which leads to jaundice, a condition that can become severe. People who suffer from G6PD deficiency must avoid certain foods and medicines containing chemicals that can trigger damage their glutathione-deficient red blood cells.
Heterogeneous Catalysts
A heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase.
Heterogeneous catalysis has at least four steps:
1. Adsorption of the reactant onto the surface of the catalyst
2. Activation of the adsorbed reactant
3. Reaction of the adsorbed reactant
4. Diffusion of the product from the surface into the gas or liquid phase (desorption).
Any one of these steps may be slow and thus may serve as the rate determining step. In general, however, in the presence of the catalyst, the overall rate of the reaction is faster than it would be if the reactants were in the gas or liquid phase.
Figure $6$ illustrates the steps that chemists believe to occur in the reaction of compounds containing a carbon–carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon–carbon double bonds) to produce saturated fats and oils (which contain only carbon–carbon single bonds).
Other significant industrial processes that involve the use of heterogeneous catalysts include the preparation of sulfuric acid, the preparation of ammonia, the oxidation of ammonia to nitric acid, and the synthesis of methanol, CH3OH. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (Figure $7$).
Automobile Catalytic Converters
Scientists developed catalytic converters to reduce the amount of toxic emissions produced by burning gasoline in internal combustion engines. Catalytic converters take advantage of all five factors that affect the speed of chemical reactions to ensure that exhaust emissions are as safe as possible.
By utilizing a carefully selected blend of catalytically active metals, it is possible to effect complete combustion of all carbon-containing compounds to carbon dioxide while also reducing the output of nitrogen oxides. This is particularly impressive when we consider that one step involves adding more oxygen to the molecule and the other involves removing the oxygen (Figure $6$).
Most modern, three-way catalytic converters possess a surface impregnated with a platinum-rhodium catalyst, which catalyzes the conversion nitric oxide into dinitrogen and oxygen as well as the conversion of carbon monoxide and hydrocarbons such as octane into carbon dioxide and water vapor:
$\ce{2NO2}(g)⟶\ce{N2}(g)+\ce{2O2}(g)\ [5pt] \ce{2CO}(g)+\ce{O2}(g)⟶\ce{2CO2}(g)\ [5pt] \ce{2C8H18}(g)+\ce{25O2}(g)⟶\ce{16CO2}(g)+\ce{18H2O}(g) \nonumber$
In order to be as efficient as possible, most catalytic converters are preheated by an electric heater. This ensures that the metals in the catalyst are fully active even before the automobile exhaust is hot enough to maintain appropriate reaction temperatures.
Enzyme Structure and Function
The study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important compounds, particularly those that are involved in cellular metabolism. Different classes of enzymes perform a variety of functions, as shown in Table $1$.
ATP hydrolysis.”" data-quail-id="131" data-mt-width="1016">
Table $1$: Classes of Enzymes and Their Functions
Class Function
oxidoreductases redox reactions
transferases transfer of functional groups
hydrolases hydrolysis reactions
lyases group elimination to form double bonds
isomerases isomerization
ligases bond formation with ATP hydrolysis
Enzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme’s active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction (Figure $7$).
Summary
Catalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy. Catalysts can be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than the reactants).
Footnotes
1. “The Nobel Prize in Chemistry 1995,” Nobel Prize.org, accessed February 18, 2015, Nobel Prizes Chemistry [www.nobelprize.org].
Glossary
heterogeneous catalyst
catalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur
homogeneous catalyst
catalyst present in the same phase as the reactants | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.8%3A_Catalysis.txt |
12.1: Chemical Reaction Rates
Q12.1.1
What is the difference between average rate, initial rate, and instantaneous rate?
Solution
First, a general reaction rate must be defined to know what any variation of a rate is. The reaction rate is defined as the measure of the change in concentration of the reactants or products per unit time. The rate of a chemical reaction is not a constant and rather changes continuously, and can be influenced by temperature. Rate of a reaction can be defined as the disappearance of any reactant or appearance of any product. Thus, an average rate is the average reaction rate over a given period of time in the reaction, the instantaneous rate is the reaction rate at a specific given moment during the reaction, and the initial rate is the instantaneous rate at the very start of the reaction (when the product begins to form).
The instantaneous rate of a reaction can be denoted as $\lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t} \nonumber$
Q12.1.2
Ozone decomposes to oxygen according to the equation $\ce{2O3}(g)⟶\ce{3O2}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O3 and the formation of oxygen.
Solution
For the general reaction, aA ---> bB, the rate of the reaction can be expressed in terms of the disappearance of A or the appearance of B over a certain time period as follows.
$- \dfrac{1}{a}\dfrac{\Delta [A]}{\Delta t} = - \dfrac{1}{b}\dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{\Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{\Delta [D]}{\Delta t}$
We want the rate of a reaction to be positive, but the change in the concentration of a reactant, A, will be negative because it is being used up to be transformed into product, B. Therefore, when expressing the rate of the reaction in terms of the change in the concentration of A, it is important to add a negative sign in front to ensure the overall rate positive.
Lastly, the rate must be normalized according to the stoichiometry of the reaction. In the decomposition of ozone to oxygen, two moles of ozone form three moles of oxygen gas. This means that the increase in oxygen gas will be 1.5 times as great as the decrease in ozone. Because the rate of the reaction should be able to describe both species, we divide the change in concentration by its stoichiometric coefficient in the balanced reaction equation to deal with this issue.
Therefore, the rate of the reaction of the decomposition of ozone into oxygen gas can be described as follows:
$Rate=-\frac{Δ[O3]}{2ΔT}=\frac{Δ[O2]}{3ΔT}$
Answer
$Rate=-\frac{Δ[O3]}{2ΔT}=\frac{Δ[O2]}{3ΔT}\] Q12.1.3 In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction $\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3. Solution In this problem we are asked to write the equation that relates rate expressions in terms of disappearance of the reactants of the equation and in terms of the formation of the product. A reaction rate gives insight to how rate is affected as a function of concentration of the substances in the equation. Rates can often be expressed on graphs of concentration vs time expressed in change (${\Delta}$) of concentration and time and in a short enough time interval, the instantaneous rate can be approximated. If we were to analyze the reaction given, the graph would demonstrate that Cl2 decreases, that F2 decreases 3 times as quickly, and then ClF3 increases at a rate doubles. The reactants are being used and converted to product so they decrease while products increase. For this problem, we can apply the general formula of a rate to the specific aspects of a problem where the general form follows: $aA+bB⟶cC+dD\nonumber$. And the rate can then be written as $rate=-\frac {1}{a}\frac{{\Delta}[A]}{{\Delta}t}$ $=-\frac {1}{b}\frac{{\Delta}[B]}{{\Delta}t}$ $=\frac {1}{c}\frac{{\Delta}[C]}{{\Delta}t}$ $=\frac {1}{d}\frac{{\Delta}[D]}{{\Delta}t}.$ Here the negative signs are used to keep the convention of expressing rates as positive numbers. In this specific case we use the stoichiometry to get the specific rates of disappearance and formation (back to what was said in the first paragraph). So, the problem just involves referring the to the equation and its balanced coefficients. Based upon the equation we see that Cl2 is a reactant and has no coefficient, F2 has a coefficient of 3 and is also used up, and then ClF3 is a product that increases two-fold with a coefficient of 2. So, the rate here can be written as: $rate=-\frac{{\Delta}[Cl_2]}{{\Delta}t}=-\frac {1}{3}\frac{{\Delta}[F_2]}{{\Delta}t}=\frac {1}{2}\frac{{\Delta}[ClF_3]}{{\Delta}t}\nonumber$ Answer $\ce{rate}=+\dfrac{1}{2}\dfrac{Δ[\ce{CIF3}]}{Δt}=−\dfrac{Δ[\ce{Cl2}]}{Δt}=−\dfrac{1}{3}\dfrac{Δ[\ce{F2}]}{Δt}\nonumber$ Q12.1.4 A study of the rate of dimerization of C4H6 gave the data shown in the table: $\ce{2C4H6⟶C8H12}\nonumber$ Time (s) 0 1600 3200 4800 6200 [C4H6] (M) 1.00 × 10−2 5.04 × 10−3 3.37 × 10−3 2.53 × 10−3 2.08 × 10−3 1. Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s. 2. Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C4H6]. What are the units of this rate? 3. Determine the average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b). Solution 1.) The average rate of dimerization is the change in concentration of a reactant per unit time. In this case it would be: $rate$ $of$ $dimerization=-\frac{\Delta [C_4H_6]}{\Delta t}$ Rate of dimerization between 0 s and 1600 s: $rate$ $of$ $dimerization=-\frac{5.04×10^{-3}M-1.00×10^{-2}M}{1600 s-0 s}$ $rate$ $of$ $dimerization=3.10 × 10^{-6} \frac{M}{s}$ Rate of dimerization between 1600 s and 3200 s: $rate$ $of$ $dimerization=-\frac{3.37×10^{-3}M-5.04×10^{-3}M}{3200 s-1600 s}$ $rate$ $of$ $dimerization=1.04 × 10^{-6} \frac{M}{s}$ 2.) The instantaneous rate of dimerization at 3200 s can be found by graphing time versus [C4H6]. Because you want to find the rate of dimerization at 3200 s, you need to find the slope between 1600 s and 3200 s and also 3200 s and 4800 s. For the slope between 1600 s and 3200 s use the points (1600 s, 5.04 x 10-3 M) and (3200 s, 3.37 x 10-3 M) $\frac{3.37×10^{-3}M-5.04×10^{-3}M}{3200 s-1600 s}$ $\frac{-0.00167 M}{1600 s}$ $-1.04×10^{-6}\frac{M}{s}$ For the slope between 3200 s and 4800 s use the points (3200s, 3.37 x 10-3 M) and (4800s, 2.53 x 10-3 M) $\frac{2.53×10^{-3}M-3.37×10^{-3}M}{4800 s-3200 s}$ $\frac{-8.4×10^{-4} M}{1600 s}$ $-5.25×10^{-7}\frac{M}{s}$ Take the two slopes you just found and find the average of them to get the instantaneous rate of dimerization. $\frac{-1.04×10^{-6}\frac{M}{s}+-5.25×x10^{-7}\frac{M}{s}}{2}$ $\frac{-1.565×10^{-6}\frac{M}{s}}{2}$ $-7.83×10^-7\frac{M}{s}$ The instantaneous rate of dimerization is $-7.83×10^-7\frac{M}{s}$ and the units of this rate is $\frac{M}{s}$. 3.) The average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s can be found by using our answers from part a and b. If you look back up at the original equation, you could see that C4H6 and C8H12 are related in a two to one ratio. For every two moles of C4H6 used, there is one mole of C8H12 produced. For this reaction, the average rate of dimerization and the average rate of formation can be linked through this equation: $\frac{-1}{2}\frac{\Delta [C_4H_6]}{\Delta t}=\frac{\Delta [C_8H_{12}]}{\Delta t}$ Notice that reactant side is negative because the reactants are being used up in the reaction. So, for the average rate of formation of C8H12 at 1600 s, use the rate of dimerization between 0 s and 1600 s we found earlier and plug into the equation: $\frac{-1}{2}×3.10 × 10^{-6} \frac{M}{s}=\frac{\Delta [C_8H_{12}]}{\Delta t}$ $\frac{\Delta [C_8H_{12}]}{\Delta t}=1.55×10^{-6}\frac{M}{s}$ The average rate of formation for C8H12 at 1600 s is $1.55×10^{-6}\frac{M}{s}$. The rate of formation will be positive because products are being formed. The instantaneous rate of formation for C8H12 can be linked to the instantaneous rate of dimerization by this equation: $\frac{-1}{2}\frac{d[C_4H_6]}{dt}=\frac{d[C_8H_{12}]}{dt}$ So, for the instantaneous rate of formation for C8H12 at 3200 s, use the value of instantaneous rate of dimerization at 3200 s found earlier and plug into the equation: $\frac{-1}{2}×-7.83×10^-7\frac{M}{s}=\frac{d[C_8H_{12}]}{dt}$ $\frac{d[C_8H_{12}]}{dt}=-3.92×10^{-7}\frac{M}{s}$ The instantaneous rate of formation for C8H12 at 3200 s is $-3.92×10^-7\frac{M}{s}$ Answer 1. $3.10 × 10^{-6} \frac{M}{s}$ and $1.04 × 10^{-6} \frac{M}{s}$ 2. $-7.83×10^-7\frac{M}{s}$ and $\frac{M}{s}$ 3. $-3.92×10^-7\frac{M}{s}$ Q12.1.5 A study of the rate of the reaction represented as $2A⟶B$ gave the following data: Time (s) 0.0 5.0 10.0 15.0 20.0 25.0 35.0 [A] (M) 1.00 0.952 0.625 0.465 0.370 0.308 0.230 1. Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s. 2. Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate? 3. Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s. Solution Equations: $\frac{-\bigtriangleup A}{\bigtriangleup time}$ and Rate=$\frac{-\bigtriangleup A}{2\bigtriangleup time}=\frac{\bigtriangleup B}{time}$ Solve: 1.)The change in A from 0s to 10s is .625-1=-.375 so $\frac{-\bigtriangleup A}{\bigtriangleup time}$=.375/10= 0.0374 M/s Similarly, the change in A from 10 to 20 seconds is .370-.625=-.255 so $\frac{-\bigtriangleup A}{\bigtriangleup time}$=.255/20-10= 0.0255M/s 2.) We can estimate the rate law graphing the points against different order equations to determine the right order. Zero Order: $\frac{d[A]}{dt}=-k\nonumber$ $\int_{A_{\circ}}^{A}d[A]=-k\int_{0}^{t}dt\nonumber$ $[A]=-kt+[A_{\circ}]\nonumber$ First Order: $\frac{d[A]}{dt}=-k[A]\nonumber$ $\int_{A_{\circ}}^{A}\frac{d[A]}{[A]}=-kdt\nonumber$ $Ln(A)=-kt+Ln(A_{\circ})\nonumber$ Second Order: $\frac{d[A]}{dt}=-k[A]^{2}\nonumber$ $\int_{A\circ}^{A}\frac{d[A]}{[A]^{2}}=-k\int_{0}^{t}dt\nonumber$ $\frac{1}{[A]}=kt+\frac{1}{[A_{\circ}]}\nonumber$ Now that we have found the linear from of each order we will plot the points vs an [A] y-axis, a Ln(A) y-axis, and a 1/[A] y-axis. whichever of the plots has the most linear points will give us a good idea of the order and the slope will be the k value. Here we notice that the second order is most linear so we conclude the Rate to be.. $\frac{-d[A]}{2dt}=k[A]^{2}\nonumber$ At 15 seconds [A]=.465 and from the slope of the graph we find k=.116.so if we plug this data in and multiply both sides by 2 to get rid of the 2 in the denominator on the left side of the equation we find that the rate of disappearance of A is .05 M/s where the units are equivalent to [mol*L-1*s-1] 3.) Using the equation $\frac{-\bigtriangleup A}{2\bigtriangleup time}=\frac{\bigtriangleup B}{time}$ we divide the rates in part a and b in half to get .0188 M/s from 0 to 10 seconds and .025 M/s for the estimated instantaneous rate at 15s. Answer (a) average rate, 0 − 10 s = 0.0375 mol L−1 s−1; average rate, 12 − 18 s = 0.0225 mol L−1 s−1; (b) instantaneous rate, 15 s = 0.0500 mol L−1 s−1; (c) average rate for B formation = 0.0188 mol L−1 s−1; instantaneous rate for B formation = 0.0250 mol L−1 s−1 Q12.1.6 Consider the following reaction in aqueous solution: $\ce{5Br-}(aq)+\ce{BrO3-}(aq)+\ce{6H+}(aq)⟶\ce{3Br2}(aq)+\ce{3H2O}(l)\nonumber$ If the rate of disappearance of Br(aq) at a particular moment during the reaction is 3.5 × 10−4 M s−1, what is the rate of appearance of Br2(aq) at that moment? Solution Step 1. Define the rate of the reaction. Recall: For the general reaction: aA + bB → cC+ dD $rate =- \frac{\Delta[A]}{a\Delta{t}}=- \frac{\Delta[B]}{b\Delta{t}}= \frac{\Delta[C]}{c\Delta{t}}=\frac{\Delta[D]}{d\Delta{t}}$ So, for the reaction: $5Br^−(aq)+BrO^−_3(aq)+6H^+→3Br_2(aq)+3H_2O(l)$ The rate would be: $rate =- \frac{\Delta[Br^-]}{5\Delta{t}}=- \frac{\Delta[BrO^-_3]}{\Delta{t}}= -\frac{\Delta[H^+]}{6\Delta{t}}=\frac{\Delta[Br_2]}{3\Delta{t}}=\frac{H_2O}{3\Delta{t}}$ Step 2. Since we are given the rate for the disappearance of $Br^-$(aq) is $3.5x10^-4 Ms^{-1}$, and we want to find the rate of appearance of $Br_2$(aq). Therefore we set the two rates equal to each other. $rate =- \frac{\Delta[Br^-]}{5\Delta{t}}= \frac{\Delta[Br_2]}{3\Delta{t}}$ And,$-\frac{\Delta[Br^-]}{\Delta{t}}= -3.5x10^{-4} Ms^{-1}$ So, $3.5x10^{-4} Ms^{-1}$ = $\frac{5}{3}\frac{\Delta[Br_2]}{\Delta{t}}$ Step 3. Now solve the equation. $\frac{(3.5x10^{-4})(3)}{5} = \frac{\Delta[Br_2]}{\Delta{t}}$ $\frac{\Delta[Br_2]}{\Delta{t}} = 2.1 x 10^{-4} Ms^{-1}$ Answer $\frac{\Delta[Br_2]}{\Delta{t}} = 2.1 x 10^{-4} Ms^{-1}$ 12.2: Factors Affecting Reaction Rates Q12.2.1 Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium. Solution Molarity of Hydrochloric Acid • Reaction rates are affected by the frequency at which molecules collide. High Molarity=High Concentration which means more molecules are available to collide thus a faster reaction that one with a low molarity of HCl at a fixed volume. Temperature of Solution • Higher temperatures increase the rate of reaction because molecules move faster thus colliding more frequently • increasing temperatures allows for more particles to move past activation energy barrier to start the reaction Size of pieces of Magnesium • reaction rate is dependent on solid reactant size; smaller pieces increases the chance of collision because they enable a greater surface area thus faster reaction rate Q12.2.2 Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference. 1. What happens when the angle of the collision is changed? 2. Explain how this is relevant to rate of reaction. Solution According to the collision theory, there are many factors that cause a reaction to happen, with three of the factors being how often the molecules or atoms collide, the molecules' or atoms' orientations, and if there is sufficient energy for the reaction to happen. So, if the angle of the plunger is changed, the atom that is shot (a lone Oxygen atom in this case) will hit the other molecule (CO in this case) at a different spot and at a different angle, therefore changing the orientation and the number of proper collisions will most likely not cause for a reaction to happen. Thanks to the simulation, we can see that this is true: depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other (no reaction happens). In this particular case, the rate of the reaction will decrease because, by changing the angle, the molecules or atoms won't collide with the correct orientation or as often with the correct orientation. Q12.2.3 In the PhET Reactions & Rates interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature? S12.2.3 Based on the Collision Theory, a reaction will only occur if the molecules collide with proper orientation and with sufficient energy required for the reaction to occur. The minimum energy the molecules must collide with is called the activation energy (energy of transition state). Increasing the concentration of reactants increases the probability that reactants will collide in the correct orientation since there are more reactants in the same volume of space. Therefore, increasing the concentration of reactants would increase the rate of the reaction. Decreasing the concentration of reactants would decrease the rate of reaction because the overall number of possible collisions would decrease. Temperature is directly related the the kinetic energy of molecules and activation energy $E_a$ is the minimum energy required for a reaction to occur and doesn't change for a reaction. Increasing the temperature increases the kinetic energy of the reactants meaning the reactants will move faster and collide with each other more frequently. Therefore, increasing the temperature increase the rate of the reaction. Decreasing the temperature decreases the rate of reaction since the molecules will have less kinetic energy, move slower, and therefore collide with each other less frequently. Q12.2.4 In the PhET Reactions & Rates interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options. 1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow? 2. Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction. Solution a. On the simulation, we select the default setting and the reaction A+BC. In the default setting, we see frequent collisions, a low initial temperature, and a total average energy lower than the energy of activation. The collision theory states that the rate of a reaction is directly proportional to (the fraction of molecules with required orientation), (fractions of collisions with required energy), and (collision frequency). Although we see moving and frequently colliding reactants, the rate of the forward reaction is actually slow because it takes a long time for the products, AB and C, to start appearing. This is mainly because the fractions of collisions with required energy is low, coming from the average energy of the molecules being lower than the energy of activation. b. The reaction proceeds at an even faster rate. Again, the collision theory states that the rate of a reaction is directly proportional to (the fraction of molecules with required orientation), (fractions of collisions with required energy), and (collision frequency). Because molecules have a higher amount of energy, they have more kinetic energy. With an increased kinetic energy, the molecules not only collide more but also increase in the fraction of collision. However, the forward reaction and the backward reaction both proceed at a fast rate, so both happen almost simultaneously. It takes a shorter time for both reactions to happen. With both of the reactions adding up together overall, there is eventually a state of equilibrium. The process at which equilibrium is reached, however, is faster. Therefore, the amount of products of A+BC stays the same after a while. 12.3: Rate Laws Q12.3.1 How do the rate of a reaction and its rate constant differ? S12.3.1 The rate of a reaction or reaction rate is the change in the concentration of either the reactant or the product over a period of time. If the concentrations change, the rate also changes. Rate for A → B: $rate=\frac{\Delta[B]}{\Delta t}=-\frac{\Delta[A]}{\Delta t}$ The rate constant (k) is a proportionality constant that relates the reaction rates to reactants. If the concentrations change, the rate constant does not change. For a reaction with the general equation: $aA+bB→cC+dD$ the experimentally determined rate law usually has the following form: $rate=k[A]^m[B]^n$ Q12.3.2 Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions: 1. What is the order of the reaction with respect to that reactant? 2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant? Solution (a) 2; (b) 1 Q12.3.3 Tripling the concentration of a reactant increases the rate of a reaction nine times. With this knowledge, answer the following questions: 1. What is the order of the reaction with respect to that reactant? 2. Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times. What is the order of the reaction with respect to that reactant? Q12.3.4 How much and in what direction will each of the following affect the rate of the reaction: $\ce{CO}(g)+\ce{NO2}(g)⟶\ce{CO2}(g)+\ce{NO}(g)$ if the rate law for the reaction is $\ce{rate}=k[\ce{NO2}]^2$? 1. Decreasing the pressure of NO2 from 0.50 atm to 0.250 atm. 2. Increasing the concentration of CO from 0.01 M to 0.03 M. Solution (a) The process reduces the rate by a factor of 4. (b) Since CO does not appear in the rate law, the rate is not affected. Q12.3.5 How will each of the following affect the rate of the reaction: $\ce{CO}(g)+\ce{NO2}(g)⟶\ce{CO2}(g)+\ce{NO}(g)$ if the rate law for the reaction is $\ce{rate}=k[\ce{NO2}][\ce{CO}]$ ? 1. Increasing the pressure of NO2 from 0.1 atm to 0.3 atm 2. Increasing the concentration of CO from 0.02 M to 0.06 M. Q12.3.6 Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction $\ce{NO + O3⟶NO2 + O2}$ is first order with respect to both NO and O3 with a rate constant of 2.20 × 107 L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 × 10−6 M and [O3] = 5.9 × 10−7 M? Solution 4.3 × 10−5 mol/L/s Q12.3.7 Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: $\ce{^{32}_{15}P⟶^{32}_{16}S + e-}\nonumber$ Rate = 4.85 × 10−2 $\mathrm{day^{-1}\:[^{32}P]}$ What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M? Q12.3.8 The rate constant for the radioactive decay of 14C is 1.21 × 10−4 year−1. The products of the decay are nitrogen atoms and electrons (beta particles): $\ce{^6_{14}C⟶^{6}_{14}N + e-}\nonumber$ $\ce{rate}=k[\ce{^6_{14}C}]\nonumber$ What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 × 10−9 M? Solution 7.9 × 10−13 mol/L/year Q12.3.9 What is the instantaneous rate of production of N atoms Q12.3.8 in a sample with a carbon-14 content of 1.5 × 10−9 M? Q12.3.10 The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 × 10−8 L/mol/s. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 × 10−4 M? Q12.3.11 Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men: [C2H5OH] (M) 4.4 × 10−2 3.3 × 10−2 2.2 × 10−2 Rate (mol/L/h) 2.0 × 10−2 2.0 × 10−2 2.0 × 10−2 Determine the rate equation, the rate constant, and the overall order for this reaction. Solution rate = k; k = 2.0 × 10−2 mol/L/h (about 0.9 g/L/h for the average male); The reaction is zero order. Q12.3.12 Under certain conditions the decomposition of ammonia on a metal surface gives the following data: [NH3] (M) 1.0 × 10−3 2.0 × 10−3 3.0 × 10−3 Rate (mol/L/h1) 1.5 × 10−6 1.5 × 10−6 1.5 × 10−6 Determine the rate equation, the rate constant, and the overall order for this reaction. Q12.3.13 Nitrosyl chloride, NOCl, decomposes to NO and Cl2. $\ce{2NOCl}(g)⟶\ce{2NO}(g)+\ce{Cl2}(g)\nonumber$ Determine the rate equation, the rate constant, and the overall order for this reaction from the following data: [NOCl] (M) 0.10 0.20 0.30 Rate (mol/L/h) 8.0 × 10−10 3.2 × 10−9 7.2 × 10−9 Solution Before we can figure out the rate constant first we must first determine the basic rate equation and rate order. The basic rate equation for this reaction, where n is the rate order of NOCl and k is the rate constant, is $rate = k[NOCl]^n\nonumber$ since NOCl is the reactant in the reaction. In order to figure out the order of the reaction we must find the order of [NOCl] as it is the only reactant in the reaction. To do this we must examine how the rate of the reaction changes as the concentration of NOCl changes. As [NOCl] doubles in concentration from 0.10 M to 0.20 M the rate goes from 8.0 x 10-10 to 3.2 x 10-9 (3.2 x 10-9(mol/L/h))/(8.0 x 10-10(mol/L/h)) = 4 so we conclude that as [NOCl] doubles, the rate goes up by 4. Since 22= 4 we can say that the order of [NOCl] is 2 so our updated rate law is $rate = k[NOCl]^2\nonumber$ Now that we have the order, we can substitute the first experimental values from the given table to find the rate constant, k (8.0 x 10-10(mol/L/h)) = k(0.10 M)2 so $k= \dfrac{8.0 \times 10^{-10}}{ (0.10\, M)^2} = 8 \times 10^{-8} M^{-1} sec^{-1}\nonumber$ We were able to find the units of k using rate order, when the rate order is 2 units of k are M-1 x sec-1 So the rate equation is: rate = k[NOCl]2, it is second order, and k = 8 x 10-8 M-1 x sec-1 Overall rate law : $rate = \underbrace{(8 \times 10^{-8})}_{\text{1/(M x sec)}} [NOCl]^2\nonumber$ Answer rate = k[NOCl]2; k = 8.0 × 10−8 L/mol/s; second order Q12.3.14 From the following data, determine the rate equation, the rate constant, and the order with respect to A for the reaction $A⟶2C$. [A] (M) 1.33 × 10−2 2.66 × 10−2 3.99 × 10−2 Rate (mol/L/h) 3.80 × 10−7 1.52 × 10−6 3.42 × 10−6 Solution A. Using the experimental data, we can compare the effects of changing [A] on the rate of reaction by relating ratios of [A] to ratios of rates $\frac{2.66 \times 10^{-2}}{1.33 \times 10^{-2}} = 2\nonumber$ and $\frac{1.52 \times 10^{-6}}{3.8 \times 10^{-7}} = 4\nonumber$ B. From this we know that doubling the concentration of A will result in quadrupling the rate of reaction. The order of this reaction is 2. C. We can now write the rate equation since we know the order: $rate=k[A]^2\nonumber$ D. By plugging in one set of experimental data into our rate equation we can solve for the rate constant, k: $3.8 \times 10^{-7} = k \times (1.33 \times 10^{-2})^{2}\nonumber$ $k = \frac{3.8 \times 10^{-7}}{1.769 \times 10^{-4}}\nonumber$ $k= .00215 M^{-1}s^{-1}\nonumber$ Answer $k= .00215 M^{-1}s^{-1}$ 2nd Order Q12.3.15 Nitrogen(II) oxide reacts with chlorine according to the equation: $\ce{2NO}(g)+\ce{Cl2}(g)⟶\ce{2NOCl}(g)\nonumber$ The following initial rates of reaction have been observed for certain reactant concentrations: [NO] (mol/L1) [Cl2] (mol/L) Rate (mol/L/h) 0.50 0.50 1.14 1.00 0.50 4.56 1.00 1.00 9.12 What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant? Solution For the general equation, $aA + bB \rightarrow cC + dD$ The rate can be written as $rate = k[A]^{m}[B]^{n}$ where k is the rate constant, and m and n are the reaction orders. For our equation $2NO(g) + Cl_{2}(g) \rightarrow 2NOCl(g)$ the $rate = k[NO]^{m}[Cl_{2}]^{n}$ Now, we need to find the reaction orders. Reaction orders can only be found through experimental values. We can compare two reactions where one of the reactants has the same concentration for both trials, and solve for the reaction order. $\frac{rate_{1}}{rate_{2}}=\frac{[NO]_{1}^{m}[Cl_{2}]_{1}^{n}}{[NO]_{2}^{m}[Cl_{2}]_{2}^{n}}$ We can use the data in the table provided. If we plug in the values for rows 1 and 2, we see that the values for the concentration of Cl will cancel, leaving just the rates and the concentrations of NO. $\frac{1.14}{4.56}=\frac{[0.5]^{m}}{[1.0]^{m}}$ We can now solve for m, and we find that m =2. This means that the reaction order for [NO] is 2. Now we must find the value of n. To do so, we can use the same equation but with the values from rows 2 and 3. This time, the concentration of NO will cancel out. $\frac{4.56}{9.12}=\frac{[0.5]^{n}}{[1.0]^{n}}$ When we solve for n, we find that n = 1. This means that the reaction order for [Cl2] is 1. We are one step closer to finishing our rate equation. $rate = k[NO]^{2}[Cl_{2}]$ Finally, we can solve for the rate constant. To do this, we can use one of the trials of the experiment, and plug in the values for the rate, and concentrations of reactants, then solve for k. $1.14 mol/L/h = k[0.5 mol/L]^{2}[0.5mol/L]$ $k=9.12L^{2}mol^{-2}h^{-1}$ So, our final rate equation is: $rate = (9.12 L^{2} mol^{-2}h^{-1})[NO]^{2}[Cl_{2}]$ *A common mistake is forgetting units. Make sure to track your units throughout the process of determining your rate constant. Be careful because the units will change relative to the reaction order. Answer rate = k[NO]2[Cl]2; k = 9.12 L2 mol−2 h−1; second order in NO; first order in Cl2 Q12.3.17 Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: $\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g)\nonumber$ Determine the rate equation, the rate constant, and the orders with respect to each reactant from the following data: [NO] (M) 0.30 0.60 0.60 [H2] (M) 0.35 0.35 0.70 Rate (mol/L/s) 2.835 × 10−3 1.134 × 10−2 2.268 × 10−2 Solution Determine the rate equation, the rate constant, and the orders with respect to each reactant. The rate constant and the orders can be determined through the differential rate law. The general form of the differential rate law is given below: aA + bB + cC ==> products where A, B, and C are the concentrations of the reactants, k is the rate constant, and n,m, and p refer to the order of each reactant. To find the orders of each reactant, we see that when [NO] doubles but [H2] doesn't change, the rate quadruples, meaning that [NO] is a second order reaction ([NO]2). When [H2] doubles but [NO] doesn't change, the rate doubles, meaning that [H2] is a first order reaction. So the rate law would look something like this: Rate = k[NO]2[H2] We can use this rate law to determine the value of the rate constant. Plug in the data for reactant concentration and rate from one of the trials to solve for k the rate constant. In this case, we chose to use the data from trial 1 from the second column of the data table. 2.835x10-3 = k[0.3]2[0.35] k = .09 M-2/s-1 Q12.3.18 For the reaction $A⟶B+C$, the following data were obtained at 30 °C: [A] (M) 0.230 0.356 0.557 Rate (mol/L/s) 4.17 × 10−4 9.99 × 10−4 2.44 × 10−3 1. What is the order of the reaction with respect to [A], and what is the rate equation? 2. What is the rate constant? Solution 1. The rate equation for an $n$ order reaction is given as $\frac{dr}{dt}={k}{[A]^n}$. Where $[A]$ is the concentration in M, and $\frac{dr}{dt}$ is the rate in M/s. We can then use each set of data points, plug its values into the rate equation and solve for $n$. Note you can use any of the data points as long as the concentration corresponds to its rate. Rate equation 1: $4.17 \times {10}^{-4}={k}{[0.230]^n}$ Rate equation 2: $9.99 \times {10}^{-4}={k}{[0.356]^n}$ We divide Rate equation 1 by Rate equation 2 in order to cancel out k, the rate constant. ${\frac{4.17 \times {10}^{-4}}{9.99 \times {10}^{-4}}} = {\frac{k[0.230]^n}{k[0.356]^n}}$ ${0.417}={0.646^n}$ Now the only unknown we have is $n$. Using logarithm rules one can solve for it. $ln{\: 0.417}={n \cdot ln{\: 0.646}}$ $\frac{ln{\: 0.417}}{ln{\:0.646}}=n=2$ The rate equation is second order with respect to A and is written as $\frac{dr}{dt}={k}{[A]^2}$. 2. We can solve for $k$ by plugging in any data point into our rate equation $\frac{dr}{dt}={k}{[A]^2}$. Using the first data points for instance $[A]=0.230 \:\frac{mol}{L}$ and $\frac{dr}{dt} = 4.17 \times {10}^{-4} \:\frac{mol}{L \cdot s}$] we get the equation $4.17 \times {10}^{-4} \:\frac{mol}{L \cdot s}={k}{[0.230 \:\frac{mol}{L}]^2}$ Which solves for $k=7.88 \times {10}^{-3} \frac{L}{mol \cdot s}$ Since we know this is a second order reaction the appropriate units for $k$ can also be written as $\frac{1}{M \cdot s}$ Answer (a) The rate equation is second order in A and is written as rate = k[A]2. (b) k = 7.88 × 10−13 L mol−1 s−1 Q12.3.19 For the reaction $Q⟶W+X$, the following data were obtained at 30 °C: [Q]initial (M) 0.170 0.212 0.357 Rate (mol/L/s) 6.68 × 10−3 1.04 × 10−2 2.94 × 10−2 1. What is the order of the reaction with respect to [Q], and what is the rate equation? 2. What is the rate constant? Solution What is the order of the reaction with respect to [Q], and what is the rate equation? • Order reaction: 2 because when you use the ratio trial 3:2, it will look like this: • ($\dfrac{2.94*10^{-2}}{1.04*10^{-2}}$) = ($\dfrac{0.357^{x}}{0.212^{x}}$) • 2.82 = 1.7x • x = 2 so the order of reaction is 2 • Rate reaction equation: Rate=k[Q]2 What is the rate constant? • To find the rate constant (k) simply plug and calculate one of the trials into the rate equation • 1.04 x 10-2=k[0.212]2 • k=0.231 $M^{-1}s^{-1}$ Answer Order: 2 k=0.231 $M^{-1}s^{-1}$ Q12.3.20 The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6.2 × 10−4 min−1. $\ce{2N2O5⟶4NO2 + O2}\nonumber$ What is the rate of the reaction when [N2O5] = 0.40 M? Solution Step 1: The first step is to write the rate law. We know the general formula for for a first-order rate law. It is as follows: Rate=k[A] Step 2: We now plug in [N2O5] in for [A] in our general rate law. We also plug in our rate constant (k), which was given to us. Now our equation looks as follows: Rate=(6.2x10-4 min-1)[N2O5] Step 3: We now plug in our given molarity. [N2O5]=0.4 M. Now our equation looks as follows: Rate=(6.2x10-4 min-1)(0.4 M) Step 4: We now solve our equation. Rate=(6.2x10-4 min-1)(0.4 M)= 2.48x10-4 M/min. Step 5: Use significant figures and unit conversion to round 2.48x10-4 M/min to 2.5 × 10−4 (moles)L-1min-1 Answer (a) 2.5 × 10−4 mol/L/min Q12.3.21 The annual production of HNO3 in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel. 1. $\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g)$ 2. $\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)$ 3. $\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g)$ The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 × 10−6 L2/mol2/s. Solution To determine the rate law for an equation we need to look at its slow step. Since both equation a and c are fast, equation b can be considered the slow step of the reaction. The slow step is also considered the rate determining step of the system. Hence, The rate determining step is the second step because it's the slow step. rate of production of $NO_2 = k [A]^m [B]^n$ $rate = k [NO]^2 [O_2]^1~M/s$ $rate = (5.8*10^{-6}) [0.75]^2 [0.5]^1 ~M/s$ $rate = 1.6*10^{-6}~M/s$ Answer $rate = 1.6*10^{-6}~M/s$ Q12.3.22 The following data have been determined for the reaction: $\ce{I- + OCl- ⟶ IO- + Cl-}\nonumber$ 1 2 3 $\mathrm{[I^-]_{initial}}$ (M) 0.10 0.20 0.30 $\mathrm{[OCl^-]_{initial}}$ (M) 0.050 0.050 0.010 Rate (mol/L/s) 3.05 × 10−4 6.20 × 10−4 1.83 × 10−4 Determine the rate equation and the rate constant for this reaction. Solution Using the reactants, we can form the rate law of the reaction:$ r=k[OCl^-]^n[I^-]^m \]
From there, we need to use the data to determine the order of both $[OCl^-]$ and $[I^-]$. In doing so, we need to compare $r_1$ to $r_2$ such that:
$\frac {r_1}{r_2} = \frac {(0.10^m)(0.050^n)}{(0.20^m)(0.050^n)} = \frac {3.05 \times 10^{-4}}{6.20 \times 10^{-4}}$
$0.5^m = 0.5$
$m = 1$
We can "cross out" the concentration of $[OCl^-]$ because it has the same concentration in both of the trials used.
Now that we know m ($[I^-]$) has a first order of 1.
We cannot "cross out" $[I^-]$ to find $[OCl^-]$ because no two trials have the same concentration. In order to solve for n we will plug in 1 for m.
$\frac {r_1}{r_3} = \frac {(0.10^{1})(0.050^n)}{(0.30^{1})(0.010^n)} = \frac {3.05 \times 10^{-4}}{1.83 \times 10^{-4}}$
$\frac {1}{3} (5^{n}) = 1.6666667$
$5^{n} = 5$
$n = 1$
Since we know that orders of both n and m are equal to one, we can not substitute them into the rate law equation along with the respective concentrations (from either the first, second, or third reaction) and solve for the rate constant, k.
$r=k[OCl^-]^n[I^-]^m$
$3.05 * 10^{-4}= k[0.05]^1[0.10]^1$
$k = 6.1 * 10^{-2} \frac {L}{mol \times s}$
Thus the overall rate law is: $r = (6.1 * 10^{-2} \frac {L}{mol \times s})[OCl^-][I^-] \] The units for K depend on the overall order of the reaction. To find the overall order we add m and n together. By doing this we find an overall order of 2. This is why the units for K are$ \frac {L}{mol \times s} \]
Answer
rate = k[I][OCl−1]; k = 6.1 × 10−2 L mol −1 s−1
Q12.3.23
In the reaction
$2NO + Cl_2 → 2NOCl\nonumber$
the reactants and products are gases at the temperature of the reaction. The following rate data were measured for three experiments:
Initial p{NO} Initial p{Cl2} Initial rate
(atm) (atm) (moles of A consumed atm sec-1)
0.50 0.50 5.1 x 10-3
1.0 1.0 4.0 x 10-2
0.50 1.0 1.0 x 10-2
1. From these data, write the rate equation for this gas reaction. What order is the reaction in NO, Cl2, and overall?
2. Calculate the specific rate constant for this reaction.
Solution
a. The rate equation can be determined by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction $A+B\rightarrow products$, for example, we need to determine k and the exponents m and n in the following equation:
$rate=k[A]^m[B]^n\nonumber$
To do this, the initial concentration of B can be kept constant while varying the initial concentration of A and calculating the initial reaction rate. This information would deduce the reaction order with respect to A. The same process can be done to find the reaction order with respect to B.
In this particular example,
$\frac{rate_2}{rate_3}=\frac{k[A_2]^m[B_2]^n}{k[A_3]^m[B_3]^n}\nonumber$
So taking the values from the table,
$\frac{4.0*10^{-2}}{1.0*10^{-2}}=\frac{k[1.0]^m[1.0]^n}{k[0.5]^m[1.0]^n}\nonumber$
and by canceling like terms, you are left with
$\frac{4.0*10^{-2}}{1.0*10^{-2}}=\frac{[1.0]^m}{[0.5]^m}\nonumber$
Now, solve for m
$4=2^m\Longrightarrow m=2$ Because m=2, the reaction with respect to $NO$ is 2. $NO$ is second order.
You can repeat the same process to find n.
$\frac{rate_3}{rate_1}=\frac{k[A_3]^m[B_3]^n}{k[A_1]^m[B_1]^n}\nonumber$
Taking the values from the table,
$\frac{1.0*10^{-2}}{5.1*10^{-3}}=\frac{k[0.5]^m[1.0]^n}{k[0.5]^m[0.5]^n}\nonumber$
and by canceling like terms, you are left with
$\frac{1.0*10^{-2}}{5.1*10^{-3}}=\frac{[1.0]^n}{[0.5]^n}\nonumber$
Now this time, solve for n
$2=2^n\Longrightarrow n=1$
Because n=1, the reaction with respect to $Cl_2$ is 1. $Cl_2$ is first order.
So the rate equation is$rate=k[NO]^2[Cl_2]^1\nonumber$
To find the overall rate order, you simply add the orders together. Second order + first order makes the overall reaction third order.
b. The rate constant is calculated by inserting the data from any row of the table into the experimentally determined rate law and solving for k. For a third order reaction, the units of k are $frac{1}{atm^2*sec}$. Using Experiment 1,
$rate=k[NO]^2[Cl_2]^1\Longrightarrow 5.1*10^{-3} \frac{atm}{sec}=k[0.5m atm]^2[0.5 atm]^1\nonumber$
$k=0.0408 \frac{1}{atm^2*sec}\nonumber$
Answer
$NO$ is second order.
$Cl_2$ is first order.
Overall reaction order is three.
b)
$k=0.0408\; atm^{-2}*sec^{-1}$
12.4: Integrated Rate Laws
Q12.4.1
Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of A at varying times.
Solution
To determine the order of a reaction when given the data series, one must graph the data how it is, graph it as natural log of [A], and graph it as 1/[A]. Whichever method yields a straight line will determine the order. Respective of the methods of graphing above, if a straight line is yielded by the first graphing method its a 0 order, if by the second method it's a 1st order, and the third graphing method its a 2nd order. When the order of the graph is known, a series of equations, given in the above image, can be used with the various points on the graph to determine the value of k. We can see that we need an initial value of A and a final value of A, and both of these would be given by the data.
Zeroth order when plotting initial concentrating versus final concentration you have a negative linear slope.
$[A] = [A]_0 − kt\nonumber$
First order when plotting ln[initial concentration] versus ln[ final concentration] you have a negative linear slope.
$\ln[A] = \ln[A]_0 − kt\nonumber$
Second order when plotting the 1/[initial concentration] versus 1/[final concentration] you have a positive linear slope.
$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt\nonumber$
Q12.4.2
Use the data provided to graphically determine the order and rate constant of the following reaction: $\ce{SO2Cl2 ⟶ SO2 + Cl2}$
Time (s) 0 5.00 × 103 1.00 × 104 1.50 × 104 2.50 × 104 3.00 × 104 4.00 × 104
[SO2Cl2] (M) 0.100 0.0896 0.0802 0.0719 0.0577 0.0517 0.0415
Solution
Use the data to graphically determine the order and rate constant of the following reaction.
In order to determine the rate law for a reaction from a set of data consisting of concentration (or the values of some function of concentration) versus time, make three graphs of the data based on the integrated rate laws of each order reaction.
[concentration] versus time (linear for a zero order reaction) ln [concentration] versus time (linear for a 1st order reaction) 1 / [concentration] versus time (linear for a 2nd order reaction)
slope= -2.0 x 10-5
k = 2.0 x 10-5
The graph that is linear indicates the order of the reaction. Then, you can find the correct rate equation:
zero order reaction rate = k (k = - slope of line)
1st order reaction rate = k[A] (k = - slope of line)
2nd order reaction rate = k[A]2 (k = slope of line)
In this graph, ln(concentration) vs time is linear, indicating that the reaction is first order.
k=-slope of line
Answer
Plotting a graph of ln[SO2Cl2] versus t reveals a linear trend; therefore we know this is a first-order reaction:
k = −2.20 × 105 s−1
Q12.4.3
Use the data provided in a graphical method to determine the order and rate constant of the following reaction:
$2P⟶Q+W\nonumber$
Time (s) 9.0 13.0 18.0 22.0 25.0
[P] (M) 1.077 × 10−3 1.068 × 10−3 1.055 × 10−3 1.046 × 10−3 1.039 × 10−3
Solution
Add texts here. Do not delete this text first.
Q12.4.4
Pure ozone decomposes slowly to oxygen, $\ce{2O3}(g)⟶\ce{3O2}(g)$. Use the data provided in a graphical method and determine the order and rate constant of the reaction.
Time (h) 0 2.0 × 103 7.6 × 103 1.23 × 104 1.70 × 104 1.70 × 104
[O3] (M) 1.00 × 10−5 4.98 × 10−6 2.07 × 10−6 1.39 × 10−6 1.22 × 10−6 1.05 × 10−6
Solution
To determine the order and rate constant, you need to graph the data for zero order, first order, and second order by plotting concentration versus time- [A] vs. time, natural logarithm (ln) of [A] vs. time, and 1/[A] vs. time respectively. The order of the reaction is determined by identifying which of these three graphs produces a straight line. The rate constant k is represented by the slope of the graph. The graphs with their respective data values would be
Time (s) 9.0 13.0 18.0 22.0 25.0
[P] (M) 1.077 × 10−3 1.068 × 10−3 1.055 × 10−3 1.046 × 10−3 1.039 × 10−3
Time (s) 9.0 13.0 18.0 22.0 25.0
ln [P] (M) -6.83358 -6.84197 -6.85421 -6.86278 -6.8695
Time (s) 9.0 13.0 18.0 22.0 25.0
1/[P] (M) 928.5051 936.3296 947.8673 956.0229 962.4639
Since each graph yields a straight line the order and rate constant of the reaction cannot be determined.
To identify how the concentrations changes a function of time, requires solving the appropriate differential equation (i.e., the differential rate law).
The zero-order rate law predicts in a linear decay of concentration with time
The 1st-order rate law predicts in an exponential decay of concentration with time
The 2nd-order rate law predicts in an reciprocal decay of concentration with time
The plot is not linear, so the reaction is not zero order.
The plot is not linear, so the reaction is not first order.
The plot is nicely linear, so the reaction is second order.
To a second order equation, $1/[A] \ = k*t + 1/[A_0]$
Thus, the value of K is the slope of the graph Time vs $\frac{1}{\ce{O3}}$,
k = 50.3*10^6 L mol−1 h−1
Answer
The plot is nicely linear, so the reaction is second order.
k = 50.1 L mol−1 h−1
Q12.4.5
From the given data, use a graphical method to determine the order and rate constant of the following reaction:
$2X⟶Y+Z$
Time (s) 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0
[X] (M) 0.0990 0.0497 0.0332 0.0249 0.0200 0.0166 0.0143 0.0125
Solution
In order to determine the order of the reaction we need to plot the data using three different graphs. All three graphs will have time in seconds as the x-axis, but the y-axis is what will differ. One graph will plot concentration versus time, the second will plot natural log of concentration versus time, and the other will plot 1/concentration versus times. Whichever graph results in a line, we know that must be the order of the reaction. If we get a line using the first graph, it will be zero order, if it is a line for the second graph it will be first order, and if it is a line for the third graph it will be a second order reaction. Now lets plot the data to determine the order.
We can clearly see that the third graph, which plots 1/M versus time, is a straight line while the other two are slightly curved. Therefore, we can determine that the rate of this reaction is second order. This also tells us that the units of the rate constant which should be M-2s-1 for a second order reaction.
To determine the rate constant, called k, we simple need to figure out the slope of the third graph since that is the order of this reaction. To find the slope of the line, we take two points and subtract the y values and then divide them by the difference of the x values. This is how to do it:
Use the points (5, 10.101) and (40, 80).
Now use these to get the slop, aka the rate constant: (80-10.101)/(40-5) = 1.997 = k
So the rate constant for this second order reaction is 1.997 M-1s-1.
Q12.4.6
What is the half-life for the first-order decay of phosphorus-32? $\ce{(^{32}_{15}P⟶^{32}_{16}S + e- )}$ The rate constant for the decay is 4.85 × 10−2 day−1.
Solution
This is a first order reaction, so we can use our half life equation below:
$t_{1/2}=\frac{0.693}{k}\nonumber$
The rate constant is given to us in units per day. All we have to do, is to plug it into the equation.
$t_{1/2}=\frac{0.693}{4.85*10^{-2}}\nonumber$
$=14.3\; days\nonumber$A12.4.6
14.3 d
Q12.4.7
What is the half-life for the first-order decay of carbon-14? $\ce{(^6_{14}C⟶^7_{14}N + e- )}$ The rate constant for the decay is 1.21 × 10−4 year−1.
Solution
To find the half life, we need to use the first-order half-life equation. All half life reactions undergo first order reactions.
The half-life equation for first order is $t_{1/2}=ln2/k \nonumber$with k being the rate constant. The rate constant for carbon-14 was given as $1.21 × 10^{-4} year^{−1}$.
Plug it in the equation. $t_{1/2}=ln2/(1.21 × 10^{−4} year^{−1})\nonumber$ and solve for $t_{1/2}$.
When you calculate it, the half life for carbon-14 is 5.73*103
Answer
The half-life for carbon-14 is calculated to be 5.73*103
Q12.4.8
What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 × 10−8 L/mol/s.
Solution
The half-life of a reaction, t1/2, is the amount of time that is required for a reactant concentration to decrease by half compared to its initial concentration. When solving for the half-life of a reaction, we should first consider the order of reaction to determine it's rate law. In this case, we are told that this reaction is second-order, so we know that the integrated rate law is given as:
$\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0}\nonumber$
Isolating for time, we find that:
$t_{1/2} = \dfrac{1}{k[A]_0}\nonumber$
Now it is just a matter of substituting the information we have been given to calculate $t_{1/2}$, where the rate constant, ${k}$, is equal to 8.0 × 10−8 L/mol/s and initial concentration, ${[A]_0}$, is equal to 0.15M:
$t_{1/2} = \dfrac{1}{(8.0×10^{-8})(0.15)} = {8.33×10^7 seconds}\nonumber$
Answer
8.33 × 107 s
Q12.4.9
What is the half-life for the decomposition of O3 when the concentration of O3 is 2.35 × 10−6 M? The rate constant for this second-order reaction is 50.4 L/mol/h.
Solution
Add texts here. Do not delete this text first.
Since the reaction is second order, its half-life is
$t_{1/2}=\dfrac{1}{(50.4M^{-1}/h)[2.35×10^{-6}M]}\nonumber$
So, half-life is 8443 hours.
Q12.4.10
The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L/mol/s. If the initial concentration is 0.500 mol/L, what is the value of t1/2?
Solution
As mentioned in the question the reaction of compound A will result in the formation of compounds C and D. This reaction was found to be second-order in A. Therefore, we should use the second order equation for half-life which relates the rate constant and initial concentrations to the half-life:
$t_{\frac{1}{2}}=\frac{1}{k[A]_{0}}\nonumber$
Since we were given k (rate constant) and Initial concentration of A, we have everything needed to calculate the half life of A.
$k=0.5\frac{\frac{L}{mol}}{s}\nonumber$
$[A]_{0}=2.42\frac{mol}{L}\nonumber$
When we plug in the given information notice that the units cancel out to seconds.
$t_{\frac{1}{2}}=\frac{1}{\frac{2.42Lmol^{-}}{s}[0.500\frac{mol}{L}]}=0.826 s\nonumber$
Answer
0.826 s
Q12.4.11
The half-life of a reaction of compound A to give compounds D and E is 8.50 minutes when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?
Solution
Organize the given variables:
(half-life of A) $t_{1/2}=8.50min$
(initial concentration of A) $[A]_{0}=0.150mol/L$
(target concentration of A) $[A]=0.0300mol/L$
Find the the rate constant k, using the half-life formulas for each respective order. After finding k, use the integrated rate law respective to each order and the initial and target concentrations of A to find the time it took for the concentration to drop.
(a) first order with respect to A
(half-life) $t_{1/2}=\frac{ln(2)}{k}=\frac{0.693}{k}$
(rearranged for k) $k=\frac{0.693}{t_{1/2}}$
(plug in t1/2 = 8.50 min) $k=\frac{0.693}{8.50min}=0.0815min^{-1}$
(integrated rate law) $ln[A]=-kt+ln[A]_{0}$
(rearranged for t) $ln(\frac{[A]}{[A]_{0}})=-kt$
$-ln(\frac{[A]}{[A]_{0}})=kt$
$ln(\frac{[A]}{[A]_{0}})^{-1}=kt$
$ln(\frac{[A]_{0}}{[A]})=kt$
$t=\frac{ln(\frac{[A]_{0}}{[A]})}{k}$
(plug in variables) $t=\frac{ln(\frac{0.150mol/L}{0.0300mol/L})}{0.0815min^{-1}}=\frac{ln(5.00)}{0.0815min^{-1}}=19.7min$
(b) second order with respect to A
(half-life) $t_{1/2}=\frac{1}{k[A]_{0}}$
(rearranged for k) $k=\frac{1}{t_{1/2}[A]_{0}}$
(plug in variables) $k=\frac{1}{(8.50min)(0.150mol/L)}=\frac{1}{1.275min\cdot mol/L}=0.784L/mol\cdot min$
(integrated rate law) $\frac{1}{[A]}=kt+\frac{1}{[A]_{0}}$
(rearranged for t) $\frac{1}{[A]}-\frac{1}{[A]_{0}}=kt$
$t=\frac{1}{k}(\frac{1}{[A]}-\frac{1}{[A]_{0}})$
(plug in variables) $t=\frac{1}{0.784L/mol\cdot min}(\frac{1}{0.0300mol/L}-\frac{1}{0.150mol/L})=\frac{1}{0.784L/mol\cdot min}(\frac{80}{3}L/mol)=34.0min$
Answer
a) 19.7 min
b) 34.0 min
Q12.4.12
Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 × 104 g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 × 10−6 g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.
[Penicillin] (M) Rate (mol/L/min)
2.0 × 10−6 1.0 × 10−10
3.0 × 10−6 1.5 × 10−10
4.0 × 10−6 2.0 × 10−10
Solution
The first step is to solve for the order or the reaction. This can be done by setting up two expressions which equate the rate to the rate constant times the molar concentration of penicillin raised to the power of it's order. Once we have both expressions set up, we can divide them to cancel out k (rate constant) and use a basic logarithm to solve for the exponent, which is the order. It will look like this.
rate(mol/L/min)=k[M]x
(1.0 x 10-10)=k[2.0 x 10-6]x
(1.5 x 10-10)=k[3.0 x 10-6]x
Dividing the two equations results in the expression:
(2/3)=(2/3)x
*A single ratio equation can also be set up to solve for the reaction order:
*$\frac{rate_{1}}{rate_{2}}=\frac{k[Penicillin]_{1}^{x}}{k[Penicillin]_{2}^{x}}\nonumber$
*We then solve for x in a similar fashion.
*$\frac{1.0x10^{-10}}{1.5x10^{-10}}=\frac{[2.0x10^{-6}]^{x}}{[3.0x10^{-6}]^{x}}\nonumber$
We can now use the natural logarithm to solve for x, or simply and intuitively see that in order for the equation to work, x must be equal to one. Thus, the reaction is of the first order.
Now that we have the order of the reaction, we can proceed to solve for the value of the rate constant. Substituting x=1 into our first equation yields the expression:
(1 x 10-10)=k[2.0 x 10-6]1
k=(1 x 10-10)/(2 x 10-6)
k= (5 x 10-5) min-1
We have a unit of min-1 because we divided (mol/L/min) by molarity, which is in (mol/L), yielding a unit of min-1.
We were given two important pieces of information to finish the problem. It is stated that the enzyme has a molecular weight of 3 × 104 g/mol, and that we have a one liter solution that contains (0.15 x 10-6 g) of penicillinase. Dividing the amount of grams by the molecular weight yields 5 x 10-12 moles.
(0.15 x 10-6) g / (3 x 104) g/mol = (5 x 10-12) mol
Now that we have the amount of moles, we can divide our rate constant by this value.
(5 x 10-5) min-1 / (5 x 10-12) mol = (1 x 107) mol-1 min -1
Answer
The reaction is first order with k = 1.0 × 107 mol−1 min−1
Q12.4.13
Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?
Solution
This problem is asking us for the percentage of radioactivity remaining after a certain time for both isotopes after 48 hours. We must identify an equation that will help us solve this and we can determine that we can determine this information using the first order equation.
This equation Ln(N/No)= -kt tells that the Natural log of the fraction remaining is equal to the rate constant times time. To determine the rate constant, we can also compute .693 over the half-life given in the information.
For Technetium-99 we can determine the rate constant by plugging into the second equation: .693/6 hrs= .1155 h-1
Now that we have the rate constant we can plug in : Ln(N/No)=-(.1155h-1)(48h) so Ln(N/No)=-5.544 and if we take the inverse of the natural log, we get (N/No)=3.9x10 -3 and if we multiply this by 100, we get .39% remaining.
We can do this same process for Thallium-201 and plugin: .693/73 hrs= .009493151 h-1 and when we plug this into the first order equation we get:
Ln(N/No)=-(.009493h-1)(48h) so Ln(N/No)=-.45567248 and when we take the inverse of the natural log, we get (N/No)=.6340 and when multiplied by 100, we get 63.40% remaining which makes sense since its half-life is 73 hours and only 48 hours have passed, half of the amount has yet to be consumed.
Answer
Technetium-99: 0.39%
Thallium-201: 63.40%
Q12.4.14
There are two molecules with the formula C3H6. Propene, $\ce{CH_3CH=CH_2}$, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:
When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of 5.95 × 10−4 s−1. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C?
Solution
Use the equation $t{_1}{_/}{_2} = \frac{ln2} k\nonumber$ since this is a first-order reaction. You can tell that this is a first order reaction due to the units of measurement of the rate constant, which is s-1. Different orders of reactions lead to different rate constants, and a rate constant of s-1 will always be first order.
Plug into the equation, and you get half life = 1164.95 seconds. To convert this to hours, we would divide this number by 3600 seconds/hour, to get 0.324 hours.
Use the integrated first order rate law $ln\frac{[A]}{[A]_0} = -kt\nonumber$. In this equation, [A]0 represents the initial amount of compound present at time 0, while [A] represents the amount of compound that is left after the reaction has occurred. Therefore, the fraction $\frac{[A]}{[A]_0}\nonumber$ is equal to the fraction of cyclopropane that remains after a certain amount of time, in this case, 0.75 hours.
Substitute x for the fraction of $\frac{[A]}{[A]_0}\nonumber$ into the integrated rate law: $ln\frac{[A]}{[A]_0} = -kt\nonumber$ $ln(x) = -5.95x10^{-4}(0.75)\nonumber$ $x=e^{(-0.000595)(0.75)}\nonumber$ = 0.20058 = 20%.
So, the half life is 0.324 hours, and 20% of the cyclopropane will remain as 80% will have formed propene.
Answer
0.324 hours. ; 20% remains
Q12.4.16
Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the nuclear equation is $\ce{^{18}_9F ⟶ _8^{18}O + ^0_{1}e^+}$.) Physicians use 18F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.
1. What is the rate constant for the decomposition of fluorine-18?
2. If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?
3. How long does it takFe for 99.99% of the 18F to decay?
Solution
a) The nuclear decay of an isotope of an element is represented by the first order equation:
ln(N/N0) = −kt
Where t is time, N0 is the initial amount of the substance, N is the amount of the substance after time t, and k is the rate constant. We can rearrange the equation and isolate k so that we could solve for the rate constant:
k = [-ln(N/N0)] / t
We are given that fluorine-18 has a half-life of 109.7 minutes. Since we have the half-life, we can choose an arbitrary value for N0 and use half of that value for N. In this case, we choose 100 for N0 and 50 for N. Now we can plug in those values into the equation above and solve for k.
k = [-ln(50/100)] / 109.7
k = 0.6931 / 109.7 = 0.006319 min-1
The rate constant for this reaction is 0.006319 min-1.
b) For this problem, we are able to use the same equation from part a:
ln(N/N0) = −kt
However, this time we are given the amount of time elapsed instead of the half-life, and we are asked to determine the percent of fluorine-18 radioactivity remaining after that time. In this problem, we must plug in values for N0, k (determined from part a), and t.
But first, since we are given the elapsed time in hours, we must convert it into minutes:
5.59 hours x (60 minutes / 1 hours) = 335.4 minutes
This gives us the value for t. We also have values for k (0.006319 min-1) and N0 (again an arbitrary number.) Now we can plug values into the original equation, giving us:
ln(N/100) = −(0.006319)(335.4)
We solve this equation by taking the exponential of both sides:
eln(N/100) = e−(0.006319)(335.4)
where eln equals 1 and now we can just solve for N:
N/100 = e−(0.006319)(335.4)
N = [e−(0.006319)(335.4)] x 100 = 12.0
Since 100 was used as the initial amount and 12.0 was determined as the remaining amount, 12.0 can be used as the percentage of remaining amount of radioactivity of fluorine-18. Thus the percent of fluorine-18 radioactivity remaining after 5.59 hours is 12.0%.
c) This part of the question is much like the previous two parts, but this time we are given the initial amount of radioactivity, the final amount of radioactivity and we are asked do determine how long it took for that amount of radioactivity to decay. We are able to use the same equation:
ln(N/N0) = −kt
However, now we are given N and N0 and we have already determined k from before. We are told that 99.99% of the radioactivity has decayed, so we can use 100 and 0.01 for N0 and N respectively. We plug these values in to the equation, solve for t, and get
ln(0.01/1000) = −0.006319t
-9.21 = −0.006319t
t = 1458 minutes
Therefore, its takes 1458 minutes for 99.99% of the radioactivity to decay.
Answer
a) 0.006319 min-1
b) 12.0%
c) 1458 minutes
Q12.4.17
Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for $\dfrac{1}{64}$ of the initial dose to remain in the athlete’s body?
Solution
252 days
for first order reaction: t1/2 = 0.693 / k
k = 0.693 / 42
k = 0.0165
for first order reaction: [A] = [A]0 e-kt
1/64 initial means that: [A] = 1/64 [A]0
therefore: 1/64 [A]0 = [A]0 e-0.0165t
t = 252 days
Q12.4.18
Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.
Solution
In order to find out what year King Richard III died, set [A]/[A0] (the percent of carbon-14 still contained) equal to 0.5time(t)/half life (t1/2) or use the equation N(t) = N0e-rt.
Using the first equation:
$A/A_{0}$ = $0.5^{t/t_{1/2}}$ plug in the given numbers $.9379 = 0.5^{t/5730}$ and solve for t.
$ln.9379$ = $(t/5730)(ln0.5)$ (using the rule of logs)
$-.0641$ = $(t/5730)(-.693)$
$-367.36$ = $-.693t$
$t = 530.1 years$
Using $N(t) = N_{0}e^{-rt}$ this problem is solved by the following:
$1/2 = e^{-5730r}$
$r = 0.000121$
Now that we know what r is, we can use this value in our original formula and solve for t, the amount of years that have passed.
This time, we use 93.78, the percent of the carbon-14 remaining as N(t) and 100 as the original, N0.
$93.78 = 100e^{-0.000121t}$
$t = 530.7$ years
Another way of doing this is by using these two equations:
λ = $\dfrac{0.693}{t_{1/2}}$ and $\dfrac{n_{t}}{n_{0}}$ = -λt
$n_{t}$ = concentration at time t (93.79)
$n_{0}$ = initial concentration (100)
First solve for lambda or the decay constant by plugging in the half life.
Then plug in lambda and the other numbers into the second equation, and solve for t- which should equal to 530.1 years as well.
If we want to find out what year King Richard III died, we take the current year, 2017, and subtract 530 years. Doing this, we find that King Richard III died in the year 1487.
Answer
King Richard III died in the year 1487
Q12.4.19
Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:
Initial [C3H5N3O9] (M) 4.88 3.52 2.29 1.81 5.33 4.05 2.95 1.72
t (s) 300 300 300 300 180 180 180 180
% Decomposed 52.0 52.9 53.2 53.9 34.6 35.9 36.0 35.4
Solution
First we need to understand what the question is asking for: the average rate constant. The average rate constant is the variable "k" when discussing kinetics and it can be defined as the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances. Knowing that we need to find K in this first order reaction, we can look to formulas that include "k," initial and final concentrations $[A]_o and [A]_t$, and half life time "t." Since this is a first order reaction, we can look to the first order equations, and doing that we find one that includes the variables given in the question: $\ln[A]_t=-kt+\ln[A]_o\nonumber$
For the first reaction, we have an initial concentration of 4.88 M, and a percentage decomposed. To find the final concentration, we must multiply the initial concentration by the percentage decomposed to know how much decomposed, and subtract that from the original to find out how much is left: 4.88M x 0.52= 2.54 M and 4.88M-2.54M=2.34M
Now, we have the variables we need, and we plug it into the equation above:
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[2.34M]=-k(300s)+\ln[4.88M]$
k=${-(\ln[2.34M]-\ln[4.88M])}\over 300$
$k=2.45x10^{-3}$
Since it asks for the rate constant of each experiment, we now must do the same procedure for each data set to find the rate constant:
Second experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[1.66M]=-k(300s)+\ln[3.52M]$
k=${-(\ln[1.66M]-\ln[3.52M])}\over 300$
$k=2.51x10^{-3}$
Third experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[1.07M]=-k(300s)+\ln[2.29M]$
k=${-(\ln[1.07M]-\ln[2.29M])}\over 300$
$k=2.54x10^{-3}$
Fourth experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[0.834M]=-k(300s)+\ln[1.81M]$
k=${-(\ln[0.834M]-\ln[1.81M])}\over 300$
$k=2.58x10^{-3}$
Fifth Experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[3.49M]=-k(180s)+\ln[5.33M]$
k=${-(\ln[3.49M]-\ln[5.33M])}\over 180$
$k=2.35x10^{-3}$
Sixth Experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[2.60M]=-k(180s)+\ln[4.05M]$
k=${-(\ln[2.60M]-\ln[4.05M])}\over 180$
$k=2.46x10^{-3}$
Seventh Experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[1.89M]=-k(180s)+\ln[2.95M]$
k=${-(\ln[1.89M]-\ln[2.95M])}\over 180$
$k=2.47x10^{-3}$
Eighth experiment
$\ln[A]_t=-kt+\ln[A]_o$
$\ln[1.11M]=-k(180s)+\ln[1.72M]$
k=${-(\ln[1.11M]-\ln[1.72M])}\over 180$
$k=2.43x10^{-3}$
Answer
[A]0 (M) k × 103 (s−1)
4.88 2.45
3.52 2.51
2.29 2.54
1.81 2.58
5.33 2.35
4.05 2.44
2.95 2.47
1.72 2.43
Q12.4.20
For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene $\ce{(CH2=CH–CH=CH2)}$ has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:
The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 × 10−4 s−1 at 150 °C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 °C with an initial pressure of 55 torr.
Solution
Since this is a first order reaction, the integrated rate law is: $[A_{t}]=[A_{0}]e^{-kt}$
Partial Pressure: Use the integrated rate law to find the partial pressure at 30 minutes:
Use $A_0$ = 55 torr, t = 30 min, and k = $2.0 * 10^{-4}s^{-1}$ to solve the integrated rate law equation:
$[A_{30}]=(55 torr)*e^{-(2.0x10^{-4}\frac{1}{sec})(30min\cdot\frac{60sec}{1 min})}$
Solve this equation to get:
$[A_{30}]=(55 torr)*e^{-0.36}$
$A_{30}]$ = 38.37 torr.
Initial Concentration: Find the initial concentration using the ideal gas law.
The ideal gas law is given by $PV = nRT → n = \frac{PV}{RT}$. Use this form of the gas law to solve for the initial concentration n.
Use V = 0.53L, R = 0.08206 $\frac{L*atm}{mol*L}$, T = 423.15 K, and P = $\frac{1 atm}{760}$ = 0.07237 atm .
Solve the ideal gas equation using these values:
$n=\frac{(55torr)(0.53L)}{(0.08206\frac{L*atm}{mol*K})(423.15K)} = 0.00110$ moles cyclobutene.
Now find the initial concentration of cyclobutene $A_0$ using the equation $[A_0] = \frac{n}{V}$:
$A_0 = \frac{n}{V} = \frac{0.00110 moles}{0.53 L} = 0.00208 M$
Concentration at 30 minutes: Find the concentration of cyclobutene at 30 minutes by using the integrated rate law given above, using time t = 30 minutes, or 1800 seconds.
$[A_{30}]=(0.00208M)e^{-0.36}= 0.00145M$
So at 30 minutes, the cyclobutene concentration is 0.00145 M, and the partial pressure is 38.37 torr.
Answer
Partial Pressure: 38.37 torr.
Concentration: 0.00145 M
12.5: Collision Theory
Q12.5.1
Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
Solution
The two factors that may prevent a collision from producing a chemical reaction are:
1. Kinetic energy of the molecule
In order for chemical reactions to occur, molecules require enough velocity to overcome the minimum activation energy needed to break the old bonds and form new bonds with other molecules. At higher temperatures, the molecules possess the minimum amount of kinetic energy needed which ensures the collisions will be energetic enough to lead to a reaction.
2. The orientation of molecules during the collision
Two molecules have to collide in the right orientation in order for the reaction to occur. Molecules have to orient properly for another molecule to collide at the right activation state.
Q12.5.2
When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
Solution
There has to be contact between reactants for a reaction to occur. The more the reactants collide, the more often reactions can occur. Factors that determine reaction rates include concentration of reactants, temperature, physical states of reactants, surface area, and the use of a catalyst. The reaction rate usually increases as the concentration of a reactant increases. Increasing the temperature increases the average kinetic energy of molecules, causing them to collide more frequently, which increases the reaction rate. When two reactants are in the same fluid phase, their particles collide more frequently, which increases the reaction rate. If the surface area of a reactant is increased, more particles are exposed to the other reactant therefore more collisions occur and the rate of reaction increases. A catalyst participates in a chemical reaction and increases the reaction rate without changing itself.
Q12.5.3
What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
Solution
Activation energy is the energy barrier that must be overcome in order for a reaction to occur. To get the molecules into a state that allows them to break and form bonds, the molecules must be contorted (deformed, or bent) into an unstable state called the transition state. The transition state is a high-energy state, and some amount of energy – the activation energy – must be added in order for the molecule reach it. Because the transition state is unstable, reactant molecules don’t stay there long, but quickly proceed to the next step of the chemical reaction.The activated complex is the highest energy of the transition state of the reaction.
Q12.5.5
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
Solution
This method is based on the Arrhenius equation which can be used to show the effect of a change of temperature on the rate constant, and therefore on the rate of reaction.
The rate constant is different from reaction rat in that the reaction rate is the measure of how fast or slow a chemical reaction takes place while a rate constant is a constant that shows the relationship between the reaction rate and the concentrations of the reactants or products.
For example, for the reaction $A + B \rightarrow C$, the rate law would be:
$rate = k[A]^a[B]^b$
k = rate constant
[A] = concentration of reactant A
a = order of reaction with respect to A
[B] = concentration of reactant B
b = order of reaction with respect to B
However, the rate constant remains constant only if you are changing the concentration of the reactants. If you change the temperature or the catalyst of the reaction, the rate constant will change and this is demonstrated by the Arrhenius equation:
$k = Ae^\frac{-E_a}{RT}$
$ln \left(\frac{k_1}{k_2}\right) = \left(\frac{-E_a}{R}\right)\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
k = rate constant
A = frequency factor
$E_a$ = activation energy
e = exponential function, $e^x$
R = gas constant
T = temperature (K)
In other words, the activation energy of a reaction, $E_a$, from a series of data that includes the rate of reaction, k, at varying temperatures can be determined by graphing it on a plot of $\ln k$ versus $\frac{1}{T}$. You can then use the slope of the graph you have plotted to solve for $E_a$ by setting the slope equal to $\frac{-E_a}{R}$.
Q12.5.6
How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
S12.5.6
Collision theory states that the rates of chemical reactions depend on the fraction of molecules with the correct orientation, fraction of collisions with required energy, and the collision frequency. Because the fraction of collisions with required energy is a function of temperature, as temperature increases, the fraction of collisions with required energy also increases. The kinetic energy of reactants also increases with temperature which means molecules will collide more often increasing collisions frequency. With increased fraction of collisions with required energy and collisions frequency, the rate of chemical reaction increases. We see mathematically, from the Arrhenius equation, that temperature and the rate constant are related.
$k=Ae^{\frac {E_a}{RT}}$
where k is the rate constant, A is a specific constant, R is 8.3145 J/K, Ea is the reaction-specific activation energy in J, and T is temperature in K. We see from the equation that k is very sensitive to changes in the temperature.
Q12.5.7
The rate of a certain reaction doubles for every 10 °C rise in temperature.
1. How much faster does the reaction proceed at 45 °C than at 25 °C?
2. How much faster does the reaction proceed at 95 °C than at 25 °C?
Solution
By finding the difference in temperature, 45 °C - 25 °C, we get 20 °C. Since the rate of the reaction doubles every 10 °C increase in temperature and the rate of the reaction experienced a 20 °C increase in temperature, we see that the reaction rate doubled twice (22 = 4). As a result, the reaction proceeds four times faster.
Following the same process as in part a, we get the difference in temperature to be 70 °C. Since the rate of the reaction doubles every 10 °C increase in temperature and the system experienced a 70 °C change, we see that the reaction doubled seven times (27 = 128). We can see the reaction proceeds 128 times faster.
(a) 4-times faster (b) 128-times faster
Q12.5.8
In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher?
S12.5.8
First off, it is important to recognize that this decomposition reaction is a first-order reaction, which can be written as follows: $\mathrm2NaClO_3\to2NaCl + 3O_2$
Understanding this, it is important to be able to then be able to recognize which equation would be most useful given the initial conditions presented by the question. Since we are dealing with time, percentage of material left, and temperature, the only viable equation that could relate all of this would be the Arrhenius Equation, which is written as follows: $\mathrm \ln(\frac{k_2}{k_1}) = \frac {Ea}{R}({\frac1{t_1}}-{\frac{1}{t_{2}}})$
However, this problem does not give us enough information such as what the activation energy is or the initial temperature in order to mathematically solve this problem. Additionally, the problem tells us to approximate how long the decomposition would take, which means we are asked to answer this question conceptually based on our knowledge of thermodynamics and reaction rates. As a general rule of thumb, we know that for every 10˚C rise in temperature the rate of reaction doubles. Since the question tells us that there is a 20˚C rise in temperature we can deduce that the reaction rate doubles twice, as per the general rule mentioned before. This means the overall reaction rate for this decomposition would quadruple, or would be 4 times faster than the reaction rate at the initial temperature.
We can gut check this answer by recalling how an increase in the average kinetic energy (temperature) decreases the time it takes for the reaction to take place and increase the reaction rate. Thus, if we increase the temperature we should have a faster reaction rate.
Q12.5.9
The rate constant at 325 °C for the decomposition reaction $\ce{C4H8⟶2C2H4}$ is 6.1 × 10−8 s−1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
Solution
S12.5.9
Using the Arrhenius equation allows me to find the frequency factor, A.
k=Ae-Ea/RT
k, Ea, R, and T are all known values. k, Ea, and T are given in the problem as 6.1x10-8, 261 kJ, and 598 K, respectively.
So, plugging them into the equation gives:
6.1x10-8 s-1=Ae(-261000 J)/(8.3145 J/mol)(598 K)
Take e(-261000 J)/(8.3145 J/mol)(598) and get 1.59 x 10-23. Divide k, 6.1 x 10-8, by 1.59 x 10-23 and get A=3.9 x 1015s-1
A12.5.9
$\mathrm{3.9×10^{15}\:s^{−1}}$
Q12.5.10
The rate constant for the decomposition of acetaldehyde (CH3CHO), to methane (CH4), and carbon monoxide (CO), in the gas phase is 1.1 × 10−2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.
S12.5.10
The equation for relating the rate constant and activation energy of a reaction is the Arrhenius equation:
$k = Ae^ {-\frac{E_a}{RT}}$
When given two rate constants at two different temperatures but for the same reaction, the Arrhenius equation can be rewritten as:
$ln (\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$
In this problem, all the variables are given except for the Ea (activation energy).
k1 = 1.1 × 10−2 L/mol/s
T1 = 703 K
k2 = 4.95 L/mol/s
T2 = 865 K
R = 8.314 J/(mol K) (Ideal Gas Constant)
Now plug in all these values into the equation, and solve for Ea.
$ln (\frac{4.95\frac{L}{mol×s}}{1.1 × 10^{-2}\frac{L}{mol×s}}) = \frac{E_a}{8.314 × 10^{-3}\frac{kJ}{mol×K}} (\frac{1}{703} - \frac{1}{865})$
Ea = 190 kJ (2 sig figs)
Q12.5.11
An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
Solution
43.0 kJ/mol
Q12.5.12
In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
1. the change in free energy per second
2. the change in temperature per second
3. the number of collisions per second
4. the number of product molecules
Q12.5.13
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:
Temperature (K) k (M−1 s−1)
555 6.23 × 10−7
575 2.42 × 10−6
645 1.44 × 10−4
700 2.01 × 10−3
What is the value of the activation energy (in kJ/mol) for this reaction?
Solution
177 kJ/mol
Q12.5.14
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
T (K) k (s−1)
293 0.054
298 0.100
Q12.5.15
The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
$\ce{C12H22O11 + H2O ⟶ C6H12O6 + C6H12O6}\nonumber$
follows a first-order rate equation for the disappearance of sucrose: Rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
1. In neutral solution, k = 2.1 × 10−11 s−1 at 27 °C and 8.5 × 10−11 s−1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10−7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
Solution
Ea = 108 kJ
A = 2.0 × 108 s−1
k = 3.2 × 10−10 s−1
(b) 1.81 × 108 h or 7.6 × 106 day. (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.
Q12.5.16
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first $A+BC⟶AB+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
Q12.5.17
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first $A+BC⟶AB+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?
Solution
The A atom has enough energy to react with BC; however, the different angles at which it bounces off of BC without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.
12.6: Reaction Mechanisms
Q12.6.1
Why are elementary reactions involving three or more reactants very uncommon?
Q12.6.2
In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction $A+B⟶C$ ? Can we predict the effect if the reaction is known to be an elementary reaction?
Solution
Add texts here. Do not delete this text first.
No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate equation. Yes. If the reaction is an elementary reaction, then doubling the concentration of A doubles the rate.
Q12.6.3
Phosgene, COCl2, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by:
step 1: Cl + CO → COCl
step 2: COCl + Cl2→ COCl2 + Cl
1. Write the overall reaction equation.
2. Identify any reaction intermediates.
3. Identify any intermediates.
Q12.6.4
Define these terms:
1. unimolecular reaction
2. bimolecular reaction
3. elementary reaction
4. overall reaction
Q12.6.5
What is the rate equation for the elementary termolecular reaction $A+2B⟶\ce{products}$? For $3A⟶\ce{products}$?
Solution
Add texts here. Do not delete this text first.
We are given that both of these reactions are elementary termolecular. The molecularity of a reaction refers to the number of reactant particles that react together with the proper and energy and orientation. Termolecular reactions have three atoms to collide simultaneously. As it is termolecular, and there are no additional reactants aside from the three given in each reaction, there are no intermediate reactions. The rate law for elementary reactions is determined by the stoichiometry of the reaction without needed experimental data.
The basic rate form for the elementary step is what follows:
$rate= {k} \cdot {reactant \ 1}^{i} \cdot {reactant \ 2}^{ii} \cdot ...$ Where i and ii are the stochiometric coefficient from reactant 1 and 2 respectively.
For: $3A \rightarrow products$
${k} \cdot {A}^3 = rate$
For: $A + 2B \rightarrow products$
${k} \cdot {[A]} \cdot {[B]}^2 = rate$
Note that the order of these reactions are both three.
Answer
Add texts here. Do not delete this text first.
Rate = k[A][B]2; Rate = k[A]3
Q12.6.6
Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?
(a) $\ce{Cl2 + CO ⟶ Cl2CO}$
$\ce{rate}=k\ce{[Cl2]^{3/2}[CO]}$
(b) $\ce{PCl3 + Cl2 ⟶ PCl5}$
$\ce{rate}=k\ce{[PCl3][Cl2]}$
(c) $\ce{2NO + H2 ⟶ N2 + H2O}$
$\ce{rate}=k\ce{[NO][H2]}$
(d) $\ce{2NO + O2 ⟶ 2NO2}$
$\ce{rate}=k\ce{[NO]^2[O2]}$
(e) $\ce{NO + O3 ⟶ NO2 + O2}$
$\ce{rate}=k\ce{[NO][O3]}$
Solution
Add texts here. Do not delete this text first.
An elementary reaction is a chemical reaction in which the reactants directly form products in a single step. In another words, the rate law for the overall reaction is same as experimentally found rate law. Out of 5 options, option (b),(d), and (e) are such reactions
Q12.6.7
Write the rate equation for each of the following elementary reactions:
1. $\ce{O3 \xrightarrow{sunlight} O2 + O}$
2. $\ce{O3 + Cl ⟶ O2 + ClO}$
3. $\ce{ClO + O⟶ Cl + O2}$
4. $\ce{O3 + NO ⟶ NO2 + O2}$
5. $\ce{NO2 + O ⟶ NO + O2}$
Solution
Add texts here. Do not delete this text first.
Rate equations are dependent on the reactants and not the products.
The rate law of a reaction can be found using a rate constant (which is found experimentally), and the initial concentrations of reactants.
A general solution for the equation
$aA + bB \rightarrow cC + dD$
is $rate = k[A]^{m}[B]^{n}$ where m and n are reaction orders.
However, reaction orders are found experimentally, and since we do not have experimental data for these reactions, we can disregard that part of the equation.
To find the rate laws, all we have to do is plug the reactants into the rate formula. This is only due to the case that these are elementary reactions. Further reading on elementary reactions can be found on Libre Texts.
a. O3 ⟶ O2 + O
To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".
Rate = k[O3]
b. O3 + Cl ⟶ O2 + ClO
To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".
Rate = k[O3][Cl]
c. ClO + O ⟶ Cl + O2
To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".
Rate = k[ClO][O]
d. O3 + NO ⟶ NO2 + O2
To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".
Rate = k[O3][NO]
e. NO2 + O ⟶ NO + O2
To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".
Rate = k[NO2][O]
Answer
Add texts here. Do not delete this text first.
(a) Rate1 = k[O3]; (b) Rate2 = k[O3][Cl]; (c) Rate3 = k[ClO][O]; (d) Rate2 = k[O3][NO]; (e) Rate3 = k[NO2][O]
Q12.6.8
Nitrogen(II) oxide, NO, reacts with hydrogen, H2, according to the following equation:
$\ce{2NO + 2H2 ⟶ N2 + 2H2O}\nonumber$
What would the rate law be if the mechanism for this reaction were:
$\ce{2NO + H2 ⟶ N2 + H2O2\:(slow)}\nonumber$
$\ce{H2O2 + H2 ⟶ 2H2O\:(fast)}\nonumber$
The rate law of the mechanism is determined by the slow step of the reaction. Since the slow step is an elementary step, the rate law can be drawn from the coefficients of the chemical equation. So therefore, the rate law is as follows: rate=k[NO]2[H2]. Since both NO and H2 are reactants in the overall reaction (therefore are not intermediates in the reaction), no further steps have to be done to determine the rate law.
Q12.6.9
Consider the reaction
CH4 + Cl2 → CH3Cl + HCl (occurs under light)
The mechanism is a chain reaction involving Cl atoms and CH3 radicals. Which of the following steps does not terminate this chain reaction?
1. CH3 + Cl → CH3CI
2. CH3 + HCl → CH4 + Cl
3. CH3 + CH3 → C2H2
4. Cl + Cl → Cl2
Solution
Add texts here. Do not delete this text first.
Chain reactions involve reactions that create products necessary for more reactions to occur. In this case, a reaction step will continue the chain reaction if a radical is generated. Radicals are highly reactive particles, so more reactions in the chain will take place as long as they are present. The chlorine is considered a free radical as it has an unpaired electron; for this reason it is very reactive and propagates a chain reaction. It does so by taking an electron from a stable molecule and making that molecule reactive, and that molecule goes on to react with stable species, and in that manner a long series of "chain" reactions are initiated. A chlorine radical will continue the chain by completing the following reaction:
${Cl \cdot}+{CH_4} \rightarrow {CH_3 \cdot}+{HCl}$
The ${CH_3}$ generated by this reaction can then react with other species, continuing to propagate the chain reaction.
Option 1 is incorrect because the only species it produces is ${CH_3Cl}$, a product in the overall reaction that is unreactive. This terminates the chain reaction because it fails to produce any $Cl$ or $CH_3$ radicals that are necessary for further propagating the overall reaction.
Option 2 is the correct answer because it produces a $Cl$ radical. This $Cl$ radical can continue the chain by colliding with $CH_4$ molecules.
Option 3 is incorrect because it fails to produce a radical capable of continuing the chain.
Option 4 is incorrect because it produces $Cl_2$, a molecule that does not react unless additional light is supplied. Therefore, this step breaks the chain.
Answer
Add texts here. Do not delete this text first.
Answer: Option 2: ${CH_3}+{HCl} \rightarrow {CH_4}+{Cl}$
Q12.6.10
Experiments were conducted to study the rate of the reaction represented by this equation.
$\ce{2NO}(g)+\ce{2H2}(g)⟶\ce{N2}(g)+\ce{2H2O}(g)\nonumber$
Initial concentrations and rates of reaction are given here.
Experiment Initial Concentration [NO] (mol/L) Initial Concentration, [H2] (mol/L) Initial Rate of Formation of N2 (mol/L min)
1 0.0060 0.0010 1.8 × 10−4
2 0.0060 0.0020 3.6 × 10−4
3 0.0010 0.0060 0.30 × 10−4
4 0.0020 0.0060 1.2 × 10−4
Consider the following questions:
1. Determine the order for each of the reactants, NO and H2, from the data given and show your reasoning.
2. Write the overall rate law for the reaction.
3. Calculate the value of the rate constant, k, for the reaction. Include units.
4. For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H2 had been consumed.
5. The following sequence of elementary steps is a proposed mechanism for the reaction.
Step 1: $\ce{NO + NO ⇌ N2O2}$
Step 2: $\ce{N2O2 + H2 ⇌ H2O + N2O}$
Step 3: $\ce{N2O + H2 ⇌ N2 + H2O}$
Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.
S12.6.10
1. i) Find the order for [NO] by using experiment 3 and 4 where [H2] is constant
Notice that [NO] doubles from experiment 3 to 4 and the rate quadruples. So the order for [NO] is 2
ii) Find the order for [H2] by using experiment 1 and 2 where [NO] is constant
Notice that [H2] doubles from experiment 1 to 2 and the rate doubles as well. So the order for [H2] is 1
2. Put in the order for each product as the exponents for the corresponding reactant.
$rate = k [NO]^2 [H_2]$
3. Put in the concentrations and the rate from one of the experiments into the rate law and solve for k. (Here, experiement 1 is used but any of them will work)
$rate = k [NO]^2 [H_2]$ $.00018 = k [.006]^2 [.001]$ $k = 5000 M^{-2}s^{-1}$
4. Plug in values for experiment 2 into the rate law equation and solve for the concentration of NO
$.00036=5000[NO]^2[.001]$ $[NO]^2= 7.2 x 10^{-5}$ $[NO] = .0085 M$
5. Write the rate laws for each step and then see which matches the rate law we found in question 2. The rate determining step (the slow step) is the one that gives the rate for the overall reaction. Because of this, only those concentrations will influence the overall reaction, contrary to what we would believe if we just looked at the overall reaction.
Step 1: $NO + NO \rightleftharpoons N_2O_2$
$rate =k_1[NO]^2$ This rate law is not the same as the one we calculate in question 2 so this can not be the rate determining step
Step 2: $N_2O_2+H_2 \rightleftharpoons N_2O + N_2O$
$rate = k_2[N_2O_2][H_2]$
Since $N_2O_2$ is an intermediate you must replace it in the rate law equation. Intermediates can not be in the rate law because they do not appear in the overall reaction. Here you can take the reverse of equation 1 (k-1) and substitute the other side (the reactants of equation 1) for the intermediate in the rate law equation.
$rate_1 = rate_{-1}\nonumber$
$k_1[NO]^2 = k_{-1}[N_2O_2]\nonumber$
$[N_2O_2] = \frac{k_1[NO]^2}{k_{-1}}\nonumber$
$rate= \frac{k_2k_{1}[NO]^2[H_2]}{k_{-1}}$
Overall: $rate={k[NO]^2[H_2]}$ This is the same so it is the rate determining step.
So $N_2O_2+H_2 \rightleftharpoons N_2O + N_2O$ is the rate determining step step.
Answer
Add texts here. Do not delete this text first.
(a) NO: 2
$\ce {H2}$ : 1
(b) Rate = k [NO]2[H2];
(c) k = 5.0 × 103 mol−2 L−2 min−1;
(d) 0.0050 mol/L;
(e) Step II is the rate-determining step.
Q12.6.11
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:
• $\ce{Cl2}(g)⇌\ce{2Cl}(g)$ (fast, k1 represents the forward rate constant, k−1 the reverse rate constant)
• $\ce{CO}(g)+\ce{Cl}(g)⟶\ce{COCl}(g)$ (slow, k2 the rate constant)
• $\ce{COCl}(g)+\ce{Cl}(g)⟶\ce{COCl2}(g)$ (fast, k3 the rate constant)
1. Write the overall reaction.
2. Identify all intermediates.
3. Write the rate law for each elementary reaction.
4. Write the overall rate law expression.
Solution
Add texts here. Do not delete this text first.
1. To write the overall reaction you have to identify the intermediates and leave them out. The easiest way to do this is to write out all the products and reactants and cross out anything that is on both sides.
Cl2(g) + CO(g) + 2Cl(g) +COCl(g) 2Cl(g) + COCl(g) + COCl2(g)
In this you will cross out the 2Cl(g) molecules and the COCl(g). What is left after that is the overall reaction.
Cl2(g) + CO(g) + COCl2(g)
2. For part two you will just list the intermediates that you crossed out.
Cl and COCl are intermediates
3. Each rate law will be the rate equal to the rate constant times the concentrations of the reactants
reaction 1
(forward) rate=k1[Cl2] ( reverse) rate=k-1[Cl]
reaction 2
rate=k2[CO][Cl]
Reaction 3
rate=k3[COCl][Cl]
4. The overall rate law is based off the slowest step (step #2), since it is the rate determining step, but Cl is present in that rate law so we have to replace it with an equivalent that does not contain an intermediate. To do this you use the equilibrium since the rates are the same you can set up the rate laws of the forward and reverse equal to each other.
k1[Cl2] = k-1[Cl]
[Cl]= k1[Cl2]/k-1
rate=k2[CO]k1[Cl2]/k-1
rate=k°[CO][Cl2]
Steps to replacing and intermediate
1. Set the forward and reverse reaction equal to each other using separate constants
2. Solve for the intermediate using algebra
3. Plug into the rate determining formula
4. All the k's will be condensed into a K prime constant
12.7: Catalysis
Q12.7.1
Account for the increase in reaction rate brought about by a catalyst.
Q12.7.2
Compare the functions of homogeneous and heterogeneous catalysts.
Q12.7.3
Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F2, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is: $\ce{O3 \xrightarrow{sunlight} O2 + O}\ \ce{O3 + Cl ⟶ O2 + ClO}\ \ce{ClO + O ⟶ Cl + O2}\nonumber$
1. Explain why chlorine atoms are catalysts in the gas-phase transformation: $\ce{2O3⟶3O2}\nonumber$
2. Nitric oxide is also involved in the decomposition of ozone by the mechanism: $\ce{O3 \xrightarrow{sunlight} O2 + O\ O3 + NO ⟶ NO2 + O2\ NO2 + O ⟶ NO + O2}\nonumber$
Is NO a catalyst for the decomposition? Explain your answer.
Q12.7.4
For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed:
(a)
Q12.7.5
For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed:
(a)
(b)
Q12.7.6
For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:
(a)
(b)
Q12.7.7
For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:
(a)
(b)
Q12.7.8
1. Based on the diagrams in Question Q12.7.6, which of the reactions has the fastest rate? Which has the slowest rate?
2. Based on the diagrams in Question Q12.7.7, which of the reactions has the fastest rate? Which has the slowest rate? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/17%3A_Kinetics/17.E%3A_Kinetics_%28Exercises%29.txt |
The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table.
18: Representative Metals Metalloids and Nonmetals
Learning Objectives
• Classify elements
• Make predictions about the periodicity properties of the representative elements
We begin this section by examining the behaviors of representative metals in relation to their positions in the periodic table. The primary focus of this section will be the application of periodicity to the representative metals.
It is possible to divide elements into groups according to their electron configurations. The representative elements are elements where the s and p orbitals are filling. The transition elements are elements where the d orbitals (groups 3–11 on the periodic table) are filling, and the inner transition metals are the elements where the f orbitals are filling. The d orbitals fill with the elements in group 11; therefore, the elements in group 12 qualify as representative elements because the last electron enters an s orbital. Metals among the representative elements are the representative metals. Metallic character results from an element’s ability to lose its outer valence electrons and results in high thermal and electrical conductivity, among other physical and chemical properties. There are 20 nonradioactive representative metals in groups 1, 2, 3, 12, 13, 14, and 15 of the periodic table (the elements shaded in yellow in Figure $1$). The radioactive elements copernicium, flerovium, polonium, and livermorium are also metals but are beyond the scope of this chapter.
In addition to the representative metals, some of the representative elements are metalloids. A metalloid is an element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors. The remaining representative elements are nonmetals. Unlike metals, which typically form cations and ionic compounds (containing ionic bonds), nonmetals tend to form anions or molecular compounds. In general, the combination of a metal and a nonmetal produces a salt. A salt is an ionic compound consisting of cations and anions.
A salt is an ionic compound consisting of cations and anions.
Most of the representative metals do not occur naturally in an uncombined state because they readily react with water and oxygen in the air. However, it is possible to isolate elemental beryllium, magnesium, zinc, cadmium, mercury, aluminum, tin, and lead from their naturally occurring minerals and use them because they react very slowly with air. Part of the reason why these elements react slowly is that these elements react with air to form a protective coating. The formation of this protective coating is passivation. The coating is a nonreactive film of oxide or some other compound. Elemental magnesium, aluminum, zinc, and tin are important in the fabrication of many familiar items, including wire, cookware, foil, and many household and personal objects. Although beryllium, cadmium, mercury, and lead are readily available, there are limitations in their use because of their toxicity.
Group 1: The Alkali Metals
The alkali metals lithium, sodium, potassium, rubidium, cesium, and francium constitute group 1 of the periodic table. Although hydrogen is in group 1 (and also in group 17), it is a nonmetal and deserves separate consideration later in this chapter. The name alkali metal is in reference to the fact that these metals and their oxides react with water to form very basic (alkaline) solutions.
The properties of the alkali metals are similar to each other as expected for elements in the same family. The alkali metals have the largest atomic radii and the lowest first ionization energy in their periods. This combination makes it very easy to remove the single electron in the outermost (valence) shell of each. The easy loss of this valence electron means that these metals readily form stable cations with a charge of 1+. Their reactivity increases with increasing atomic number due to the ease of losing the lone valence electron (decreasing ionization energy). Since oxidation is so easy, the reverse, reduction, is difficult, which explains why it is hard to isolate the elements. The solid alkali metals are very soft; lithium, shown in Figure $2$, has the lowest density of any metal (0.5 g/cm3).
The alkali metals all react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. This means they are easier to oxidize than is hydrogen. As an example, the reaction of lithium with water is:
$\ce{2Li}(s)+\ce{2H2O}(l)⟶\ce{2LiOH}(aq)+\ce{H2}(g) \nonumber$
Alkali metals react directly with all the nonmetals (except the noble gases) to yield binary ionic compounds containing 1+ metal ions. These metals are so reactive that it is necessary to avoid contact with both moisture and oxygen in the air. Therefore, they are stored in sealed containers under mineral oil, as shown in Figure $3$, to prevent contact with air and moisture. The pure metals never exist free (uncombined) in nature due to their high reactivity. In addition, this high reactivity makes it necessary to prepare the metals by electrolysis of alkali metal compounds.
Unlike many other metals, the reactivity and softness of the alkali metals make these metals unsuitable for structural applications. However, there are applications where the reactivity of the alkali metals is an advantage. For example, the production of metals such as titanium and zirconium relies, in part, on the ability of sodium to reduce compounds of these metals. The manufacture of many organic compounds, including certain dyes, drugs, and perfumes, utilizes reduction by lithium or sodium.
Sodium and its compounds impart a bright yellow color to a flame, as seen in Figure $4$. Passing an electrical discharge through sodium vapor also produces this color. In both cases, this is an example of an emission spectrum as discussed in the chapter on electronic structure. Streetlights sometime employ sodium vapor lights because the sodium vapor penetrates fog better than most other light. This is because the fog does not scatter yellow light as much as it scatters white light. The other alkali metals and their salts also impart color to a flame. Lithium creates a bright, crimson color, whereas the others create a pale, violet color.
Video $1$: This video demonstrates the reactions of the alkali metals with water.
Group 2: The Alkaline Earth Metals
The alkaline earth metals (beryllium, magnesium, calcium, strontium, barium, and radium) constitute group 2 of the periodic table. The name alkaline metal comes from the fact that the oxides of the heavier members of the group react with water to form alkaline solutions. The nuclear charge increases when going from group 1 to group 2. Because of this charge increase, the atoms of the alkaline earth metals are smaller and have higher first ionization energies than the alkali metals within the same period. The higher ionization energy makes the alkaline earth metals less reactive than the alkali metals; however, they are still very reactive elements. Their reactivity increases, as expected, with increasing size and decreasing ionization energy. In chemical reactions, these metals readily lose both valence electrons to form compounds in which they exhibit an oxidation state of 2+. Due to their high reactivity, it is common to produce the alkaline earth metals, like the alkali metals, by electrolysis. Even though the ionization energies are low, the two metals with the highest ionization energies (beryllium and magnesium) do form compounds that exhibit some covalent characters. Like the alkali metals, the heavier alkaline earth metals impart color to a flame. As in the case of the alkali metals, this is part of the emission spectrum of these elements. Calcium and strontium produce shades of red, whereas barium produces a green color.
Magnesium is a silver-white metal that is malleable and ductile at high temperatures. Passivation decreases the reactivity of magnesium metal. Upon exposure to air, a tightly adhering layer of magnesium oxycarbonate forms on the surface of the metal and inhibits further reaction. (The carbonate comes from the reaction of carbon dioxide in the atmosphere.) Magnesium is the lightest of the widely used structural metals, which is why most magnesium production is for lightweight alloys.
Magnesium (Figure $5$), calcium, strontium, and barium react with water and air. At room temperature, barium shows the most vigorous reaction. The products of the reaction with water are hydrogen and the metal hydroxide. The formation of hydrogen gas indicates that the heavier alkaline earth metals are better reducing agents (more easily oxidized) than is hydrogen. As expected, these metals react with both acids and nonmetals to form ionic compounds. Unlike most salts of the alkali metals, many of the common salts of the alkaline earth metals are insoluble in water because of the high lattice energies of these compounds, containing a divalent metal ion.
The potent reducing power of hot magnesium is useful in preparing some metals from their oxides. Indeed, magnesium’s affinity for oxygen is so great that burning magnesium reacts with carbon dioxide, producing elemental carbon:
$\ce{2Mg}(s)+\ce{CO2}(g)⟶\ce{2MgO}(s)+\ce{C}(s) \nonumber$
For this reason, a CO2 fire extinguisher will not extinguish a magnesium fire. Additionally, the brilliant white light emitted by burning magnesium makes it useful in flares and fireworks.
Group 12 Metals
The elements in group 12 are transition elements; however, the last electron added is not a d electron, but an s electron. Since the last electron added is an s electron, these elements qualify as representative metals, or post-transition metals. The group 12 elements behave more like the alkaline earth metals than transition metals. Group 12 contains the four elements zinc, cadmium, mercury, and copernicium. Each of these elements has two electrons in its outer shell (ns2). When atoms of these metals form cations with a charge of 2+, where the two outer electrons are lost, they have pseudo-noble gas electron configurations. Mercury is sometimes an exception because it also exhibits an oxidation state of 1+ in compounds that contain a diatomic $\ce{Hg2^2+}$ ion. In their elemental forms and in compounds, cadmium and mercury are both toxic.
Zinc is the most reactive in group 12, and mercury is the least reactive. (This is the reverse of the reactivity trend of the metals of groups 1 and 2, in which reactivity increases down a group. The increase in reactivity with increasing atomic number only occurs for the metals in groups 1 and 2.) The decreasing reactivity is due to the formation of ions with a pseudo-noble gas configuration and to other factors that are beyond the scope of this discussion. The chemical behaviors of zinc and cadmium are quite similar to each other but differ from that of mercury.
Zinc and cadmium have lower reduction potentials than hydrogen, and, like the alkali metals and alkaline earth metals, they will produce hydrogen gas when they react with acids. The reaction of zinc with hydrochloric acid, shown in Figure $6$, is:
Zinc is a silvery metal that quickly tarnishes to a blue-gray appearance. This change in color is due to an adherent coating of a basic carbonate, Zn2(OH)2CO3, which passivates the metal to inhibit further corrosion. Dry cell and alkaline batteries contain a zinc anode. Brass (Cu and Zn) and some bronze (Cu, Sn, and sometimes Zn) are important zinc alloys. About half of zinc production serves to protect iron and other metals from corrosion. This protection may take the form of a sacrificial anode (also known as a galvanic anode, which is a means of providing cathodic protection for various metals) or as a thin coating on the protected metal. Galvanized steel is steel with a protective coating of zinc.
Sacrificial Anodes
A sacrificial anode, or galvanic anode, is a means of providing cathodic protection of various metals. Cathodic protection refers to the prevention of corrosion by converting the corroding metal into a cathode. As a cathode, the metal resists corrosion, which is an oxidation process. Corrosion occurs at the sacrificial anode instead of at the cathode.
The construction of such a system begins with the attachment of a more active metal (more negative reduction potential) to the metal needing protection. Attachment may be direct or via a wire. To complete the circuit, a salt bridge is necessary. This salt bridge is often seawater or ground water. Once the circuit is complete, oxidation (corrosion) occurs at the anode and not the cathode.
The commonly used sacrificial anodes are magnesium, aluminum, and zinc. Magnesium has the most negative reduction potential of the three and serves best when the salt bridge is less efficient due to a low electrolyte concentration such as in freshwater. Zinc and aluminum work better in saltwater than does magnesium. Aluminum is lighter than zinc and has a higher capacity; however, an oxide coating may passivate the aluminum. In special cases, other materials are useful. For example, iron will protect copper.
Mercury is very different from zinc and cadmium. Mercury is the only metal that is liquid at 25 °C. Many metals dissolve in mercury, forming solutions called amalgams (see the feature on Amalgams), which are alloys of mercury with one or more other metals. Mercury, shown in Figure $7$, is a nonreactive element that is more difficult to oxidize than hydrogen. Thus, it does not displace hydrogen from acids; however, it will react with strong oxidizing acids, such as nitric acid:
$\ce{Hg}(l)+\ce{HCl}(aq)⟶\textrm{no reaction} \nonumber$
$\ce{3Hg}(l)+\ce{8HNO3}(aq)⟶\ce{3Hg(NO3)2}(aq)+\ce{4H2O}(l)+\ce{2NO}(g) \nonumber$
The clear NO initially formed quickly undergoes further oxidation to the reddish brown NO2.
Most mercury compounds decompose when heated. Most mercury compounds contain mercury with a 2+-oxidation state. When there is a large excess of mercury, it is possible to form compounds containing the $\ce{Hg2^2+}$ ion. All mercury compounds are toxic, and it is necessary to exercise great care in their synthesis.
Amalgams
An amalgam is an alloy of mercury with one or more other metals. This is similar to considering steel to be an alloy of iron with other metals. Most metals will form an amalgam with mercury, with the main exceptions being iron, platinum, tungsten, and tantalum. Due to toxicity issues with mercury, there has been a significant decrease in the use of amalgams. Historically, amalgams were important in electrolytic cells and in the extraction of gold. Amalgams of the alkali metals still find use because they are strong reducing agents and easier to handle than the pure alkali metals. Prospectors had a problem when they found finely divided gold. They learned that adding mercury to their pans collected the gold into the mercury to form an amalgam for easier collection. Unfortunately, losses of small amounts of mercury over the years left many streams in California polluted with mercury. Dentists use amalgams containing silver and other metals to fill cavities. There are several reasons to use an amalgam including low cost, ease of manipulation, and longevity compared to alternate materials. Dental amalgams are approximately 50% mercury by weight, which, in recent years, has become a concern due to the toxicity of mercury.
After reviewing the best available data, the Food and Drug Administration (FDA) considers amalgam-based fillings to be safe for adults and children over six years of age. Even with multiple fillings, the mercury levels in the patients remain far below the lowest levels associated with harm. Clinical studies have found no link between dental amalgams and health problems. Health issues may not be the same in cases of children under six or pregnant women. The FDA conclusions are in line with the opinions of the Environmental Protection Agency (EPA) and Centers for Disease Control (CDC). The only health consideration noted is that some people are allergic to the amalgam or one of its components.
Group 13
Group 13 contains the metalloid boron and the metals aluminum, gallium, indium, and thallium. The lightest element, boron, is semiconducting, and its binary compounds tend to be covalent and not ionic. The remaining elements of the group are metals, but their oxides and hydroxides change characters. The oxides and hydroxides of aluminum and gallium exhibit both acidic and basic behaviors. A substance, such as these two, that will react with both acids and bases is amphoteric. This characteristic illustrates the combination of nonmetallic and metallic behaviors of these two elements. Indium and thallium oxides and hydroxides exhibit only basic behavior, in accordance with the clearly metallic character of these two elements. The melting point of gallium is unusually low (about 30 °C) and will melt in your hand.
Aluminum is amphoteric because it will react with both acids and bases. A typical reaction with an acid is:
$\ce{2Al}(s)+\ce{6HCl}(aq)⟶\ce{2AlCl3}(aq)+\ce{3H2}(g) \nonumber$
The products of the reaction of aluminum with a base depend upon the reaction conditions, with the following being one possibility:
$\ce{2Al}(s)+\ce{2NaOH}(aq)+\ce{6H2O}(l)⟶\ce{2Na[Al(OH)4]}(aq)+\ce{3H2}(g) \nonumber$
With both acids and bases, the reaction with aluminum generates hydrogen gas.
The group 13 elements have a valence shell electron configuration of ns2np1. Aluminum normally uses all of its valence electrons when it reacts, giving compounds in which it has an oxidation state of 3+. Although many of these compounds are covalent, others, such as AlF3 and Al2(SO4)3, are ionic. Aqueous solutions of aluminum salts contain the cation $\ce{[Al(H2O)6]^3+}$, abbreviated as Al3+(aq). Gallium, indium, and thallium also form ionic compounds containing M3+ ions. These three elements exhibit not only the expected oxidation state of 3+ from the three valence electrons but also an oxidation state (in this case, 1+) that is two below the expected value. This phenomenon, the inert pair effect, refers to the formation of a stable ion with an oxidation state two lower than expected for the group. The pair of electrons is the valence s orbital for those elements. In general, the inert pair effect is important for the lower p-block elements. In an aqueous solution, the Tl+(aq) ion is more stable than is Tl3+(aq). In general, these metals will react with air and water to form 3+ ions; however, thallium reacts to give thallium(I) derivatives. The metals of group 13 all react directly with nonmetals such as sulfur, phosphorus, and the halogens, forming binary compounds.
The metals of group 13 (Al, Ga, In, and Tl) are all reactive. However, passivation occurs as a tough, hard, thin film of the metal oxide forms upon exposure to air. Disruption of this film may counter the passivation, allowing the metal to react. One way to disrupt the film is to expose the passivated metal to mercury. Some of the metal dissolves in the mercury to form an amalgam, which sheds the protective oxide layer to expose the metal to further reaction. The formation of an amalgam allows the metal to react with air and water.
Video $2$: Although easily oxidized, the passivation of aluminum makes it very useful as a strong, lightweight building material. Because of the formation of an amalgam, mercury is corrosive to structural materials made of aluminum. This video demonstrates how the integrity of an aluminum beam can be destroyed by the addition of a small amount of elemental mercury.
The most important uses of aluminum are in the construction and transportation industries, and in the manufacture of aluminum cans and aluminum foil. These uses depend on the lightness, toughness, and strength of the metal, as well as its resistance to corrosion. Because aluminum is an excellent conductor of heat and resists corrosion, it is useful in the manufacture of cooking utensils.
Aluminum is a very good reducing agent and may replace other reducing agents in the isolation of certain metals from their oxides. Although more expensive than reduction by carbon, aluminum is important in the isolation of Mo, W, and Cr from their oxides.
Group 14
The metallic members of group 14 are tin, lead, and flerovium. Carbon is a typical nonmetal. The remaining elements of the group, silicon and germanium, are examples of semimetals or metalloids. Tin and lead form the stable divalent cations, Sn2+ and Pb2+, with oxidation states two below the group oxidation state of 4+. The stability of this oxidation state is a consequence of the inert pair effect. Tin and lead also form covalent compounds with a formal 4+-oxidation state. For example, SnCl4 and PbCl4 are low-boiling covalent liquids.
Tin reacts readily with nonmetals and acids to form tin(II) compounds (indicating that it is more easily oxidized than hydrogen) and with nonmetals to form either tin(II) or tin(IV) compounds (Figure $8$), depending on the stoichiometry and reaction conditions. Lead is less reactive. It is only slightly easier to oxidize than hydrogen, and oxidation normally requires a hot concentrated acid.
Many of these elements exist as allotropes. Allotropes are two or more forms of the same element in the same physical state with different chemical and physical properties. There are two common allotropes of tin. These allotropes are grey (brittle) tin and white tin. As with other allotropes, the difference between these forms of tin is in the arrangement of the atoms. White tin is stable above 13.2 °C and is malleable like other metals. At low temperatures, gray tin is the more stable form. Gray tin is brittle and tends to break down to a powder. Consequently, articles made of tin will disintegrate in cold weather, particularly if the cold spell is lengthy. The change progresses slowly from the spot of origin, and the gray tin that is first formed catalyzes further change. In a way, this effect is similar to the spread of an infection in a plant or animal body, leading people to call this process tin disease or tin pest.
The principal use of tin is in the coating of steel to form tin plate-sheet iron, which constitutes the tin in tin cans. Important tin alloys are bronze (Cu and Sn) and solder (Sn and Pb). Lead is important in the lead storage batteries in automobiles.
Group 15
Bismuth, the heaviest member of group 15, is a less reactive metal than the other representative metals. It readily gives up three of its five valence electrons to active nonmetals to form the tri-positive ion, Bi3+. It forms compounds with the group oxidation state of 5+ only when treated with strong oxidizing agents. The stability of the 3+-oxidation state is another example of the inert pair effect.
Summary
This section focuses on the periodicity of the representative elements. These are the elements where the electrons are entering the s and p orbitals. The representative elements occur in groups 1, 2, and 12–18. These elements are representative metals, metalloids, and nonmetals. The alkali metals (group 1) are very reactive, readily form ions with a charge of 1+ to form ionic compounds that are usually soluble in water, and react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. The outermost electrons of the alkaline earth metals (group 2) are more difficult to remove than the outer electron of the alkali metals, leading to the group 2 metals being less reactive than those in group 1. These elements easily form compounds in which the metals exhibit an oxidation state of 2+. Zinc, cadmium, and mercury (group 12) commonly exhibit the group oxidation state of 2+ (although mercury also exhibits an oxidation state of 1+ in compounds that contain $\ce{Hg2^2+}$). Aluminum, gallium, indium, and thallium (group 13) are easier to oxidize than is hydrogen. Aluminum, gallium, and indium occur with an oxidation state 3+ (however, thallium also commonly occurs as the Tl+ ion). Tin and lead form stable divalent cations and covalent compounds in which the metals exhibit the 4+-oxidation state.
Glossary
alkaline earth metal
any of the metals (beryllium, magnesium, calcium, strontium, barium, and radium) occupying group 2 of the periodic table; they are reactive, divalent metals that form basic oxides
allotropes
two or more forms of the same element, in the same physical state, with different chemical structures
bismuth
heaviest member of group 15; a less reactive metal than other representative metals
metal (representative)
atoms of the metallic elements of groups 1, 2, 12, 13, 14, 15, and 16, which form ionic compounds by losing electrons from their outer s or p orbitals
metalloid
element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors
passivation
metals with a protective nonreactive film of oxide or other compound that creates a barrier for chemical reactions; physical or chemical removal of the passivating film allows the metals to demonstrate their expected chemical reactivity
representative element
element where the s and p orbitals are filling
representative metal
metal among the representative elements | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.01%3A_Periodicity.txt |
Learning Objectives
• Identify natural sources of representative metals
• Describe electrolytic and chemical reduction processes used to prepare these elements from natural sources
Because of their reactivity, we do not find most representative metals as free elements in nature. However, compounds that contain ions of most representative metals are abundant. In this section, we will consider the two common techniques used to isolate the metals from these compounds—electrolysis and chemical reduction.
These metals primarily occur in minerals, with lithium found in silicate or phosphate minerals, and sodium and potassium found in salt deposits from evaporation of ancient seas and in silicates. The alkaline earth metals occur as silicates and, with the exception of beryllium, as carbonates and sulfates. Beryllium occurs as the mineral beryl, Be3Al2Si6O18, which, with certain impurities, may be either the gemstone emerald or aquamarine. Magnesium is in seawater and, along with the heavier alkaline earth metals, occurs as silicates, carbonates, and sulfates. Aluminum occurs abundantly in many types of clay and in bauxite, an impure aluminum oxide hydroxide. The principle tin ore is the oxide cassiterite, SnO2, and the principle lead and thallium ores are the sulfides or the products of weathering of the sulfides. The remaining representative metals occur as impurities in zinc or aluminum ores.
Electrolysis
Ions of metals in of groups 1 and 2, along with aluminum, are very difficult to reduce; therefore, it is necessary to prepare these elements by electrolysis, an important process discussed in the chapter on electrochemistry. Briefly, electrolysis involves using electrical energy to drive unfavorable chemical reactions to completion; it is useful in the isolation of reactive metals in their pure forms. Sodium, aluminum, and magnesium are typical examples.
The Preparation of Sodium
The most important method for the production of sodium is the electrolysis of molten sodium chloride; the set-up is a Downs cell, shown in Figure $1$. The reaction involved in this process is:
$\ce{2NaCl}(l)\:\mathrm{\underset{600\:°C}{\xrightarrow{electrolysis}}}\:\ce{2Na}(l)+\ce{Cl2}(g) \nonumber$
The electrolysis cell contains molten sodium chloride (melting point 801 °C), to which calcium chloride has been added to lower the melting point to 600 °C (a colligative effect). The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Chloride ions migrate to the positively charged anode, lose electrons, and undergo oxidation to chlorine gas. The overall cell reaction comes from adding the following reactions:
\begin{align} &\textrm{at the cathode: }\ce{2Na+}+\ce{2e-}⟶\ce{2Na}(l)\ &\textrm{at the anode: }\ce{2Cl-}⟶\ce{Cl2}(g)+\ce{2e-}\ &\textrm{overall change: }\ce{2Na+}+\ce{2Cl-}⟶\ce{2Na}(l)+\ce{Cl2}(g) \end{align} \nonumber
Separation of the molten sodium and chlorine prevents recombination. The liquid sodium, which is less dense than molten sodium chloride, floats to the surface and flows into a collector. The gaseous chlorine goes to storage tanks. Chlorine is also a valuable product.
The Preparation of Aluminum
The preparation of aluminum utilizes a process invented in 1886 by Charles M. Hall, who began to work on the problem while a student at Oberlin College in Ohio. Paul L. T. Héroult discovered the process independently a month or two later in France. In honor to the two inventors, this electrolysis cell is known as the Hall–Héroult cell. The Hall–Héroult cell is an electrolysis cell for the production of aluminum. Figure $2$ illustrates the Hall–Héroult cell.
The production of aluminum begins with the purification of bauxite, the most common source of aluminum. The reaction of bauxite, AlO(OH), with hot sodium hydroxide forms soluble sodium aluminate, while clay and other impurities remain undissolved:
$\ce{AlO(OH)}(s)+\ce{NaOH}(aq)+\ce{H2O}(l)⟶\ce{Na[Al(OH)4]}(aq) \nonumber$
After the removal of the impurities by filtration, the addition of acid to the aluminate leads to the reprecipitation of aluminum hydroxide:
$\ce{Na[Al(OH)4]}(aq)+\ce{H3O+}(aq)⟶\ce{Al(OH)3}(s)+\ce{Na+}(aq)+\ce{2H2O}(l) \nonumber$
The next step is to remove the precipitated aluminum hydroxide by filtration. Heating the hydroxide produces aluminum oxide, Al2O3, which dissolves in a molten mixture of cryolite, Na3AlF6, and calcium fluoride, CaF2. Electrolysis of this solution takes place in a cell like that shown in Figure $2$. Reduction of aluminum ions to the metal occurs at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode.
The Preparation of Magnesium
Magnesium is the other metal that is isolated in large quantities by electrolysis. Seawater, which contains approximately 0.5% magnesium chloride, serves as the major source of magnesium. Addition of calcium hydroxide to seawater precipitates magnesium hydroxide. The addition of hydrochloric acid to magnesium hydroxide, followed by evaporation of the resultant aqueous solution, leaves pure magnesium chloride. The electrolysis of molten magnesium chloride forms liquid magnesium and chlorine gas:
$\ce{MgCl2}(aq)+\ce{Ca(OH)2}(aq)⟶\ce{Mg(OH)2}(s)+\ce{CaCl2}(aq) \nonumber$
$\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)⟶\ce{MgCl2}(aq)+\ce{2H2O}(l) \nonumber$
$\ce{MgCl2}(l)⟶\ce{Mg}(l)+\ce{Cl2}(g) \nonumber$
Some production facilities have moved away from electrolysis completely. In the next section, we will see how the Pidgeon process leads to the chemical reduction of magnesium.
Chemical Reduction
It is possible to isolate many of the representative metals by chemical reduction using other elements as reducing agents. In general, chemical reduction is much less expensive than electrolysis, and for this reason, chemical reduction is the method of choice for the isolation of these elements. For example, it is possible to produce potassium, rubidium, and cesium by chemical reduction, as it is possible to reduce the molten chlorides of these metals with sodium metal. This may be surprising given that these metals are more reactive than sodium; however, the metals formed are more volatile than sodium and can be distilled for collection. The removal of the metal vapor leads to a shift in the equilibrium to produce more metal (via the Le Chatelier’s principle). The production of magnesium, zinc, and tin provide additional examples of chemical reduction.
The Preparation of Magnesium
The Pidgeon process involves the reaction of magnesium oxide with elemental silicon at high temperatures to form pure magnesium:
$\ce{Si}(s)+\ce{2MgO}(s)\xrightarrow{Δ}\ce{SiO2}(s)+\ce{2Mg}(g) \nonumber$
Although this reaction is unfavorable in terms of thermodynamics, the removal of the magnesium vapor produced takes advantage of Le Chatelier’s principle to continue the forward progress of the reaction. Over 75% of the world’s production of magnesium, primarily in China, comes from this process.
The Preparation of Zinc
Zinc ores usually contain zinc sulfide, zinc oxide, or zinc carbonate. After separation of these compounds from the ores, heating in air converts the ore to zinc oxide by one of the following reactions:
$\ce{2ZnS}(s)+\ce{3O2}(g)\xrightarrow{Δ}\ce{2ZnO}(s)+\ce{2SO2}(g) \nonumber$
$\ce{ZnCO3}(s)\xrightarrow{Δ}\ce{ZnO}(s)+\ce{CO2}(g) \nonumber$
Carbon, in the form of coal, reduces the zinc oxide to form zinc vapor:
$\ce{ZnO}(s)+\ce{C}(s)⟶\ce{Zn}(g)+\ce{CO}(g) \nonumber$
The zinc can be distilled (boiling point 907 °C) and condensed. This zinc contains impurities of cadmium (767 °C), iron (2862 °C), lead (1750 °C), and arsenic (613 °C). Careful redistillation produces pure zinc. Arsenic and cadmium are distilled from the zinc because they have lower boiling points. At higher temperatures, the zinc is distilled from the other impurities, mainly lead and iron.
The Preparation of Tin
The ready reduction of tin(IV) oxide by the hot coals of a campfire accounts for the knowledge of tin in the ancient world. In the modern process, the roasting of tin ores containing SnO2 removes contaminants such as arsenic and sulfur as volatile oxides. Treatment of the remaining material with hydrochloric acid removes the oxides of other metals. Heating the purified ore with carbon at temperature above 1000 °C produces tin:
$\ce{SnO2}(s)+\ce{2C}(s)\xrightarrow{Δ}\ce{Sn}(s)+\ce{2CO}(g) \nonumber$
The molten tin collects at the bottom of the furnace and is drawn off and cast into blocks.
Summary
Because of their chemical reactivity, it is necessary to produce the representative metals in their pure forms by reduction from naturally occurring compounds. Electrolysis is important in the production of sodium, potassium, and aluminum. Chemical reduction is the primary method for the isolation of magnesium, zinc, and tin. Similar procedures are important for the other representative metals.
Glossary
chemical reduction
method of preparing a representative metal using a reducing agent
Downs cell
electrochemical cell used for the commercial preparation of metallic sodium (and chlorine) from molten sodium chloride
Hall–Héroult cell
electrolysis apparatus used to isolate pure aluminum metal from a solution of alumina in molten cryolite
Pidgeon process
chemical reduction process used to produce magnesium through the thermal reaction of magnesium oxide with silicon | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.02%3A_Occurrence_and_Preparation_of_the_Representative_Metals.txt |
Learning Objectives
• Describe the general preparation, properties, and uses of the metalloids
• Describe the preparation, properties, and compounds of boron and silicon
A series of six elements called the metalloids separate the metals from the nonmetals in the periodic table. The metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. These elements look metallic; however, they do not conduct electricity as well as metals so they are semiconductors. They are semiconductors because their electrons are more tightly bound to their nuclei than are those of metallic conductors. Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form monatomic anions. This intermediate behavior is in part due to their intermediate electronegativity values. In this section, we will briefly discuss the chemical behavior of metalloids and deal with two of these elements—boron and silicon—in more detail.
The metalloid boron exhibits many similarities to its neighbor carbon and its diagonal neighbor silicon. All three elements form covalent compounds. However, boron has one distinct difference in that its 2s22p1 outer electron structure gives it one less valence electron than it has valence orbitals. Although boron exhibits an oxidation state of 3+ in most of its stable compounds, this electron deficiency provides boron with the ability to form other, sometimes fractional, oxidation states, which occur, for example, in the boron hydrides.
Silicon has the valence shell electron configuration 3s23p2, and it commonly forms tetrahedral structures in which it is sp3 hybridized with a formal oxidation state of 4+. The major differences between the chemistry of carbon and silicon result from the relative strength of the carbon-carbon bond, carbon’s ability to form stable bonds to itself, and the presence of the empty 3d valence-shell orbitals in silicon. Silicon’s empty d orbitals and boron’s empty p orbital enable tetrahedral silicon compounds and trigonal planar boron compounds to act as Lewis acids. Carbon, on the other hand, has no available valence shell orbitals; tetrahedral carbon compounds cannot act as Lewis acids. Germanium is very similar to silicon in its chemical behavior.
Arsenic and antimony generally form compounds in which an oxidation state of 3+ or 5+ is exhibited; however, arsenic can form arsenides with an oxidation state of 3−. These elements tarnish only slightly in dry air but readily oxidize when warmed.
Tellurium combines directly with most elements. The most stable tellurium compounds are the tellurides—salts of Te2− formed with active metals and lanthanides—and compounds with oxygen, fluorine, and chlorine, in which tellurium normally exhibits an oxidation state 2+ or 4+. Although tellurium(VI) compounds are known (for example, TeF6), there is a marked resistance to oxidation to this maximum group oxidation state.
Structures of the Metalloids
Covalent bonding is the key to the crystal structures of the metalloids. In this regard, these elements resemble nonmetals in their behavior.
Elemental silicon, germanium, arsenic, antimony, and tellurium are lustrous, metallic-looking solids. Silicon and germanium crystallize with a diamond structure. Each atom within the crystal has covalent bonds to four neighboring atoms at the corners of a regular tetrahedron. Single crystals of silicon and germanium are giant, three-dimensional molecules. There are several allotropes of arsenic with the most stable being layer like and containing puckered sheets of arsenic atoms. Each arsenic atom forms covalent bonds to three other atoms within the sheet. The crystal structure of antimony is similar to that of arsenic, both shown in Figure $1$. The structures of arsenic and antimony are similar to the structure of graphite, covered later in this chapter. Tellurium forms crystals that contain infinite spiral chains of tellurium atoms. Each atom in the chain bonds to two other atoms.
Video $1$: Explore a cubic diamond crystal structure.
Pure crystalline boron is transparent. The crystals consist of icosahedra, as shown in Figure $2$, with a boron atom at each corner. In the most common form of boron, the icosahedra pack together in a manner similar to the cubic closest packing of spheres. All boron-boron bonds within each icosahedron are identical and are approximately 176 pm in length. In the different forms of boron, there are different arrangements and connections between the icosahedra.
The name silicon is derived from the Latin word for flint, silex. The metalloid silicon readily forms compounds containing Si-O-Si bonds, which are of prime importance in the mineral world. This bonding capability is in contrast to the nonmetal carbon, whose ability to form carbon-carbon bonds gives it prime importance in the plant and animal worlds.
Occurrence, Preparation, and Compounds of Boron and Silicon
Boron constitutes less than 0.001% by weight of the earth’s crust. In nature, it only occurs in compounds with oxygen. Boron is widely distributed in volcanic regions as boric acid, B(OH)3, and in dry lake regions, including the desert areas of California, as borates and salts of boron oxyacids, such as borax, Na2B4O7⋅10H2O.
Elemental boron is chemically inert at room temperature, reacting with only fluorine and oxygen to form boron trifluoride, BF3, and boric oxide, B2O3, respectively. At higher temperatures, boron reacts with all nonmetals, except tellurium and the noble gases, and with nearly all metals; it oxidizes to B2O3 when heated with concentrated nitric or sulfuric acid. Boron does not react with nonoxidizing acids. Many boron compounds react readily with water to give boric acid, B(OH)3 (sometimes written as H3BO3).
Reduction of boric oxide with magnesium powder forms boron (95–98.5% pure) as a brown, amorphous powder:
$\ce{B2O3}(s)+\ce{3Mg}(s)⟶\ce{2B}(s)+\ce{3MgO}(s) \nonumber$
An amorphous substance is a material that appears to be a solid, but does not have a long-range order like a true solid. Treatment with hydrochloric acid removes the magnesium oxide. Further purification of the boron begins with conversion of the impure boron into boron trichloride. The next step is to heat a mixture of boron trichloride and hydrogen:
$\ce{2BCl3}(g)+\ce{3H2}(g)\:\mathrm{\xrightarrow{1500\:°C}}\:\ce{2B}(s)+\ce{6HCl}(g) \hspace{20px} ΔH°=\mathrm{253.7\: kJ} \nonumber$
Silicon makes up nearly one-fourth of the mass of the earth’s crust—second in abundance only to oxygen. The crust is composed almost entirely of minerals in which the silicon atoms are at the center of the silicon-oxygen tetrahedron, which connect in a variety of ways to produce, among other things, chains, layers, and three-dimensional frameworks. These minerals constitute the bulk of most common rocks, soil, and clays. In addition, materials such as bricks, ceramics, and glasses contain silicon compounds.
It is possible to produce silicon by the high-temperature reduction of silicon dioxide with strong reducing agents, such as carbon and magnesium:
$\ce{SiO2}(s)+\ce{2C}(s)\xrightarrow{Δ}\ce{Si}(s)+\ce{2CO}(g) \nonumber$
$\ce{SiO2}(s)+\ce{2Mg}(s)\xrightarrow{Δ}\ce{Si}(s)+\ce{2MgO}(s) \nonumber$
Extremely pure silicon is necessary for the manufacture of semiconductor electronic devices. This process begins with the conversion of impure silicon into silicon tetrahalides, or silane (SiH4), followed by decomposition at high temperatures. Zone refining, illustrated in Figure $3$, completes the purification. In this method, a rod of silicon is heated at one end by a heat source that produces a thin cross-section of molten silicon. Slowly lowering the rod through the heat source moves the molten zone from one end of the rod to other. As this thin, molten region moves, impurities in the silicon dissolve in the liquid silicon and move with the molten region. Ultimately, the impurities move to one end of the rod, which is then cut off.
This highly purified silicon, containing no more than one part impurity per million parts of silicon, is the most important element in the computer industry. Pure silicon is necessary in semiconductor electronic devices such as transistors, computer chips, and solar cells.
Like some metals, passivation of silicon occurs due the formation of a very thin film of oxide (primarily silicon dioxide, SiO2). Silicon dioxide is soluble in hot aqueous base; thus, strong bases destroy the passivation. Removal of the passivation layer allows the base to dissolve the silicon, forming hydrogen gas and silicate anions. For example:
$\ce{Si}(s)+\ce{4OH-}(aq)⟶\ce{SiO4^4-}(aq)+\ce{2H2}(g) \nonumber$
Silicon reacts with halogens at high temperatures, forming volatile tetrahalides, such as SiF4.
Unlike carbon, silicon does not readily form double or triple bonds. Silicon compounds of the general formula SiX4, where X is a highly electronegative group, can act as Lewis acids to form six-coordinate silicon. For example, silicon tetrafluoride, SiF4, reacts with sodium fluoride to yield Na2[SiF6], which contains the octahedral $\ce{[SiF6]^2-}$ ion in which silicon is sp3d2 hybridized:
$\ce{2NaF}(s)+\ce{SiF4}(g)⟶\ce{Na2SiF6}(s) \nonumber$
Antimony reacts readily with stoichiometric amounts of fluorine, chlorine, bromine, or iodine, yielding trihalides or, with excess fluorine or chlorine, forming the pentahalides SbF5 and SbCl5. Depending on the stoichiometry, it forms antimony(III) sulfide, Sb2S3, or antimony(V) sulfide when heated with sulfur. As expected, the metallic nature of the element is greater than that of arsenic, which lies immediately above it in group 15.
Boron and Silicon Halides
Boron trihalides—BF3, BCl3, BBr3, and BI3—can be prepared by the direct reaction of the elements. These nonpolar molecules contain boron with sp2 hybridization and a trigonal planar molecular geometry. The fluoride and chloride compounds are colorless gasses, the bromide is a liquid, and the iodide is a white crystalline solid.
Except for boron trifluoride, the boron trihalides readily hydrolyze in water to form boric acid and the corresponding hydrohalic acid. Boron trichloride reacts according to the equation:
$\ce{BCl3}(g)+\ce{3H2O}(l)⟶\ce{B(OH)3}(aq)+\ce{3HCl}(aq) \nonumber$
Boron trifluoride reacts with hydrofluoric acid, to yield a solution of fluoroboric acid, HBF4:
$\ce{BF3}(aq)+\ce{HF}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{BF4-}(aq) \nonumber$
In this reaction, the BF3 molecule acts as the Lewis acid (electron pair acceptor) and accepts a pair of electrons from a fluoride ion:
All the tetrahalides of silicon, SiX4, have been prepared. Silicon tetrachloride can be prepared by direct chlorination at elevated temperatures or by heating silicon dioxide with chlorine and carbon:
$\ce{SiO2}(s)+\ce{2C}(s)+\ce{2Cl2}(g)\xrightarrow{Δ}\ce{SiCl4}(g)+\ce{2CO}(g) \nonumber$
Silicon tetrachloride is a covalent tetrahedral molecule, which is a nonpolar, low-boiling (57 °C), colorless liquid. It is possible to prepare silicon tetrafluoride by the reaction of silicon dioxide with hydrofluoric acid:
$\ce{SiO2}(s)+\ce{4HF}(g)⟶\ce{SiF4}(g)+\ce{2H2O}(l) \hspace{20px} ΔH°=\mathrm{−191.2\: kJ} \nonumber$
Hydrofluoric acid is the only common acid that will react with silicon dioxide or silicates. This reaction occurs because the silicon-fluorine bond is the only bond that silicon forms that is stronger than the silicon-oxygen bond. For this reason, it is possible to store all common acids, other than hydrofluoric acid, in glass containers.
Except for silicon tetrafluoride, silicon halides are extremely sensitive to water. Upon exposure to water, SiCl4 reacts rapidly with hydroxide groups, replacing all four chlorine atoms to produce unstable orthosilicic acid, Si(OH)4 or H4SiO4, which slowly decomposes into SiO2.
Boron and Silicon Oxides and Derivatives
Boron burns at 700 °C in oxygen, forming boric oxide, B2O3. Boric oxide is necessary for the production of heat-resistant borosilicate glass, like that shown in Figure $4$ and certain optical glasses. Boric oxide dissolves in hot water to form boric acid, B(OH)3:
$\ce{B2O3}(s)+\ce{3H2O}(l)⟶\ce{2B(OH)3}(aq) \nonumber$
The boron atom in B(OH)3 is sp2 hybridized and is located at the center of an equilateral triangle with oxygen atoms at the corners. In solid B(OH)3, hydrogen bonding holds these triangular units together. Boric acid, shown in Figure $5$, is a very weak acid that does not act as a proton donor but rather as a Lewis acid, accepting an unshared pair of electrons from the Lewis base OH:
$\ce{B(OH)3}(aq)+\ce{2H2O}(l)⇌\ce{B(OH)4-}(aq)+\ce{H3O+}(aq) \hspace{20px} K_\ce{a}=5.8×10^{−10} \nonumber$
Heating boric acid to 100 °C causes molecules of water to split out between pairs of adjacent –OH groups to form metaboric acid, HBO2. At about 150 °C, additional B-O-B linkages form, connecting the BO3 groups together with shared oxygen atoms to form tetraboric acid, H2B4O7. Complete water loss, at still higher temperatures, results in boric oxide.
Borates are salts of the oxyacids of boron. Borates result from the reactions of a base with an oxyacid or from the fusion of boric acid or boric oxide with a metal oxide or hydroxide. Borate anions range from the simple trigonal planar $\ce{BO3^3-}$ ion to complex species containing chains and rings of three- and four-coordinated boron atoms. The structures of the anions found in CaB2O4, K[B5O6(OH)4]⋅2H2O (commonly written KB5O8⋅4H2O) and Na2[B4O5(OH)4]⋅8H2O (commonly written Na2B4O7⋅10H2O) are shown in Figure $6$. Commercially, the most important borate is borax, Na2[B4O5(OH)4]⋅8H2O, which is an important component of some laundry detergents. Most of the supply of borax comes directly from dry lakes, such as Searles Lake in California, or is prepared from kernite, Na2B4O7⋅4H2O.
Silicon dioxide, silica, occurs in both crystalline and amorphous forms. The usual crystalline form of silicon dioxide is quartz, a hard, brittle, clear, colorless solid. It is useful in many ways—for architectural decorations, semiprecious jewels, and frequency control in radio transmitters. Silica takes many crystalline forms, or polymorphs, in nature. Trace amounts of Fe3+ in quartz give amethyst its characteristic purple color. The term quartz is also used for articles such as tubing and lenses that are manufactured from amorphous silica. Opal is a naturally occurring form of amorphous silica.
The contrast in structure and physical properties between silicon dioxide and carbon dioxide is interesting, as illustrated in Figure $7$. Solid carbon dioxide (dry ice) contains single CO2 molecules with each of the two oxygen atoms attached to the carbon atom by double bonds. Very weak intermolecular forces hold the molecules together in the crystal. The volatility of dry ice reflect these weak forces between molecules. In contrast, silicon dioxide is a covalent network solid. In silicon dioxide, each silicon atom links to four oxygen atoms by single bonds directed toward the corners of a regular tetrahedron, and SiO4 tetrahedra share oxygen atoms. This arrangement gives a three dimensional, continuous, silicon-oxygen network. A quartz crystal is a macromolecule of silicon dioxide. The difference between these two compounds is the ability of the group 14 elements to form strong π bonds. Second-period elements, such as carbon, form very strong π bonds, which is why carbon dioxide forms small molecules with strong double bonds. Elements below the second period, such as silicon, do not form π bonds as readily as second-period elements, and when they do form, the π bonds are weaker than those formed by second-period elements. For this reason, silicon dioxide does not contain π bonds but only σ bonds.
At 1600 °C, quartz melts to yield a viscous liquid. When the liquid cools, it does not crystallize readily but usually supercools and forms a glass, also called silica. The SiO4 tetrahedra in glassy silica have a random arrangement characteristic of supercooled liquids, and the glass has some very useful properties. Silica is highly transparent to both visible and ultraviolet light. For this reason, it is important in the manufacture of lamps that give radiation rich in ultraviolet light and in certain optical instruments that operate with ultraviolet light. The coefficient of expansion of silica glass is very low; therefore, rapid temperature changes do not cause it to fracture. CorningWare and other ceramic cookware contain amorphous silica.
Silicates are salts containing anions composed of silicon and oxygen. In nearly all silicates, sp3-hybridized silicon atoms occur at the centers of tetrahedra with oxygen at the corners. There is a variation in the silicon-to-oxygen ratio that occurs because silicon-oxygen tetrahedra may exist as discrete, independent units or may share oxygen atoms at corners in a variety of ways. In addition, the presence of a variety of cations gives rise to the large number of silicate minerals. Many ceramics are composed of silicates. By including small amounts of other compounds, it is possible to modify the physical properties of the silicate materials to produce ceramics with useful characteristics.
Summary
The elements boron, silicon, germanium, arsenic, antimony, and tellurium separate the metals from the nonmetals in the periodic table. These elements, called metalloids or sometimes semimetals, exhibit properties characteristic of both metals and nonmetals. The structures of these elements are similar in many ways to those of nonmetals, but the elements are electrical semiconductors.
Glossary
amorphous
solid material such as a glass that does not have a regular repeating component to its three-dimensional structure; a solid but not a crystal
borate
compound containing boron-oxygen bonds, typically with clusters or chains as a part of the chemical structure
polymorph
variation in crystalline structure that results in different physical properties for the resulting compound
silicate
compound containing silicon-oxygen bonds, with silicate tetrahedra connected in rings, sheets, or three-dimensional networks, depending on the other elements involved in the formation of the compounds | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.03%3A_Structure_and_General_Properties_of_the_Metalloids.txt |
Learning Objectives
• Describe structure and properties of nonmetals
The nonmetals are elements located in the upper right portion of the periodic table. Their properties and behavior are quite different from those of metals on the left side. Under normal conditions, more than half of the nonmetals are gases, one is a liquid, and the rest include some of the softest and hardest of solids. The nonmetals exhibit a rich variety of chemical behaviors. They include the most reactive and least reactive of elements, and they form many different ionic and covalent compounds. This section presents an overview of the properties and chemical behaviors of the nonmetals, as well as the chemistry of specific elements. Many of these nonmetals are important in biological systems.
In many cases, trends in electronegativity enable us to predict the type of bonding and the physical states in compounds involving the nonmetals. We know that electronegativity decreases as we move down a given group and increases as we move from left to right across a period. The nonmetals have higher electronegativities than do metals, and compounds formed between metals and nonmetals are generally ionic in nature because of the large differences in electronegativity between them. The metals form cations, the nonmetals form anions, and the resulting compounds are solids under normal conditions. On the other hand, compounds formed between two or more nonmetals have small differences in electronegativity between the atoms, and covalent bonding—sharing of electrons—results. These substances tend to be molecular in nature and are gases, liquids, or volatile solids at room temperature and pressure.
In normal chemical processes, nonmetals do not form monatomic positive ions (cations) because their ionization energies are too high. All monatomic nonmetal ions are anions; examples include the chloride ion, Cl, the nitride ion, N3−, and the selenide ion, Se2. The common oxidation states that the nonmetals exhibit in their ionic and covalent compounds are shown in Figure $1$. Remember that an element exhibits a positive oxidation state when combined with a more electronegative element and that it exhibits a negative oxidation state when combined with a less electronegative element.
The first member of each nonmetal group exhibits different behaviors, in many respects, from the other group members. The reasons for this include smaller size, greater ionization energy, and (most important) the fact that the first member of each group has only four valence orbitals (one 2s and three 2p) available for bonding, whereas other group members have empty d orbitals in their valence shells, making possible five, six, or even more bonds around the central atom. For example, nitrogen forms only NF3, whereas phosphorus forms both PF3 and PF5.
Remember that an element exhibits a positive oxidation state when combined with a more electronegative element and that it exhibits a negative oxidation state when combined with a less electronegative element.
Another difference between the first group member and subsequent members is the greater ability of the first member to form π bonds. This is primarily a function of the smaller size of the first member of each group, which allows better overlap of atomic orbitals. Nonmetals, other than the first member of each group, rarely form π bonds to nonmetals that are the first member of a group. For example, sulfur-oxygen π bonds are well known, whereas sulfur does not normally form stable π bonds to itself.
The variety of oxidation states displayed by most of the nonmetals means that many of their chemical reactions involve changes in oxidation state through oxidation-reduction reactions. There are five general aspects of the oxidation-reduction chemistry:
1. Nonmetals oxidize most metals. The oxidation state of the metal becomes positive as it undergoes oxidation and that of the nonmetal becomes negative as it undergoes reduction. For example:
$4 \underset{0}{\ce{Fe}} \left( s \right) + 3 \underset{0}{\ce{O_2}} \left( g \right) \rightarrow 2 \underset{+3}{\ce{Fe_2}} \underset{-2}{\ce{O_3}} \left( s \right) \nonumber$
2. With the exception of nitrogen and carbon, which are poor oxidizing agents, a more electronegative nonmetal oxidizes a less electronegative nonmetal or the anion of the nonmetal:
$\underset{0}{\ce{S}} \left( s \right) + \underset{0}{\ce{O_2}} \left( g \right) \rightarrow 2 \underset{+4}{\ce{S}} \underset{-2}{\ce{O_2}} \left( s \right) \nonumber$
$\underset{0}{\ce{Cl_2}} \left( g \right) + 2 \ce{I^-} \left( aq \right) \rightarrow \underset{0}{\ce{I_2}} \left( s \right) + 2 \ce{Cl^-} \left( aq \right) \nonumber$
• Fluorine and oxygen are the strongest oxidizing agents within their respective groups; each oxidizes all the elements that lie below it in the group. Within any period, the strongest oxidizing agent is in group 17. A nonmetal often oxidizes an element that lies to its left in the same period. For example:
$2 \underset{0}{\ce{As}} \left( s \right) + 3 \underset{0}{\ce{Br_2}} \left( l \right) \rightarrow 2 \underset{+3}{\ce{As}} \underset{-1}{\ce{Br_3}} \left( s \right) \nonumber$
• The stronger a nonmetal is as an oxidizing agent, the more difficult it is to oxidize the anion formed by the nonmetal. This means that the most stable negative ions are formed by elements at the top of the group or in group 17 of the period.
• Fluorine and oxygen are the strongest oxidizing elements known. Fluorine does not form compounds in which it exhibits positive oxidation states; oxygen exhibits a positive oxidation state only when combined with fluorine. For example: $2 \underset{0}{\ce{F_2}} \left( g \right) + 2 \ce{OH^-} \left( aq \right) \rightarrow \underset{+2}{\ce{O}} \underset{-1}{\ce{F_2}} \left( g \right) + 2 \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \nonumber$
With the exception of most of the noble gases, all nonmetals form compounds with oxygen, yielding covalent oxides. Most of these oxides are acidic, that is, they react with water to form oxyacids. Recall from the acid-base chapter that an oxyacid is an acid consisting of hydrogen, oxygen, and some other element. Notable exceptions are carbon monoxide, CO, nitrous oxide, N2O, and nitric oxide, NO. There are three characteristics of these acidic oxides:
1. Oxides such as SO2 and N2O5, in which the nonmetal exhibits one of its common oxidation states, are acid anhydrides and react with water to form acids with no change in oxidation state. The product is an oxyacid. For example: $\ce{SO2}(g)+\ce{H2O}(l)⟶\ce{H2SO3}(aq) \nonumber$ $\ce{N2O5}(s)+\ce{H2O}(l)⟶\ce{2HNO3}(aq) \nonumber$
2. Those oxides such as NO2 and ClO2, in which the nonmetal does not exhibit one of its common oxidation states, also react with water. In these reactions, the nonmetal is both oxidized and reduced. For example: $3 \underset{+4}{\ce{N}} \ce{O_2} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow 2 \ce{H} \underset{+5}{\ce{N}} \ce{O_3} \left( aq \right) + \underset{+2}{\ce{N}} \ce{O} \left( g \right) \nonumber$ Reactions in which the same element is both oxidized and reduced are called disproportionation reactions.
3. The acid strength increases as the electronegativity of the central atom increases. To learn more, see the discussion in the chapter on acid-base chemistry.
The binary hydrogen compounds of the nonmetals also exhibit an acidic behavior in water, although only HCl, HBr, and HI are strong acids. The acid strength of the nonmetal hydrogen compounds increases from left to right across a period and down a group. For example, ammonia, NH3, is a weaker acid than is water, H2O, which is weaker than is hydrogen fluoride, HF. Water, H2O, is also a weaker acid than is hydrogen sulfide, H2S, which is weaker than is hydrogen selenide, H2Se. Weaker acidic character implies greater basic character.
Structures of the Nonmetals
The structures of the nonmetals differ dramatically from those of metals. Metals crystallize in closely packed arrays that do not contain molecules or covalent bonds. Nonmetal structures contain covalent bonds, and many nonmetals consist of individual molecules. The electrons in nonmetals are localized in covalent bonds, whereas in a metal, there is delocalization of the electrons throughout the solid.
The noble gases are all monatomic, whereas the other nonmetal gases—hydrogen, nitrogen, oxygen, fluorine, and chlorine—normally exist as the diatomic molecules H2, N2, O2, F2, and Cl2. The other halogens are also diatomic; Br2 is a liquid and I2 exists as a solid under normal conditions. The changes in state as one moves down the halogen family offer excellent examples of the increasing strength of intermolecular London forces with increasing molecular mass and increasing polarizability.
Oxygen has two allotropes: O2, dioxygen, and O3, ozone. Phosphorus has three common allotropes, commonly referred to by their colors: white, red, and black. Sulfur has several allotropes. There are also many carbon allotropes. Most people know of diamond, graphite, and charcoal, but fewer people know of the recent discovery of fullerenes, carbon nanotubes, and graphene.
Descriptions of the physical properties of three nonmetals that are characteristic of molecular solids follow.
Carbon
Carbon occurs in the uncombined (elemental) state in many forms, such as diamond, graphite, charcoal, coke, carbon black, graphene, and fullerene.
Diamond, shown in Figure $2$, is a very hard crystalline material that is colorless and transparent when pure. Each atom forms four single bonds to four other atoms at the corners of a tetrahedron (sp3 hybridization); this makes the diamond a giant molecule. Carbon-carbon single bonds are very strong, and, because they extend throughout the crystal to form a three-dimensional network, the crystals are very hard and have high melting points (~4400 °C).
Graphite, also shown in Figure $2$, is a soft, slippery, grayish-black solid that conducts electricity. These properties relate to its structure, which consists of layers of carbon atoms, with each atom surrounded by three other carbon atoms in a trigonal planar arrangement. Each carbon atom in graphite forms three σ bonds, one to each of its nearest neighbors, by means of sp2-hybrid orbitals. The unhybridized p orbital on each carbon atom will overlap unhybridized orbitals on adjacent carbon atoms in the same layer to form π bonds. Many resonance forms are necessary to describe the electronic structure of a graphite layer; Figure $3$ illustrates two of these forms.
Atoms within a graphite layer are bonded together tightly by the σ and π bonds; however, the forces between layers are weak. London dispersion forces hold the layers together. To learn more, see the discussion of these weak forces in the chapter on liquids and solids. The weak forces between layers give graphite the soft, flaky character that makes it useful as the so-called “lead” in pencils and the slippery character that makes it useful as a lubricant. The loosely held electrons in the resonating π bonds can move throughout the solid and are responsible for the electrical conductivity of graphite.
Other forms of elemental carbon include carbon black, charcoal, and coke. Carbon black is an amorphous form of carbon prepared by the incomplete combustion of natural gas, CH4. It is possible to produce charcoal and coke by heating wood and coal, respectively, at high temperatures in the absence of air.
Recently, new forms of elemental carbon molecules have been identified in the soot generated by a smoky flame and in the vapor produced when graphite is heated to very high temperatures in a vacuum or in helium. One of these new forms, first isolated by Professor Richard Smalley and coworkers at Rice University, consists of icosahedral (soccer-ball-shaped) molecules that contain 60 carbon atoms, C60. This is buckminsterfullerene (often called bucky balls) after the architect Buckminster Fuller, who designed domed structures, which have a similar appearance (Figure $4$).
Nanotubes and Graphene
Graphene and carbon nanotubes are two recently discovered allotropes of carbon. Both of the forms bear some relationship to graphite. Graphene is a single layer of graphite (one atom thick), as illustrated in Figure $2$d, whereas carbon nanotubes roll the layer into a small tube, as illustrated in Figure $5$.
Graphene is a very strong, lightweight, and efficient conductor of heat and electricity discovered in 2003. As in graphite, the carbon atoms form a layer of six-membered rings with sp2-hybridized carbon atoms at the corners. Resonance stabilizes the system and leads to its conductivity. Unlike graphite, there is no stacking of the layers to give a three-dimensional structure. Andre Geim and Kostya Novoselov at the University of Manchester won the 2010 Nobel Prize in Physics for their pioneering work characterizing graphene.
The simplest procedure for preparing graphene is to use a piece of adhesive tape to remove a single layer of graphene from the surface of a piece of graphite. This method works because there are only weak London dispersion forces between the layers in graphite. Alternative methods are to deposit a single layer of carbon atoms on the surface of some other material (ruthenium, iridium, or copper) or to synthesize it at the surface of silicon carbide via the sublimation of silicon.
There currently are no commercial applications of graphene. However, its unusual properties, such as high electron mobility and thermal conductivity, should make it suitable for the manufacture of many advanced electronic devices and for thermal management applications.
Carbon nanotubes are carbon allotropes, which have a cylindrical structure. Like graphite and graphene, nanotubes consist of rings of sp2-hybridized carbon atoms. Unlike graphite and graphene, which occur in layers, the layers wrap into a tube and bond together to produce a stable structure. The walls of the tube may be one atom or multiple atoms thick.
Carbon nanotubes are extremely strong materials that are harder than diamond. Depending upon the shape of the nanotube, it may be a conductor or semiconductor. For some applications, the conducting form is preferable, whereas other applications utilize the semiconducting form.
The basis for the synthesis of carbon nanotubes is the generation of carbon atoms in a vacuum. It is possible to produce carbon atoms by an electrical discharge through graphite, vaporization of graphite with a laser, and the decomposition of a carbon compound.
The strength of carbon nanotubes will eventually lead to some of their most exciting applications, as a thread produced from several nanotubes will support enormous weight. However, the current applications only employ bulk nanotubes. The addition of nanotubes to polymers improves the mechanical, thermal, and electrical properties of the bulk material. There are currently nanotubes in some bicycle parts, skis, baseball bats, fishing rods, and surfboards.
Phosphorus
The name phosphorus comes from the Greek words meaning light bringing. When phosphorus was first isolated, scientists noted that it glowed in the dark and burned when exposed to air. Phosphorus is the only member of its group that does not occur in the uncombined state in nature; it exists in many allotropic forms. We will consider two of those forms: white phosphorus and red phosphorus.
White phosphorus is a white, waxy solid that melts at 44.2 °C and boils at 280 °C. It is insoluble in water, is very soluble in carbon disulfide, and bursts into flame in air. As a solid, as a liquid, as a gas, and in solution, white phosphorus exists as P4 molecules with four phosphorus atoms at the corners of a regular tetrahedron (Figure $5$). Each phosphorus atom covalently bonds to the other three atoms in the molecule by single covalent bonds. White phosphorus is the most reactive allotrope and is very toxic.
Heating white phosphorus to 270–300 °C in the absence of air yields red phosphorus. Red phosphorus is denser, has a higher melting point (~600 °C), is much less reactive, is essentially nontoxic, and is easier and safer to handle than is white phosphorus. Its structure is highly polymeric and appears to contain three-dimensional networks of P4 tetrahedra joined by P-P single bonds. Red phosphorus is insoluble in solvents that dissolve white phosphorus. When red phosphorus is heated, P4 molecules sublime from the solid.
Sulfur
The allotropy of sulfur is far greater and more complex than that of any other element. Sulfur is the brimstone referred to in the Bible and other places, and references to sulfur occur throughout recorded history—right up to the relatively recent discovery that it is a component of the atmospheres of Venus and of Io, a moon of Jupiter. The most common and most stable allotrope of sulfur is yellow, rhombic sulfur, so named because of the shape of its crystals. Rhombic sulfur is the form to which all other allotropes revert at room temperature. Crystals of rhombic sulfur melt at 113 °C. Cooling this liquid gives long needles of monoclinic sulfur. This form is stable from 96 °C to the melting point, 119 °C. At room temperature, it gradually reverts to the rhombic form.
Both rhombic sulfur and monoclinic sulfur contain S8 molecules in which atoms form eight-membered, puckered rings that resemble crowns (Figure $6$). Each sulfur atom is bonded to each of its two neighbors in the ring by covalent S-S single bonds.
When rhombic sulfur melts, the straw-colored liquid is quite mobile; its viscosity is low because S8 molecules are essentially spherical and offer relatively little resistance as they move past each other. As the temperature rises, S-S bonds in the rings break, and polymeric chains of sulfur atoms result. These chains combine end to end, forming still longer chains that tangle with one another. The liquid gradually darkens in color and becomes so viscous that finally (at about 230 °C) it does not pour easily. The dangling atoms at the ends of the chains of sulfur atoms are responsible for the dark red color because their electronic structure differs from those of sulfur atoms that have bonds to two adjacent sulfur atoms. This causes them to absorb light differently and results in a different visible color. Cooling the liquid rapidly produces a rubberlike amorphous mass, called plastic sulfur.
Sulfur boils at 445 °C and forms a vapor consisting of S2, S6, and S8 molecules; at about 1000 °C, the vapor density corresponds to the formula S2, which is a paramagnetic molecule like O2 with a similar electronic structure and a weak sulfur-sulfur double bond.
As seen in this discussion, an important feature of the structural behavior of the nonmetals is that the elements usually occur with eight electrons in their valence shells. If necessary, the elements form enough covalent bonds to supplement the electrons already present to possess an octet. For example, members of group 15 have five valence electrons and require only three additional electrons to fill their valence shells. These elements form three covalent bonds in their free state: triple bonds in the N2 molecule or single bonds to three different atoms in arsenic and phosphorus. The elements of group 16 require only two additional electrons. Oxygen forms a double bond in the O2 molecule, and sulfur, selenium, and tellurium form two single bonds in various rings and chains. The halogens form diatomic molecules in which each atom is involved in only one bond. This provides the electron required necessary to complete the octet on the halogen atom. The noble gases do not form covalent bonds to other noble gas atoms because they already have a filled outer shell.
Summary
Nonmetals have structures that are very different from those of the metals, primarily because they have greater electronegativity and electrons that are more tightly bound to individual atoms. Most nonmetal oxides are acid anhydrides, meaning that they react with water to form acidic solutions. Molecular structures are common for most of the nonmetals, and several have multiple allotropes with varying physical properties.
Glossary
acid anhydride
compound that reacts with water to form an acid or acidic solution
disproportionation reaction
chemical reaction where a single reactant is simultaneously reduced and oxidized; it is both the reducing agent and the oxidizing agent | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.04%3A_Structure_and_General_Properties_of_the_Nonmetals.txt |
Learning Objectives
• Describe the properties, preparation, and compounds of hydrogen
Hydrogen is the most abundant element in the universe. The sun and other stars are composed largely of hydrogen. Astronomers estimate that 90% of the atoms in the universe are hydrogen atoms. Hydrogen is a component of more compounds than any other element. Water is the most abundant compound of hydrogen found on earth. Hydrogen is an important part of petroleum, many minerals, cellulose and starch, sugar, fats, oils, alcohols, acids, and thousands of other substances.
At ordinary temperatures, hydrogen is a colorless, odorless, tasteless, and nonpoisonous gas consisting of the diatomic molecule H2. Hydrogen is composed of three isotopes, and unlike other elements, these isotopes have different names and chemical symbols: protium, 1H, deuterium, 2H (or “D”), and tritium 3H (or “T”). In a naturally occurring sample of hydrogen, there is one atom of deuterium for every 7000 H atoms and one atom of radioactive tritium for every 1018 H atoms. The chemical properties of the different isotopes are very similar because they have identical electron structures, but they differ in some physical properties because of their differing atomic masses. Elemental deuterium and tritium have lower vapor pressure than ordinary hydrogen. Consequently, when liquid hydrogen evaporates, the heavier isotopes are concentrated in the last portions to evaporate. Electrolysis of heavy water, D2O, yields deuterium. Most tritium originates from nuclear reactions.
Preparation of Hydrogen
Elemental hydrogen must be prepared from compounds by breaking chemical bonds. The most common methods of preparing hydrogen follow.
From Steam and Carbon or Hydrocarbons
Water is the cheapest and most abundant source of hydrogen. Passing steam over coke (an impure form of elemental carbon) at 1000 °C produces a mixture of carbon monoxide and hydrogen known as water gas:
$\ce{C}(s)+\ce{H2O}(g)\:\mathrm{\xrightarrow{1000\:°C}}\:\underset{\Large\mathrm{water\: gas}}{\ce{CO}(g)+\ce{H2}(g)} \nonumber$
Water gas is as an industrial fuel. It is possible to produce additional hydrogen by mixing the water gas with steam in the presence of a catalyst to convert the CO to CO2. This reaction is the water gas shift reaction.
It is also possible to prepare a mixture of hydrogen and carbon monoxide by passing hydrocarbons from natural gas or petroleum and steam over a nickel-based catalyst. Propane is an example of a hydrocarbon reactant:
$\ce{C3H8}(g)+\ce{3H2O}(g)\:\mathrm{\underset{catalyst}{\xrightarrow{900\:°C}}}\:\ce{3CO}(g)+\ce{7H2}(g) \nonumber$
Electrolysis
Hydrogen forms when direct current electricity passes through water (electrolysis) containing an electrolyte such as H2SO4, (Figure $1$.) Bubbles of hydrogen form at the cathode, and oxygen evolves at the anode. The net reaction is:
$\ce{2H2O}(l)+\ce{electrical\: energy}⟶\ce{2H2}(g)+\ce{O2}(g) \nonumber$
Reaction of Metals with Acids
This is the most convenient laboratory method of producing hydrogen. Metals with lower reduction potentials reduce the hydrogen ion in dilute acids to produce hydrogen gas and metal salts. For example, as shown in Figure $2$, iron in dilute hydrochloric acid produces hydrogen gas and iron(II) chloride:
$\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{2Cl-}(aq)⟶\ce{Fe^2+}(aq)+\ce{2Cl-}(aq)+\ce{H2}(g)+\ce{2H2O}(l) \nonumber$
Reaction of Ionic Metal Hydrides with Water
It is possible to produce hydrogen from the reaction of hydrides of the active metals, which contain the very strongly basic H anion, with water:
$\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca^2+}(aq)+\ce{2OH-}(aq)+\ce{2H2}(g) \nonumber$
Metal hydrides are expensive but convenient sources of hydrogen, especially where space and weight are important factors. They are important in the inflation of life jackets, life rafts, and military balloons.
Reactions
Under normal conditions, hydrogen is relatively inactive chemically, but when heated, it enters into many chemical reactions.
Two thirds of the world’s hydrogen production is devoted to the manufacture of ammonia, which is a fertilizer and used in the manufacture of nitric acid. Large quantities of hydrogen are also important in the process of hydrogenation, discussed in the chapter on organic chemistry.
It is possible to use hydrogen as a nonpolluting fuel. The reaction of hydrogen with oxygen is a very exothermic reaction, releasing 286 kJ of energy per mole of water formed. Hydrogen burns without explosion under controlled conditions. The oxygen-hydrogen torch, because of the high heat of combustion of hydrogen, can achieve temperatures up to 2800 °C. The hot flame of this torch is useful in cutting thick sheets of many metals. Liquid hydrogen is also an important rocket fuel (Figure $3$).
An uncombined hydrogen atom consists of a nucleus and one valence electron in the 1s orbital. The n = 1 valence shell has a capacity for two electrons, and hydrogen can rightfully occupy two locations in the periodic table. It is possible to consider hydrogen a group 1 element because hydrogen can lose an electron to form the cation, H+. It is also possible to consider hydrogen to be a group 17 element because it needs only one electron to fill its valence orbital to form a hydride ion, H, or it can share an electron to form a single, covalent bond. In reality, hydrogen is a unique element that almost deserves its own location in the periodic table.
Reactions with Elements
When heated, hydrogen reacts with the metals of group 1 and with Ca, Sr, and Ba (the more active metals in group 2). The compounds formed are crystalline, ionic hydrides that contain the hydride anion, H, a strong reducing agent and a strong base, which reacts vigorously with water and other acids to form hydrogen gas.
The reactions of hydrogen with nonmetals generally produce acidic hydrogen compounds with hydrogen in the 1+ oxidation state. The reactions become more exothermic and vigorous as the electronegativity of the nonmetal increases. Hydrogen reacts with nitrogen and sulfur only when heated, but it reacts explosively with fluorine (forming HF) and, under some conditions, with chlorine (forming HCl). A mixture of hydrogen and oxygen explodes if ignited. Because of the explosive nature of the reaction, it is necessary to exercise caution when handling hydrogen (or any other combustible gas) to avoid the formation of an explosive mixture in a confined space. Although most hydrides of the nonmetals are acidic, ammonia and phosphine (PH3) are very, very weak acids and generally function as bases. There is a summary of these reactions of hydrogen with the elements in Table $1$.
Table $1$: Chemical Reactions of Hydrogen with Other Elements
General Equation Comments
$\ce{MH\: or\: MH2⟶MOH\: or\: M(OH)2 + H2}$ ionic hydrides with group 1 and Ca, Sr, and Ba
$\ce{H2 + C⟶ (no\: reaction)}$
$\ce{3H2 + N2 ⟶ 2NH3}$ requires high pressure and temperature; low yield
$\ce{2H2 + O2 ⟶ 2H2O}$ exothermic and potentially explosive
$\ce{H2 + S ⟶ H2S}$ requires heating; low yield
$\ce{H2 + X2 ⟶ 2HX}$ X = F, Cl, Br, and I; explosive with F2; low yield with I2
Reaction with Compounds
Hydrogen reduces the heated oxides of many metals, with the formation of the metal and water vapor. For example, passing hydrogen over heated CuO forms copper and water. Hydrogen may also reduce the metal ions in some metal oxides to lower oxidation states:
$\ce{H2}(g)+\ce{MnO2}(s)\xrightarrow{Δ}\ce{MnO}(s)+\ce{H2O}(g) \nonumber$
Hydrogen Compounds
Other than the noble gases, each of the nonmetals forms compounds with hydrogen. For brevity, we will discuss only a few hydrogen compounds of the nonmetals here.
Nitrogen Hydrogen Compounds
Ammonia, NH3, forms naturally when any nitrogen-containing organic material decomposes in the absence of air. The laboratory preparation of ammonia is by the reaction of an ammonium salt with a strong base such as sodium hydroxide. The acid-base reaction with the weakly acidic ammonium ion gives ammonia, illustrated in Figure $4$. Ammonia also forms when ionic nitrides react with water. The nitride ion is a much stronger base than the hydroxide ion:
$\ce{Mg3N2}(s)+\ce{6H2O}(l)⟶\ce{3Mg(OH)2}(s)+\ce{2NH3}(g) \nonumber$
The commercial production of ammonia is by the direct combination of the elements in the Haber process:
$\ce{N2}(g)+\ce{3H2}(g)\xrightleftharpoons{\ce{catalyst}}\ce{2NH3}(g) \hspace{20px} ΔH°=\mathrm{−92\: kJ} \nonumber$
Ammonia is a colorless gas with a sharp, pungent odor. Smelling salts utilize this powerful odor. Gaseous ammonia readily liquefies to give a colorless liquid that boils at −33 °C. Due to intermolecular hydrogen bonding, the enthalpy of vaporization of liquid ammonia is higher than that of any other liquid except water, so ammonia is useful as a refrigerant. Ammonia is quite soluble in water (658 L at STP dissolves in 1 L H2O).
The chemical properties of ammonia are as follows:
1. Ammonia acts as a Brønsted base, as discussed in the chapter on acid-base chemistry. The ammonium ion is similar in size to the potassium ion; compounds of the two ions exhibit many similarities in their structures and solubilities.
2. Ammonia can display acidic behavior, although it is a much weaker acid than water. Like other acids, ammonia reacts with metals, although it is so weak that high temperatures are necessary. Hydrogen and (depending on the stoichiometry) amides (salts of $\ce{NH2-}$), imides (salts of NH2), or nitrides (salts of N3−) form.
3. The nitrogen atom in ammonia has its lowest possible oxidation state (3−) and thus is not susceptible to reduction. However, it can be oxidized. Ammonia burns in air, giving NO and water. Hot ammonia and the ammonium ion are active reducing agents. Of particular interest are the oxidations of ammonium ion by nitrite ion, $\ce{NO2-}$, to yield pure nitrogen and by nitrate ion to yield nitrous oxide, N2O.
4. There are a number of compounds that we can consider derivatives of ammonia through the replacement of one or more hydrogen atoms with some other atom or group of atoms. Inorganic derivations include chloramine, NH2Cl, and hydrazine, N2H4:
Chloramine, NH2Cl, results from the reaction of sodium hypochlorite, NaOCl, with ammonia in basic solution. In the presence of a large excess of ammonia at low temperature, the chloramine reacts further to produce hydrazine, N2H4:
$\ce{NH3}(aq)+\ce{OCl-}(aq)⟶\ce{NH2Cl}(aq)+\ce{OH-}(aq) \nonumber$
$\ce{NH2Cl}(aq)+\ce{NH3}(aq)+\ce{OH-}(aq)⟶\ce{N2H4}(aq)+\ce{Cl-}(aq)+\ce{H2O}(l) \nonumber$
Anhydrous hydrazine is relatively stable in spite of its positive free energy of formation:
$\ce{N2}(g)+\ce{2H2}(g)⟶\ce{N2H4}(l) \hspace{20px} ΔG^\circ_\ce{f}=\mathrm{149.2\:kJ\:mol^{−1}} \nonumber$
Hydrazine is a fuming, colorless liquid that has some physical properties remarkably similar to those of H2O (it melts at 2 °C, boils at 113.5 °C, and has a density at 25 °C of 1.00 g/mL). It burns rapidly and completely in air with substantial evolution of heat:
$\ce{N2H4}(l)+\ce{O2}(g)⟶\ce{N2}(g)+\ce{2H2O}(l) \hspace{20px} ΔH°=\mathrm{−621.5\:kJ\:mol^{−1}} \nonumber$
Like ammonia, hydrazine is both a Brønsted base and a Lewis base, although it is weaker than ammonia. It reacts with strong acids and forms two series of salts that contain the $\ce{N2H5+}$ and $\ce{N2H6^2+}$ ions, respectively. Some rockets use hydrazine as a fuel.
Phosphorus Hydrogen Compounds
The most important hydride of phosphorus is phosphine, PH3, a gaseous analog of ammonia in terms of both formula and structure. Unlike ammonia, it is not possible to form phosphine by direct union of the elements. There are two methods for the preparation of phosphine. One method is by the action of an acid on an ionic phosphide. The other method is the disproportionation of white phosphorus with hot concentrated base to produce phosphine and the hydrogen phosphite ion:
$\ce{AlP}(s)+\ce{3H3O+}(aq)⟶\ce{PH3}(g)+\ce{Al^3+}(aq)+\ce{3H2O}(l) \nonumber$
$\ce{P4}(s)+\ce{4OH-}(aq)+\ce{2H2O}(l)⟶\ce{2HPO3^2-}(aq)+\ce{2PH3}(g) \nonumber$
Phosphine is a colorless, very poisonous gas, which has an odor like that of decaying fish. Heat easily decomposes phosphine $(\ce{4PH3}⟶\ce{P4}+\ce{6H2})$, and the compound burns in air. The major uses of phosphine are as a fumigant for grains and in semiconductor processing. Like ammonia, gaseous phosphine unites with gaseous hydrogen halides, forming phosphonium compounds like PH4Cl and PH4I. Phosphine is a much weaker base than ammonia; therefore, these compounds decompose in water, and the insoluble PH3 escapes from solution.
Sulfur Hydrogen Compounds
Hydrogen sulfide, H2S, is a colorless gas that is responsible for the offensive odor of rotten eggs and of many hot springs. Hydrogen sulfide is as toxic as hydrogen cyanide; therefore, it is necessary to exercise great care in handling it. Hydrogen sulfide is particularly deceptive because it paralyzes the olfactory nerves; after a short exposure, one does not smell it.
The production of hydrogen sulfide by the direct reaction of the elements (H2 + S) is unsatisfactory because the yield is low. A more effective preparation method is the reaction of a metal sulfide with a dilute acid. For example:
$\ce{FeS}(s)+\ce{2H3O+}(aq)⟶\ce{Fe^2+}(aq)+\ce{H2S}(g)+\ce{2H2O}(l) \nonumber$
It is easy to oxidize the sulfur in metal sulfides and in hydrogen sulfide, making metal sulfides and H2S good reducing agents. In acidic solutions, hydrogen sulfide reduces Fe3+ to Fe2+, $\ce{MnO4-}$ to Mn2+, $\ce{Cr2O7^2-}$ to Cr3+, and HNO3 to NO2. The sulfur in H2S usually oxidizes to elemental sulfur, unless a large excess of the oxidizing agent is present. In which case, the sulfide may oxidize to $\ce{SO3^2-}$ or $\ce{SO4^2-}$ (or to SO2 or SO3 in the absence of water):
$\ce{2H2S}(g)+\ce{O2}(g)⟶\ce{2S}(s)+\ce{2H2O}(l) \nonumber$
This oxidation process leads to the removal of the hydrogen sulfide found in many sources of natural gas. The deposits of sulfur in volcanic regions may be the result of the oxidation of H2S present in volcanic gases.
Hydrogen sulfide is a weak diprotic acid that dissolves in water to form hydrosulfuric acid. The acid ionizes in two stages, yielding hydrogen sulfide ions, HS, in the first stage and sulfide ions, S2−, in the second. Since hydrogen sulfide is a weak acid, aqueous solutions of soluble sulfides and hydrogen sulfides are basic:
$\ce{S^2-}(aq)+\ce{H2O}(l)⇌\ce{HS-}(aq)+\ce{OH-}(aq) \nonumber$
$\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H2S}(g)+\ce{OH-}(aq) \nonumber$
Halogen Hydrogen Compounds
Binary compounds containing only hydrogen and a halogen are hydrogen halides. At room temperature, the pure hydrogen halides HF, HCl, HBr, and HI are gases.
In general, it is possible to prepare the halides by the general techniques used to prepare other acids. Fluorine, chlorine, and bromine react directly with hydrogen to form the respective hydrogen halide. This is a commercially important reaction for preparing hydrogen chloride and hydrogen bromide.
The acid-base reaction between a nonvolatile strong acid and a metal halide will yield a hydrogen halide. The escape of the gaseous hydrogen halide drives the reaction to completion. For example, the usual method of preparing hydrogen fluoride is by heating a mixture of calcium fluoride, CaF2, and concentrated sulfuric acid:
$\ce{CaF2}(s)+\ce{H2SO4}(aq)⟶\ce{CaSO4}(s)+\ce{2HF}(g) \nonumber$
Gaseous hydrogen fluoride is also a by-product in the preparation of phosphate fertilizers by the reaction of fluoroapatite, Ca5(PO4)3F, with sulfuric acid. The reaction of concentrated sulfuric acid with a chloride salt produces hydrogen chloride both commercially and in the laboratory.
In most cases, sodium chloride is the chloride of choice because it is the least expensive chloride. Hydrogen bromide and hydrogen iodide cannot be prepared using sulfuric acid because this acid is an oxidizing agent capable of oxidizing both bromide and iodide. However, it is possible to prepare both hydrogen bromide and hydrogen iodide using an acid such as phosphoric acid because it is a weaker oxidizing agent. For example:
$\ce{H3PO4}(l)+\ce{Br-}(aq)⟶\ce{HBr}(g)+\ce{H2PO4-}(aq) \nonumber$
All of the hydrogen halides are very soluble in water, forming hydrohalic acids. With the exception of hydrogen fluoride, which has a strong hydrogen-fluoride bond, they are strong acids. Reactions of hydrohalic acids with metals, metal hydroxides, oxides, or carbonates produce salts of the halides. Most chloride salts are soluble in water. AgCl, PbCl2, and Hg2Cl2 are the commonly encountered exceptions.
The halide ions give the substances the properties associated with X(aq). The heavier halide ions (Cl, Br, and I) can act as reducing agents, and the lighter halogens or other oxidizing agents will oxidize them:
$\ce{Cl2}(aq)+\ce{2e-}⟶\ce{2Cl-}(aq) \hspace{20px} E°=\mathrm{1.36\:V}\ \ce{Br2}(aq)+\ce{2e-}⟶\ce{2Br-}(aq) \hspace{20px} E°=\mathrm{1.09\:V}\ \ce{I2}(aq)+\ce{2e-}⟶\ce{2I-}(aq) \hspace{20px} E°=\mathrm{0.54\:V} \nonumber$
For example, bromine oxidizes iodine:
$\ce{Br2}(aq)+\ce{2HI}(aq)⟶\ce{2HBr}(aq)+\ce{I2}(aq) \hspace{20px} E°=\mathrm{0.55\:V} \nonumber$
Hydrofluoric acid is unique in its reactions with sand (silicon dioxide) and with glass, which is a mixture of silicates:
$\ce{SiO2}(s)+\ce{4HF}(aq)⟶\ce{SiF4}(g)+\ce{2H2O}(l) \nonumber$
$\ce{CaSiO3}(s)+\ce{6HF}(aq)⟶\ce{CaF2}(s)+\ce{SiF4}(g)+\ce{3H2O}(l) \nonumber$
The volatile silicon tetrafluoride escapes from these reactions. Because hydrogen fluoride attacks glass, it can frost or etch glass and is used to etch markings on thermometers, burets, and other glassware.
The largest use for hydrogen fluoride is in production of hydrochlorofluorocarbons for refrigerants, in plastics, and in propellants. The second largest use is in the manufacture of cryolite, Na3AlF6, which is important in the production of aluminum. The acid is also important in the production of other inorganic fluorides (such as BF3), which serve as catalysts in the industrial synthesis of certain organic compounds.
Hydrochloric acid is relatively inexpensive. It is an important and versatile acid in industry and is important for the manufacture of metal chlorides, dyes, glue, glucose, and various other chemicals. A considerable amount is also important for the activation of oil wells and as pickle liquor—an acid used to remove oxide coating from iron or steel that is to be galvanized, tinned, or enameled. The amounts of hydrobromic acid and hydroiodic acid used commercially are insignificant by comparison.
Summary
Hydrogen is the most abundant element in the universe and its chemistry is truly unique. Although it has some chemical reactivity that is similar to that of the alkali metals, hydrogen has many of the same chemical properties of a nonmetal with a relatively low electronegativity. It forms ionic hydrides with active metals, covalent compounds in which it has an oxidation state of 1− with less electronegative elements, and covalent compounds in which it has an oxidation state of 1+ with more electronegative nonmetals. It reacts explosively with oxygen, fluorine, and chlorine, less readily with bromine, and much less readily with iodine, sulfur, and nitrogen. Hydrogen reduces the oxides of metals with lower reduction potentials than chromium to form the metal and water. The hydrogen halides are all acidic when dissolved in water.
Glossary
Haber process
main industrial process used to produce ammonia from nitrogen and hydrogen; involves the use of an iron catalyst and elevated temperatures and pressures
hydrogen halide
binary compound formed between hydrogen and the halogens: HF, HCl, HBr, and HI
hydrogenation
addition of hydrogen (H2) to reduce a compound | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.05%3A_Occurrence_Preparation_and_Compounds_of_Hydrogen.txt |
Learning Objectives
• Describe the preparation, properties, and uses of some representative metal carbonates
The chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates—compounds that contain the carbonate anions, $\ce{CO3^2-}$. The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates—compounds that contain the hydrogen carbonate anion, $\ce{HCO3-}$, also known as the bicarbonate anion.
With the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include:
\begin{align} \ce{Na2O}(s)+\ce{CO2}(g) &⟶ \ce{Na2CO3}(s)\[4pt] \ce{Ca(OH)2}(s)+\ce{CO2}(g) &⟶\ce{CaCO3}(s)+\ce{H2O}(l) \end{align} \nonumber
The carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are:
\begin{align} \ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{CaCO3}(s) \[4pt] \ce{Pb^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{PbCO3}(s) \end{align} \nonumber
Pearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as Al3+ or Sn4+ behave differently in this reaction as carbon dioxide and the corresponding oxide form instead of the carbonate.
Alkali metal hydrogen carbonates such as NaHCO3 and CsHCO3 form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide:
$\ce{OH-}(aq)+\ce{CO2}(aq)⟶\ce{HCO3-}(aq) \nonumber$
It is possible to isolate the solids by evaporation of the water from the solution.
Although they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when CaCO3 dissolves in water containing dissolved carbon dioxide:
$\ce{CaCO3}(s)+\ce{CO2}(aq)+\ce{H2O}(l)⟶\ce{Ca^2+}(aq)+\ce{2HCO3-}(aq) \nonumber$
Hydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure $1$, form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate.
The two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, Na3(CO3)(HCO3)(H2O)2. Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces Na2CO3:
$\ce{2Na3(CO3)(HCO3)(H2O)2}(s)⟶\ce{3Na2CO3}(s)+\ce{5H2O}(l)+\ce{CO2}(g) \nonumber$
Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction:
$\ce{CO3^2-}(aq)+\ce{H2O}(l)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber$
Carbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid (stomach acid), as shown in Figure $2$, illustrates the reaction:
$\ce{CaCO3}(s)+\ce{2HCl}(aq)⟶\ce{CaCl2}(aq)+\ce{CO2}(g)+\ce{H2O}(l) \nonumber$
Other applications of carbonates include glass making—where carbonate ions serve as a source of oxide ions—and synthesis of oxides.
Hydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water:
$\ce{KHCO3}(aq)+\ce{KOH}(aq)⟶\ce{K2CO3}(aq)+\ce{H2O}(l) \nonumber$
With acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda (bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate (cream of tartar), KHC4H4O6. As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide:
$\ce{HC4H4O6-}(aq)+\ce{HCO3-}(aq)⟶\ce{C4H4O6^2-}(aq)+\ce{CO2}(g)+\ce{H2O}(l) \nonumber$
Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods.
Summary
The usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone (CaCO3), the antacid Tums (CaCO3), and baking soda (NaHCO3) are common examples. Carbonates and hydrogen carbonates decompose in the presence of acids and most decompose on heating.
Glossary
bicarbonate anion
salt of the hydrogen carbonate ion, $\ce{HCO3-}$
carbonate
salt of the anion $\ce{CO3^2-}$; often formed by the reaction of carbon dioxide with bases
hydrogen carbonate
salt of carbonic acid, H2CO3 (containing the anion $\ce{HCO3-}$) in which one hydrogen atom has been replaced; an acid carbonate; also known as bicarbonate ion | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.06%3A_Occurrence_Preparation_and_Properties_of_Carbonates.txt |
Learning Objectives
• Describe the properties, preparation, and uses of nitrogen
Most pure nitrogen comes from the fractional distillation of liquid air. The atmosphere consists of 78% nitrogen by volume. This means there are more than 20 million tons of nitrogen over every square mile of the earth’s surface. Nitrogen is a component of proteins and of the genetic material (DNA/RNA) of all plants and animals.
Under ordinary conditions, nitrogen is a colorless, odorless, and tasteless gas. It boils at 77 K and freezes at 63 K. Liquid nitrogen is a useful coolant because it is inexpensive and has a low boiling point. Nitrogen is very unreactive because of the very strong triple bond between the nitrogen atoms. The only common reactions at room temperature occur with lithium to form Li3N, with certain transition metal complexes, and with hydrogen or oxygen in nitrogen-fixing bacteria. The general lack of reactivity of nitrogen makes the remarkable ability of some bacteria to synthesize nitrogen compounds using atmospheric nitrogen gas as the source one of the most exciting chemical events on our planet. This process is one type of nitrogen fixation. In this case, nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. Nitrogen fixation also occurs when lightning passes through air, causing molecular nitrogen to react with oxygen to form nitrogen oxides, which are then carried down to the soil.
Nitrogen Fixation
All living organisms require nitrogen compounds for survival. Unfortunately, most of these organisms cannot absorb nitrogen from its most abundant source—the atmosphere. Atmospheric nitrogen consists of N2 molecules, which are very unreactive due to the strong nitrogen-nitrogen triple bond. However, a few organisms can overcome this problem through a process known as nitrogen fixation, illustrated in Figure $1$.
Nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. To date, the only known kind of biological organisms capable of nitrogen fixation are microorganisms. These organisms employ enzymes called nitrogenases, which contain iron and molybdenum. Many of these microorganisms live in a symbiotic relationship with plants, with the best-known example being the presence of rhizobia in the root nodules of legumes.
Large volumes of atmospheric nitrogen are necessary for making ammonia—the principal starting material used for preparation of large quantities of other nitrogen-containing compounds. Most other uses for elemental nitrogen depend on its inactivity. It is helpful when a chemical process requires an inert atmosphere. Canned foods and luncheon meats cannot oxidize in a pure nitrogen atmosphere, so they retain a better flavor and color, and spoil less rapidly, when sealed in nitrogen instead of air. This technology allows fresh produce to be available year-round, regardless of growing season.
There are compounds with nitrogen in all of its oxidation states from 3− to 5+. Much of the chemistry of nitrogen involves oxidation-reduction reactions. Some active metals (such as alkali metals and alkaline earth metals) can reduce nitrogen to form metal nitrides. In the remainder of this section, we will examine nitrogen-oxygen chemistry.
There are well-characterized nitrogen oxides in which nitrogen exhibits each of its positive oxidation numbers from 1+ to 5+. When ammonium nitrate is carefully heated, nitrous oxide (dinitrogen oxide) and water vapor form. Stronger heating generates nitrogen gas, oxygen gas, and water vapor. No one should ever attempt this reaction—it can be very explosive. In 1947, there was a major ammonium nitrate explosion in Texas City, Texas, and, in 2013, there was another major explosion in West, Texas. In the last 100 years, there were nearly 30 similar disasters worldwide, resulting in the loss of numerous lives. In this oxidation-reduction reaction, the nitrogen in the nitrate ion oxidizes the nitrogen in the ammonium ion. Nitrous oxide, shown in Figure $2$, is a colorless gas possessing a mild, pleasing odor and a sweet taste. It finds application as an anesthetic for minor operations, especially in dentistry, under the name “laughing gas.”
Low yields of nitric oxide, NO, form when heating nitrogen and oxygen together. NO also forms when lightning passes through air during thunderstorms. Burning ammonia is the commercial method of preparing nitric oxide. In the laboratory, the reduction of nitric acid is the best method for preparing nitric oxide. When copper reacts with dilute nitric acid, nitric oxide is the principal reduction product:
$\ce{3Cu}(s)+\ce{8HNO3}(aq)⟶\ce{2NO}(g)+\ce{3Cu(NO3)2}(aq)+\ce{4H2O}(l) \nonumber$
Gaseous nitric oxide is the most thermally stable of the nitrogen oxides and is the simplest known thermally stable molecule with an unpaired electron. It is one of the air pollutants generated by internal combustion engines, resulting from the reaction of atmospheric nitrogen and oxygen during the combustion process.
At room temperature, nitric oxide is a colorless gas consisting of diatomic molecules. As is often the case with molecules that contain an unpaired electron, two molecules combine to form a dimer by pairing their unpaired electrons to form a bond. Liquid and solid NO both contain N2O2 dimers, like that shown in Figure $3$. Most substances with unpaired electrons exhibit color by absorbing visible light; however, NO is colorless because the absorption of light is not in the visible region of the spectrum.
Cooling a mixture of equal parts nitric oxide and nitrogen dioxide to −21 °C produces dinitrogen trioxide, a blue liquid consisting of N2O3 molecules (Figure $4$). Dinitrogen trioxide exists only in the liquid and solid states. When heated, it reverts to a mixture of NO and NO2.
It is possible to prepare nitrogen dioxide in the laboratory by heating the nitrate of a heavy metal, or by the reduction of concentrated nitric acid with copper metal, as shown in Figure $5$. Commercially, it is possible to prepare nitrogen dioxide by oxidizing nitric oxide with air.
The nitrogen dioxide molecule (Figure $6$) contains an unpaired electron, which is responsible for its color and paramagnetism. It is also responsible for the dimerization of NO2. At low pressures or at high temperatures, nitrogen dioxide has a deep brown color that is due to the presence of the NO2 molecule. At low temperatures, the color almost entirely disappears as dinitrogen tetraoxide, N2O4, forms. At room temperature, an equilibrium exists:
$\ce{2NO2}(g)⇌\ce{N2O4}(g) \hspace{20px} K_P=6.86 \nonumber$
Dinitrogen pentaoxide, N2O5 (Figure $7$), is a white solid that is formed by the dehydration of nitric acid by phosphorus(V) oxide (tetraphosphorus decoxide):
$\ce{P4O10}(s)+\ce{4HNO3}(l)⟶\ce{4HPO3}(s)+\ce{2N2O5}(s) \nonumber$
It is unstable above room temperature, decomposing to N2O4 and O2.
The oxides of nitrogen(III), nitrogen(IV), and nitrogen(V) react with water and form nitrogen-containing oxyacids. Nitrogen(III) oxide, N2O3, is the anhydride of nitrous acid; HNO2 forms when N2O3 reacts with water. There are no stable oxyacids containing nitrogen with an oxidation state of 4+; therefore, nitrogen(IV) oxide, NO2, disproportionates in one of two ways when it reacts with water. In cold water, a mixture of HNO2 and HNO3 forms. At higher temperatures, HNO3 and NO will form. Nitrogen(V) oxide, N2O5, is the anhydride of nitric acid; HNO3 is produced when N2O5 reacts with water:
$\ce{N2O5}(s)+\ce{H2O}(l)⟶\ce{2HNO3}(aq) \nonumber$
The nitrogen oxides exhibit extensive oxidation-reduction behavior. Nitrous oxide resembles oxygen in its behavior when heated with combustible substances. N2O is a strong oxidizing agent that decomposes when heated to form nitrogen and oxygen. Because one-third of the gas liberated is oxygen, nitrous oxide supports combustion better than air (one-fifth oxygen). A glowing splinter bursts into flame when thrust into a bottle of this gas. Nitric oxide acts both as an oxidizing agent and as a reducing agent. For example:
$\textrm{oxidizing agent: }\ce{P4}(s)+\ce{6NO}(g)⟶\ce{P4O6}(s)+\ce{3N2}(g) \nonumber$
$\textrm{reducing agent: }\ce{Cl2}(g)+\ce{2NO}(g)⟶\ce{2ClNO}(g) \nonumber$
Nitrogen dioxide (or dinitrogen tetraoxide) is a good oxidizing agent. For example:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber$
$\ce{NO2}(g)+\ce{2HCl}(aq)⟶\ce{NO}(g)+\ce{Cl2}(g)+\ce{H2O}(l) \nonumber$
Summary
Nitrogen exhibits oxidation states ranging from 3− to 5+. Because of the stability of the N≡N triple bond, it requires a great deal of energy to make compounds from molecular nitrogen. Active metals such as the alkali metals and alkaline earth metals can reduce nitrogen to form metal nitrides. Nitrogen oxides and nitrogen hydrides are also important substances.
Glossary
nitrogen fixation
formation of nitrogen compounds from molecular nitrogen | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.07%3A_Occurrence_Preparation_and_Properties_of_Nitrogen.txt |
Learning Objectives
• Describe the properties, preparation, and uses of phosphorus
The industrial preparation of phosphorus is by heating calcium phosphate, obtained from phosphate rock, with sand and coke:
$\ce{2Ca3(PO4)2}(s)+\ce{6SiO2}(s)+\ce{10C}(s) \xrightarrow{Δ} \ce{6CaSiO3}(l)+\ce{10CO}(g)+\ce{P4}(g) \nonumber$
The phosphorus distills out of the furnace and is condensed into a solid or burned to form P4O10. The preparation of many other phosphorus compounds begins with P4O10. The acids and phosphates are useful as fertilizers and in the chemical industry. Other uses are in the manufacture of special alloys such as ferrophosphorus and phosphor bronze. Phosphorus is important in making pesticides, matches, and some plastics. Phosphorus is an active nonmetal. In compounds, phosphorus usually occurs in oxidation states of 3−, 3+, and 5+. Phosphorus exhibits oxidation numbers that are unusual for a group 15 element in compounds that contain phosphorus-phosphorus bonds; examples include diphosphorus tetrahydride, H2P-PH2, and tetraphosphorus trisulfide, P4S3, illustrated in Figure $1$.
Phosphorus Oxygen Compounds
Phosphorus forms two common oxides, phosphorus(III) oxide (or tetraphosphorus hexaoxide), P4O6, and phosphorus(V) oxide (or tetraphosphorus decaoxide), P4O10, both shown in Figure $2$. Phosphorus(III) oxide is a white crystalline solid with a garlic-like odor. Its vapor is very poisonous. It oxidizes slowly in air and inflames when heated to 70 °C, forming P4O10. Phosphorus(III) oxide dissolves slowly in cold water to form phosphorous acid, H3PO3.
Phosphorus(V) oxide, P4O10, is a white powder that is prepared by burning phosphorus in excess oxygen. Its enthalpy of formation is very high (−2984 kJ), and it is quite stable and a very poor oxidizing agent. Dropping P4O10 into water produces a hissing sound, heat, and orthophosphoric acid:
$\ce{P4O10}(s)+\ce{6H2O}(l)⟶\ce{4H3PO4}(aq) \nonumber$
Because of its great affinity for water, phosphorus(V) oxide is an excellent drying agent for gases and solvents, and for removing water from many compounds.
Phosphorus Halogen Compounds
Phosphorus will react directly with the halogens, forming trihalides, PX3, and pentahalides, PX5. The trihalides are much more stable than the corresponding nitrogen trihalides; nitrogen pentahalides do not form because of nitrogen’s inability to form more than four bonds.
The chlorides PCl3 and PCl5, both shown in Figure $3$, are the most important halides of phosphorus. Phosphorus trichloride is a colorless liquid that is prepared by passing chlorine over molten phosphorus. Phosphorus pentachloride is an off-white solid that is prepared by oxidizing the trichloride with excess chlorine. The pentachloride sublimes when warmed and forms an equilibrium with the trichloride and chlorine when heated.
Like most other nonmetal halides, both phosphorus chlorides react with an excess of water and yield hydrogen chloride and an oxyacid: PCl3 yields phosphorous acid H3PO3 and PCl5 yields phosphoric acid, H3PO4.
The pentahalides of phosphorus are Lewis acids because of the empty valence d orbitals of phosphorus. These compounds readily react with halide ions (Lewis bases) to give the anion $\ce{PX6-}$. Whereas phosphorus pentafluoride is a molecular compound in all states, X-ray studies show that solid phosphorus pentachloride is an ionic compound, $\ce{[PCl4+][PCl6- ]}$, as are phosphorus pentabromide, $\ce{[PBr4+]}$[Br], and phosphorus pentaiodide, $\ce{[PI4+]}$[I].
Summary
Phosphorus (group 15) commonly exhibits oxidation states of 3− with active metals and of 3+ and 5+ with more electronegative nonmetals. The halogens and oxygen will oxidize phosphorus. The oxides are phosphorus(V) oxide, P4O10, and phosphorus(III) oxide, P4O6. The two common methods for preparing orthophosphoric acid, H3PO4, are either the reaction of a phosphate with sulfuric acid or the reaction of water with phosphorus(V) oxide. Orthophosphoric acid is a triprotic acid that forms three types of salts. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.08%3A_Occurrence_Preparation_and_Properties_of_Phosphorus.txt |
Learning Objectives
• Describe the properties, preparation, and compounds of oxygen
• Describe the preparation, properties, and uses of some representative metal oxides, peroxides, and hydroxides
Oxygen is the most abundant element on the earth’s crust. The earth’s surface is composed of the crust, atmosphere, and hydrosphere. About 50% of the mass of the earth’s crust consists of oxygen (combined with other elements, principally silicon). Oxygen occurs as O2 molecules and, to a limited extent, as O3 (ozone) molecules in air. It forms about 20% of the mass of the air. About 89% of water by mass consists of combined oxygen. In combination with carbon, hydrogen, and nitrogen, oxygen is a large part of plants and animals.
Oxygen is a colorless, odorless, and tasteless gas at ordinary temperatures. It is slightly denser than air. Although it is only slightly soluble in water (49 mL of gas dissolves in 1 L at STP), oxygen’s solubility is very important to aquatic life.
Most of the oxygen isolated commercially comes from air and the remainder from the electrolysis of water. The separation of oxygen from air begins with cooling and compressing the air until it liquefies. As liquid air warms, oxygen with its higher boiling point (90 K) separates from nitrogen, which has a lower boiling point (77 K). It is possible to separate the other components of air at the same time based on differences in their boiling points.
Oxygen is essential in combustion processes such as the burning of fuels. Plants and animals use the oxygen from the air in respiration. The administration of oxygen-enriched air is an important medical practice when a patient is receiving an inadequate supply of oxygen because of shock, pneumonia, or some other illness.
The chemical industry employs oxygen for oxidizing many substances. A significant amount of oxygen produced commercially is important in the removal of carbon from iron during steel production. Large quantities of pure oxygen are also necessary in metal fabrication and in the cutting and welding of metals with oxyhydrogen and oxyacetylene torches.
Liquid oxygen is important to the space industry. It is an oxidizing agent in rocket engines. It is also the source of gaseous oxygen for life support in space.
As we know, oxygen is very important to life. The energy required for the maintenance of normal body functions in human beings and in other organisms comes from the slow oxidation of chemical compounds. Oxygen is the final oxidizing agent in these reactions. In humans, oxygen passes from the lungs into the blood, where it combines with hemoglobin, producing oxyhemoglobin. In this form, blood transports the oxygen to tissues, where it is transferred to the tissues. The ultimate products are carbon dioxide and water. The blood carries the carbon dioxide through the veins to the lungs, where the blood releases the carbon dioxide and collects another supply of oxygen. Digestion and assimilation of food regenerate the materials consumed by oxidation in the body; the energy liberated is the same as if the food burned outside the body.
Green plants continually replenish the oxygen in the atmosphere by a process called photosynthesis. The products of photosynthesis may vary, but, in general, the process converts carbon dioxide and water into glucose (a sugar) and oxygen using the energy of light:
\begin{alignat}{3} &\ce{6CO2}(g) \:+\: &&\ce{6H2O}(l) \:\mathrm{\underset{light}{\xrightarrow{chlorophyll}}}\: &&\ce{C6H12O6}(aq) \:+\: &&\ce{6O2}(g)\ &\mathrm{carbon\ dioxide} &&\ce{water} &&\ce{glucose} &&\ce{oxygen} \end{alignat} \nonumber
Thus, the oxygen that became carbon dioxide and water by the metabolic processes in plants and animals returns to the atmosphere by photosynthesis.
When dry oxygen is passed between two electrically charged plates, ozone (O3, illustrated in Figure $1$), an allotrope of oxygen possessing a distinctive odor, forms. The formation of ozone from oxygen is an endothermic reaction, in which the energy comes from an electrical discharge, heat, or ultraviolet light:
$\ce{3O2}(g)\xrightarrow{\ce{electric\: discharge}}\ce{2O3}(g) \hspace{20px} ΔH°=\mathrm{287\: kJ} \nonumber$
The sharp odor associated with sparking electrical equipment is due, in part, to ozone.
Ozone forms naturally in the upper atmosphere by the action of ultraviolet light from the sun on the oxygen there. Most atmospheric ozone occurs in the stratosphere, a layer of the atmosphere extending from about 10 to 50 kilometers above the earth’s surface. This ozone acts as a barrier to harmful ultraviolet light from the sun by absorbing it via a chemical decomposition reaction:
$\ce{O3}(g)\xrightarrow{\ce{ultraviolet\: light}}\ce{O}(g)+\ce{O2}(g) \nonumber$
The reactive oxygen atoms recombine with molecular oxygen to complete the ozone cycle. The presence of stratospheric ozone decreases the frequency of skin cancer and other damaging effects of ultraviolet radiation. It has been clearly demonstrated that chlorofluorocarbons, CFCs (known commercially as Freons), which were present as aerosol propellants in spray cans and as refrigerants, caused depletion of ozone in the stratosphere. This occurred because ultraviolet light also causes CFCs to decompose, producing atomic chlorine. The chlorine atoms react with ozone molecules, resulting in a net removal of O3 molecules from stratosphere. This process is explored in detail in our coverage of chemical kinetics. There is a worldwide effort to reduce the amount of CFCs used commercially, and the ozone hole is already beginning to decrease in size as atmospheric concentrations of atomic chlorine decrease. While ozone in the stratosphere helps protect us, ozone in the troposphere is a problem. This ozone is a toxic component of photochemical smog.
The uses of ozone depend on its reactivity with other substances. It can be used as a bleaching agent for oils, waxes, fabrics, and starch: It oxidizes the colored compounds in these substances to colorless compounds. It is an alternative to chlorine as a disinfectant for water.
Reactions
Elemental oxygen is a strong oxidizing agent. It reacts with most other elements and many compounds.
Reaction with Elements
Oxygen reacts directly at room temperature or at elevated temperatures with all other elements except the noble gases, the halogens, and few second- and third-row transition metals of low reactivity (those with higher reduction potentials than copper). Rust is an example of the reaction of oxygen with iron. The more active metals form peroxides or superoxides. Less active metals and the nonmetals give oxides. Two examples of these reactions are:
$\ce{2Mg}(s)+\ce{O2}(g)⟶\ce{2MgO}(s) \nonumber$
$\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s) \nonumber$
The oxides of halogens, at least one of the noble gases, and metals with higher reduction potentials than copper do not form by the direct action of the elements with oxygen.
Reaction with Compounds
Elemental oxygen also reacts with some compounds. If it is possible to oxidize any of the elements in a given compound, further oxidation by oxygen can occur. For example, hydrogen sulfide, H2S, contains sulfur with an oxidation state of 2−. Because the sulfur does not exhibit its maximum oxidation state, we would expect H2S to react with oxygen. It does, yielding water and sulfur dioxide. The reaction is:
$\ce{2H2S}(g)+\ce{3O2}(g)⟶\ce{2H2O}(l)+\ce{2SO2}(g) \nonumber$
It is also possible to oxidize oxides such as CO and P4O6 that contain an element with a lower oxidation state. The ease with which elemental oxygen picks up electrons is mirrored by the difficulty of removing electrons from oxygen in most oxides. Of the elements, only the very reactive fluorine can oxidize oxides to form oxygen gas.
Oxides, Peroxides, and Hydroxides
Compounds of the representative metals with oxygen fall into three categories: (1) oxides, containing oxide ions, O2−; (2) peroxides, containing peroxides ions, $\ce{O2^2-}$, with oxygen-oxygen covalent single bonds and a very limited number of superoxides, containing superoxide ions, $\ce{O2-}$, with oxygen-oxygen covalent bonds that have a bond order of $\frac{3}{2}$, In addition, there are (3) hydroxides, containing hydroxide ions, OH. All representative metals form oxides. Some of the metals of group 2 also form peroxides, MO2, and the metals of group 1 also form peroxides, M2O2, and superoxides, MO2.
Oxides
It is possible to produce the oxides of most representative metals by heating the corresponding hydroxides (forming the oxide and gaseous water) or carbonates (forming the oxide and gaseous CO2). Equations for example reactions are:
$\ce{2Al(OH)3}(s)\xrightarrow{Δ}\ce{Al2O3}(s)+\ce{3H2O}(g) \nonumber$
$\ce{CaCO3}(s)\xrightarrow{Δ}\ce{CaO}(s)+\ce{CO2}(g) \nonumber$
However, alkali metal salts generally are very stable and do not decompose easily when heated. Alkali metal oxides result from the oxidation-reduction reactions created by heating nitrates or hydroxides with the metals. Equations for sample reactions are:
$\ce{2KNO3}(s)+\ce{10K}(s)\xrightarrow{Δ}\ce{6K2O}(s)+\ce{N2}(g) \nonumber$
$\ce{2LiOH}(s)+\ce{2Li}(s)\xrightarrow{Δ}\ce{2Li2O}(s)+\ce{H2}(g) \nonumber$
With the exception of mercury(II) oxide, it is possible to produce the oxides of the metals of groups 2–15 by burning the corresponding metal in air. The heaviest member of each group, the member for which the inert pair effect is most pronounced, forms an oxide in which the oxidation state of the metal ion is two less than the group oxidation state (inert pair effect). Thus, Tl2O, PbO, and Bi2O3 form when burning thallium, lead, and bismuth, respectively. The oxides of the lighter members of each group exhibit the group oxidation state. For example, SnO2 forms from burning tin. Mercury(II) oxide, HgO, forms slowly when mercury is warmed below 500 °C; it decomposes at higher temperatures.
Burning the members of groups 1 and 2 in air is not a suitable way to form the oxides of these elements. These metals are reactive enough to combine with nitrogen in the air, so they form mixtures of oxides and ionic nitrides. Several also form peroxides or superoxides when heated in air.
Ionic oxides all contain the oxide ion, a very powerful hydrogen ion acceptor. With the exception of the very insoluble aluminum oxide, Al2O3, tin(IV), SnO2, and lead(IV), PbO2, the oxides of the representative metals react with acids to form salts. Some equations for these reactions are:
$\ce{Na2O}+\ce{2HNO3}(aq)⟶\ce{2NaNO3}(aq)+\ce{H2O}(l) \nonumber$
$\ce{CaO}(s)+\ce{2HCL}(aq)⟶\ce{CaCl2}(aq)+\ce{H2O}(l) \nonumber$
$\ce{SnO}(s)+\ce{2HClO4}(aq)⟶\ce{Sn(ClO4)2}(aq)+\ce{H2O}(l) \nonumber$
The oxides of the metals of groups 1 and 2 and of thallium(I) oxide react with water and form hydroxides. Examples of such reactions are:
$\ce{Na2O}(s)+\ce{H2O}(l)⟶\ce{NaOH}(aq) \nonumber$
$\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(aq) \nonumber$
$\ce{Tl2O}(s)+\ce{H2O}(aq)⟶\ce{2TlOH}(aq) \nonumber$
The oxides of the alkali metals have little industrial utility, unlike magnesium oxide, calcium oxide, and aluminum oxide. Magnesium oxide is important in making firebrick, crucibles, furnace linings, and thermal insulation—applications that require chemical and thermal stability. Calcium oxide, sometimes called quicklime or lime in the industrial market, is very reactive, and its principal uses reflect its reactivity. Pure calcium oxide emits an intense white light when heated to a high temperature (as illustrated in Figure $2$:). Blocks of calcium oxide heated by gas flames were the stage lights in theaters before electricity was available. This is the source of the phrase “in the limelight.”
Calcium oxide and calcium hydroxide are inexpensive bases used extensively in chemical processing, although most of the useful products prepared from them do not contain calcium. Calcium oxide, CaO, is made by heating calcium carbonate, CaCO3, which is widely and inexpensively available as limestone or oyster shells:
$\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g) \nonumber$
Although this decomposition reaction is reversible, it is possible to obtain a 100% yield of CaO by allowing the CO2 to escape. It is possible to prepare calcium hydroxide by the familiar acid-base reaction of a soluble metal oxide with water:
$\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s) \nonumber$
Both CaO and Ca(OH)2 are useful as bases; they accept protons and neutralize acids.
Alumina (Al2O3) occurs in nature as the mineral corundum, a very hard substance used as an abrasive for grinding and polishing. Corundum is important to the jewelry trade as ruby and sapphire. The color of ruby is due to the presence of a small amount of chromium; other impurities produce the wide variety of colors possible for sapphires. Artificial rubies and sapphires are now manufactured by melting aluminum oxide (melting point = 2050 °C) with small amounts of oxides to produce the desired colors and cooling the melt in such a way as to produce large crystals. Ruby lasers use synthetic ruby crystals.
Zinc oxide, ZnO, was a useful white paint pigment; however, pollutants tend to discolor the compound. The compound is also important in the manufacture of automobile tires and other rubber goods, and in the preparation of medicinal ointments. For example, zinc-oxide-based sunscreens, as shown in Figure $3$, help prevent sunburn. The zinc oxide in these sunscreens is present in the form of very small grains known as nanoparticles. Lead dioxide is a constituent of charged lead storage batteries. Lead(IV) tends to revert to the more stable lead(II) ion by gaining two electrons, so lead dioxide is a powerful oxidizing agent.
Peroxides and Superoxides
Peroxides and superoxides are strong oxidizers and are important in chemical processes. Hydrogen peroxide, H2O2, prepared from metal peroxides, is an important bleach and disinfectant. Peroxides and superoxides form when the metal or metal oxides of groups 1 and 2 react with pure oxygen at elevated temperatures. Sodium peroxide and the peroxides of calcium, strontium, and barium form by heating the corresponding metal or metal oxide in pure oxygen:
$\ce{2Na}(s)+\ce{O2}(g)\xrightarrow{Δ}\ce{Na2O2}(s) \nonumber$
$\ce{2Na2O}(s)+\ce{O2}(g)\xrightarrow{Δ}\ce{2Na2O2}(s) \nonumber$
$\ce{2SrO}(s)+\ce{O2}(g)\xrightarrow{Δ}\ce{2SrO2}(s) \nonumber$
The peroxides of potassium, rubidium, and cesium can be prepared by heating the metal or its oxide in a carefully controlled amount of oxygen:
$\ce{2K}(s)+\ce{O2}(g)⟶\ce{K2O2}(s) \hspace{20px} \mathrm{(2\:mol\: K\: per\: mol\: O_2)} \nonumber$
With an excess of oxygen, the superoxides KO2, RbO2, and CsO2 form. For example:
$\ce{K}(s)+\ce{O2}(g)⟶\ce{KO2}(s) \hspace{20px} \ce{(1\: mol\: K\: per\: mol\: O_2)} \nonumber$
The stability of the peroxides and superoxides of the alkali metals increases as the size of the cation increases.
Hydroxides
Hydroxides are compounds that contain the OH ion. It is possible to prepare these compounds by two general types of reactions. Soluble metal hydroxides can be produced by the reaction of the metal or metal oxide with water. Insoluble metal hydroxides form when a solution of a soluble salt of the metal combines with a solution containing hydroxide ions.
With the exception of beryllium and magnesium, the metals of groups 1 and 2 react with water to form hydroxides and hydrogen gas. Examples of such reactions include:
$\ce{2Li}(s)+\ce{2H2O}(l)⟶\ce{2LiOH}(aq)+\ce{H2}(g) \nonumber$
$\ce{Ca}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{H2}(g) \nonumber$
However, these reactions can be violent and dangerous; therefore, it is preferable to produce soluble metal hydroxides by the reaction of the respective oxide with water:
$\ce{Li2O}(s)+\ce{H2O}(l)⟶\ce{2LiOH}(aq) \nonumber$
$\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(aq) \nonumber$
Most metal oxides are base anhydrides. This is obvious for the soluble oxides because they form metal hydroxides. Most other metal oxides are insoluble and do not form hydroxides in water; however, they are still base anhydrides because they will react with acids.
It is possible to prepare the insoluble hydroxides of beryllium, magnesium, and other representative metals by the addition of sodium hydroxide to a solution of a salt of the respective metal. The net ionic equations for the reactions involving a magnesium salt, an aluminum salt, and a zinc salt are:
$\ce{Mg^2+}(aq)+\ce{2OH-}(aq)⟶\ce{Mg(OH)2}(s) \nonumber$
$\ce{Al^3+}(aq)+\ce{3OH-}(aq)⟶\ce{Al(OH)3}(s) \nonumber$
$\ce{Zn^2+}(aq)+\ce{2OH-}(aq)⟶\ce{Zn(OH)2}(s) \nonumber$
An excess of hydroxide must be avoided when preparing aluminum, gallium, zinc, and tin(II) hydroxides, or the hydroxides will dissolve with the formation of the corresponding complex ions: $\ce{Al(OH)4-}$, $\ce{Ga(OH)4-}$, $\ce{Zn(OH)4^2-}$, and $\ce{Sn(OH)3-}$ (Figure $4$). The important aspect of complex ions for this chapter is that they form by a Lewis acid-base reaction with the metal being the Lewis acid.
Industry uses large quantities of sodium hydroxide as a cheap, strong base. Sodium chloride is the starting material for the production of NaOH because NaCl is a less expensive starting material than the oxide. Sodium hydroxide is among the top 10 chemicals in production in the United States, and this production was almost entirely by electrolysis of solutions of sodium chloride. This process is the chlor-alkali process, and it is the primary method for producing chlorine.
Sodium hydroxide is an ionic compound and melts without decomposition. It is very soluble in water, giving off a great deal of heat and forming very basic solutions: 40 grams of sodium hydroxide dissolves in only 60 grams of water at 25 °C. Sodium hydroxide is employed in the production of other sodium compounds and is used to neutralize acidic solutions during the production of other chemicals such as petrochemicals and polymers.
Many of the applications of hydroxides are for the neutralization of acids (such as the antacid shown in Figure $5$) and for the preparation of oxides by thermal decomposition. An aqueous suspension of magnesium hydroxide constitutes the antacid milk of magnesia. Because of its ready availability (from the reaction of water with calcium oxide prepared by the decomposition of limestone, CaCO3), low cost, and activity, calcium hydroxide is used extensively in commercial applications needing a cheap, strong base. The reaction of hydroxides with appropriate acids is also used to prepare salts.
The Chlor-Alkali Process
Although they are very different chemically, there is a link between chlorine and sodium hydroxide because there is an important electrochemical process that produces the two chemicals simultaneously. The process known as the chlor-alkali process, utilizes sodium chloride, which occurs in large deposits in many parts of the world. This is an electrochemical process to oxidize chloride ion to chlorine and generate sodium hydroxide.
Passing a direct current of electricity through a solution of NaCl causes the chloride ions to migrate to the positive electrode where oxidation to gaseous chlorine occurs when the ion gives up an electron to the electrode:
$\ce{2Cl-}(aq)⟶\ce{Cl2}(g)+\ce{2e-} \hspace{20px} \textrm{(at the positive electrode)} \nonumber$
The electrons produced travel through the outside electrical circuit to the negative electrode. Although the positive sodium ions migrate toward this negative electrode, metallic sodium does not form because sodium ions are too difficult to reduce under the conditions used. (Recall that metallic sodium is active enough to react with water and hence, even if produced, would immediately react with water to produce sodium ions again.) Instead, water molecules pick up electrons from the electrode and undergo reduction to form hydrogen gas and hydroxide ions:
$\ce{2H2O}(l)+\ce{2e-}\textrm{ (from the negative electrode)}⟶\ce{H2}(g)+\ce{2OH-}(aq) \nonumber$
The overall result is the conversion of the aqueous solution of NaCl to an aqueous solution of NaOH, gaseous Cl2, and gaseous H2:
$\ce{2Na+}(aq)+\ce{2Cl-}(aq)+\ce{2H2O}(l)\xrightarrow{\ce{electrolysis}}\ce{2Na+}(aq)+\ce{2OH-}(aq)+\ce{Cl2}(g)+\ce{H2}(g) \nonumber$
Nonmetal Oxygen Compounds
Most nonmetals react with oxygen to form nonmetal oxides. Depending on the available oxidation states for the element, a variety of oxides might form. Fluorine will combine with oxygen to form fluorides such as OF2, where the oxygen has a 2+-oxidation state.
Sulfur Oxygen Compounds
The two common oxides of sulfur are sulfur dioxide, SO2, and sulfur trioxide, SO3. The odor of burning sulfur comes from sulfur dioxide. Sulfur dioxide, shown in Figure $6$, occurs in volcanic gases and in the atmosphere near industrial plants that burn fuel containing sulfur compounds.
Commercial production of sulfur dioxide is from either burning sulfur or roasting sulfide ores such as ZnS, FeS2, and Cu2S in air. (Roasting, which forms the metal oxide, is the first step in the separation of many metals from their ores.) A convenient method for preparing sulfur dioxide in the laboratory is by the action of a strong acid on either sulfite salts containing the $\ce{SO3^2-}$ ion or hydrogen sulfite salts containing $\ce{HSO3-}$. Sulfurous acid, H2SO3, forms first, but quickly decomposes into sulfur dioxide and water. Sulfur dioxide also forms when many reducing agents react with hot, concentrated sulfuric acid. Sulfur trioxide forms slowly when heating sulfur dioxide and oxygen together, and the reaction is exothermic:
$\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} ΔH°=\mathrm{−197.8\:kJ} \nonumber$
Sulfur dioxide is a gas at room temperature, and the SO2 molecule is bent. Sulfur trioxide melts at 17 °C and boils at 43 °C. In the vapor state, its molecules are single SO3 units (Figure $7$), but in the solid state, SO3 exists in several polymeric forms.
The sulfur oxides react as Lewis acids with many oxides and hydroxides in Lewis acid-base reactions, with the formation of sulfites or hydrogen sulfites, and sulfates or hydrogen sulfates, respectively.
Halogen Oxygen Compounds
The halogens do not react directly with oxygen, but it is possible to prepare binary oxygen-halogen compounds by the reactions of the halogens with oxygen-containing compounds. Oxygen compounds with chlorine, bromine, and iodine are oxides because oxygen is the more electronegative element in these compounds. On the other hand, fluorine compounds with oxygen are fluorides because fluorine is the more electronegative element.
As a class, the oxides are extremely reactive and unstable, and their chemistry has little practical importance. Dichlorine oxide, formally called dichlorine monoxide, and chlorine dioxide, both shown in Figure $8$, are the only commercially important compounds. They are important as bleaching agents (for use with pulp and flour) and for water treatment.
Nonmetal Oxyacids and Their Salts
Nonmetal oxides form acids when allowed to react with water; these are acid anhydrides. The resulting oxyanions can form salts with various metal ions.
Nitrogen Oxyacids and Salts
Nitrogen pentaoxide, N2O5, and NO2 react with water to form nitric acid, HNO3. Alchemists, as early as the eighth century, knew nitric acid (shown in Figure $9$) as aqua fortis (meaning "strong water"). The acid was useful in the separation of gold from silver because it dissolves silver but not gold. Traces of nitric acid occur in the atmosphere after thunderstorms, and its salts are widely distributed in nature. There are tremendous deposits of Chile saltpeter, NaNO3, in the desert region near the boundary of Chile and Peru. Bengal saltpeter, KNO3, occurs in India and in other countries of the Far East.
In the laboratory, it is possible to produce nitric acid by heating a nitrate salt (such as sodium or potassium nitrate) with concentrated sulfuric acid:
$\ce{NaNO3}(s)+\ce{H2SO4}(l)\xrightarrow{Δ}\ce{NaHSO4}(s)+\ce{HNO3}(g) \nonumber$
The Ostwald process is the commercial method for producing nitric acid. This process involves the oxidation of ammonia to nitric oxide, NO; oxidation of nitric oxide to nitrogen dioxide, NO2; and further oxidation and hydration of nitrogen dioxide to form nitric acid:
$\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g) \nonumber$
$\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g) \nonumber$
$\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g) \nonumber$
or
$\ce{4NO2}(g)+\ce{O2}(g)+\ce{2H2O}(g)⟶\ce{4HNO3}(l) \nonumber$
Pure nitric acid is a colorless liquid. However, it is often yellow or brown in color because NO2 forms as the acid decomposes. Nitric acid is stable in aqueous solution; solutions containing 68% of the acid are commercially available concentrated nitric acid. It is both a strong oxidizing agent and a strong acid.
The action of nitric acid on a metal rarely produces H2 (by reduction of H+) in more than small amounts. Instead, the reduction of nitrogen occurs. The products formed depend on the concentration of the acid, the activity of the metal, and the temperature. Normally, a mixture of nitrates, nitrogen oxides, and various reduction products form. Less active metals such as copper, silver, and lead reduce concentrated nitric acid primarily to nitrogen dioxide. The reaction of dilute nitric acid with copper produces NO. In each case, the nitrate salts of the metals crystallize upon evaporation of the resultant solutions.
Nonmetallic elements, such as sulfur, carbon, iodine, and phosphorus, undergo oxidation by concentrated nitric acid to their oxides or oxyacids, with the formation of NO2:
$\ce{S}(s)+\ce{6HNO3}(aq)⟶\ce{H2SO4}(aq)+\ce{6NO2}(g)+\ce{2H2O}(l) \nonumber$
$\ce{C}(s)+\ce{4HNO3}(aq)⟶\ce{CO2}(g)+\ce{4NO2}(g)+\ce{2H2O}(l) \nonumber$
Nitric acid oxidizes many compounds; for example, concentrated nitric acid readily oxidizes hydrochloric acid to chlorine and chlorine dioxide. A mixture of one part concentrated nitric acid and three parts concentrated hydrochloric acid (called aqua regia, which means royal water) reacts vigorously with metals. This mixture is particularly useful in dissolving gold, platinum, and other metals that are more difficult to oxidize than hydrogen. A simplified equation to represent the action of aqua regia on gold is:
$\ce{Au}(s)+\ce{4HCl}(aq)+\ce{3HNO3}(aq)⟶\ce{HAuCl4}(aq)+\ce{3NO2}(g)+\ce{3H2O}(l) \nonumber$
Video $1$: Although gold is generally unreactive, you can watch a video of the complex mixture of compounds present in aqua regia dissolving it into solution.
Nitrates, salts of nitric acid, form when metals, oxides, hydroxides, or carbonates react with nitric acid. Most nitrates are soluble in water; indeed, one of the significant uses of nitric acid is to prepare soluble metal nitrates.
Nitric acid finds extensive use in the laboratory and in chemical industries as a strong acid and strong oxidizing agent. It is important in the manufacture of explosives, dyes, plastics, and drugs. Salts of nitric acid (nitrates) are valuable as fertilizers. Gunpowder is a mixture of potassium nitrate, sulfur, and charcoal.
The reaction of N2O3 with water gives a pale blue solution of nitrous acid, HNO2. However, HNO2 (shown in Figure $10$) is easier to prepare by the addition of an acid to a solution of nitrite; nitrous acid is a weak acid, so the nitrite ion is basic in aqueous solution:
$\ce{NO2-}(aq)+\ce{H3O+}(aq)⟶\ce{HNO2}(aq)+\ce{H2O}(l) \nonumber$
Nitrous acid is very unstable and exists only in solution. It disproportionates slowly at room temperature (rapidly when heated) into nitric acid and nitric oxide. Nitrous acid is an active oxidizing agent with strong reducing agents, and strong oxidizing agents oxidize it to nitric acid.
Sodium nitrite, NaNO2, is an additive to meats such as hot dogs and cold cuts. The nitrite ion has two functions. It limits the growth of bacteria that can cause food poisoning, and it prolongs the meat’s retention of its red color. The addition of sodium nitrite to meat products is controversial because nitrous acid reacts with certain organic compounds to form a class of compounds known as nitrosamines. Nitrosamines produce cancer in laboratory animals. This has prompted the FDA to limit the amount of NaNO2 in foods.
The nitrites are much more stable than the acid, but nitrites, like nitrates, can explode. Nitrites, like nitrates, are also soluble in water (AgNO2 is only slightly soluble).
Phosphorus Oxyacids and Salts
Pure orthophosphoric acid, H3PO4 (Figure $11$), forms colorless, deliquescent crystals that melt at 42 °C. The common name of this compound is phosphoric acid, and is commercially available as a viscous 82% solution known as syrupy phosphoric acid. One use of phosphoric acid is as an additive to many soft drinks.
One commercial method of preparing orthophosphoric acid is to treat calcium phosphate rock with concentrated sulfuric acid:
$\ce{Ca3(PO4)2}(s)+\ce{3H2SO4}(aq)⟶\ce{2H3PO4}(aq)+\ce{3CaSO4}(s) \nonumber$
Dilution of the products with water, followed by filtration to remove calcium sulfate, gives a dilute acid solution contaminated with calcium dihydrogen phosphate, Ca(H2PO4)2, and other compounds associated with calcium phosphate rock. It is possible to prepare pure orthophosphoric acid by dissolving P4O10 in water.
The action of water on P4O6, PCl3, PBr3, or PI3 forms phosphorous acid, H3PO3 (shown in Figure $12$). The best method for preparing pure phosphorous acid is by hydrolyzing phosphorus trichloride:
$\ce{PCl3}(l)+\ce{3H2O}(l)⟶\ce{H3PO3}(aq)+\ce{3HCl}(g) \nonumber$
Heating the resulting solution expels the hydrogen chloride and leads to the evaporation of water. When sufficient water evaporates, white crystals of phosphorous acid will appear upon cooling. The crystals are deliquescent, very soluble in water, and have an odor like that of garlic. The solid melts at 70.1 °C and decomposes at about 200 °C by disproportionation into phosphine and orthophosphoric acid:
$\ce{4H3PO3}(l)⟶\ce{PH3}(g)+\ce{3H3PO4}(l) \nonumber$
Phosphorous acid forms only two series of salts, which contain the dihydrogen phosphite ion, $\ce{H2PO3-}$, or the hydrogen phosphate ion, $\ce{HPO3^2-}$, respectively. It is not possible to replace the third atom of hydrogen because it is not very acidic, as it is not easy to ionize the P-H bond.
Sulfur Oxyacids and Salts
The preparation of sulfuric acid, H2SO4 (Figure $13$), begins with the oxidation of sulfur to sulfur trioxide and then converting the trioxide to sulfuric acid. Pure sulfuric acid is a colorless, oily liquid that freezes at 10.5 °C. It fumes when heated because the acid decomposes to water and sulfur trioxide. The heating process causes the loss of more sulfur trioxide than water, until reaching a concentration of 98.33% acid. Acid of this concentration boils at 338 °C without further change in concentration (a constant boiling solution) and is commercially concentrated H2SO4. The amount of sulfuric acid used in industry exceeds that of any other manufactured compound.
The strong affinity of concentrated sulfuric acid for water makes it a good dehydrating agent. It is possible to dry gases and immiscible liquids that do not react with the acid by passing them through the acid.
Sulfuric acid is a strong diprotic acid that ionizes in two stages. In aqueous solution, the first stage is essentially complete. The secondary ionization is not nearly so complete, and $\ce{HSO4-}$ is a moderately strong acid (about 25% ionized in solution of a $\ce{HSO4-}$ salt: Ka = 1.2 × 10−2).
Being a diprotic acid, sulfuric acid forms both sulfates, such as Na2SO4, and hydrogen sulfates, such as NaHSO4. Most sulfates are soluble in water; however, the sulfates of barium, strontium, calcium, and lead are only slightly soluble in water.
Among the important sulfates are Na2SO4⋅10H2O and Epsom salts, MgSO4⋅7H2O. Because the $\ce{HSO4-}$ ion is an acid, hydrogen sulfates, such as NaHSO4, exhibit acidic behavior, and this compound is the primary ingredient in some household cleansers.
Hot, concentrated sulfuric acid is an oxidizing agent. Depending on its concentration, the temperature, and the strength of the reducing agent, sulfuric acid oxidizes many compounds and, in the process, undergoes reduction to SO2, $\ce{HSO3-}$, $\ce{SO3^2-}$, S, H2S, or S2−.
Sulfur dioxide dissolves in water to form a solution of sulfurous acid, as expected for the oxide of a nonmetal. Sulfurous acid is unstable, and it is not possible to isolate anhydrous H2SO3. Heating a solution of sulfurous acid expels the sulfur dioxide. Like other diprotic acids, sulfurous acid ionizes in two steps: The hydrogen sulfite ion, $\ce{HSO3-}$, and the sulfite ion, $\ce{SO3^2-}$, form. Sulfurous acid is a moderately strong acid. Ionization is about 25% in the first stage, but it is much less in the second (Ka1 = 1.2 × 10−2 and Ka2 = 6.2 × 10−8).
In order to prepare solid sulfite and hydrogen sulfite salts, it is necessary to add a stoichiometric amount of a base to a sulfurous acid solution and then evaporate the water. These salts also form from the reaction of SO2 with oxides and hydroxides. Heating solid sodium hydrogen sulfite forms sodium sulfite, sulfur dioxide, and water:
$\ce{2NaHSO3}(s)\xrightarrow{Δ}\ce{Na2SO3}(s)+\ce{SO2}(g)+\ce{H2O}(l) \nonumber$
Strong oxidizing agents can oxidize sulfurous acid. Oxygen in the air oxidizes it slowly to the more stable sulfuric acid:
$\ce{2H2SO3}(aq)+\ce{O2}(g)+\ce{2H2O}(l)\xrightarrow{Δ}\ce{2H3O+}(aq)+\ce{2HSO4-}(aq) \nonumber$
Solutions of sulfites are also very susceptible to air oxidation to produce sulfates. Thus, solutions of sulfites always contain sulfates after exposure to air.
Halogen Oxyacids and Their Salts
The compounds HXO, HXO2, HXO3, and HXO4, where X represents Cl, Br, or I, are the hypohalous, halous, halic, and perhalic acids, respectively. The strengths of these acids increase from the hypohalous acids, which are very weak acids, to the perhalic acids, which are very strong. Table $1$ lists the known acids, and, where known, their pKa values are given in parentheses.
Table $1$: Oxyacids of the Halogens
Name Fluorine Chlorine Bromine Iodine
hypohalous HOF HOCl (7.5) HOBr (8.7) HOI (11)
halous HClO2 (2.0)
halic HClO3 HBrO3 HIO3 (0.8)
perhalic HClO4 HBrO4 HIO4 (1.6)
paraperhalic H5IO6 (1.6)
The only known oxyacid of fluorine is the very unstable hypofluorous acid, HOF, which is prepared by the reaction of gaseous fluorine with ice:
$\ce{F2}(g)+\ce{H2O}(s)⟶\ce{HOF}(g)+\ce{HF}(g) \nonumber$
The compound is very unstable and decomposes above −40 °C. This compound does not ionize in water, and there are no known salts. It is uncertain whether the name hypofluorous acid is even appropriate for HOF; a more appropriate name might be hydrogen hypofluorite.
The reactions of chlorine and bromine with water are analogous to that of fluorine with ice, but these reactions do not go to completion, and mixtures of the halogen and the respective hypohalous and hydrohalic acids result. Other than HOF, the hypohalous acids only exist in solution. The hypohalous acids are all very weak acids; however, HOCl is a stronger acid than HOBr, which, in turn, is stronger than HOI.
The addition of base to solutions of the hypohalous acids produces solutions of salts containing the basic hypohalite ions, OX. It is possible to isolate these salts as solids. All of the hypohalites are unstable with respect to disproportionation in solution, but the reaction is slow for hypochlorite. Hypobromite and hypoiodite disproportionate rapidly, even in the cold:
$\ce{3XO-}(aq)⟶\ce{2X-}(aq)+\ce{XO3-}(aq) \nonumber$
Sodium hypochlorite is an inexpensive bleach (Clorox) and germicide. The commercial preparation involves the electrolysis of cold, dilute, aqueous sodium chloride solutions under conditions where the resulting chlorine and hydroxide ion can react. The net reaction is:
$\ce{Cl-}(aq)+\ce{H2O}(l)\xrightarrow{\ce{electrical\: energy}}\ce{ClO-}(aq)+\ce{H2}(g) \nonumber$
The only definitely known halous acid is chlorous acid, HClO2, obtained by the reaction of barium chlorite with dilute sulfuric acid:
$\ce{Ba(ClO2)2}(aq)+\ce{H2SO4}(aq)⟶\ce{BaSO4}(s)+\ce{2HClO2}(aq) \nonumber$
Filtering the insoluble barium sulfate leaves a solution of HClO2. Chlorous acid is not stable; it slowly decomposes in solution to yield chlorine dioxide, hydrochloric acid, and water. Chlorous acid reacts with bases to give salts containing the chlorite ion (shown in Figure $14$). Sodium chlorite finds an extensive application in the bleaching of paper because it is a strong oxidizing agent and does not damage the paper.
Chloric acid, HClO3, and bromic acid, HBrO3, are stable only in solution. The reaction of iodine with concentrated nitric acid produces stable white iodic acid, HIO3:
$\ce{I2}(s)+\ce{10HNO3}(aq)⟶\ce{2HIO3}(s)+\ce{10NO2}(g)+\ce{4H2O}(l) \nonumber$
It is possible to obtain the lighter halic acids from their barium salts by reaction with dilute sulfuric acid. The reaction is analogous to that used to prepare chlorous acid. All of the halic acids are strong acids and very active oxidizing agents. The acids react with bases to form salts containing chlorate ions (shown in Figure $15$). Another preparative method is the electrochemical oxidation of a hot solution of a metal halide to form the appropriate metal chlorates. Sodium chlorate is a weed killer; potassium chlorate is used as an oxidizing agent.
Perchloric acid, HClO4, forms when treating a perchlorate, such as potassium perchlorate, with sulfuric acid under reduced pressure. The HClO4 can be distilled from the mixture:
$\ce{KClO4}(s)+\ce{H2SO4}(aq)⟶\ce{HClO4}(g)+\ce{KHSO4}(s) \nonumber$
Dilute aqueous solutions of perchloric acid are quite stable thermally, but concentrations above 60% are unstable and dangerous. Perchloric acid and its salts are powerful oxidizing agents, as the very electronegative chlorine is more stable in a lower oxidation state than 7+. Serious explosions have occurred when heating concentrated solutions with easily oxidized substances. However, its reactions as an oxidizing agent are slow when perchloric acid is cold and dilute. The acid is among the strongest of all acids. Most salts containing the perchlorate ion (Figure $16$) are soluble. It is possible to prepare them from reactions of bases with perchloric acid and, commercially, by the electrolysis of hot solutions of their chlorides.
Perbromate salts are difficult to prepare, and the best syntheses currently involve the oxidation of bromates in basic solution with fluorine gas followed by acidification. There are few, if any, commercial uses of this acid or its salts.
There are several different acids containing iodine in the 7+-oxidation state; they include metaperiodic acid, HIO4, and paraperiodic acid, H5IO6. These acids are strong oxidizing agents and react with bases to form the appropriate salts.
Summary
Oxygen is one of the most reactive elements. This reactivity, coupled with its abundance, makes the chemistry of oxygen very rich and well understood. Compounds of the representative metals with oxygen exist in three categories (1) oxides, (2) peroxides and superoxides, and (3) hydroxides. Heating the corresponding hydroxides, nitrates, or carbonates is the most common method for producing oxides. Heating the metal or metal oxide in oxygen may lead to the formation of peroxides and superoxides. The soluble oxides dissolve in water to form solutions of hydroxides. Most metals oxides are base anhydrides and react with acids. The hydroxides of the representative metals react with acids in acid-base reactions to form salts and water. The hydroxides have many commercial uses.
All nonmetals except fluorine form multiple oxides. Nearly all of the nonmetal oxides are acid anhydrides. The acidity of oxyacids requires that the hydrogen atoms bond to the oxygen atoms in the molecule rather than to the other nonmetal atom. Generally, the strength of the oxyacid increases with the number of oxygen atoms bonded to the nonmetal atom and not to a hydrogen.
Glossary
base anhydride
metal oxide that behaves as a base towards acids
chlor-alkali process
electrolysis process for the synthesis of chlorine and sodium hydroxide
hydrogen sulfate
$\ce{HSO4-}$ ion
hydrogen sulfite
$\ce{HSO3-}$ ion
hydroxide
compound of a metal with the hydroxide ion OH or the group −OH
nitrate
$\ce{NO3-}$ ion; salt of nitric acid
Ostwald process
industrial process used to convert ammonia into nitric acid
oxide
binary compound of oxygen with another element or group, typically containing O2− ions or the group –O– or =O
ozone
allotrope of oxygen; O3
peroxide
molecule containing two oxygen atoms bonded together or as the anion, $\ce{O2^2-}$
photosynthesis
process whereby light energy promotes the reaction of water and carbon dioxide to form carbohydrates and oxygen; this allows photosynthetic organisms to store energy
sulfate
$\ce{SO4^2-}$ ion
sulfite
$\ce{SO3^2-}$ ion
superoxide
oxide containing the anion $\ce{O2-}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.09%3A_Occurrence_Preparation_and_Compounds_of_Oxygen.txt |
Learning Objectives
• Describe the properties, preparation, and uses of sulfur
Sulfur exists in nature as elemental deposits as well as sulfides of iron, zinc, lead, and copper, and sulfates of sodium, calcium, barium, and magnesium. Hydrogen sulfide is often a component of natural gas and occurs in many volcanic gases, like those shown in Figure $1$. Sulfur is a constituent of many proteins and is essential for life.
The Frasch process, illustrated in Figure $2$, is important in the mining of free sulfur from enormous underground deposits in Texas and Louisiana. Superheated water (170 °C and 10 atm pressure) is forced down the outermost of three concentric pipes to the underground deposit. The hot water melts the sulfur. The innermost pipe conducts compressed air into the liquid sulfur. The air forces the liquid sulfur, mixed with air, to flow up through the outlet pipe. Transferring the mixture to large settling vats allows the solid sulfur to separate upon cooling. This sulfur is 99.5% to 99.9% pure and requires no purification for most uses.
Larger amounts of sulfur also come from hydrogen sulfide recovered during the purification of natural gas.
Sulfur exists in several allotropic forms. The stable form at room temperature contains eight-membered rings, and so the true formula is S8. However, chemists commonly use S to simplify the coefficients in chemical equations; we will follow this practice in this book.
Like oxygen, which is also a member of group 16, sulfur exhibits a distinctly nonmetallic behavior. It oxidizes metals, giving a variety of binary sulfides in which sulfur exhibits a negative oxidation state (2−). Elemental sulfur oxidizes less electronegative nonmetals, and more electronegative nonmetals, such as oxygen and the halogens, will oxidize it. Other strong oxidizing agents also oxidize sulfur. For example, concentrated nitric acid oxidizes sulfur to the sulfate ion, with the concurrent formation of nitrogen(IV) oxide:
$\ce{S}(s)+\ce{6HNO3}(aq)⟶\ce{2H3O+}(aq)+\ce{SO4^2-}(aq)+\ce{6NO2}(g) \nonumber$
The chemistry of sulfur with an oxidation state of 2− is similar to that of oxygen. Unlike oxygen, however, sulfur forms many compounds in which it exhibits positive oxidation states.
Summary
Sulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, S2−, in which it has as oxidation state of 2−. Sulfur reacts with most nonmetals.
Glossary
Frasch process
important in the mining of free sulfur from enormous underground deposits
18.11: Occurrence Preparation and Properties of Halogens
Learning Objectives
• Describe the preparation, properties, and uses of halogens
• Describe the properties, preparation, and uses of halogen compounds
The elements in group 17 are the halogens. These are the elements fluorine, chlorine, bromine, iodine, and astatine. These elements are too reactive to occur freely in nature, but their compounds are widely distributed. Chlorides are the most abundant; although fluorides, bromides, and iodides are less common, they are reasonably available. In this section, we will examine the occurrence, preparation, and properties of halogens. Next, we will examine halogen compounds with the representative metals followed by an examination of the interhalogens. This section will conclude with some applications of halogens.
Occurrence and Preparation
All of the halogens occur in seawater as halide ions. The concentration of the chloride ion is 0.54 M; that of the other halides is less than 10–4 M. Fluoride also occurs in minerals such as CaF2, Ca(PO4)3F, and Na3AlF6. Chloride also occurs in the Great Salt Lake and the Dead Sea, and in extensive salt beds that contain NaCl, KCl, or MgCl2. Part of the chlorine in your body is present as hydrochloric acid, which is a component of stomach acid. Bromine compounds occur in the Dead Sea and underground brines. Iodine compounds are found in small quantities in Chile saltpeter, underground brines, and sea kelp. Iodine is essential to the function of the thyroid gland.
The best sources of halogens (except iodine) are halide salts. It is possible to oxidize the halide ions to free diatomic halogen molecules by various methods, depending on the ease of oxidation of the halide ion. Fluoride is the most difficult to oxidize, whereas iodide is the easiest.
The major method for preparing fluorine is electrolytic oxidation. The most common electrolysis procedure is to use a molten mixture of potassium hydrogen fluoride, KHF2, and anhydrous hydrogen fluoride. Electrolysis causes HF to decompose, forming fluorine gas at the anode and hydrogen at the cathode. It is necessary to keep the two gases separated to prevent their explosive recombination to reform hydrogen fluoride.
Most commercial chlorine comes from the electrolysis of the chloride ion in aqueous solutions of sodium chloride; this is the chlor-alkali process discussed previously. Chlorine is also a product of the electrolytic production of metals such as sodium, calcium, and magnesium from their fused chlorides. It is also possible to prepare chlorine by the chemical oxidation of the chloride ion in acid solution with strong oxidizing agents such as manganese dioxide ($\ce{MnO2}$) or sodium dichromate ($\ce{Na2Cr2O7}$). The reaction with manganese dioxide is:
$\ce{MnO}_{2(s)}+\ce{2Cl^-}_{(aq)}+\ce{4H_3O^+}_{(aq)}⟶\ce{Mn}^{2+}_{(aq)}+\ce{Cl}_{2(g)}+\ce{6H_2O}_{(l)}\nonumber$
The commercial preparation of bromine involves the oxidation of bromide ion by chlorine:
$\ce{2Br^-}_{(aq)}+\ce{Cl}_{2(g)}⟶\ce{Br}_{2(l)}+\ce{2Cl^-}_{(aq)} \nonumber$
Chlorine is a stronger oxidizing agent than bromine. This method is important for the production of essentially all domestic bromine.
Some iodine comes from the oxidation of iodine chloride, $\ce{ICl}$, or iodic acid, $\ce{HlO_3}$. The commercial preparation of iodine utilizes the reduction of sodium iodate, $\ce{NaIO_3}$, an impurity in deposits of Chile saltpeter, with sodium hydrogen sulfite:
$\ce{2IO^-}_{3(aq)}+\ce{5HSO^-}_{3(aq)}⟶\ce{3HSO^-}_{4(aq)}+\ce{2SO^{2-}}_{4(aq)}+\ce{H_2O}_{(l)}+\ce{I}_{2(s)} \nonumber$
Properties of the Halogens
Fluorine is a pale yellow gas, chlorine is a greenish-yellow gas, bromine is a deep reddish-brown liquid, and iodine is a grayish-black crystalline solid. Liquid bromine has a high vapor pressure, and the reddish vapor is readily visible in Figure $1$. Iodine crystals have a noticeable vapor pressure. When gently heated, these crystals sublime and form a beautiful deep violet vapor.
Bromine is only slightly soluble in water, but it is miscible in all proportions in less polar (or nonpolar) solvents such as chloroform, carbon tetrachloride, and carbon disulfide, forming solutions that vary from yellow to reddish-brown, depending on the concentration.
Iodine is soluble in chloroform, carbon tetrachloride, carbon disulfide, and many hydrocarbons, giving violet solutions of I2 molecules. Iodine dissolves only slightly in water, giving brown solutions. It is quite soluble in aqueous solutions of iodides, with which it forms brown solutions. These brown solutions result because iodine molecules have empty valence d orbitals and can act as weak Lewis acids towards the iodide ion. The equation for the reversible reaction of iodine (Lewis acid) with the iodide ion (Lewis base) to form triiodide ion, $\ce{I3-}$, is:
$\ce{I}_{2(s)}+\ce{I^-}_{(aq)}⟶\ce{I^-}_{3(aq)} \label{4}$
The easier it is to oxidize the halide ion, the more difficult it is for the halogen to act as an oxidizing agent. Fluorine generally oxidizes an element to its highest oxidation state, whereas the heavier halogens may not. For example, when excess fluorine reacts with sulfur, $\ce{SF_6}$ forms. Chlorine gives $\ce{SCl_2}$ and bromine, $\ce{S_2Br_2}$. Iodine does not react with sulfur.
Fluorine is the most powerful oxidizing agent of the known elements. It spontaneously oxidizes most other elements; therefore, the reverse reaction, the oxidation of fluorides, is very difficult to accomplish. Fluorine reacts directly and forms binary fluorides with all of the elements except the lighter noble gases (He, Ne, and Ar). Fluorine is such a strong oxidizing agent that many substances ignite on contact with it. Drops of water inflame in fluorine and form $\ce{O_2}$, $\ce{OF_2}$, $\ce{H_2O_2}$, $\ce{O_3}$, and $\ce{HF}$. Wood and asbestos ignite and burn in fluorine gas. Most hot metals burn vigorously in fluorine. However, it is possible to handle fluorine in copper, iron, or nickel containers because an adherent film of the fluoride salt passivates their surfaces. Fluorine is the only element that reacts directly with the noble gas xenon.
Although it is a strong oxidizing agent, chlorine is less active than fluorine. Mixing chlorine and hydrogen in the dark makes the reaction between them to be imperceptibly slow. Exposure of the mixture to light causes the two to react explosively. Chlorine is also less active towards metals than fluorine, and oxidation reactions usually require higher temperatures. Molten sodium ignites in chlorine. Chlorine attacks most nonmetals (C, N2, and O2 are notable exceptions), forming covalent molecular compounds. Chlorine generally reacts with compounds that contain only carbon and hydrogen (hydrocarbons) by adding to multiple bonds or by substitution.
In cold water, chlorine undergoes a disproportionation reaction:
$\ce{Cl}_{2(aq)}+\ce{2H_2O}_{(l)}⟶\ce{HOCl}_{(aq)}+\ce{H_3O^+}_{(aq)}+\ce{Cl^-}_{(aq)} \nonumber$
Half the chlorine atoms oxidize to the 1+ oxidation state (hypochlorous acid), and the other half reduce to the 1− oxidation state (chloride ion). This disproportionation is incomplete, so chlorine water is an equilibrium mixture of chlorine molecules, hypochlorous acid molecules, hydronium ions, and chloride ions. When exposed to light, this solution undergoes a photochemical decomposition:
$\ce{2HOCl}_{(aq)}+\ce{2H_2O}_{(l)}\xrightarrow{\ce{sunlight}} \ce{2H_3O^+}_{(aq)}+\ce{2Cl^-}_{(aq)}+\ce{O}_{2(g)} \nonumber$
The nonmetal chlorine is more electronegative than any other element except fluorine, oxygen, and nitrogen. In general, very electronegative elements are good oxidizing agents; therefore, we would expect elemental chlorine to oxidize all of the other elements except for these three (and the nonreactive noble gases). Its oxidizing property, in fact, is responsible for its principal use. For example, phosphorus(V) chloride, an important intermediate in the preparation of insecticides and chemical weapons, is manufactured by oxidizing the phosphorus with chlorine:
$\ce{P}_{4(s)}+\ce{10Cl}_{2(g)}⟶\ce{4PCl}_{5(l)} \nonumber$
A great deal of chlorine is also used to oxidize, and thus to destroy, organic or biological materials in water purification and in bleaching.
The chemical properties of bromine are similar to those of chlorine, although bromine is the weaker oxidizing agent and its reactivity is less than that of chlorine.
Iodine is the least reactive of the halogens. It is the weakest oxidizing agent, and the iodide ion is the most easily oxidized halide ion. Iodine reacts with metals, but heating is often required. It does not oxidize other halide ions.
Compared with the other halogens, iodine reacts only slightly with water. Traces of iodine in water react with a mixture of starch and iodide ion, forming a deep blue color. This reaction is a very sensitive test for the presence of iodine in water.
Halides of the Representative Metals
Thousands of salts of the representative metals have been prepared. The binary halides are an important subclass of salts. A salt is an ionic compound composed of cations and anions, other than hydroxide or oxide ions. In general, it is possible to prepare these salts from the metals or from oxides, hydroxides, or carbonates. We will illustrate the general types of reactions for preparing salts through reactions used to prepare binary halides.
The binary compounds of a metal with the halogens are the halides. Most binary halides are ionic. However, mercury, the elements of group 13 with oxidation states of 3+, tin(IV), and lead(IV) form covalent binary halides. The direct reaction of a metal and a halogen produce the halide of the metal. Examples of these oxidation-reduction reactions include:
$\ce{Cd}_{(s)}+\ce{Cl}_{2(g)}⟶\ce{CdCl}_{2(s)} \nonumber$
$\ce{2Ga}_{(l)}+\ce{3Br}_{2(l)}⟶\ce{2GaBr}_{3(s)} \nonumber$
Video $1$: Reactions of the alkali metals with elemental halogens are very exothermic and often quite violent. Under controlled conditions, they provide exciting demonstrations for budding students of chemistry. You can view the initial heating of the sodium that removes the coating of sodium hydroxide, sodium peroxide, and residual mineral oil to expose the reactive surface. The reaction with chlorine gas then proceeds very nicely.
If a metal can exhibit two oxidation states, it may be necessary to control the stoichiometry in order to obtain the halide with the lower oxidation state. For example, preparation of tin(II) chloride requires a 1:1 ratio of Sn to $\ce{Cl_2}$, whereas preparation of tin(IV) chloride requires a 1:2 ratio:
$\ce{Sn}_{(s)}+\ce{Cl}_{2(g)}⟶\ce{SnCl}_{2(s)} \nonumber$
$\ce{Sn}_{(s)}+\ce{2Cl}_{2(g)}⟶\ce{SnCl}_{4(l)}\nonumber$
The active representative metals—those that are easier to oxidize than hydrogen—react with gaseous hydrogen halides to produce metal halides and hydrogen. The reaction of zinc with hydrogen fluoride is:
$\ce{Zn}_{(s)}+\ce{2HF}_{(g)}⟶\ce{ZnF}_{2(s)}+\ce{H}_{2(g)} \nonumber$
The active representative metals also react with solutions of hydrogen halides to form hydrogen and solutions of the corresponding halides. Examples of such reactions include:
$\ce{Cd}_{(s)}+\ce{2HBr}_{(aq)}⟶\ce{CdBr}_{2(aq)}+\ce{H}_{2(g)} \nonumber$
$\ce{Sn}_{(s)}+\ce{2HI}_{(aq)}⟶\ce{SnI}_{2(aq)}+\ce{H}_{2(g)} \nonumber$
Hydroxides, carbonates, and some oxides react with solutions of the hydrogen halides to form solutions of halide salts. It is possible to prepare additional salts by the reaction of these hydroxides, carbonates, and oxides with aqueous solution of other acids:
$\ce{CaCo}_{3(s)}+\ce{2HCl}_{(aq)}⟶\ce{CaCl}_{2(aq)}+\ce{CO}_{2(g)}+\ce{H_2O}_{(l)} \nonumber$
$\ce{TlOH}_{(aq)}+\ce{HF}_{(aq)}⟶\ce{TlF}_{(aq)}+\ce{H_2O}_{(l)} \nonumber$
A few halides and many of the other salts of the representative metals are insoluble. It is possible to prepare these soluble salts by metathesis reactions that occur when solutions of soluble salts are mixed (Figure $2$). Metathesis reactions are examined in the chapter on the stoichiometry of chemical reactions.
Several halides occur in large quantities in nature. The ocean and underground brines contain many halides. For example, magnesium chloride in the ocean is the source of magnesium ions used in the production of magnesium. Large underground deposits of sodium chloride, like the salt mine shown in Figure $3$, occur in many parts of the world. These deposits serve as the source of sodium and chlorine in almost all other compounds containing these elements. The chlor-alkali process is one example.
Interhalogens
Compounds formed from two or more different halogens are interhalogens. Interhalogen molecules consist of one atom of the heavier halogen bonded by single bonds to an odd number of atoms of the lighter halogen. The structures of IF3, IF5, and IF7 are illustrated in Figure $4$ Formulas for other interhalogens, each of which comes from the reaction of the respective halogens, are in Table $1$.
Note from Table $1$ that fluorine is able to oxidize iodine to its maximum oxidation state, 7+, whereas bromine and chlorine, which are more difficult to oxidize, achieve only the 5+-oxidation state. A 7+-oxidation state is the limit for the halogens. Because smaller halogens are grouped about a larger one, the maximum number of smaller atoms possible increases as the radius of the larger atom increases. Many of these compounds are unstable, and most are extremely reactive. The interhalogens react like their component halides; halogen fluorides, for example, are stronger oxidizing agents than are halogen chlorides.
Table $1$: Interhalogens
YX YX3 YX5 YX7
BrCl(g)
BrF(g) BrF3(l) BrF5(l)
ClF(g) ClF3(g) ClF5(g)
IBr(s)
ICl(l) ICl3(s)
IF(s) IF3(s) IF5(l) IF7(g)
The ionic polyhalides of the alkali metals, such as KI3, KICl2, KICl4, CsIBr2, and CsBrCl2, which contain an anion composed of at least three halogen atoms, are closely related to the interhalogens. As seen previously, the formation of the polyhalide anion $\ce{I_3^-}$ is responsible for the solubility of iodine in aqueous solutions containing an iodide ion.
Applications
The fluoride ion and fluorine compounds have many important uses. Compounds of carbon, hydrogen, and fluorine are replacing Freons (compounds of carbon, chlorine, and fluorine) as refrigerants. Teflon is a polymer composed of –CF2CF2– units. Fluoride ion is added to water supplies and to some toothpastes as SnF2 or NaF to fight tooth decay. Fluoride partially converts teeth from Ca5(PO4)3(OH) into Ca5(PO4)3F.
Chlorine is important to bleach wood pulp and cotton cloth. The chlorine reacts with water to form hypochlorous acid, which oxidizes colored substances to colorless ones. Large quantities of chlorine are important in chlorinating hydrocarbons (replacing hydrogen with chlorine) to produce compounds such as tetrachloride (CCl4), chloroform (CHCl3), and ethyl chloride (C2H5Cl), and in the production of polyvinyl chloride (PVC) and other polymers. Chlorine is also important to kill the bacteria in community water supplies.
Bromine is important in the production of certain dyes, and sodium and potassium bromides are used as sedatives. At one time, light-sensitive silver bromide was a component of photographic film.
Iodine in alcohol solution with potassium iodide is an antiseptic (tincture of iodine). Iodide salts are essential for the proper functioning of the thyroid gland; an iodine deficiency may lead to the development of a goiter. Iodized table salt contains 0.023% potassium iodide. Silver iodide is useful in the seeding of clouds to induce rain; it was important in the production of photographic film and iodoform, CHI3, is an antiseptic.
Summary
The halogens form halides with less electronegative elements. Halides of the metals vary from ionic to covalent; halides of nonmetals are covalent. Interhalogens form by the combination of two or more different halogens. All of the representative metals react directly with elemental halogens or with solutions of the hydrohalic acids (HF, HCl, HBr, and HI) to produce representative metal halides. Other laboratory preparations involve the addition of aqueous hydrohalic acids to compounds that contain such basic anions, such as hydroxides, oxides, or carbonates. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.10%3A_Occurrence_Preparation_and_Properties_of_Sulfur.txt |
Learning Objectives
• Describe the properties, preparation, and uses of the noble gases
The elements in group 18 are the noble gases (helium, neon, argon, krypton, xenon, and radon). They earned the name “noble” because they were assumed to be nonreactive since they have filled valence shells. In 1962, Dr. Neil Bartlett at the University of British Columbia proved this assumption to be false.
These elements are present in the atmosphere in small amounts. Some natural gas contains 1–2% helium by mass. Helium is isolated from natural gas by liquefying the condensable components, leaving only helium as a gas. The United States possesses most of the world’s commercial supply of this element in its helium-bearing gas fields. Argon, neon, krypton, and xenon come from the fractional distillation of liquid air. Radon comes from other radioactive elements. More recently, it was observed that this radioactive gas is present in very small amounts in soils and minerals. Its accumulation in well-insulated, tightly sealed buildings, however, constitutes a health hazard, primarily lung cancer.
The boiling points and melting points of the noble gases are extremely low relative to those of other substances of comparable atomic or molecular masses. This is because only weak London dispersion forces are present, and these forces can hold the atoms together only when molecular motion is very slight, as it is at very low temperatures. Helium is the only substance known that does not solidify on cooling at normal pressure. It remains liquid close to absolute zero (0.001 K) at ordinary pressures, but it solidifies under elevated pressure.
Helium is used for filling balloons and lighter-than-air craft because it does not burn, making it safer to use than hydrogen. Helium at high pressures is not a narcotic like nitrogen. Thus, mixtures of oxygen and helium are important for divers working under high pressures. Using a helium-oxygen mixture avoids the disoriented mental state known as nitrogen narcosis, the so-called rapture of the deep. Helium is important as an inert atmosphere for the melting and welding of easily oxidizable metals and for many chemical processes that are sensitive to air.
Liquid helium (boiling point, 4.2 K) is an important coolant to reach the low temperatures necessary for cryogenic research, and it is essential for achieving the low temperatures necessary to produce superconduction in traditional superconducting materials used in powerful magnets and other devices. This cooling ability is necessary for the magnets used for magnetic resonance imaging, a common medical diagnostic procedure. The other common coolant is liquid nitrogen (boiling point, 77 K), which is significantly cheaper.
Neon is a component of neon lamps and signs. Passing an electric spark through a tube containing neon at low pressure generates the familiar red glow of neon. It is possible to change the color of the light by mixing argon or mercury vapor with the neon or by utilizing glass tubes of a special color.
Argon was useful in the manufacture of gas-filled electric light bulbs, where its lower heat conductivity and chemical inertness made it preferable to nitrogen for inhibiting the vaporization of the tungsten filament and prolonging the life of the bulb. Fluorescent tubes commonly contain a mixture of argon and mercury vapor. Argon is the third most abundant gas in dry air.
Krypton-xenon flash tubes are used to take high-speed photographs. An electric discharge through such a tube gives a very intense light that lasts only $\dfrac{1}{50,000}$ of a second. Krypton forms a difluoride, KrF2, which is thermally unstable at room temperature.
Stable compounds of xenon form when xenon reacts with fluorine. Xenon difluoride, XeF2, forms after heating an excess of xenon gas with fluorine gas and then cooling. The material forms colorless crystals, which are stable at room temperature in a dry atmosphere. Xenon tetrafluoride, XeF4, (Figure $1$) and xenon hexafluoride, XeF6, are prepared in an analogous manner, with a stoichiometric amount of fluorine and an excess of fluorine, respectively. Compounds with oxygen are prepared by replacing fluorine atoms in the xenon fluorides with oxygen.
When XeF6 reacts with water, a solution of XeO3 results and the xenon remains in the 6+-oxidation state:
$\ce{XeF6}(s)+\ce{3H2O}(l)⟶\ce{XeO3}(aq)+\ce{6HF}(aq) \nonumber$
Dry, solid xenon trioxide, XeO3, is extremely explosive—it will spontaneously detonate. Both XeF6 and XeO3 disproportionate in basic solution, producing xenon, oxygen, and salts of the perxenate ion, $\ce{XeO6^4-}$, in which xenon reaches its maximum oxidation sate of 8+.
Radon apparently forms RnF2—evidence of this compound comes from radiochemical tracer techniques.
Unstable compounds of argon form at low temperatures, but stable compounds of helium and neon are not known.
Summary
The most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds.
Glossary
halide
compound containing an anion of a group 17 element in the 1− oxidation state (fluoride, F; chloride, Cl; bromide, Br; and iodide, I)
interhalogen
compound formed from two or more different halogens | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.12%3A_Occurrence_Preparation_and_Properties_of_the_Noble_Gases.txt |
18.1: Periodicity
How do alkali metals differ from alkaline earth metals in atomic structure and general properties?
The alkali metals all have a single s electron in their outermost shell. In contrast, the alkaline earth metals have a completed s subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period.
Why does the reactivity of the alkali metals decrease from cesium to lithium?
Predict the formulas for the nine compounds that may form when each species in column 1 of Table reacts with each species in column 2.
1 2
Na I
Sr Se
Al O
$\ce{Na + I2 ⟶ 2NaI\ 2Na + Se ⟶ Na2Se\ 2Na + O2 ⟶ Na2O2}$
$\ce{Sr + I2⟶SrI2\ Sr + Se⟶SeSe\ 2Sr + O2⟶2SrO}$
$\ce{2Al + 3I2⟶2AlI3\ 2Al + 3Se⟶Al2Se3\ 4Al + 3O2⟶2Al2O3}$
Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples.
1. (a) the most metallic of the elements Al, Be, and Ba
2. (b) the most covalent of the compounds NaCl, CaCl2, and BeCl2
3. (c) the lowest first ionization energy among the elements Rb, K, and Li
4. (d) the smallest among Al, Al+, and Al3+
5. (e) the largest among Cs+, Ba2+, and Xe
Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other?
The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of $\mathrm{\dfrac{35.7\: g}{100\: mL}}$ compared with $\mathrm{\dfrac{53.8\: g}{100\: mL}}$ for SrCl2. Heating to 100 °C provides an easy test, since the solubility of NaCl is $\mathrm{\dfrac{39.12\: g}{100\: mL}}$, but that of SrCl2 is $\mathrm{\dfrac{100.8\: g}{100\: mL}}$. Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl2) that this method would be viable and perhaps the easiest and least expensive test to perform.
The reaction of quicklime, CaO, with water produces slaked lime, Ca(OH)2, which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic:
$\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s) \hspace{20px} ΔH=\mathrm{−350\: kJ\:mol^{−1}}$
1. (a) What is the enthalpy of reaction per gram of quicklime that reacts?
2. (b) How much heat, in kilojoules, is associated with the production of 1 ton of slaked lime?
Write a balanced equation for the reaction of elemental strontium with each of the following:
1. (a) oxygen
2. (b) hydrogen bromide
3. (c) hydrogen
4. (d) phosphorus
5. (e) water
(a) $\ce{2Sr}(s)+\ce{O2}(g)⟶\ce{2SrO}(s)$; (b) $\ce{Sr}(s)+\ce{2HBr}(g)⟶\ce{SrBr2}(s)+\ce{H2}(g)$; (c) $\ce{Sr}(s)+\ce{H2}(g)⟶\ce{SrH2}(s)$; (d) $\ce{6Sr}(s)+\ce{P4}(s)⟶\ce{2Sr3P2}(s)$; (e) $\ce{Sr}(s)+\ce{2H2O}(l)⟶\ce{Sr(OH)2}(aq)+\ce{H2}(g)$
How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate?
What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl2.) How many pounds of fish is this?
11 lb
The elements sodium, aluminum, and chlorine are in the same period.
1. (a) Which has the greatest electronegativity?
2. (b) Which of the atoms is smallest?
3. (c) Write the Lewis structure for the simplest covalent compound that can form between aluminum and chlorine.
4. (d) Will the oxide of each element be acidic, basic, or amphoteric?
Does metallic tin react with HCl?
Yes, tin reacts with hydrochloric acid to produce hydrogen gas.
What is tin pest, also known as tin disease?
Compare the nature of the bonds in PbCl2 to that of the bonds in PbCl4.
In PbCl2, the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl4, the bonding is covalent, as evidenced by it being an unstable liquid at room temperature.
Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare?
18.2: Occurrence and Preparation of the Representative Metals
Write an equation for the reduction of cesium chloride by elemental calcium at high temperature.
$\ce{2CsCl}(l)+\ce{Ca}(g)\:\mathrm{\overset{countercurrent \ fractionating \ tower}{\xrightarrow{\hspace{40px}}}}\:\ce{2Cs}(g)+\ce{CaCl2}(l)$
Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal?
Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples.
Cathode (reduction): $\ce{2Li+} + \ce{2e-}⟶\ce{2Li}(l)$; Anode (oxidation): $\ce{2Cl-}⟶\ce{Cl2}(g)+\ce{2e-}$; Overall reaction: $\ce{2Li+}+\ce{2Cl-}⟶\ce{2Li}(l)+\ce{Cl2}(g)$
The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine.
Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples.
What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water?
0.5035 g H2
How many grams of oxygen gas are necessary to react completely with 3.01 × 1021 atoms of magnesium to yield magnesium oxide?
Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction?
Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning.
Why is it possible for an active metal like aluminum to be useful as a structural metal?
Describe the production of metallic aluminum by electrolytic reduction.
Extract from ore: $\ce{AlO(OH)}(s)+\ce{NaOH}(aq)+\ce{H2O}(l)⟶\ce{Na[Al(OH)4]}(aq)$
Recover: $\ce{2Na[Al(OH)4]}(s)+\ce{H2SO4}(aq)⟶\ce{2Al(OH)3}(s)+\ce{Na2SO4}(aq)+\ce{2H2O}(l)$
Sinter: $\ce{2Al(OH)3}(s)⟶\ce{Al2O3}(s)+\ce{3H2O}(g)$
Dissolve in Na3AlF6(l) and electrolyze: $\ce{Al^3+}+\ce{3e-}⟶\ce{Al}(s)$
What is the common ore of tin and how is tin separated from it?
A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H2SnO3, is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample?
25.83%
Consider the production of 100 kg of sodium metal using a current of 50,000 A, assuming a 100% yield.
(a) How long will it take to produce the 100 kg of sodium metal?
(b) What volume of chlorine at 25 °C and 1.00 atm forms?
What mass of magnesium forms when 100,000 A is passed through a MgCl2 melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield?
39 kg
18.3: Structure and General Properties of the Metalloids
Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples.
1. (a) GeH4
2. (b) SbF3
3. (c) Te(OH)6
4. (d) H2Te
5. (e) GeF2
6. (f) TeCl4
7. (g) $\ce{SiF6^2-}$
8. (h) SbCl5
9. (i) TeF6
Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding.
1. (a) H3BPH3
2. (b) $\ce{BF4-}$
3. (c) BBr3
4. (d) B(CH3)3
5. (e) B(OH)3
(a) H3BPH3:
;
(b) $\ce{BF4-}$:
;
(c) BBr3:
;
(d) B(CH3)3:
;
(e) B(OH)3:
Describe the hybridization of boron and the molecular structure about the boron in each of the following:
1. (a) H3BPH3
2. (b) $\ce{BF4-}$
3. (c) BBr3
4. (d) B(CH3)3
5. (e) B(OH)3
Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure.
1s22s22p63s23p23d0.
Write a Lewis structure for each of the following molecules and ions:
1. (a) (CH3)3SiH
2. (b) $\ce{SiO4^4-}$
3. (c) Si2H6
4. (d) Si(OH)4
5. (e) $\ce{SiF6^2-}$
Describe the hybridization of silicon and the molecular structure of the following molecules and ions:
1. (a) (CH3)3SiH
2. (b) $\ce{SiO4^4-}$
3. (c) Si2H6
4. (d) Si(OH)4
5. (e) $\ce{SiF6^2-}$
(a) (CH3)3SiH: sp3 bonding about Si; the structure is tetrahedral; (b) $\ce{SiO4^4-}$: sp3 bonding about Si; the structure is tetrahedral; (c) Si2H6: sp3 bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH)4: sp3 bonding about Si; the structure is tetrahedral; (e) $\ce{SiF6^2-}$: sp3d2 bonding about Si; the structure is octahedral
Describe the hybridization and the bonding of a silicon atom in elemental silicon.
Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding.
(a) SiH4
(b) Si2H6
(c) SiCl3H
(d) SiF4
(e) SiCl2F2
(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar
Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide.
Name each of the following compounds:
1. (a) TeO2
2. (b) Sb2S3
3. (c) GeF4
4. (d) SiH4
5. (e) GeH4
(a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride
Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature):
1. (a) F2
2. (b) O2
3. (c) S
4. (d) Se
5. (e) Br2
Why is boron limited to a maximum coordination number of four in its compounds?
Boron has only s and p orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no d orbitals are available in boron.
Write a formula for each of the following compounds:
1. (a) silicon dioxide
2. (b) silicon tetraiodide
3. (c) silane
4. (d) silicon carbide
5. (e) magnesium silicide
From the data given in Appendix I , determine the standard enthalpy change and the standard free energy change for each of the following reactions:
1. (a) $\ce{BF3}(g)+\ce{3H2O}(l)⟶\ce{B(OH)3}(s)+\ce{3HF}(g)$
2. (b) $\ce{BCl3}(g)+\ce{3H2O}(l)⟶\ce{B(OH)3}(s)+\ce{3HCl}(g)$
3. (c) $\ce{B2H6}(g)+\ce{6H2O}(l)⟶\ce{2B(OH)3}(s)+\ce{6H2}(g)$
(a) ΔH° = 87 kJ; ΔG° = 44 kJ; (b) ΔH° = −109.9 kJ; Δ = −154.7 kJ; (c) ΔH° = −510 kJ; ΔG° = −601.5 kJ
A hydride of silicon prepared by the reaction of Mg2Si with acid exerted a pressure of 306 torr at 26 °C in a bulb with a volume of 57.0 mL. If the mass of the hydride was 0.0861 g, what is its molecular mass? What is the molecular formula for the hydride?
Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it?
A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond.
18.4: Structure and General Properties of the Nonmetals
Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure?
Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P4 molecules instead of P2 molecules?
In the N2 molecule, the nitrogen atoms have an σ bond and two π bonds holding the two atoms together. The presence of three strong bonds makes N2 a very stable molecule. Phosphorus is a third-period element, and as such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming three σ bonds.
Write balanced chemical equations for the reaction of the following acid anhydrides with water:
1. (a) SO3
2. (b) N2O3
3. (c) Cl2O7
4. (d) P4O10
5. (e) NO2
Determine the oxidation number of each element in each of the following compounds:
1. (a) HCN
2. (b) OF2
3. (c) AsCl3
(a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1−
Determine the oxidation state of sulfur in each of the following:
1. (a) SO3
2. (b) SO2
3. (c) $\ce{SO3^2-}$
Arrange the following in order of increasing electronegativity: F; Cl; O; and S.
S < Cl < O < F
Why does white phosphorus consist of tetrahedral P4 molecules while nitrogen consists of diatomic N2 molecules?
18.5: Occurrence, Preparation, and Compounds of Hydrogen
Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals?
The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself.
The reaction of calcium hydride, CaH2, with water can be characterized as a Lewis acid-base reaction:
$\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)$
Identify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidation-reduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction.
In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why?
Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form.
What mass of CaH2 is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is:
$\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)$
What mass of hydrogen gas results from the reaction of 8.5 g of KH with water?
$\ce{KH + H2O ⟶ KOH + H2}$
0.43 g H2
18.6: Occurrence, Preparation, and Properties of Carbonates
Carbon forms the $\ce{CO3^2-}$ ion, yet silicon does not form an analogous $\ce{SiO3^2-}$ ion. Why?
Complete and balance the following chemical equations:
(a) hardening of plaster containing slaked lime
$\ce{Ca(OH)2 + CO2 ⟶}$
(b) removal of sulfur dioxide from the flue gas of power plants
$\ce{CaO + SO2 ⟶}$
(c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise
$\ce{NaHCO3 + NaH2PO4 ⟶}$
(a) $\ce{Ca(OH)2}(aq)+\ce{CO2}(g)⟶\ce{CaCO3}(s)+\ce{H2O}(l)$; (b) $\ce{CaO}(s)+\ce{SO2}(g)⟶\ce{CaSO3}(s)$;
(c) $\ce{2NaHCO3}(s)+\ce{NaH2PO4}(aq)⟶\ce{Na3PO4}(aq)+\ce{2CO2}(g)+\ce{2H2O}(l)$
Heating a sample of Na2CO3xH2O weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous Na2CO3. What is the formula of the hydrated compound?
18.7: Occurrence, Preparation, and Properties of Nitrogen
Write the Lewis structures for each of the following:
1. (a) NH2−
2. (b) N2F4
3. (c) $\ce{NH2-}$
4. (d) NF3
5. (e) $\ce{N3-}$
(a) NH2−:
; (b) N2F4: ; (c) $\ce{NH2-}$: ; (d) NF3: ; (e) $\ce{N3-}$:
For each of the following, indicate the hybridization of the nitrogen atom (for $\ce{N3-}$, the central nitrogen).
1. (a) N2F4
2. (b) $\ce{NH2-}$
3. (c) NF3
4. (d) $\ce{N3-}$
Explain how ammonia can function both as a Brønsted base and as a Lewis base.
Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate.
Brønsted base: $\ce{NH3 + H3O+ ⟶ NH4+ + H2O}$ Lewis base: $\ce{2NH3 + Ag+ ⟶ [H3N−Ag−NH3]+}$
Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples.
1. (a) NCl3
2. (b) ClNO
3. (c) N2O5
4. (d) N2O3
5. (e) $\ce{NO2-}$
6. (f) N2O4
7. (g) N2O
8. (h) $\ce{NO3-}$
9. (i) HNO2
10. (j) HNO3
For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples.
(a) NO2
(b) $\ce{NO2-}$
(c) $\ce{NO2+}$
(a) NO2:
Nitrogen is sp2 hybridized. The molecule has a bent geometry with an ONO bond angle of approximately 120°. (b) $\ce{NO2-}$: Nitrogen is sp2 hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than 120°. (c) $\ce{NO2+}$:
Nitrogen is sp hybridized. The molecule has a linear geometry with an ONO bond angle of 180°.
How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce?
Although PF5 and AsF5 are stable, nitrogen does not form NF5 molecules. Explain this difference among members of the same group.
Nitrogen cannot form a NF5 molecule because it does not have d orbitals to bond with the additional two fluorine atoms.
The equivalence point for the titration of a 25.00-mL sample of CsOH solution with 0.1062 M HNO3 is at 35.27 mL. What is the concentration of the CsOH solution?
18.8: Occurrence, Preparation, and Properties of Phosphorus
Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.
1. (a) PH3
2. (b) $\ce{PH4+}$
3. (c) P2H4
4. (d) $\ce{PO4^3-}$
5. (e) PF5
(a)
;
(b)
;
(c)
;
(d)
;
(e)
Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry.
1. (a) PH3
2. (b) $\ce{PH4+}$
3. (c) P2H4
4. (d) $\ce{PO4^3-}$
Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.)
1. (a) $\ce{P4 + Al⟶}$
2. (b) $\ce{P4 + Na⟶}$
3. (c) $\ce{P4 + F2⟶}$
4. (d) $\ce{P4 + Cl2⟶}$
5. (e) $\ce{P4 + O2⟶}$
6. (f) $\ce{P4O6 + O2⟶}$
(a) $\ce{P4}(s)+\ce{4Al}(s)⟶\ce{4AlP}(s)$; (b) $\ce{P4}(s)+\ce{12Na}(s)⟶\ce{4Na3P}(s)$; (c) $\ce{P4}(s)+\ce{10F2}(g)⟶\ce{4PF5}(l)$; (d) $\ce{P4}(s)+\ce{6Cl2}(g)⟶\ce{4PCl3}(l)$ or $\ce{P4}(s)+\ce{10Cl2}(g)⟶\ce{4PCl5}(l)$; (e) $\ce{P4}(s)+\ce{3O2}(g)⟶\ce{P4O6}(s)$ or $\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s)$; (f) $\ce{P4O6}(s)+\ce{2O2}(g)⟶\ce{P4O10}(s)$
Describe the hybridization of phosphorus in each of the following compounds: P4O10, P4O6, PH4I (an ionic compound), PBr3, H3PO4, H3PO3, PH3, and P2H4. You may wish to review the chapter on advanced theories of covalent bonding.
What volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl3 is an excess of water? Note that when H3PO3 is titrated under these conditions, only one proton of the acid molecule reacts.
291 mL
How much POCl3 can form from 25.0 g of PCl5 and the appropriate amount of H2O?
How many tons of Ca3(PO4)2 are necessary to prepare 5.0 tons of phosphorus if the yield is 90%?
28 tons
Write equations showing the stepwise ionization of phosphorous acid.
Draw the Lewis structures and describe the geometry for the following:
1. (a) $\ce{PF4+}$
2. (b) PF5
3. (c) $\ce{PF6-}$
4. (d) POF3
(a)
;
(b)
;
(c)
;
(d)
Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms?
Assign an oxidation state to phosphorus in each of the following:
1. (a) NaH2PO3
2. (b) PF5
3. (c) P4O6
4. (d) K3PO4
5. (e) Na3P
6. (f) Na4P2O7
(a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+
Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus.
1. (a) Write the empirical formula of phosphorus(V) oxide.
2. (b) What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280.
3. (c) Write balanced equations for the production of phosphorus(V) oxide and phosphoric acid.
4. (d) Determine the mass of phosphorus required to make 1.00 × 104 kg of phosphoric acid, assuming a yield of 98.85%.
18.9: Occurrence, Preparation, and Compounds of Oxygen
Predict the product of burning francium in air.
FrO2
Using equations, describe the reaction of water with potassium and with potassium oxide.
Write balanced chemical equations for the following reactions:
1. (a) zinc metal heated in a stream of oxygen gas
2. (b) zinc carbonate heated until loss of mass stops
3. (c) zinc carbonate added to a solution of acetic acid, CH3CO2H
4. (d) zinc added to a solution of hydrobromic acid
(a) $\ce{2Zn}(s)+\ce{O2}(g)⟶\ce{2ZnO}(s)$; (b) $\ce{ZnCO3}(s)⟶\ce{ZnO}(s)+\ce{CO2}(g)$; (c) $\ce{ZnCO3}(s)+\ce{2CH3COOH}(aq)⟶\ce{Zn(CH3COO)2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$; (d) $\ce{Zn}(s)+\ce{2HBr}(aq)⟶\ce{ZnBr2}(aq)+\ce{H2}(g)$
Write balanced chemical equations for the following reactions:
1. (a) cadmium burned in air
2. (b) elemental cadmium added to a solution of hydrochloric acid
3. (c) cadmium hydroxide added to a solution of acetic acid, CH3CO2H
Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations.
$\ce{Al(OH)3}(s)+\ce{3H+}(aq)⟶\ce{Al^3+}+\ce{3H2O}(l)$; $\ce{Al(OH)3}(s)+\ce{OH-}⟶\ce{[Al(OH)4]-}(aq)$
Write balanced chemical equations for the following reactions:
1. (a) metallic aluminum burned in air
2. (b) elemental aluminum heated in an atmosphere of chlorine
3. (c) aluminum heated in hydrogen bromide gas
4. (d) aluminum hydroxide added to a solution of nitric acid
Write balanced chemical equations for the following reactions:
1. (a) sodium oxide added to water
2. (b) cesium carbonate added to an excess of an aqueous solution of HF
3. (c) aluminum oxide added to an aqueous solution of HClO4
4. (d) a solution of sodium carbonate added to solution of barium nitrate
5. (e) titanium metal produced from the reaction of titanium tetrachloride with elemental sodium
(a) $\ce{Na2O}(s)+\ce{H2O}(l)⟶\ce{2NaOH}(aq)$; (b) $\ce{Cs2CO3}(s)+\ce{2HF}(aq)⟶\ce{2CsF}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$; (c) $\ce{Al2O3}(s)+\ce{6HClO4}(aq)⟶\ce{2Al(ClO4)3}(aq)+\ce{3H2O}(l)$; (d) $\ce{Na2CO3}(aq)+\ce{Ba(NO3)2}(aq)⟶\ce{2NaNO3}(aq)+\ce{BaCO3}(s)$; (e) $\ce{TiCl4}(l)+\ce{4Na}(s)⟶\ce{Ti}(s)+\ce{4NaCl}(s)$
What volume of 0.250 M H2SO4 solution is required to neutralize a solution that contains 5.00 g of CaCO3?
Which is the stronger acid, HClO4 or HBrO4? Why?
HClO4 is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid.
Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state.
1. (a) Mg
2. (b) Rb
3. (c) Ga
4. (d) C2H2
5. (e) CO
Which is the stronger acid, H2SO4 or H2SeO4? Why? You may wish to review the chapter on acid-base equilibria.
As H2SO4 and H2SeO4 are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H2SO4 is the stronger acid.
18.10: Occurrence, Preparation, and Properties of Sulfur
Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid.
Give the hybridization and oxidation state for sulfur in SO2, in SO3, and in H2SO4.
SO2, sp2 4+; SO3, sp2, 6+; H2SO4, sp3, 6+
Which is the stronger acid, NaHSO3 or NaHSO4?
Determine the oxidation state of sulfur in SF6, SO2F2, and KHS.
SF6: S = 6+; SO2F2: S = 6+; KHS: S = 2−
Which is a stronger acid, sulfurous acid or sulfuric acid? Why?
Oxygen forms double bonds in O2, but sulfur forms single bonds in S8. Why?
Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen.
Give the Lewis structure of each of the following:
1. (a) SF4
2. (b) K2SO4
3. (c) SO2Cl2
4. (d) H2SO3
5. (e) SO3
Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent.
There are many possible answers including:
$\ce{Cu}(s)+\ce{2H2SO4}(l)⟶\ce{CuSO4}(aq)+\ce{SO2}(g)+\ce{2H2O}(l)$
$\ce{C}(s)+\ce{2H2SO4}(l)⟶\ce{CO2}(g)+\ce{2SO2}(g)+\ce{2H2O}(l)$
Explain why sulfuric acid, H2SO4, which is a covalent molecule, dissolves in water and produces a solution that contains ions.
How many grams of Epsom salts (MgSO4⋅7H2O) will form from 5.0 kg of magnesium?
5.1 × 104 g
18.11: Occurrence, Preparation, and Properties of Halogens
What does it mean to say that mercury(II) halides are weak electrolytes?
Why is SnCl4 not classified as a salt?
SnCl4 is not a salt because it is covalently bonded. A salt must have ionic bonds.
The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions:
(a) reaction of a weak base and a strong acid
$\ce{NH3 + HClO4⟶}$
(b) preparation of a soluble silver salt for silver plating
$\ce{Ag2CO3 + HNO3⟶}$
(c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride
$\ce{SrCl2}(aq)+\ce{H2O}(l)\xrightarrow{\ce{electrolysis}}$
Which is the stronger acid, HClO3 or HBrO3? Why?
In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO3 is stronger than HBrO3; Cl is more electronegative than Br.
What is the hybridization of iodine in IF3 and IF5?
Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.
(a) IF5
(b) $\ce{I3-}$
(c) PCl5
(d) SeF4
(e) ClF3
(a)
;
(b)
;
(c)
;
(d)
;
(e)
Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties?
Name each of the following compounds:
(a) BrF3
(b) NaBrO3
(c) PBr5
(d) NaClO4
(e) KClO
(a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite
Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid.
What is the oxidation state of the halogen in each of the following?
(a) H5IO6
(b) $\ce{IO4-}$
(c) ClO2
(d) ICl3
(e) F2
(a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0
Physiological saline concentration—that is, the sodium chloride concentration in our bodies—is approximately 0.16 M. A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought?
18.12: Occurrence, Preparation, and Properties of the Noble Gases
Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.
1. (a) XeF2
2. (b) XeF4
3. (c) XeO3
4. (d) XeO4
5. (e) XeOF4
(a) sp3d hybridized; (b) sp3d2 hybridized; (c) sp3 hybridized; (d) sp3 hybridized; (e) sp3d2 hybridized;
What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry.
1. (a) XeF2
2. (b) XeF4
3. (c) XeO3
4. (d) XeO4
5. (e) XeOF4
Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry.
1. (a) XeF2
2. (b) XeF4
3. (c) XeO3
4. (d) XeO4
5. (e) XeOF4
(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar
What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry.
1. (a) XeO2F2
2. (b) KrF2
3. (c) $\ce{XeF3+}$
4. (d) $\ce{XeO6^4-}$
5. (e) XeO3
A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon.
The empirical formula is XeF6, and the balanced reactions are:
$\ce{Xe}(g)+\ce{3F2}(g)\xrightarrow{Δ}\ce{XeF6}(s)$
$\ce{XeF6}(s)+\ce{3H2}(g)⟶\ce{6HF}(g)+\ce{Xe}(g)$
Basic solutions of Na4XeO6 are powerful oxidants. What mass of Mn(NO3)2•6H2O reacts with 125.0 mL of a 0.1717 M basic solution of Na4XeO6 that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.E%3A_Representative_Metals_Metalloids_and_Nonmetals_%28Exercises%29.txt |
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. These include the d-block (groups 3–11) and f-block element elements. The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.
19: Transition Metals and Coordination Chemistry
Learning Objectives
• Outline the general approach for the isolation of transition metals from natural sources
• Describe typical physical and chemical properties of the transition metals
• Identify simple compound classes for transition metals and describe their chemical properties
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure $2$, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.
The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.
The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series.
Example $1$: Valence Electrons in Transition Metals
Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration:
1. cerium(III)
2. lead(II)
3. Ti2+
4. Am3+
5. Pd2+
For the examples that are transition metals, determine to which series they belong.
Solution
For ions, the s-valence electrons are lost prior to the d or f electrons.
1. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series.
2. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element.
3. titanium(II) [Ar]3d2; first transition series
4. americium(III) [Rn]5f6; actinide
5. palladium(II) [Kr]4d8; second transition series
Exercise $1$
Check Your Learning Give an example of an ion from the first transition series with no d electrons.
Answer
V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+.
Uses of Lanthanides in Devices
Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver (4.5 × 10−5% versus 0.79 × 10−5% by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together.
The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure $3$). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines.
As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to$470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials.
The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series.
Properties of the Transition Elements
Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (Table P1), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.
Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as $\ce{MoO4^2-}$ and $\ce{ReO4-}$.
Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.
Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts.
Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure $4$. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).
For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air.
The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+.
Example $2$: Activity of the Transition Metals
Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?
Solution
First, we need to look up the reduction half reactions (Table P1) for each oxide in the specified oxidation state:
$\ce{Cr2O7^2- + 14H+ + 6e- ⟶ 2Cr^3+ + 7H2O} \hspace{20px} \mathrm{+1.33\: V} \nonumber$
$\ce{MnO4- + 8H+ + 5e- ⟶ Mn^2+ + H2O} \hspace{20px} \mathrm{+1.51\: V} \nonumber$
$\ce{TiO2 + 4H+ + 2e- ⟶ Ti^2+ + 2H2O} \hspace{20px} \mathrm{−0.50\: V} \nonumber$
A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set.
Exercise $2$
Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from (Table P1).
Answer
$\ce{Co}(s)+\ce{2HCl}⟶\ce{H2}+\ce{CoCl2}(aq)$; no reaction because Pt(s) will not be oxidized by H+
Preparation of the Transition Elements
Ancient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure $5$). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust (Fe2O3). The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common.
Generally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal.
In general, it is not difficult to reduce ions of the d-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the f-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium.
We shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the d-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining.
1. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metal-bearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal.
2. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag—a substance with a low melting point that can be readily separated from the molten metal.
3. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals.
Isolation of Iron
The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.
The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure $6$) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.
Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:
$\ce{CO2}(g)+\ce{C}(s)⟶\ce{2CO}(g) \nonumber$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $6$.
The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:
$\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l) \nonumber$
Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure $7$).
Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle.
Isolation of Copper
The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:
$\ce{CaCO3}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)+\ce{CO2}(g) \nonumber$
$\ce{FeO}(s)+\ce{SiO2}(s)⟶\ce{FeSiO3}(l) \nonumber$
In these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion).
Reduction of the Cu2S that remains after smelting is accomplished by blowing air through the molten material. The air converts part of the Cu2S into Cu2O. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper:
$\ce{2Cu2S}(l)+\ce{3O2}(g)⟶\ce{2Cu2O}(l)+\ce{2SO2}(g) \nonumber$
$\ce{2Cu2O}(l)+\ce{Cu2S}(l)⟶\ce{6Cu}(l)+\ce{SO2}(g) \nonumber$
The copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure $8$). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry).
Isolation of Silver
Silver sometimes occurs in large nuggets (Figure $9$) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores.
Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, $\ce{[Ag(CN)2]-}$, from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:
$\ce{4Ag}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{4OH-}(aq) \nonumber$
$\ce{2Ag2S}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{2S}(s)+\ce{4OH-}(aq) \nonumber$
$\ce{AgCl}(s)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)+\ce{Cl-}(aq) \nonumber$
The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:
$\ce{2[Ag(CN)2]-}(aq)+\ce{Zn}(s)⟶\ce{2Ag}(s)+\ce{[Zn(CN)4]^2-}(aq) \nonumber$
Example $3$: Refining Redox
One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:
$\ce{4Ag}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{4OH-}(aq) \nonumber$
Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:
$\ce{4Ag}(s)+\ce{8CN-}(aq)⟶\ce{4[Ag(CN)2]-}(aq)? \nonumber$
Solution
The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state.
Exercise $3$
During the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron?
Answer
The carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0).
Transition Metal Compounds
The bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows.
Halides
Anhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example:
$\ce{2Fe}(s)+\ce{3Cl2}(g)⟶\ce{2FeCl3}(s) \nonumber$
Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state:
$\ce{Fe}(s)+\ce{2FeCl3}(s)⟶\ce{3FeCl2}(s) \nonumber$
The stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds.
In general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are:
$\ce{NiCO3}(s)+\ce{2HF}(aq)⟶\ce{NiF2}(aq)+\ce{H2O}(l)+\ce{CO2}(g) \nonumber$
$\ce{Co(OH)2}(s)+\ce{2HBr}(aq)⟶\ce{CoBr2}(aq)+\ce{2H2O}(l) \nonumber$
Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:
$\ce{Cr}(s)+\ce{2HCl}(aq)⟶\ce{CrCl2}(aq)+\ce{H2}(g) \nonumber$
The polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride (TiCl2 and TiCl3) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride (TiCl4) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier d-block elements have significant covalent characteristics.
The covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides:
$\ce{SiCl4}(l)+\ce{2H2O}(l)⟶\ce{SiO2}(s)+\ce{4HCl}(aq) \nonumber$
$\ce{TiCl4}(l)+\ce{2H2O}(l)⟶\ce{TiO2}(s)+\ce{4HCl}(aq) \nonumber$
Oxides
As with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent.
The oxides of the first transition series can be prepared by heating the metals in air. These oxides are Sc2O3, TiO2, V2O5, Cr2O3, Mn3O4, Fe3O4, Co3O4, NiO, and CuO.
Alternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide:
$\ce{FeC2O4}(s)⟶\ce{FeO}(s)+\ce{CO}(g)+\ce{CO2}(g) \nonumber$
$\ce{Co(OH)2}(s)⟶\ce{CoO}(s)+\ce{H2O}(g) \nonumber$
With the exception of CrO3 and Mn2O7, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are primarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid:
$\ce{CoO}(s)+\ce{2HNO3}(aq)⟶\ce{Co(NO3)2}(aq)+\ce{H2O}(l) \nonumber$
$\ce{Sc2O3}(s)+\ce{6HCl}(aq)⟶\ce{2ScCl3}(aq)+\ce{3H2O}(l) \nonumber$
The oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions $\ce{VO4^3-}$, $\ce{CrO4^2-}$, and $\ce{MnO4-}$. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by:
$\ce{CrO3}(s)+\ce{2Na+}(aq)+\ce{2OH-}(aq)⟶\ce{2Na+}(aq)+\ce{CrO4^2-}(aq)+\ce{H2O}(l) \nonumber$
Chromium(VI) oxide and manganese(VII) oxide react with water to form the acids H2CrO4 and HMnO4, respectively.
Hydroxides
When a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is:
$\ce{Co^2+}(aq)+\ce{2OH-}(aq)⟶\ce{Co(OH)2}(s) \nonumber$
In this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration:
$\ce{4Fe^3+}(aq)+\ce{6OH-}(aq)+\ce{nH2O}(l)⟶\ce{2Fe2O3⋅(n + 3)H2O}(s) \nonumber$
These substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal.
Carbonates
Many of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation:
$\ce{Ni^2+}(aq)+\ce{CO3^2-}⟶\ce{NiCO3}(s) \nonumber$
The reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides.
Other Salts
In many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements.
A variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide:
$\ce{2Sc}(s)+\ce{6HBr}(aq)⟶\ce{2ScBr3}(aq)+\ce{3H2}(g) \nonumber$
The common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example:
$\ce{Ni(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2ClO4-}(aq)⟶\ce{Ni^2+}(aq)+\ce{2ClO4-}(aq)+\ce{4H2O}(l) \nonumber$
Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example:
$\ce{Ba^2+}(aq)+\ce{2Cl-}(aq)+\ce{2K+}(aq)+\ce{CrO4^2-}(aq)⟶\ce{BaCrO4}(s)+\ce{2K+}(aq)+\ce{2Cl-}(aq) \nonumber$
In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements.
High Temperature Superconductors
A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.
Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.
One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7.
The new materials become superconducting at temperatures close to 90 K (Figure $10$), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K). Not only are liquid nitrogen-cooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium.
Although the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize. Superconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008.
Researchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure $11$).
Summary
The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.
Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity.
Glossary
actinide series
(also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103
coordination compound
stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons
d-block element
one of the elements in groups 3–11 with valence electrons in d orbitals
f-block element
(also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table
first transition series
transition elements in the fourth period of the periodic table (first row of the d-block), atomic numbers 21–29
fourth transition series
transition elements in the seventh period of the periodic table (fourth row of the d-block), atomic numbers 89 and 104–111
hydrometallurgy
process in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal
lanthanide series
(also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71
platinum metals
group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties
rare earth element
collection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult
second transition series
transition elements in the fifth period of the periodic table (second row of the d-block), atomic numbers 39–47
smelting
process of extracting a pure metal from a molten ore
steel
material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses
superconductor
material that conducts electricity with no resistance
third transition series
transition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.1%3A_Properties_of_Transition_Metals_and_Their_Compounds.txt |
Learning Objectives
• List the defining traits of coordination compounds
• Describe the structures of complexes containing monodentate and polydentate ligands
• Use standard nomenclature rules to name coordination compounds
• Explain and provide examples of geometric and optical isomerism
• Identify several natural and technological occurrences of coordination compounds
The hemoglobin in your blood, the chlorophyll in green plants, vitamin \(B_{12}\), and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure \(1\)). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.
Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form CH4. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure \(2\)). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals—atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.
The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure \(3\)). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.
Many other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, H2NCH2CH2NH2) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure \(4\)). Both of the atoms can coordinate to a single metal center. In the complex [Co(en)3]3+, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.
Any ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or “many teeth”) because it can bite into the metal center with more than one bond. The term chelate (pronounced “KEY-late”) from the Greek for “claw” is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab’s claw would hold a marble. Figure \(4\) showed one example of a chelate and the heme complex in hemoglobin is another important example (Figure \(5\)). It contains a polydentate ligand with four donor atoms that coordinate to iron.
Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as NH3, Cl, and H2O, are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, H2NCH2CH2NH2, and the anion of the acid glycine, \(\ce{NH2CH2CO2-}\) (Figure \(6\)) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The heme ligand (Figure \(5\)) is a tetradentate ligand.
The Naming of Complexes
The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:
1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.
2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group (e.g., Table \(1\). For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua (H2O), amine (NH3), carbonyl (CO), and nitrosyl (NO). For example, name [Pt(NH3)2Cl4] as diaminetetrachloroplatinum(IV).
3. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri- (for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris- (for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups (C5H4N).
Table \(1\): Examples of Anionic Ligands
Anionic Ligand Name
F fluoro
Cl chloro
Br bromo
I iodo
CN cyano
\(\ce{NO3-}\) nitrato
OH hydroxo
O2– oxo
\(\ce{C2O4^2-}\) oxalato
\(\ce{CO2^2-}\) carbonato
When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Tables \(2\), \(3\), and \(3\)). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state.
Table \(2\): Select Coordination Complexes based on total Charge
Examples in Which the Complex Is Cation
[Co(NH3)6]Cl3 hexaaminecobalt(III) chloride
[Pt(NH3)4Cl2]2+ tetraaminedichloroplatinum(IV) ion
[Ag(NH3)2]+ diaminesilver(I) ion
[Cr(H2O)4Cl2]Cl tetraaquadichlorochromium(III) chloride
[Co(H2NCH2CH2NH2)3]2(SO4)3 tris(ethylenediamine)cobalt(III) sulfate
Examples in Which the Complex Is Neutral
[Pt(NH3)2Cl4] diaminetetrachloroplatinum(IV)
[Ni(H2NCH2CH2NH2)2Cl2] dichlorobis(ethylenediamine)nickel(II)
Examples in Which the Complex Is an Anion
[PtCl6]2− hexachloroplatinate(IV) ion
Na2[SnCl6] sodium hexachlorostannate(IV)
Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H2O)4Cl2]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Example \(1\): Coordination Numbers and Oxidation States
Determine the name of the following complexes and give the coordination number of the central metal atom.
1. Na2[PtCl6]
2. K3[Fe(C2O4)3]
3. [Co(NH3)5Cl]Cl2
Solution
1. There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6]2−. There are six anionic chloride ligands, so −2 = −6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six.
2. The coordination sphere has a charge of 3− (based on the potassium) and the oxalate ligands each have a charge of 2−, so the metal oxidation state is given by −3 = −6 + x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six.
3. In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1−. The oxidation state is found by +2 = −1 + x and is 3+, so the complex is pentaaminechlorocobalt(III) chloride and the coordination number is six.
Exercise \(1\)
The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number.
Answer
K[Ag(CN)2]; coordination number two
The Structures of Complexes
The most common structures of the complexes in coordination compounds are octahedral, tetrahedral, and square planar (Figure \(7\)). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table \(3\) compares coordination numbers to the molecular geometry:
Table \(3\): Coordination Numbers and Molecular Geometry
Coordination Number Molecular Geometry Example
2 linear [Ag(NH3)2]+
3 trigonal planar [Cu(CN)3]2−
4 tetrahedral(d0 or d10), low oxidation states for M [Ni(CO)4]
4 square planar (d8) [NiCl4]2−
5 trigonal bipyramidal [CoCl5]2−
5 square pyramidal [VO(CN)4]2−
6 octahedral [CoCl6]3−
7 pentagonal bipyramid [ZrF7]3−
8 square antiprism [ReF8]2−
8 dodecahedron [Mo(CN)8]4−
9 and above more complicated structures [ReH9]2−
Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding d-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure \(8\). The chloride and nitrate anions in [Co(H2O)6]Cl2 and [Cr(en)3](NO3)3, and the potassium cations in K2[PtCl6], are outside the brackets and are not bonded to the metal ion.
For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as [Zn(CN)4]2− (Figure \(9\)), each of the ligand pairs forms an angle of 109.5°. In square planar complexes, such as [Pt(NH3)2Cl2], each ligand has two other ligands at 90° angles (called the cis positions) and one additional ligand at an 180° angle, in the trans position.
Isomerism in Complexes
Isomers are different chemical species that have the same chemical formula. Transition metals often form geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral [Co(NH3)4Cl2]+ ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure \(1\)). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.
Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH3)4Cl2]NO3 differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two [Co(NH3)4Cl2]NO3 isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar.
Example \(2\): Geometric Isomers
Identify which geometric isomer of [Pt(NH3)2Cl2] is shown in Figure \(9\)b. Draw the other geometric isomer and give its full name.
Solution
In the Figure \(9\)b, the two chlorine ligands occupy cis positions. The other form is shown in below. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diaminedichloroplatinum(II).
The trans isomer of [Pt(NH3)2Cl2] has each ligand directly across from an adjacent ligand.
Exercise \(2\)
Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II).
Answer
Another important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en)3]n+ [in which Mn+ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure \(11\). These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en)3]n+ and not the other.
The [Co(en)2Cl2]+ ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure \(12\)).
Linkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN− can be bound through the sulfur or nitrogen atom, affording two distinct compounds ([Co(NH3)5SCN]2+ or [Co(NH3)5NCS]2+).
Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl6][Br] and [CoCl5Br][Cl].
Coordination Complexes in Nature and Technology
Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure \(13\)). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.
Transition Metal Catalysts
One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (Figure \(14\)). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.
Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure \(13\)) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints.
The structure of heme (Figure \(15\)), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is Fe2+; oxidation of the iron to Fe3+ prevents oxygen transport.
Complexing agents often are used for water softening because they tie up such ions as Ca2+, Mg2+, and Fe2+, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, (HO2CCH2)2NCH2CH2N(CH2CO2H)2, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure \(16\)). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.
Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), HSCH2CH(SH)CH2OH, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure \(17\)). Another polydentate ligand, enterobactin, which is isolated from certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley’s anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.
Example \(3\): Chelation Therapy
Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure \(18\). Identify which atoms in this molecule could act as donor atoms.
Solution
All of the oxygen and sulfur atoms have lone pairs of electrons that can be used to coordinate to a metal center, so there are six possible donor atoms. Geometrically, only two of these atoms can be coordinated to a metal at once. The most common binding mode involves the coordination of one sulfur atom and one oxygen atom, forming a five-member ring with the metal.
Exercise \(3\)
Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations.1 Identify at least two biologically important metals that could be disrupted by chelation therapy.
Answer
Ca, Fe, Zn, and Cu
Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as [Ag(CN)2] and [Au(CN)2] are used extensively in the electroplating industry.
In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diaminedichloroplatinum(II), [Pt(NH3)2(Cl)2], and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diamine (NH3)2 portion is retained with other groups, replacing the dichloro [(Cl)2] portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin.
Summary
The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use.
Footnotes
1. National Council against Health Fraud, NCAHF Policy Statement on Chelation Therapy, (Peabody, MA, 2002).
Glossary
bidentate ligand
ligand that coordinates to one central metal through coordinate bonds from two different atoms
central metal
ion or atom to which one or more ligands is attached through coordinate covalent bonds
chelate
complex formed from a polydentate ligand attached to a central metal
chelating ligand
ligand that attaches to a central metal ion by bonds from two or more donor atoms
cis configuration
configuration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule
coordination compound
substance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions
coordination number
number of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form
coordination sphere
central metal atom or ion plus the attached ligands of a complex
donor atom
atom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal
ionization isomer
(or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere
ligand
ion or neutral molecule attached to the central metal ion in a coordination compound
linkage isomer
coordination compound that possesses a ligand that can bind to the transition metal in two different ways (CN vs. NC)
monodentate
ligand that attaches to a central metal through just one coordinate covalent bond
optical isomer
(also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers
polydentate ligand
ligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)
trans configuration
configuration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.2%3A_Coordination_Chemistry_of_Transition_Metals.txt |
Learning Objectives
• Outline the basic premise of crystal field theory (CFT)
• Identify molecular geometries associated with various d-orbital splitting patterns
• Predict electron configurations of split d orbitals for selected transition metal atoms or ions
• Explain spectral and magnetic properties in terms of CFT concepts
The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the d orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three p orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes.
Crystal Field Theory
To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized d orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.
CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.
All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized d orbitals in an octahedral complex. The five d orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure $1$. In an octahedral complex, the six ligands coordinate along the axes.
In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five d orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the d orbitals are no longer the same.
In octahedral complexes, the lobes in two of the five d orbitals, the $d_{z^2}$ and $d_{x^2−y^2}$ orbitals, point toward the ligands (Figure $1$). These two orbitals are called the eg orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the dxy, dxz, and dyz orbitals, have lobes that point between the ligands and are called the t2g orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the d orbitals of the metal ion. However, the repulsions between the electrons in the eg orbitals (the $d_{z^2}$ and $d_{x^2−y^2}$ orbitals) and the ligands are greater than the repulsions between the electrons in the t2g orbitals (the dzy, dxz, and dyz orbitals) and the ligands. This is because the lobes of the eg orbitals point directly at the ligands, whereas the lobes of the t2g orbitals point between them. Thus, electrons in the eg orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the t2g orbitals. The difference in energy may be represented as shown in Figure $2$.
The difference in energy between the eg and the t2g orbitals is called the crystal field splitting and is symbolized by Δoct, where oct stands for octahedral.
The magnitude of Δoct depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using 3d, 4d, or 5d orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series, a short version of which is given here:
$\large \underset{\textrm{a few ligands of the spectrochemical series, in order of increasing field strength of the ligand}}{\xrightarrow{\ce{I- <Br- <Cl- <F- <H2O<C2O4^2- <NH3<\mathit{en}<NO2- <CN-}}} \nonumber$
In this series, ligands on the left cause small crystal field splittings and are weak-field ligands, whereas those on the right cause larger splittings and are strong-field ligands. Thus, the Δoct value for an octahedral complex with iodide ligands (I) is much smaller than the Δoct value for the same metal with cyanide ligands (CN).
Electrons in the d orbitals follow the aufbau (“filling up”) principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy (P). Electrons will always singly occupy each orbital in a degenerate set before pairing. P is similar in magnitude to Δoct. When electrons fill the d orbitals, the relative magnitudes of Δoct and P determine which orbitals will be occupied.
In [Fe(CN)6]4−, the strong field of six cyanide ligands produces a large Δoct. Under these conditions, the electrons require less energy to pair than they require to be excited to the eg orbitals (Δoct > P). The six 3d electrons of the Fe2+ ion pair in the three t2g orbitals (Figure $3$). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.
In [Fe(H2O)6]2+, on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δoct < P). Because it requires less energy for the electrons to occupy the eg orbitals than to pair together, there will be an electron in each of the five 3d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H2O)6]2+, there will be one pair of electrons and four unpaired electrons (Figure $3$). Complexes such as the [Fe(H2O)6]2+ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.
A similar line of reasoning shows why the [Fe(CN)6]3− ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H2O)6]3+ and [FeF6]3− ions are high-spin complexes with five unpaired electrons.
Example $1$: High- and Low-Spin Complexes
Predict the number of unpaired electrons.
1. K3[CrI6]
2. [Cu(en)2(H2O)2]Cl2
3. Na3[Co(NO2)6]
Solution
The complexes are octahedral.
1. Cr3+ has a d3 configuration. These electrons will all be unpaired.
2. Cu2+ is d9, so there will be one unpaired electron.
3. Co3+ has d6 valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the t2g orbitals, leaving 0 unpaired.
Exercise $1$
The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which d-electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes?
Answer
d4, d5, d6, and d7
Example $2$: CFT for Other Geometries
CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the eg set point directly at the ligands. For tetrahedral complexes, the d orbitals remain in place, but now we have only four ligands located between the axes (Figure $4$). None of the orbitals points directly at the tetrahedral ligands. However, the eg set (along the Cartesian axes) overlaps with the ligands less than does the t2g set. By analogy with the octahedral case, predict the energy diagram for the d orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral eg set becomes a tetrahedral e set, and the octahedral t2g set becomes a t2 set.
Solution
Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so Δtet is usually small $\left(Δ_\ce{tet}=\dfrac{4}{9}Δ_\ce{oct}\right)$:
Exercise $2$
Explain how many unpaired electrons a tetrahedral d4 ion will have.
Answer
4; because Δtet is small, all tetrahedral complexes are high spin and the electrons go into the t2 orbitals before pairing
The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the z-axis. This changes the distribution of the d orbitals, as orbitals on or near the z-axis become more stable, and those on or near the x- or y-axes become less stable. This results in the octahedral t2g and the eg sets splitting and gives a more complicated pattern with no simple Δoct. The basic pattern is:
Magnetic Moments of Molecules and Ions
Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O2 that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N2 and ions such as Na+ and [Fe(CN)6]4− that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.
When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin d6 [Fe(CN)6]4− confirms that iron is diamagnetic, whereas high-spin d6 [Fe(H2O)6]2+ has four unpaired electrons with a magnetic moment that confirms this arrangement.
Colors of Transition Metal Complexes
When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed.
The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the [Cu(NH3)4]2+ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure $5$).
Example $3$: Colors of Complexes
The octahedral complex [Ti(H2O)6]3+ has a single d electron. To excite this electron from the ground state t2g orbital to the eg orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δoct and occurs at 499 nm. Calculate the value of Δoct in Joules and predict what color the solution will appear.
Solution
Using Planck's equation (refer to the section on electromagnetic energy), we calculate:
$v=\dfrac{c}{λ}\mathrm{\:so\:\dfrac{3.00×10^8\: m/s}{\dfrac{499\: nm×1\: m}{10^9\:nm}}=6.01×10^{14}\:Hz} \nonumber$
$E=hnu\mathrm{\:so\:6.63×10^{−34}\:\textrm{J⋅s}×6.01×10^{14}\:Hz=3.99×10^{−19}\:Joules/ion} \nonumber$
Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple.
Exercise $3$
A complex that appears green, absorbs photons of what wavelengths?
Answer
red, 620–800 nm
Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure $6$, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.
The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex [Fe(H2O)6]SO4 appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure $7$). In contrast, the low-spin iron(II) complex K4[Fe(CN)6] appears pale yellow because it absorbs higher-energy violet photons.
In general, strong-field ligands cause a large split in the energies of d orbitals of the central metal atom (large Δoct). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.
A coordination compound of the Cu+ ion has a d10 configuration, and all the eg orbitals are filled. To excite an electron to a higher level, such as the 4p orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN)2], for example, is colorless. On the other hand, octahedral Cu2+ complexes have a vacancy in the eg orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu2+ complexes are almost always colored—blue, blue-green violet, or yellow (Figure $8$). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes.
Summary
Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δoct) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the t2g orbitals are completely filled before any electrons occupy the eg orbitals. Weak-field ligands favor formation of high-spin complexes. The t2g and the eg orbitals are singly occupied before any are doubly occupied.
Glossary
crystal field splitting (Δoct)
difference in energy between the t2g and eg sets or t and e sets of orbitals
crystal field theory
model that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding
eg orbitals
set of two d orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the t2g orbitals
geometric isomers
isomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties
high-spin complex
complex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals
low-spin complex
complex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals
pairing energy (P)
energy required to place two electrons with opposite spins into a single orbital
spectrochemical series
ranking of ligands according to the magnitude of the crystal field splitting they induce
strong-field ligand
ligand that causes larger crystal field splittings
t2g orbitals
set of three d orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the eg orbitals according to CFT
weak-field ligand
ligand that causes small crystal field splittings | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.3%3A_Optical_and_Magnetic_Properties_of_Coordination_Compounds.txt |
19.1: Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
Q19.1.1
Write the electron configurations for each of the following elements:
1. Sc
2. Ti
3. Cr
4. Fe
5. Ru
S19.1.1
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and sub-shells. The electron configuration of each element is unique to its position on the periodic table where the energy level is determined by the period and the number of electrons is given by the atomic number of the element. There are four different types of orbitals (s, p, d, and f) which have different shapes and each orbital can hold a maximum of 2 electrons, but the p, d and f orbitals have different sub-levels, meaning that they are able to hold more electrons.
The periodic table is broken up into groups which we can use to determine orbitals and thus, write electron configurations:
Group 1 & 2: S orbital
Group 13 - 18: P orbital
Group 3 - 12: D orbital
Lanthanide & Actinides: F orbital
Each orbital (s, p, d, f) has a maximum number of electrons it can hold. An easy way to remember the electron maximum of each is to look at the periodic table and count the number of periods in each collection of groups.
Group 1 & 2: 2 (2 electrons total = 1 orbital x max of 2 electrons = 2 electrons)
Group 13 - 18: 6 (6 electrons total = 3 orbitals x 2 electrons max = 6 electrons)
Group 3 - 12: 10 (10 electrons total = 5 orbitals x 2 electrons max = 10 electrons)
Lanthanide & Actinides: 14 (14 electrons total = 7 orbitals x 2 electrons max = 14 electrons)
Electron fills the orbitals in a specific pattern that affects the order in which the long-hand versions are written:
Electron filling pattern: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f
An easier and faster way to write electron configurations is to use noble gas configurations as short-cuts. We are able to do this because the electron configurations of the noble gases always have all filled orbitals.
He: 1s22s2
Ne: 1s22s22p6
Ar: 1s22s22p63s23p6
Kr: 1s22s22p63s23p64s23d104p6
Xe: 1s22s22p63s23p64s23d104p65s24d105p6
Rn: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p6
The most common noble gas configuration used is Ar. When you want to use the noble gas configuration short-cut, you place the noble gas's symbol inside of brackets:
[Ar]
and then write it preceding the rest of the configuration, which is solely the orbitals the proceed after that of the noble gas.
a. Sc
Let's start off by identifying where Scandium sits on the periodic table: row 4, group 3. This identification is the critical basis we need to write its electron configuration.
By looking at Scandium's atomic number, 21, it gives us both the number of protons and the number of electrons. At the end of writing its electron configuration, the electrons should add up to 21.
At row 4, group 3 Sc, is a transition metal; meaning that its electron configuration will include the D orbital.
Now, we can begin to assign the 21 electrons of Sc to orbitals. As you assign electrons to their orbitals, you move right across the periodic table.
Its first 2 electrons are in the 1s orbital which is denoted as
1s2
where the "1" preceding the s denotes the fact that it is of row one, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 21-2=19 more electrons to assign.
Its next 2 electrons are in the 2s orbital which is denoted as
2s2
where the "2" preceding the s indicates that it is of row two, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 19-2=17 more electrons to assign.
Its next 6 electrons are in the 2p orbital which is denoted as
2p6
where the "2" preceding the p indicates that it is of row two, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 17-6=11 more electrons to assign.
Its next 2 electrons are in the 3s orbital which is denoted as
3s2
where the "3" preceding the s indicates that it is of row three, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 11-2=9 more electrons to assign.
Its next 6 electrons are in the 3p orbital which is denoted as
3p6
where the "3" preceding the p indicates that it is of row three, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 9-6=3 more electrons to assign.
Its next 2 electrons are in the 4s orbital which is denoted as
4s2
where the "4" preceding the s indicates that it is of row four, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 3-2=1 more electron to assign.
Its last electron would be alone in the 3 d orbital which is denoted as
3d1
where the "3" preceding the d indicates that, even though it is technically of row 4, by disregarding the first row of H and He, this is the third row and it has an exponent of 1 because there is only 1 electron to be placed in the d orbital. Now we have assigned all of the electrons to the appropriate orbitals and sub-orbitals, so that the final, entire electron configuration is written as:
1s22s22p63s23p64s23d1
This is the long-hand version of its electron configuration.
So for Sc, its short-hand version of its electron configuration would therefore be:
[Ar] 4s23d1
b. Ti
Start off by identifying where Titanium sits on the periodic table: row 4, group 4, meaning it has 22 electrons total. Titanium is one element to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 2 electrons remaining, so therefore the final orbital will be denoted as:
3d2
If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Titanium will be:
1s22s22p63s23p64s23d2
So for Ti, its short-hand version of its electron configuration would therefore be:
[Ar] 4s23d2
c. Cr
Start off by identifying where Chromium sits on the periodic table: row 4, group 6, that means it has a total of 24 electrons. But first, Cr, along with Mo, Nb, Ru, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 24 electrons that its configuration would look like:
1s22s22p63s23p64s23d4
which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that there are 4 electrons in its 3 d orbital. The most stable configurations are half-filled (d5) and full orbitals (d10), so the elements with electrons resulting in ending with the d4 or d9 are so unstable that we write its stable form instead, where an electron from the preceding s orbital will be moved to fill the d orbital, resulting in a stable orbital.
If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Chromium will be:
1s22s22p63s23p64s13d5
So for Cr, its short-hand version of its electron configuration would therefore be:
[Ar] 4s13d5
d. Fe
Start off by identifying where Iron sits on the periodic table: row 4, group 8, meaning it has 26 electrons total. This is 5 elements to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 6 electrons remaining, so therefore the final orbital will be denoted as:
3d6
If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Iron will be:
1s22s22p63s23p64s23d6
So for Fe, its short-hand version of its electron configuration would therefore be:
[Ar] 4s23d6
e. Ru
Start off by identifying where Ruthenium sits on the periodic table: row 5, group 8, that means it has a total of 44 electrons. But first, as stated earlier, Ru, along with Cr, Mo, Nb, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 44 electrons that its configuration would look like:
1s22s22p63s23p64s23d104p65s2 4d6
which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that, even though there are 4 paired electrons, there are also 4 electrons unpaired. This results in a very unstable configuration, so to restore stability, we have to use a configuration that has the most paired electrons, which would be to take an electron from the s orbital and place it in the d orbital to create:
5s14d7
If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Ru will be:
1s22s22p63s23p64s23d104p65s14d7
So for Cr, its short-hand version of its electron configuration would therefore be:
[Kr] 5s14d7
A19.1.1
1. Sc: [Ar]4s23d1
2. Ti: [Ar]4s23d2
3. Cr: [Ar]4s13d5
4. Fe: [Ar]4s23d6
5. Ru: [Kr]5s14d7 (anomalous configuration)
Q19.1.2
Write the electron configurations for each of the following elements and its ions:
1. Ti
2. Ti2+
3. Ti3+
4. Ti4+
S19.1.2
Electrons are distributed into molecular orbitals, the $s, p, d, and f$ blocks. An orbital will have a number in front of it and a letter that corresponds to the block. The s block holds two electrons, the p block holds six, the d block holds ten, and the f block holds fourteen. So, based on the number of electrons an atom has, the molecular orbitals are filled up in a certain way. The order of the orbitals is $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p$. An exponent will be put after the letter for each orbital to signify how many electrons are in that orbital. Noble gas notation can also be used by putting the noble gas prior to the element you are writing the configuration for, and then proceed by writing the orbitals filled after the noble gas. Metal ions of the d-block will have the two electrons removed from the s block prior to any electrons being removed from the proceeding d-block.
Solutions:
1. $Ti$
Titanium has an atomic number of 22, meaning it has 22 electrons. The noble gas prior to Titanium is Argon. Looking at row 4 of the periodic table, Titanium still has 4 electrons to be placed in orbitals since Argon has 18 electrons that are already placed. The remaining electrons will fill the $4s$ orbital and the remaining two electrons will go into the $3d$ orbital. [Ar]4s23d2
2. $Ti^{+2}$
This is an ion with a plus 2 charge, meaning 2 electrons have been removed. The electrons will be removed from the $4s$ orbital and the 2 remaining electrons will be placed in the $3d$ orbital. Like number 1, the prior noble gas is Argon. [Ar]3d2
3. $Ti^{+3}$
This is an ion with a plus 3 charge, meaning 3 electrons have been removed. The first 2 electrons will be removed from the $4s$ orbital, and the third will be taken from the $3d$ orbital, and the 1 remaining electron will be placed in the 3d orbital. Like number 1, the prior noble gas is Argon. [Ar]3d1
4. $Ti^{+4}$
This is an ion with a plus 4 charge, meaning 4 electrons have been removed. The first 2 electrons will be removed from the $4s$ orbital and the second 2 will be removed from the $3d$ orbital. This results in the ion having the same electron configuration as Argon. [Ar]
Answers:
1. $[Ar]4s^23d^2$
2. $[Ar]3d^2$
3. $[Ar]3d^1$
4. $[Ar]$
A19.1.2
1. $[Ar]4s^2 3d^2$
2. $[Ar]3d^2$
3. $[Ar]3d^1$
4. $[Ar]$
Q19.1.3
Write the electron configurations for each of the following elements and its 3+ ions:
1. La
2. Sm
3. Lu
S19.1.3
In order to write the electron configuration, we begin by finding the element on the periodic table. Since La, Sm, and Lu are all a period below the noble gas Xenon, we can abbreviate ${1s^2}{2s^2}{2p^6}{3s^2}{3p^6}{3d^{10}}{4s^2}{4p^6}{4d^{10}}{5s^2}{5p^6}$ as [Xe] when writing the orbital configurations. We then find the remaining of the orbital configurations using the Aufbau Principle. For other elements not just those in period 6, the shorthand notation using noble gases would be the noble gas in the period above the given element.
1. La has three additional electrons. Two of them fill the 6s shell and the other single electron is placed on the 5d shell.
$La:$ [Xe] ${6{s}^2} {5{d}^1}$
2. Sm has eight more electrons. The 6s orbital is filled as previously and the 4f orbital receives 6 electrons because pairing electrons requires lower energy on the 4f shell than on the 5d shell.
$Sm:$ [Xe] ${6{s}^2} {4{f}^6}$
3. Lu has seventeen more electrons. Two electrons fill the 6s orbital, 14 electrons fill the 4f orbital, and extra single electron goes to the 5d orbital .
$Lu:$ [Xe] ${6s^2}{4f^{14}}{5d^1}$
To find the 3+ ion electron configuration, we remove 3 electrons from the neutral configuration, starting with the 6s orbital.
1. The ionization of La3+ removes the three extra electrons. So it reverts back to the stable Xenon configuration.
${La^{3+}:}$ [Xe]
2. The ionization of Sm3+ removes two electrons from the 6s shell and one from the outermost (4f) shell
${Sm^{3+}}:$ [Xe] ${4f^5}$
3. The ionization of Lu3+ removes its two 6s shell and one from the outermost (5d) shell, leaving only a full 4f shell
$Lu^{3+}:$ [Xe] $4f^{14}$
A19.1.3
La: [Xe]6s 25d 1, La3+: [Xe]; Sm: [Xe]6s 24f 6, Sm3+: [Xe]4f 5; Lu: [Xe]6s 24f 145d 1, Lu3+: [Xe]4f 14
Q19.1.4
Why are the lanthanoid elements not found in nature in their elemental forms?
A19.1.4
Lanthanides are rarely found in their elemental forms because they readily give their electrons to other more electronegative elements, forming compounds instead of staying in a pure elemental form. They have very similar chemical properties with one another, are often found deep within the earth, and difficult to extract. They are the inner transition elements and have partially filled d orbitals that can donate electrons. Because of this, they are very reactive and electropositive.
Q19.1.5
Which of the following elements is most likely to be used to prepare La by the reduction of La2O3: Al, C, or Fe? Why?
S19.1.5
An activity series is a list of elements in decreasing order of their reactivity. Elements on the top of the list are good reducing agents because they easily give up an electron, and elements on the bottom of the series are good oxidizing agents because they are highly electronegative would really want to accept an electron.
Step 1: Compare Aluminum, Carbon, and Iron on an activity series. Many activity series include carbon and hydrogen as references. An activity series can be found here
The activity series goes in the order (from top to bottom): Aluminum, Carbon, and Iron.
Step 2: Identify which element is the best reducing agent.
Elements on the top of the list are the best reducing agents, because they give up electrons the best.
Aluminum is the best reducing agent of the options available.
Therefore aluminum will be the best reducing agent to prepare La by the reduction of La2O3 because it is the most reactive in the series amongst the three elements.
A19.1.5
Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La(III) into La.
Q19.1.6
Which of the following is the strongest oxidizing agent: $\ce{VO4^{3-}}$, $\ce{CrO4^2-}$, or $\ce{MnO4-}$?
S19.1.6
Oxidizing agents oxidize other substances. In other words, they gain electrons or become reduced. These agents should be in their highest oxidation state. In order to determine, the strength of the compounds above as oxidizing agents, determine the oxidation numbers of each constituent elements.
$\\mathrm{VO_4^{3-}}$
We know that $\mathrm{O}$ has a -2 oxidation state and the overall charge of the ion is -3. We just need to determine Vanadate's oxidation number in this compound.
$\\mathrm{V} + \mathrm{-2(4)} = \mathrm{-3}$
$\\mathrm{V} = \mathrm{+5}$
Vanadate has an oxidation number of +5, which is its highest possible oxidation state.
$\\mathrm{CrO_4^{2-}}$
Like in the previous calculation, $\mathrm{O}$ has a -2 oxidation state. The overall charge is -2. So calculate for chromium.
$\\mathrm{Cr} + \mathrm{-2(4)} = \mathrm{-2}$
$\\mathrm{Cr} = \mathrm{+6}$
Chromium is in its highest possible oxidation state of +6 in this compound.
$\\mathrm{MnO_4^-}$
$\mathrm{O}$ has a -2 oxidation state and the overall charge is -1.
$\\mathrm{Mn} + \mathrm{-2(4)} = \mathrm{-1}$
$\\mathrm{Mn} = \mathrm{+7}$
Manganese is also in its highest oxidation state, +7.
An oxidizing agent has to be able to gain electrons which, in turn, reduces its oxidation state. Here manganese has the greatest oxidation state which allows it to experience a greater decrease in its oxidation state if needed, meaning it can gain the most electrons. So among the three compounds, $\mathrm{MnO_4^-}$ is the strongest oxidizing agent. This method assumes the metals have similar electronegativities.
Alternatively, check a redox table.
A19.1.6
$MnO_4^-$
Q19.1.7
Which of the following elements is most likely to form an oxide with the formula MO3: Zr, Nb, or Mo?
S19.1.7
Mo because Zr has an oxidation state of +4 and Nb has a oxidation state of +5 and those would not balance out the charge of 3 oxygens in the state of -2 which creates a charge of -6. Mo however has multiple oxidation states, the most common being +6 which balances out the -6 charge created by 3 oxygen ions. This is why its most likely to form an oxide with the formula MO3 or $\ce{MoO3}$.
Mo
Q19.1.8
The following reactions all occur in a blast furnace. Which of these are redox reactions?
1. $\ce{3Fe2O3}(s)+\ce{CO}(g)⟶\ce{2Fe3O4}(s)+\ce{CO2}(g)$
2. $\ce{Fe3O4}(s)+\ce{CO}(g)⟶\ce{3FeO}(s)+\ce{CO2}(g)$
3. $\ce{FeO}(s)+\ce{CO}(g)⟶\ce{Fe}(l)+\ce{CO2}(g)$
4. $\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)$
5. $\ce{C}(s)+\ce{CO2}(g)⟶\ce{2CO}(g)$
6. $\ce{CaCO3}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)$
7. $\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)$
S19.1.8
o identify redox reaction, we have to determine if have to see if the equation is an oxidation-reduction reaction-meaning that the species are changing oxidation states during the reaction, which involves the transfer of electrons between two species. If a species is losing electrons, then that species is being oxidized. If a species is gaining electrons, then that species is being reduced. A way to remember this is using the acronyms OIL RIG. Oxidation Is Loss, and Reduction Is Gain, referring to electrons. Both of these must occur for an equation to be a redox reaction. Let's see if these equations are redox reactions or not:
a. In the reactants side $\ce{Fe2O3}$, Fe is has an oxidation number of +3. In the product $\ce{Fe3O4}$, Fe has an oxidation number of +2.67. Since Fe changed from +3 to +2.67, we can say that Fe had gained electrons and therefore reduced. In the reactant, CO, carbon has an oxidation number of +2, and in $\ce{CO2}$ (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
b. In the reactant, $\ce{Fe3O4}$, Fe has an oxidation number of +2.67. In the product, FeO, Fe has an oxidation number of +2. Since the oxidation of Fe has changed from +2.67 to +2, electrons have been added therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in $\ce{CO2}$ (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
c. In the reactant side, in FeO, Fe has an oxidation number of +2 and in the products side Fe has an oxidation number of 0. Since the oxidation number of Fe changed from +2 to 0, electrons have been gained and therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in $\ce{CO2}$ (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
d. In the reactants C has an oxidation number of 0, and in the products side in $\ce{CO2}$, C has an oxidation number of +4. Since the oxidation number of C has changed from 0 to +4, we can say that C has been oxidized. In the reactants, in $\ce{O2}$ oxygen has an oxidation number of 0, and in the products CO2, oxygen has an oxidation number of -2. Since the oxidation number of oxygen has changed from 0 to -2, oxygen has been reduced. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
e. In the reactants $\ce{CO2 }$ has an oxidation number of +4, and in the products side in CO, C has an oxidation number of +2. Since carbon went from +4 to +2, carbon has been reduced. In the reactants, in $\ce{CO2 }$ oxygen has an oxidation number of -4 and in the products CO carbon has an oxidation number of -2. Since oxygen went from -4 to -2, it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this is a redox reaction.
f. In the reactants, $\ce{CaCO3}$ Ca has an oxidation number of +2, and in products CaO Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation is not a redox reaction.
g. In the products CaO Ca has an oxidation number of +2, and in the products $\ce{CaSiO3}$ Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation is not a redox reaction.
a, b, c, d, e
Q19.1.9
Why is the formation of slag useful during the smelting of iron?
S19.1.9
Slag is a substance formed as a byproduct of iron ore or iron pellets melting together in a blast furnace. Slag is also the byproduct that is formed when a desired metal has been separated from its raw ore. It is important to note that slag from steel mills is created in a manner that reduces the loss of the desired iron ore. The $\ce{CaSiO3}$slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to $\ce{O2}$, which would oxidize the $\ce{Fe}$ back to $\ce{Fe2O3}$. Since Fe has a low reduction potential of -0.440 this means it has a high oxidation potential so it would easily oxidize in the presence of O2. Creating a barrier between iron and oxygen allows the maximum product of iron to be obtained in the end of smelting.
A19.1.9
The CaSiO3 slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to O2, which would oxidize the Fe back to Fe2O3.
Q19.1.10
Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer.
S19.1.10
Manganese(VII) oxide, can be written as Mn2O7.
In relation to the Lewis acid-base theory, a Lewis acid accepts lone pair electrons, and is also known as the electron pair acceptor. Based on this theory, acidity can be measured by the element's ability to accept electron pairs. By doing the math, we find that Manganese has an oxidation state of +7 (Oxygen has an oxidation state of -2, and 2x-7=-14 or this can be shown as $-7(2)+2(x)=0$ and $x=7$ since the whole compound has a charge of zero, in order to balance the ion's charge, Mn must be +7). Therefore Mn has high capability of accepting electrons due to its high positive charge. For most metals, as the oxidation number increases, so does its acidity, because of its increased ability to accept electrons.
A19.1.10
In relation to the Lewis acid-base theory, the Lewis acid accepts lone pair electrons; thus, it is also known as the electron pair acceptor. This may be any chemical species. Acids are substances that must be lower than 7. Therefore, oxides of manganese is most likely going to become more acidic in (aq) solutions if the oxidation number increases.
Q19.1.11
Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7 is required in the titration. What percentage of the ore sample was iron?
S19.1.11
To answer this question, we must first identify the net ionic equation from the given half-reactions. We can write the oxidation and reduction half-reactions:
$\text{ oxidation:} \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}$
$\text{ reduction: }\ce{Cr2O7^2-} \rightarrow \text{ Cr}^\text{ 3+}$
We can quickly balance the oxidation half-reaction by adding the appropriate number of electrons to get
$\text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}+\text{ e}^-$
The first step in balancing the reduction half-reaction is to balance elements in the equation other than O and H. In doing so, we get
$\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}$
The second step would be to add enough water molecules to balance the oxygen.
$\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
Next, we add the correct amount of H+ to balance the hydrogen atoms.
$\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
Finally, we add enough electrons to balance charge.
$\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
The electrons involved in both half-reactions must be equal in order for us to combine the two to get the net ionic equation. This can be done by multiplying each equation by the appropriate coefficient. Scaling the oxidation half-reaction by 6, we get
$6 \text{ Fe}^\text{ 2+} \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-$
Now we can combine both half-reactions to get
$6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
The electrons cancel out, so you get:
$6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 6 \text{ Fe}^\text{ 3+}+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}$
From this we can see that the mole ratio of Cr2O72- to Fe2+ is 1:6. Given that 19.17 mL (or 0.01917 L) of 0.01 M Na2Cr2O7 was needed for titration we know that
$0.01917\text{ L} \times 0.01 \text{ M} = 1.917 \times 10^{-4} \text{ mol}$
of Na2Cr2O7 reacted. Also, since any number of moles of Na2Cr2O7 produces the same number of moles of Cr2O72- in solution
$1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Na2Cr2O7} =1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}$
We can use the mole ratio of Cr2O72- to Fe2+ to determine how many moles of iron (ii) was in the solution. The number of moles of iron (ii) is the same as the number of moles of pure iron in the sample since all of the iron was converted into iron (ii).
$1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\times \frac{6\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}}{1\text{ mol}\text{ of }\ce{Cr2O7^2-}} = 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}$
$0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} = 0.0011502\text{ mol}\text{ of } \text{ Fe}$
Now we can find the number of grams of iron that were present in the 2.5 g iron ore sample.
$0.0011502\text{ mol}\text{ of } \text{ Fe}\times\frac{55.847\text { g}}{1\text{ mol}} = 0.0642352194\text{ g}\text{ of }\text{ Fe}$
Finally, we can answer the question and find what percentage of the ore sample was iron.
$\frac{0.0642352194\text{ g}}{2.5\text{ g}} \times 100 \approx 2.57\text{%}$
So 2.57% of the ore sample was iron.
2.57%
Q19.1.12
How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe2O3 to convert that Fe2O3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume.
S19.1.12
This question uses a series of unit conversions and the $PV=nRT$ equation.
The first step is to write out the balanced chemical equation for the conversion of Fe2O3 to pure iron.
$2\;Fe_2O_3(s)\rightarrow 4\;Fe(s)+3\;O_2(g)$
Next, we need to analyze the original question to determine the value that we need to solve for. Because the question asks for a value of cubic feet, we know we need to solve for volume. We can manipulate $PV=nRT$ to solve for volume.
$V={nRT}/P$
Now determine the known variables and convert into units that will be easy to deal with.
$n = 2000\:lbs\; Fe_2O_3\frac{453.592\: grams\: Fe_2O_3}{1\: lb \:Fe_2O_3}\frac{1\: mole \:Fe_2O_3}{159.69\: grams \:Fe_2O_3}\frac{3 \;moles\: O_2}{2\; moles \:Fe_2O_3}$
$n=8521\: moles\: of \:O_2$
Convert to atm for easier calculations
$R=\frac{.0821\:L\:atm}{mol\:K}$
$T=0^{\circ}C=273\:K$
$P=760 \:torr= 1 \:atm$
Now plug the numbers into the manipulated gas law to get to an answer for V.
$V=190991.8\: liters \:of\: O_2$
From here we convert liters to cubic feet.
use the conversion
$1\;L=.0353 ft^3$
thus we have 6744.811 ft3 of O2
We then refer back to the initial question and remember that this value is only 19% of the volume of the total air. So use a simple equation to determine the total volume of air in cubic feet.
$6744.811ft^3=.19x$
x=35499 ft3 of air
35499 ft3 of air
Q19.1.13
Find the potentials of the following electrochemical cell:
Cd | Cd2+ (M = 0.10) ‖ Ni2+ (M = 0.50) | Ni
S19.1.13
Step 1 Write out your two half reactions and identify which is oxidation and which is reduction using the acronym OIL RIG to remember that oxidation is loss of electrons and reduction is gain of electrons
Cd(s)⟶Cd2+(aq)+2e- (oxidation)
Ni2+(aq)+2e-⟶Ni(s) (reduction)
Step 2 Write out complete balanced equation
Cd(s)+ Ni2+(aq)⟶Cd2+(aq)+Ni(s)
Step 3 Find Eocell
Eocell = Ecathode-Eanode
oxidation: Cd(s)⟶Cd2+(aq)+2e- Eo=-0.40V
reduction: Ni2+(aq)+2e-⟶Ni(s) Eo=-0.26V
* E values come from standard reduction potentials table given above. Also, remember anode is where oxidation happens, and cathode is where reduction happens.
Eocell=-0.26-(-.40)
Eocell=0.14V
Step 4 Find Q
Q=[products]/[reactants] (look at complete balanced equation) (remember that [x] means the concentration of x typically given in molarity and that we ignore solids or liquids)
Q=[Cd2+]/[Ni2+]
Q=0.10/0.50
Q=0.2
Step 5 Calculate E using E= Eocell-(.0592/n)logQ (n is number of moles of electrons transferred and in our case the balanced reaction transfers 2 electrons)
E= 0.14-(.0592/2)log(0.2)
E= 0.14-(-.207)
E=0.16 V
0.16 V
Q19.1.14
A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
S19.1.14
A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
Assuming that metal chloride is XCl
The balance equation for the reaction would be:
$XCl(aq)+AgNO_{3}(aq)\rightarrow XNO_{3}(aq) +AgCl(s)$
The mass of AgCl = 3.03707g
To find the moles of AgCl present:
Next, we can determine the moles of AgCl present in the reaction since 1) the mass of the precipitate is given to us and 2) this value can help us determine the moles of alkali metal chloride compound present. Given the mass of AgCl is 3.03707g in the problem and the molecular mass of AgCl per mole is 143.32g, we can solve for how many moles of AgCl is in the reaction:
$moles of Agcl=\tfrac{3.03707g}{143.32g/mol}=0.0211 mol$
Since the molar ratio of the compounds are 1:1 so the number of moles of XCl used = 0.0211 mol
We can calculate the weight of Cl- with the equation:
$0.0211 mol \times 35.5g/mol = 0.7490g$
the amount of metal present in the original compound is the weight of the compound subtracted by weight of the Cl ion:
$(2.5624- 0.7490)g= 1.8134g$
And the percentage can be calculate by
$\frac{0.7490}{2.5624}\times 100= 29.23 \%$
the molar ratio of XCl is 1:1 so then
Atomic mass of metal = $=\frac{1.8134\;g\; metal}{0.0211\; mol\; RbCl}=85.943g/mol$
So the atomic mass is 85.943 g/mol which is of Rb hence the identity of the salt is RbCl
Q19.1.15
The standard reduction potential for the reaction $\ce{[Co(H2O)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(H2O)6]^2+}(aq)$ is about 1.8 V. The reduction potential for the reaction $\ce{[Co(NH3)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(NH3)6]^2+}(aq)$ is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H2O)6]2+ and/or [Co(NH3)6]2+, can be oxidized to the corresponding cobalt(III) complex by oxygen.
S19.1.15
To calculate the cell potential, we need to know the potentials for each half reaction. After doing so, we need to determine which one is being oxidized and which one is being reduced. The one that is oxidized is the anode and the one that is reduced is the cathode. To find the cell potential, you use this formula and the reduction potential values found in a reduction potential table. If E°cell is positive, $\Delta$G is negative and the reaction is spontaneous.
cell= E°cathode - E°anode
Because it states that $[Co(H_{2}O)_{6}]^{3+}$ will be oxidized, this means it is the anode.
$O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} \rightarrow 2 H_{2}O$ +1.229 V
$O_{2}$ is being reduced, so it is the cathode.
1.229V - 1.8V= -.571 V, or -0.6 V using significant figures. This cannot happen spontaneously because E°cell is negative.
For $[Co(NH_{3})_{6}]^{3+}$, it is again being oxidized, meaning it’s the anode.
1.229-.1= 1.129 V or 1.1 V using significant figures. This reaction is spontaneous because E°cell is positive.
A19.1.15
a) E ° = −0.6 V, E ° is negative so this reduction is not spontaneous.
b) E ° = +1.1 V, E ° is positive so this reduction is spontaneous.
Q19.1.16
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
1. $\ce{MnCO3}(s)+\ce{HI}(aq)⟶$
2. $\ce{CoO}(s)+\ce{O2}(g)⟶$
3. $\ce{La}(s)+\ce{O2}(g)⟶$
4. $\ce{V}(s)+\ce{VCl4}(s)⟶$
5. $\ce{Co}(s)+xs\ce{F2}(g)⟶$
6. $\ce{CrO3}(s)+\ce{CsOH}(aq)⟶$
S19.1.16
There is a myriad of reactions that can occur, which include: single replacement, double replacement, combustion, acid-base/neutralization, decomposition or synthesis. The first step to determine the products of a reaction is to identify the type of reaction. From then on, the next steps you take to predict the products will vary based on the reaction type.
1. This reaction is a double displacement reaction, in which the cations and anions of the reactants switch places to form new compounds. Writing out the equation in terms of it's aqueous ions will help you visualize what exactly is getting moved around:
$2H^{+}(aq)+2I^{-}(aq)\rightarrow 2H^{+}(aq)+CO_{3}^{2-}(aq)+Mn^{2+}(aq)+2I^{-}(aq)$
In this case, the hydrogen cations will recombine with carbonate anions whilst manganese cations will recombine with iodide anions giving us the following equation:
$\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{H2CO3}(aq)$
This is still not the final answer however, as carbonic acid is unstable and decomposes to carbon dioxide and water under standard conditions. Taking this into account, our final equation is:
$\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$
2. This reaction is a synthesis reaction, in which two or more reactants combine to form a more complex compound. In this case we are also reacting a metal oxide with oxygen which would result in another metal oxide as the product. The resulting product would be the mixed valence oxide Co3O4 in which one cobalt atom has a +2 oxidation state whilst the other two have a +3 oxidation state. Now all is left is to balance the equation:
$\ce{6CoO}(s)+\ce{O2}(g)⟶\ce{2Co3O4}(s)$
3. Like equation 2, this reaction is also a synthesis reaction involving a metal and oxygen which should result in the formation of a metal oxide. It is a matter now of balancing the oxidation states to attain a neutral compound. Oxygen will always hold a -2 oxidation state in compounds whilst lanthanum will always exhibit a +3 oxidation state. As such, a combination of 2 lanthanum atoms with a +3 oxidation state and 3 oxygen atoms with a -2 oxidation state will give us a molecule with an overall charge of 0 (3(-2)+2(+3)=0). We know our product now, La2O3, and now just need to balance the overall equation, giving us:
$\ce{4La}(s)+\ce{3O2}(g)⟶\ce{2La2O3}(s)$
4. This reaction is slightly harder to define as it encapsulates both the properties of synthesis and decomposition reactions, wherein vanadium reacts with vanadium tetrachloride to produce vanadium trichloride. This reaction however is primarily a synthesis reaction since we are combining two reactants to produce one complex compound. With vanadium trichloride as our product, we can balance the equation:
$\ce{V}(s)+\ce{3VCl4}(s)⟶\ce{4VCl3}(s)$
5. This is another synthesis reaction combining cobalt and fluorine. This equation includes the "xs" notation in front of fluorine which is short for 'excess', meaning more fluorine than actually required is present in the reactants, ensuring the reaction goes to completion. Finding the products if a simple matter of combining the cobalt and fluorine into one molecule, which already leaves us with a balanced equation:
$\ce{Co}(s)+xs\ce{F2}(g)⟶\ce{CoF2}(s)$
6. It may not be obvious here, but the reaction we've been given here is actually an acid-base/neutralization reaction, with chromium trioxide acting as the acid and cesium hydroxide as the base. Chromium trioxide is referred to as an acidic oxide which means that it will react with water to form an acid. Note that this reaction can still proceed even if the reactants aren't in the same phases. The basic rule for acid-base/neutralization reactions is they form a salt (salt being the general term for any ionic compound formed from acid-base reactions) and water. Since we know water is one of our products, our other product must be a salt composed of cesium, chromium and oxygen. Thus, our other product should be cesium chromate, and you can now balance the equation accordingly:
$\ce{CrO3}(s)+\ce{2CsOH}(aq)⟶\ce{Cs2CrO4}(aq)+\ce{H2O}$
A19.1.16
1. $\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)$
2. $\ce{6CoO}(s)+\ce{O2}(g)⟶\ce{2Co3O4}(s)$
3. $\ce{4La}(s)+\ce{3O2}(g)⟶\ce{2La2O3}(s)$
4. $\ce{V}(s)+\ce{3VCl4}(s)⟶\ce{4VCl3}(s)$
5. $\ce{Co}(s)+xs\ce{F2}(g)⟶\ce{CoF2}(s)$
6. $\ce{CrO3}(s)+\ce{2CsOH}(aq)⟶\ce{Cs2CrO4}(aq)+\ce{H2O}$
Q19.1.17
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
1. $\ce{Fe}(s)+\ce{H2SO4}(aq)⟶$
2. $\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶$
3. $\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶$
4. $\ce{Cr}(s)+\ce{O2}(g)⟶$
5. $\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶$
6. $\ce{Ti}(s)+xs\ce{F2}(g)⟶$
S19.1.17
Predict the products of each of the following reactions.
• Fe(s)+H2SO4(aq)⟶ ?
Whenever a metal reacts with an acid, the products are salt and hydrogen. Because Fe is lower on the activity series, we know that when it reacts with an acid it will result in the formation of Hydrogen gas. To simplify the equation is:
$Metal + Acid ⟶ Salt + Hydrogen$
The salt produced will depend on the metal and in this case, the metal is iron (Fe) so the resulting equation would be:
$\ce{Fe}(s)+\ce{H2SO4}(aq)⟶ \ce{FeSO4}(aq) + \ce{H2}(g)$
This equation works out as the H2 is removed from H2SO2, resulting in a SO42- ion where Fe will take on an oxidation state of Fe+2 to form FeSO4 which will be the salt in this example.
But since FeSO4 and H2SO4 are aqueous, the reactants and products can also be written as its ions where the overall equation can be:
$\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO2^{−4}}(aq)⟶\ce{Fe2+}(aq)+\ce{SO2^{−4}}(aq)+\ce{H2}(g)+\ce{2H2O}(l)$
• FeCl3(aq)+NaOH(aq)⟶ ?
In this case, adding a metal hydroxide (NaOH) to a solution with a transition metal ion (Fe) will form a transition metal hydroxide (XOH). As iron is bonded to three chlorine atoms in the reactants side, it has the oxidation state of +3 where three hydroxide ions (OH-) are needed to balance out the charges when they are bonded in the products. The remaining ions are Na+ and Cl- where they bond together in a 1:1 ratio where there are 3 molecules of NaCl once the reaction is balanced.
The overall reaction will be:
$\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶ \ce{Fe(OH)3}(s) + \ce{3NaCl}(aq)$
NOTE: Fe(OH)3(s) is a solid as it is a rule that all all transition metal hydroxides are insoluble and a precipitate is formed.
Since NaOH(aq) and NaCl(aq) are aqueous, we can write them out in their ion forms:
$\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH^{−}}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)$
• Mn(OH)2(s)+HBr(aq)⟶ ?
This is an example of a metal hydroxide reacting with an acid where a metal salt and water will always be formed:
$Metal Hydroxide + Acid ⟶ Metal Salt + Water$
When this rule is applied to this equation, we will get the following:
$\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶ \ce{MnBr2}(aq)+\ce{2H2O}(l)$
But to follow through with this question, the aqueous solutions such as HBr(aq) and MnBr2(aq) can be re-written as:
$\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)$
• Cr(s)+O2(g)⟶ ?
This is the general reaction of a metal reacting with oxygen which will always result in a metal oxide. However, the metal oxide is determined by the oxidation state of the metal so there may be several outcomes of this reaction such as:
$\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)$
$\ce{Cr}(s)+\ce{O2}(g)⟶\ce{2CrO}(s)$
$\ce{Cr}(s)+\ce{O2}(g)⟶\ce{CrO2}(s)$
$\ce{2Cr}(s)+\ce{3O2}(g)⟶\ce{CrO3}(s)$
However, Cr2O3 is the main oxide of chromium so it can be assumed that this is the general product of this reaction.
• Mn2O3(s)+HCl(aq)⟶?
This follows the general reaction of a metal oxide and an acid will always result in a salt and water
$Metal Oxide + Acid ⟶ Salt + Water$
Using this general reaction, similar to the general reactions above, the reaction will result in:
$\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)$
However, since HCl is an aqueous solution, the overall equation can also be re-written as:
$\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)$
• Ti(s)+xsF2(g)⟶?
Titanium is able to react with the halogens where there are two oxidation state that titanium can be: +3 and +4. The following reactions follow each oxidation state accordingly:
$\ce{2Ti}(s)+\ce{3F2}(g)⟶ \ce{2TiF3}(s)$
$\ce{Ti}(s)+\ce{2F2}(g)⟶\ce{TiF4}(s)$
However, since there is the symbol "xs", this indicates that F2 is added in excess so the second reaction is favored more as it drives the reaction to completion.
OVERALL:
$\ce{Ti}(s)+\ce{xsF2}(g)⟶\ce{TiF4}(g)$
A19.1.17
1. $\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO4^2-}(aq)⟶\ce{Fe^2+}(aq)+\ce{SO4^2-}(aq)+\ce{H2}(g)+\ce{2H2O}(l)$;
2. $\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH-}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)$;
3. $\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn^2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)$;
4. $\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)$;
5. $\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)$;
6. $\ce{Ti}(s)+xs\ce{F2}(g)⟶\ce{TiF4}(g)$
Q19.1.18
Describe the electrolytic process for refining copper.
S19.1.18
By electrolysis, copper can be refined and purely made. The reason why copper needs to remove the impurities is because it helps increase the electrical conductivity in electrical wire. You can refine copper and remove the impurities through electrolysis. Pure copper is important in making electrical wire, because it creates better electrical conductivity when transferring electricity. In order for better electrical conductivity, the impurities needs to be removed and this can be done by firing the impure copper to remove the impurities, such as sulfur, oxygen, etc. and shaping them into electrical anodes that can be used in electrolysis. Then the copper electrodes are placed into an electrical cell (into separate beakers) where electrical current can pass through the beakers and onto the electrodes. Through this process, the copper is stripped off of the anode and deposited onto the cathode. This process helps remove the impurities and refine copper because all the copper has been deposited onto the cathode all in one electrode. This process increases the weight of the cathode due to copper being deposited onto the cathode. This is a prime example of how to tell if an electrode is a cathode or an anode, as stated in Q17.2.9 above.
Q19.1.19
Predict the products of the following reactions and balance the equations.
1. Zn is added to a solution of Cr2(SO4)3 in acid.
2. FeCl2 is added to a solution containing an excess of $\ce{Cr2O7^2-}$ in hydrochloric acid.
3. Cr2+ is added to $\ce{Cr2O7^2-}$ in acid solution.
4. Mn is heated with CrO3.
5. CrO is added to 2HNO3 in water.
6. FeCl3 is added to an aqueous solution of NaOH.
S19.1.19
1. $\bg_black Zn$ is added to a solution of $\bg_black Cr_2(SO_4)_3$ in acid.
• Oxidized half-reaction:$3Zn(s) \rightarrow 3Zn^{2+}(aq)+6e^-$
• Reduction half reaction:
• Overall reaction:
• Chromium will precipitate out of the solution because it has a higher reduction potential than Zinc; the reaction is a single replacement.
2. $\bg_black FeCl_2$ is added to a solution containing an excess of $\bg_black Cr_2O_7^{2-}$in hydrochloric acid.
• Dissociation reaction:$6FeCl_2(s) \rightarrow 6Fe^{2+}(aq)+12Cl^-(aq)$
• Oxidation half-reaction:$6Fe^{2+}(aq) \rightarrow 6Fe^{3+}(aq)+6e^-$
• Reduction half-reaction:$Cr_2O_7^{2-}(aq) + 14 H^+(aq) + 6 e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
• Overall reaction:
• The reduction potential for permanganate is larger so the reaction is still favorable even when the oxidation of $Fe^{2+}$ is negative.
3. $\bg_black Cr^{2+}$ is added to $\bg_black Cr_2O_7^{2-}$ in acid solution.
• Reduction half-reaction:
• Oxidation half-reaction:
• Overall reaction:$6Cr^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq) \rightarrow 8Cr^{3+}(aq)+7H_2O(l)$
• The reaction is favorable with a high positive $E^\circ$
4. $\bg_black Mn$ is heated with $\bg_black CrO_3$.
• Reduction half-reaction:$8CrO_3(aq)+24H^++24e^- \rightarrow 4Cr_2O_3(aq)+12H_2O(l)$
• Oxidation half-reaction:$9Mn(s)+12H_2O(l) \rightarrow 3Mn_3O_4(aq)+24H^+(aq)+24e^-$
• Overall reaction:$8CrO_3(aq)+9Mn(s) \rightarrow^{\Delta} 4Cr_2O_3(aq)+3Mn_3O_4(aq)$
• Heat creates a product with higher energy than both previous reactants.
5. $\bg_black CrO$ is added to $\bg_black HNO_3$ in water.
• Strong acid dissociation:$2HNO_3(aq)+2H_2O(l) \rightarrow 2H_3O^+(aq)+2NO_3^-(aq)$
• Overall reaction:$2HNO_3(aq)+CrO(aq) \rightarrow Cr^{2+}(aq)+2NO_3^-(aq)+H_2O(l)$
• This reaction works by exchange of electrons to yield Chromium ions.
6. $\bg_black FeCl_3$ is added to an aqueous solution of $\bg_black NaOH$.
• Overall reaction:$3NaOH(aq)+FeCl_3(s) \rightarrow 3Na^+(aq)+FeOH_3^-(s)+3Cl^-(aq)$
• Iron hydroxide will precipitate because the two metals will exchange anions.
A19.1.19
1. $\ce{Cr2(SO4)3}(aq)+\ce{2Zn}(s)+\ce{2H3O+}(aq)⟶\ce{2Zn^2+}(aq)+\ce{H2}(g)+\ce{2H2O}(l)+\ce{2Cr^2+}(aq)+\ce{3SO4^2-}(aq)$;
2. $\ce{4TiCl3}(s)+\ce{CrO4^2-}(aq)+\ce{8H+}(aq)⟶\ce{4Ti^4+}(aq)+\ce{Cr}(s)+\ce{4H2O}(l)+\ce{12Cl-}(aq)$;
3. In acid solution between pH 2 and pH 6, $\ce{CrO4^2-}$ forms $\ce{HrCO4-}$, which is in equilibrium with dichromate ion. The reaction is $\ce{2HCrO4-}(aq)⟶\ce{Cr2O7^2-}(aq)+\ce{H2O}(l)$. At other acidic pHs, the reaction is $\ce{3Cr^2+}(aq)+\ce{CrO4^2-}(aq)+\ce{8H3O+}(aq)⟶\ce{4Cr^3+}(aq)+\ce{12H2O}(l)$;
4. $\ce{8CrO3}(s)+\ce{9Mn}(s)\overset{Δ}{⟶}\ce{4Cr2O3}(s)+\ce{3Mn3O4}(s)$;
5. $\ce{CrO}(s)+\ce{2H3O+}(aq)+\ce{2NO3-}(aq)⟶\ce{Cr^2+}(aq)+\ce{2NO3-}(aq)+\ce{3H2O}(l)$;
6. $\ce{CrCl3}(s)+\ce{3NaOH}(aq)⟶\ce{Cr(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl-}(aq)$
Q19.1.20
What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
S19.1.20
Formula for iron(II) sulfide: $FeS$
Definition of non-oxidizing acid: A non-oxidizing acid is an acid that doesn't act the oxidizing agent. Its anion is a weaker oxidizing agent than H+, thus it can't be reduced. Examples of non-oxidizing acids: $HCl, HI, HBr, H_3PO_4, H_2SO_4$
Step 2: Choose one of the non-oxidizing acid, in this case HCl, and write the chemical reaction:
$FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)$
The gas produced when iron (II) sulfide treated with a non-oxidizing acid, HCl, is H2S (dihydrogen sulfide) gas.
Q19.1.21
Predict the products of each of the following reactions and then balance the chemical equations.
1. Fe is heated in an atmosphere of steam.
2. NaOH is added to a solution of Fe(NO3)3.
3. FeSO4 is added to an acidic solution of KMnO4.
4. Fe is added to a dilute solution of H2SO4.
5. A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
6. FeCO3 is added to a solution of HClO4.
7. Fe is heated in air.
S19.1.21
a. Steam is water ($\ce{H_{2}O}$)
We can write out the reaction as:
$\ce{Fe}$ + $\ce{H_{2}O}$ → ?
This is a single replacement reaction, so $\ce{Fe}$ replaces $\ce{H_{2}}$. So, one of the products is $\ce{Fe_{3}O_{4}}$ since it is a combination of iron(II) oxide, $\ce{FeO}$, and iron(III) oxide, $\ce{Fe_{2}O_{3}}$.
The $\ce{Fe}$ is heated in an atmosphere of steam. $\ce{H_{2}}$ becomes neutrally charged and becomes another product.
After balancing the coefficients, the final reaction is:
$\ce{3Fe}(s)$ + $\ce{4H_{2}O}(g)$ → $\ce{Fe_{3}O_{4}}(s)$ + $\ce{4H_{2}}(g)$
b. $\ce{NaOH}$ added to a solution of $\ce{Fe(NO_{3})_{3}}$ is a double replacement and precipitation reaction.
We can write out the reaction as:
$\ce{NaOH}$ + $\ce{Fe(NO_{3})_{3}}$ → ?
The $\ce{Na}$ and $\ce{Fe}$ switch to form $\ce{Fe(OH)_{3}}(s)$ and $\ce{NaNO_{3}}(aq)$.
$\ce{Fe(OH)_{3}}$ is solid because it is insoluble according to solubility rules.
After balancing the coefficients in the reaction, the final reaction is:
$\ce{Fe(NO_{3})_{3}}(aq)$ + $\ce{3NaOH}(aq)$ → $\ce{Fe(OH)_{3}}(s)$ + $\ce{NaNO_{3}}(aq)$
c. For instance, the acid used to make the acidic solution is $\ce{H_{2}SO_{4}}$, then the reaction is:
$\ce{FeSO_{4}}$ + $\ce{KMnO_{4}}$ + $\ce{H_{2}SO_{4}}$ → $\ce{Fe_{2}(SO_{4})_{3}}$ + $\ce{MnSO_{4}}$ + $\ce{H_{2}O}$ + $\ce{K_{2}SO_{4}}$
Next, the net ionic reaction has to be written to get rid of the spectator ions in the reaction, this is written as:
$\ce{Fe^{2+}}$ + $\ce{MnO_{4}^{-}}$ + $\ce{H^{+}}$ → $\ce{Fe^{3+}}$ + $\ce{Mn^{2+}}$ + $\ce{H_{2}O}$
As seen in the net ionic equation above, $\ce{Fe^{2+}}$ is oxidized to $\ce{Fe^{3+}}$ and $\ce{MnO_{4}^{-}}$ is reduced to $\ce{Mn^{2+}}$. These can be written as two half reactions:
$\ce{Fe^{2+}}$ → $\ce{Fe^{3+}}$
$\ce{MnO_{4}^{-}}$ → $\ce{Mn^{2+}}$
To balance the oxidation half reaction, one electron as to be added to the $\ce{Fe^{3+}}$, this is shown as:
$\ce{Fe^{2+}}$ → $\ce{Fe^{3+}}$ + $\ce{e^{-}}$
The reduction half reaction also has to be balanced, but with $\ce{H^{+}}$ ions and $\ce{H_{2}O}$, this is shown as:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ → $\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$
After the charge of the $\ce{Mn}$ atoms are balanced, the overall charge has to be balanced on both sides because on the reactants side, the charge is $\ce{7+}$, and the charge on the products side is $\ce{2+}$. The overall charge can be balanced by adding electrons, this is shown as:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5e^{-}}$ → $\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$
Now since both half reactions are balanced, the electrons in both half reactions have to be equal, and then the half reactions are added together. After this is done, the reaction looks like this:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5Fe^{2+}}$ + $\ce{5e^{-}}$ → $\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$ + $\ce{5Fe^{3+}}$ + $\ce{5e^{-}}$
The $\ce{5e^{-}}$ on both sides cancel out and the final balanced reaction is:
$\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5Fe^{2+}}$ →$\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$ + $\ce{5Fe^{3+}}$
d. $\ce{Fe}$ added to a dilute solution of $\ce{H_{2}SO_{4}}$ is a single replacement reaction.
The $\ce{Fe}$ is added to a dilute solution so the $\ce{H_{2}SO_{4}}$ is written as separate ions.
We can write out the reaction as:
$\ce{Fe}(s)$ + $\ce{2H^+}(aq)$ + $\ce{(SO_{4})^{2-}}(aq)$ → ?
The Fe replaces the $\ce{H^+}$ ion, and becomes an $\ce{Fe^{2+}}$ ion.
$\ce{H_{2}O}$ is also a product because the solution is dilute.
Furthermore, the $\ce{FeSO_{4}}$ also has to be separated into ions as a result of the $\ce{Fe}$ being added to a dilute solution.
After balancing all of the coefficients, the final reaction is:
$\ce{Fe}(s)$ + $\ce{(2H_{3}O)^+}(aq)$ + $\ce{(SO_{4})^{2-}}(aq)$ → $\ce{Fe^{2+}}(aq)$ + $\ce{SO_{4}^{2-}}(aq)$ + $\ce{H_{2}}(g)$ + $\ce{2H_{2}O}(l)$
Note: $\ce{H^+}$ can also be written as the the hydronium ion, $\ce{(H_{3}O)^{+}}$.
e. We initially can initially write out:
$\ce{4Fe(NO_{3})_{2}}$ + $\ce{4HNO_{3}}$ + $\ce{O_{2}}$ → ?
We write the oxygen term in the reactants because it is stated that the solution is allowed to stand in air.
We just have to analyze the possible products that can be formed and we can see that the hydrogen from nitric acid can combine with oxygen gas to form water and then combining everything together, we get the final reaction to be:
$\ce{4Fe(NO_{3})_{2}}(aq)$ + $\ce{4HNO_{3}}(aq)$ + $\ce{O_{2}}(g)$ → $\ce{2H_{2}O}(l)$ + $\ce{4Fe(NO_{3})_{3}}(aq)$
f. When $\ce{FeCO_{3}}$ is added to $\ce{HClO_{4}}$, a double replacement reaction occurs.
The $\ce{Fe^{2+}}$ ion switches spots with the $\ce{H^+}$ ion to form $\ce{Fe(ClO_{4})_{2}}$ as a product.
When the $\ce{H^+}$ ion is added to the $\ce{(CO_{3})^{2-}}$ ion, $\ce{H_{2}CO_{3}}$ is formed.
After balancing the coefficients, the final reaction is:
$\ce{FeCO_{3}}(s)$ + $\ce{HClO_{4}}(aq)$ → $\ce{Fe(ClO_{4})_{2}}(aq)$ + $\ce{H_{2}O}(l)$ + $\ce{CO_{2}}(g)$
g. Air is composed of oxygen gas, which is a diatomic molecule, so it is $\ce{O_{2}}$.
Adding $\ce{Fe}$ to $\ce{O_{2}}$ will cause a synthesis reaction to occur forming $\ce{Fe_{2}O_{3}}$.
After balancing coefficients, the final reaction is:
$\ce{3Fe}(s)$ + $\ce{2O_{2}}(g)$ → $\ce{Fe_{2}O_{3}}(s)$
A19.1.21
1. $\ce{3Fe}(s)+\ce{4H2O}(g)⟶\ce{Fe3O4}(s)+\ce{4H2}(g)$;
2. $\ce{3NaOH}(aq)+\ce{Fe(NO3)3}(aq)\xrightarrow{\ce{H2O}}\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3NO3-}(aq)$;
3. $\ce{MnO_{4}^{-}}$ + $\ce{8H^{+}}$ + $\ce{5Fe^{2+}}$ →$\ce{Mn^{2+}}$ + $\ce{4H_{2}O}$ + $\ce{5Fe^{3+}}$
4. $\ce{Fe}(s)$ + $\ce{(2H_{3}O)^+}(aq)$ + $\ce{(SO_{4})^{2-}}(aq)$ → $\ce{Fe^{2+}}(aq)$ + $\ce{SO_{4}^{2-}}(aq)$ + $\ce{H_{2}}(g)$ + $\ce{2H_{2}O}(l)$
5. $\ce{4Fe(NO_{3})_{2}}(aq)$ + $\ce{4HNO_{3}}(aq)$ + $\ce{O_{2}}(g)$ → $\ce{2H_{2}O}(l)$ + $\ce{4Fe(NO_{3})_{3}}(aq)$
6. $\ce{FeCO_{3}}(s)$ + $\ce{HClO_{4}}(aq)$ → $\ce{Fe(ClO_{4})_{2}}(aq)$ + $\ce{H_{2}O}(l)$ + $\ce{CO_{2}}(g)$
7. $\ce{3Fe}(s)$ + $\ce{2O_{2}}(g)$ → $\ce{Fe_{2}O_{3}}(s)$
Q19.1.22
Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$\ce{Co(NO3)2}(s)⟶\ce{Co2O3}(s)+\ce{NO2}(g)+\ce{O2}(g)$
S19.1.22
Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$Co(NO_3){_2(s)}⟶Co_2O{_3(s)}+NO{_2(g)}+O{_2(g)}$
In this reaction, N changes oxidation states from +5 to +4 (reduced), Co changes oxidation states from +2 to +3 (oxidized), and O changes oxidation states from -2 to 0 (also oxidized).
First, split this reaction into an oxidation and reduction half reaction set, and balance all of the elements that are not hydrogen or oxygen (we will deal with these later):
$Reduction: 2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2$
Now, for the oxidation reaction, we are only dealing with O2 on the products side. In order to balance this, we will need to add water and hydrogen to both sides:
$Oxidation: 2H_2O\rightarrow O_2+4H^+$
Balance the amount of oxygens on each side by adding the correct number of water molecules (H2O), and balance the amount of hydrogen by adding the correct number of H+ atoms:
$Reduction: 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O$
$Oxidation: 2H_2O\rightarrow O_2+4H^+$
Finally, balance the charges by adding electrons to each side of the equation. For the reduction reaction, we will add 2 electrons to balance out the 2H+, and to the oxidation reaction, we will add 4 electrons to balance out the 4H+. Remember, the goal of this step is to make sure that the charges are balanced, so we can cancel them out in the end.
$Reduction: 2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O$
$Oxidation: 2H_2O\rightarrow O_2+4H^+ +4e^-$
Multiply the reduction reaction by two, in order to balance the charges so there are 4 electrons on each side of the reaction.
$Reduction: 2(2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O)$
and combine both reactions which comes out to:
$2H_2O + 4Co(NO_3)_2 + 4H^+ \rightarrow 2CO_2O_3 + 8NO_2 + 2H_2O + O_2 + 4H^+$
Cancel out like terms:
$4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}$
Both sides have overall charges of 0 and can be checked to see if they are balanced.
A19.1.22
$4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}$
Q19.1.23
Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
S19.1.23
Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
A: Step 1: look at the question and begin to write out a general product to reactant formula for this reaction.
Step 2: try to reason out why a precipitate will form but only for a finite period of time before reforming in an aqueous substance.
Step 3: With step 2 you should have noticed that the reaction is a multiple step reaction and using the rough formula that you derived in step 1, you should try and see what the series of steps are that lead to the overall product of liquid AgCN2
In this reaction we see how NaCN is added to AgNO3 .A precipitate forms but then disappears with the addition of even more NaCN, this must mean that its an intermediate reaction which will not appear as the final product. The silver and the cyanide temporarily bond, but the bond is too weak to hold them together so they are pulled apart again when NaCN is added because a new, more stronger and stable compound is formed: [Ag(CN)2]- (aq).
The actual reaction equation when it is first taking place is $AgCl(aq)+NaCN(aq)\rightarrow AgCN(s)+NaCl(aq)$
This can be written out in the following way: as CN is added, the silver and the cyanide combine : Ag+(aq)+CN(aq)→AgCN (s)
As more CN- is added the silver and two cyanide combine to create a more stable compound: Ag+(aq)+2CN(aq)→[Ag(CN)2]- (aq)
AgCN(s) + CN- (aq) → [Ag(CN)2]- (aq)
A19.1.23
As CN is added,
$\ce{Ag+}(aq)+\ce{CN-}(aq)⟶\ce{AgCN}(s)$
As more CN is added,
$\ce{Ag+}(aq)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)$
$\ce{AgCN}(s)+\ce{CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)$
Q19.1.24
Predict which will be more stable, [CrO4]2− or [WO4]2−, and explain.
S19.1.24
According to the rules associated with Crystal Field Stabilizing Energies, stable molecules contain more electrons in the lower-energy molecular orbitals than in the high-energy molecular orbitals. In this case, both complexes have O4 as ligands, and both have a -2 charge. Therefore, you determine stability by comparing the metals. Chromium is in the 3d orbital, according to the periodic table. Tungsten (W) is in the 5d orbital. 3d is a lower energy level than 5d.Higher-level orbitals are more easily ionized, and make their base elemental form more stable. If the elemental form is more stable the oxidized form is less stable. Therefore, [CrO4]2− is more stable than [WO4]2−.
A19.1.24
[CrO4]2- is more stable because Chromium is in the 3d orbital while Tungsten is in the 4d orbital, which has a higher energy level and makes it less stable.
Q19.1.25
Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M3O4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M2O3, to permit estimation of the metal’s two oxidation states.)
1. Sc2O3
2. TiO2
3. V2O5
4. CrO3
5. MnO2
6. Fe3O4
7. Co3O4
8. NiO
9. Cu2O
S19.1.25
The first step to solving this problem is looking at the rules of Oxidizing states for various elements:
chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidation_State
The main rules that will be used in these problems will be the oxidation state rule 6 which states that oxidation state for Oxygen is (-2) and rule 2 which is that the total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species. Solving these problems requires simple algebra. The oxidation states of both elements in the compound is equal to zero, so set the unknown oxidation of the element that is not oxygen to a variable ${x}$, and the oxidation state of Oxygen equal to ${-2}$. Then multiply both oxygen states by the number of atoms of the element present. Add the values together, set the equation equal to zero and solve for ${x}$.
1. $\ce{Sc2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2{x}}={0}⟶{x}={Sc}={+3}$ $Sc^{3+}$
2. $\ce{TiO2}={2{(-2)}}+{x}={0}⟶{-4}+{x}={0}⟶{x}={Ti}={+4}$ $Ti^{4+}$
3. $\ce{V2O5}={5{(-2)}}+{2{x}}={0}⟶{-10}+{2{x}}={0}⟶{x}={V}={+5}$ $V^{5+}$
4. $\ce{CrO3}={3{(-2)}}+{x}={0}⟶{-6}+{x}={0}⟶{x}={Cr}={+6}$ $Cr^{6+}$
5. $\ce{MnO2}={2{(-2)}}+{x}={0}⟶{-4}+{x}={0}⟶{x}={Mn}={+4}$ $Mn^{4+}$
6. $\ce{Fe3O4}=\ce{FeO}·\ce{Fe2O3}=$
$\ce{FeO}={-2}+{x}={0}⟶{x}={Fe}={+2}$ $Fe^{2+}$
$\ce{Fe2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Fe}={+3}$ $Fe^{3=}$
(One Fe Atom has an oxidation state of +2 and the other 2 Fe atoms have an oxidation state of +3)
7. $\ce{Co3O4}=\ce{CoO}·\ce{Co2O3}=$
$\ce{CoO}={-2}+{x}={0}⟶{x}={Co}={+2}$ $Co^{2+}$
$\ce{Co2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Co}={+3}$ $Co^{3+}$
(One Co Atom has an oxidation state of +2 and the other 2 Co atoms have an oxidation state of +3)
8. $\ce{NiO}={-2}+{x}={0}⟶{x}={Ni}={+2}$ $Ni^{2+}$
9. $\ce{Cu2O}={-2}+{2{x}}={0}⟶{-2}+{2x}={0}⟶{x}={Cu}={+1}$ $Cu^{1+}$
A19.1.25
Sc3+; Ti4+; V5+; Cr6+; Mn4+; Fe2+ and Fe3+; Co2+ and Co3+; Ni2+; Cu+
19.2: Coordination Chemistry of Transition Metals
Q19.2.1
Indicate the coordination number for the central metal atom in each of the following coordination compounds:
1. [Pt(H2O)2Br2]
2. [Pt(NH3)(py)(Cl)(Br)] (py = pyridine, C5H5N)
3. [Zn(NH3)2Cl2]
4. [Zn(NH3)(py)(Cl)(Br)]
5. [Ni(H2O)4Cl2]
6. [Fe(en)2(CN)2]+ (en = ethylenediamine, C2H8N2)
S19.2.1
First we must identify whether or not the ligand has more than one bonded atom (bidentate/polydentate). Using the table below we are able to do this.
Ligand Number of bonded atoms
Ammine (NH3) monodentate
Aqua (H2O) monodentate
Bromo (Br) monodentate
Chloro (Cl) monodentate
Cyano (CN) monodentate
Pyridine (C5H5N) monodentate
Ethylenediamine (C2H8N2) bidentate
Now that we have identified the number of bonded atoms from each ligand, we can find the total number of atoms bonded to the central metal ion, giving us the coordination number.
1. $[Pt(H_2O)_2Br_2]$: We can identify the metal ion in the complex as Pt, platinum, as the other two are listed as ligands above and are nonmetallic. We can now use the number of ligands and their bonding atoms to find its coordination number. From the table above we see that H2O has only one bonding atom and Br as well. So for each Br atom we have one bonding atom, and we have two of these, yielding 2 bonding atoms; this is the same for H2O, giving us a total number of 4 bonding atoms, and therefore a coordination number of 4.
2. $[Pt(NH_3)(py)(Cl)(Br)]$ (py = pyridine, C5H5N): The metal ion in this complex, similarly to the first one, can be identified as Pt, platinum. The ligands can be identified as NH3, pyridine, Cl, and Br, which are all monodentate ligands and have one bonding atom each. Since we have four ligands, each with one bonding atom, the total number of bonding atoms on the metal ion is 4, therefore the complex has a coordination number of 4.
3. $[Zn(NH_3)_2Cl_2]$: The metal ion in this complex can be identified as Zn, zinc, and the ligands can be identified as NH3 and Cl. Since these two are both monodentate ligands they have one bonding atom each. Since we have a total of two NH3 and two Cl ligands, we get a total of four monodentate ligands, giving us 4 bonding atoms and a coordination number of 4.
4. $[Zn(NH_3)(py)(Cl)(Br)]$: The metal ion in this complex can be identified as Zn, zinc, and the ligands can be identified as NH3, pyridine, Cl, and Br, which are all monodentate ligands and have one bonding atom each. Since we have four ligands, each with one bonding atom, the total number of bonding atoms on the metal ion is 4, therefore the complex has a coordination number of 4.
5. $[Ni(H_2O)_4Cl_2]$: The metal ion in this complex can be identified as Ni, nickel, and we can now use the number of ligands and their bonding atoms to find its coordination number. From the table above we see that H2O has only one bonding atom and Cl as well. So for each Cl atom we have one bonding atom, and we have two of these, yielding 2 bonding atoms. H2O is the same, having only one bonding atom, but there are four of these. So this gives us a total number of 6 bonding atoms, and therefore a coordination number of 6.
6. $[Fe(en)_2(CN)_2]^+$ (en = ethylenediamine, C2H8N2): The metal ion in this complex can be identified as Fe, iron, and the ligands can be identified as (en) and CN. Since (en) is bidentate, meaning it has 2 bonding atoms, and there are two of these, the total number of bonding atoms from (en) is four. Since CN is monodentate, meaning it has one bonding atom, and there are two of these, the total number of bonding atoms from CN ligand is two. So, the total number of bonding atoms is 6, therefore the complex has a coordination number of 6.
A19.2.1
1. The 2 aqua and the 2 bromo ligands form a total of 4 coordinate covalent bonds and as a result the coordination number is 4.
2. The ammine, pyridine, chloro and bromo each form one coordinate covalent bond that gives a total of 4 and hence CN=4.
3. Two ammine and two chloro ligands give a total of 4 coordinate covalent bonds and a CN = 4.
4. One ammine, a pyrimidine, a chloro and a bromo ligand give a total of 4 covalent bonds, resulting in CN = 4.
5. Four aqua ligands and two chloro ligands form a total of 6 coordinate covalent bonds and a CN =6.
6. Ethylenediamine is a bidentate ligand that forms two coordinate covalent bonds; along with two cyano ligands, it forms a total of 6 bonds, and hence has a CN=6.
Q19.2.2
Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:
1. tetrahydroxozincate(II) ion (tetrahedral)
2. hexacyanopalladate(IV) ion
3. dichloroaurate ion (note that aurum is Latin for "gold")
4. diamminedichloroplatinum(II)
5. potassium diamminetetrachlorochromate(III)
6. hexaamminecobalt(III) hexacyanochromate(III)
7. dibromobis(ethylenediamine) cobalt(III) nitrate
S19.2.2
To determine coordination numbers we must count the total number of ligands bonded to the central metal and distinguish monodentate and polydentate ligands. To determine the formulas, we use the nomenclature rules and work backwards.
1. "tetrahydroxo" = 4 hydroxide ligands; since hydroxide is a monodentate ligand, we have a total of 4 bonds to the central metal.
Coordination Number: 4
We review the basics of nomenclature and see that "tetra" = 4 and "hydroxo" = OH-. Since the charge on zinc is 2+, which is given in the nomenclature by the Roman numerals, we can calculate the total charge on the complex to be 2-.
Formula: [Zn(OH)4]2−
2. "hexacyano" = 6 cyanide ligands; since cyanide is a monodentate ligand, we have a total of 6 bonds to the central metal.
Coordination Number: 6
We review the basics of nomenclature and see that "hexa" = 6 and "cyano" = CN-. Since the charge on Pd is 4+, which is given in the nomenclature by the Roman numerals, we can calculate the total charge on the complex to be 2-.
Formula: [Pd(CN)6]2−
3. "dichloro" = 2 chloride ligands; since chloride is a monodentate ligand, we have a total of 2 bonds to the central metal.
Coordination Number: 2
We review the basics of nomenclature and see that "di" = 2 and "chloro" = Cl-. Since the charge on Au is always 1+, we can calculate the total charge on the complex to be 1-.
Formula: [AuCl2]
4. "diammine" = 2 ammonia ligands and "dichloro" = 2 chloride ligands; since both ammonia and chloride ligands are monodentate, we have a total of 4 bonds to the central metal.
Coordination Number: 4
We review the basics of nomenclature and see that "di" = 2, "chloro" = Cl- and "ammine" = NH3. Since the charge on Pt is 2+, which is given in the nomenclature by the Roman numerals, we can calculate that the total charge is 0, so the complex is neutral.
Formula: [Pt(NH3)2Cl2]
5. "diammine" = 2 ammonia ligands and "tetrachloro" = 4 chloride ligands; since both ammonia and chloride ligands are monodentate, we have a total of 6 bonds to the central metal.
Coordination Number: 6
We review the basics of nomenclature and see that "di" = 2, "ammine" = NH3, "tetra" = 4 and "chloro" = Cl-. Since the charge on the central metal, Cr, is 3+, which is given in the nomenclature by the Roman numerals, we can calculate that the total charge of the complex is 1-. The "potassium" at the front of the nomenclature indicates that it is the corresponding cation to this anionic complex.
Formula: K[Cr(NH3)2Cl4]
6. Both of the metal complexes have "hexa" monodentate ligands, which means both have coordination numbers of 6.
Coordination Number: 6
We review the basics of nomenclature and see that "hexa" = 6, "ammine" = NH3, and "cyano" = CN-. Since the charge on Cr is 3+ and Co is 3+, which is given in the nomenclature by the Roman numerals, we find that these complexes' charges balance out.
Formula: [Co(NH3)6][Cr(CN)6]
7. "dibromo" = 2 bromide ligands "bis(ethylenediamine)" = 2 (en) ligands; bromide is a monodentate ligand while (en) is a bidentate ligand. Therefore, we have a coordination number of 6.
Coordination Number: 6
We review the basics of nomenclature and see that "di" = 2, "bromo" = Br-, "bis" = 2, and "ethylenediamine" = en. Since the charge on Co is 3+, which is given in the nomenclature by the Roman numerals, we find that the total charge of the complex is 1+. Nitrate is the corresponding anion to this cationic complex.
Formula: [Co(en)2Br2]NO3
A19.2.2
1. 4, [Zn(OH)4]2−;
2. 6, [Pd(CN)6]2−;
3. 2, [AuCl2];
4. 4, [Pt(NH3)2Cl2];
5. 6, K[Cr(NH3)2Cl4];
6. 6, [Co(NH3)6][Cr(CN)6];
7. 6, [Co(en)2Br2]NO3
Q19.2.3
Give the coordination number for each metal ion in the following compounds:
1. [Co(CO3)3]3− (note that CO32− is bidentate in this complex)
2. [Cu(NH3)4]2+
3. [Co(NH3)4Br2]2(SO4)3
4. [Pt(NH3)4][PtCl4]
5. [Cr(en)3](NO3)3
6. [Pd(NH3)2Br2] (square planar)
7. K3[Cu(Cl)5]
8. [Zn(NH3)2Cl2]
S19.2.3
You can determine a compound's coordination number based on how many ligands are bound to the central atom.
1) In this compound, Cobalt is the central atom, and it has 3 CO32- molecules attached to it. However, CO32- is a bidentate ligand, which means it binds to the central atom in two places rather than one. This means that the coordination number of [Co(CO3)3]3- is 6. A coordination number of 6 means that the structure is most likely octahedral.
2) In this compound, Copper is the central atom. 4 ammonia molecules are attached to it. This means the coordination number is 4, and the structure is likely tetrahedral.
3) For this compound, we can ignore the (SO4)3 because it is not bound to the central atom. The central atom is cobalt, and it has 4 ammonia molecules and 2 bromine molecules bound to it. The coordination number is 6.
4) There are two compounds here, indicated by the brackets. The central atom for both is platinum. One of them has 4 ammonia molecules attached, and the other has 4 chlorine atoms attached. Both complexes have a coordination number of 4.
5) We can ignore (NO3)3 for this compound. The central atom is Chromium. There are 3 ethylenediamine molecules attached to the chromium. Ethylenediamine is a bidentate ligand, so the coordination number is 6.
6) Palladium is the central atom. 2 ammonia molecules and 2 bromine atoms are bound to the palladium atom. The coordination number is 4.
7) We can ignore the K3 structure. Copper is the central atom, and there are 5 chlorine molecules attached to it. The coordination number is 5, so the structure is either trigonal bipyramidal or square pyramidal.
8) In this compound, zinc is the central atom. There are 2 ammonia molecules and 2 chlorine atoms attached. This means that the coordination number is 4.
Q19.2.4
Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.
1. [Pt(H2O)2Br2] (square planar)
2. [Pt(NH3)(py)(Cl)(Br)] (square planar, py = pyridine, C5H5N)
3. [Zn(NH3)3Cl]+ (tetrahedral)
4. [Pt(NH3)3Cl]+ (square planar)
5. [Ni(H2O)4Cl2]
6. [Co(C2O4)2Cl2]3− (note that $\ce{C2O4^2-}$ is the bidentate oxalate ion, $\ce{−O2CCO2-}$)
S19.2.4
Cis and trans are a type of geometric isomer, meaning there is a difference in the orientation in which the ligands are attached to the central metal. In cis, two of the same ligands are adjacent to one another and in trans, two of the same ligands are directly across from one another.
Optical isomers → have the ability to rotate light, optical isomers are also chiral. Only chiral complexes have optical isomers
Chiral → asymmetric, structure of its mirror image is not superimposable
Enantiomers: chiral optical isomers (compound can have multiple enantiomers)
Tetrahedral complex with 4 distinct ligands → always chiral
• For tetrahedral, if 2 ligands are the same, then it cannot be chiral, has a plane of symmetry
Solutions:
a. $[Pt(H_2O)_2Br_2]$ (square planar)
This complex has 2 kinds of ligands. The matching ligands can either be adjacent to each other and be cis, or they can be across from each other and be trans.
b. $[Pt(NH_3)(py)(Cl)(Br)]$ (square planar, py = pyridine, $C_5H_5N$)
This complex has 4 different ligands. There is no plane of symmetry in any of the enantiomers, making the structures chiral and therefore has optical isomers.
c. $[Zn(NH_3)_3Cl]^+$ (tetrahedral)
There is a plane of symmetry from $NH_3$ through $Zn$ to the other $NH_3$, therefore it is not chiral.
d. $[Pt(NH_3)_3Cl]^+$ (square planar)
There is a plane of symmetry from $NH_3$ through $Pt$ to the other $NH_3$, therefore it is not chiral.
e. $[Ni(H_2O)_2Cl_2]$
The $Cl$ ligands can either be right next to each other, or directly across from one another allowing for both cis and trans geometries.
f. $[Co(C_2O_4)_2Cl_2]^-3$ (note that $C_2O_4^-2$ is the bidentate oxalate ion, $^−O_2CCO_2^-$
There is a plane of symmetry from $Cl$ through Co to the other $Cl$ in a "trans" chlorine configuration, therefore it is not chiral in a chlorine "trans" configuration. However, there is no symmetry in the chlorine "cis" configuration, indicating multiple "cis" isomers.
A19.2.4
a. [Pt(H2O)2Br2]:
b. [Pt(NH3)(py)(Cl)(Br)]:
c. [Zn(NH3)3Cl]+ :
d. [Pt(NH3)3Cl]+ :
e. [Ni(H2O)4Cl2]:
f. [Co(C2O4)2Cl2]3−:
Q19.2.5
Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):
1. [Co(en)2(NO2)Cl]+
2. [Co(en)2Cl2]+
3. [Pt(NH3)2Cl4]
4. [Cr(en)3]3+
5. [Pt(NH3)2Cl2]
S19.2.5
We are instructed to draw all geometric isomers and optical isomers for the specified compound. Optical isomers exist when an isomer configuration is not superimposable on its mirror image. This means there are two distinct molecular shapes. Often a left and right hand are cited as an example; if you were to take your right hand and place it upon your left, you cannot make the major parts of your hand align on top of one another. The basic idea when deciding whether something is optically active is to look for a plane of symmetry--if you are able to bisect a compound in a manner that establishes symmetry, then the compound does not have an optical isomer.
Cis isomers exist when there are 2 ligands of the same species placed at 90 degree angles from each other. Trans isomers exist when there are 2 ligands of the same species placed at 180 degree angles from each other.
Problem 1
This compound is an octahedral molecule, so the six ligands (atoms in the complex that are not the central transition metal) are placed around the central atom at 90 degree angles. Two optical isomers exist for [Co(en)2(NO2)Cl]+. The second isomer is drawn by taking the mirror image of the first.
Problem 2
This compound is also an octahedral molecule. Two cis (optical) isomers and one trans isomer exist for [Co(en)2Cl2]+. The trans isomer can be drawn by placing the chlorine ligands in positions where they form a 180 degree angle with the central atom. The first cis isomer can be drawn by placing the chlorine ligands in positions where they form a 90 degree angle with the central atom. The second cis isomer can be found by mirroring the first cis isomer, like we did in problem 1.
Problem 3
This compound is also an octahedral molecule. One trans isomer and one cis isomer of [Pt(NH3)2Cl4] exist. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle with the central atom.
Problem 4
This compound is also an octahedral molecule. Two optical isomers for [Cr(en)3]3+ exist. The second optical isomer can be drawn by taking the mirror image of the first optical isomer.
Problem 5
This compound is a square planar complex, so the ligands are placed around the central atom in a plane, at 90 angles. A trans isomer and a cis isomer exist for the complex [Pt(NH3)2Cl2]. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle in the plane with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle in the plane with the central atom.
Q19.2.6
Name each of the compounds or ions given in Exercise Q19.2.3, including the oxidation state of the metal.
S19.2.6
Rules to follow for coordination complexes
1. Cations are always named before the anions.
2. Ligands are named before the metal atom or ion.
3. Ligand names are modified with an ‐o added to the root name of an anion. For neutral ligands the name of the molecule is used, with the exception of OH2, NH3, CO and NO.
4. The prefixes mono‐, di‐, tri‐, tetra‐, penta‐, and hexa‐ are used to denote the number of simple ligands.
5. The prefixes bis‐, tris‐, tetrakis‐, etc., are used for more complicated ligands or ones that already contain di‐, tri‐, etc.
6. The oxidation state of the central metal ion is designated by a Roman numeral in parentheses.
7. When more than one type of ligand is present, they are named alphabetically. Prefixes do not affect the order.
8. If the complex ion has a negative charge, the suffix –ate is added to the name of the metal.
9. In the case of complex‐ion isomerism the names cis, trans, fac, or mer may precede the formula of the complex‐ion name to indicate the spatial arrangement of the ligands. Cis means the ligands occupy adjacent coordination positions, and trans means opposite positions just as they do for organic compounds. The complexity of octahedral complexes allows for two additional geometric isomers that are peculiar to coordination complexes. Fac means facial, or that the three like ligands occupy the vertices of one face of the octahedron. Mer means meridional, or that the three like ligands occupy the vertices of a triangle one side of which includes the central metal atom or ion.
A19.2.6
1. tricarbonatocobaltate(III) ion;
2. tetraaminecopper(II) ion;
3. tetraaminedibromocobalt(III) sulfate;
4. tetraamineplatinum(II) tetrachloroplatinate(II);
5. tris-(ethylenediamine)chromium(III) nitrate;
6. diaminedibromopalladium(II);
7. potassium pentachlorocuprate(II);
8. diaminedichlorozinc(II)
Q19.2.7
Name each of the compounds or ions given in Exercise Q19.2.5.
S19.2.7
Given:
1. [Co(en)2(NO2)Cl]+
2. [Co(en)2Cl2]+
3. [Pt(NH3)2Cl4]
4. [Cr(en)3]3+
5. [Pt(NH3)2Cl2]
Wanted:
Names of the above compounds.
1. [Co(en)2(NO2)Cl]+
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Co: Cobalt
en: Ethylenediamine
NO2: Nitro
Cl: Chloro
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(en)2: bis(Ethylenediamine)
Step 3: Find the charges of the ligands. Charges can be found here.
en: 0
NO2: -1
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Co + 2(en) + (NO2) + Cl = 1
Co +2(0) + (-1) + (-1) = 1
Co = 3
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Chlorobis(ethylenediamine)nitrocobalt(III)
2. [Co(en)2Cl2]+
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Co: Cobalt
en: Ethylenediamine
Cl: Chloro
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(en)2: bis(Ethylenediamine)
Cl2: dichloro
Step 3: Find the charges of the ligands. Charges can be found here.
en: 0
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Co + 2(en) +2(Cl) = 1
Co + 2(0) + 2(-1) = 1
Co = 3
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Dichlorobis(Ethylenediamine)cobalt(III)
3. [Pt(NH3)2Cl4]
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Pt: Platinum
NH3: Ammine
Cl: Chloro
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(NH3)2: diammine
Cl4: tetrachloro
Step 3: Find the charges of the ligands. Charges can be found here.
NH3: 0
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Pt + 2(NH3) + 4(Cl) = 0
Pt + 2(0) + 4(-1) = 0
Pt = 4
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Diamminetetrachloroplatinum(IV)
4. [Cr(en)3]3+
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
Cr: Cromium
en: ethylenediamine
Step 2: Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here.
(en)3: tris(ethylenediamine)
Step 3: Find the charges of the ligands. Charges can be found here.
en: 0
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Cr + 3(en) = 3
Cr + 3(0) = 3
Cr = 3
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Tris(ethylenediamine)cromium(III)
5. [Pt(NH3)2Cl2]
Step 1: Attain the names of the ligands and metal cation. Names can be found here.
NH3: Ammine
Cl: Chloro
Pt: Platinum
Step 2: Add the appropriate pre-fixes to each ligand depending on the number.
(NH3)2: diammine
Cl2: dichloro
Step 3: Find the charges of the ligands. Charges can be found here.
NH3: 0
Cl: -1
Step 4: Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges.
Pt + 2(NH3) + 2(Cl) = 0
Pt + 2(0) + 2(-1) = 0
Pt = 2
Step 5: For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last.
Diamminedichloroplatinum(II)
A19.2.7
1. Chlorobis(ethylenediamine)nitrocobalt(III)
2. Dichlorobis(Ethylenediamine)cobalt(III)
3. Diamminetetrachloroplatinum(IV)
4. Tris(ethylenediamine)cromium(III)
5. Diamminedichloroplatinum(II)
Q19.2.8
Specify whether the following complexes have isomers.
1. tetrahedral [Ni(CO)2(Cl)2]
2. trigonal bipyramidal [Mn(CO)4NO]
3. [Pt(en)2Cl2]Cl2
S19.2.8
Isomers are compounds that have the same number of atoms, but have different structures. Structural isomers (linkage, ionization, coordination) and stereoisomers (geometric and optical) can occur with several compounds.
1. tetrahedral $\mathrm{[Ni(CO)_2(Cl)_2]}$
(Fig 1.) In this model, nickel is the dark green central atom, carbonyl ligands are the pink atom, and chloro ligands are the light green atoms.
Immediately, we can cancel out the possibility of linkage, ionization, and coordination isomers. There are no other coordination complexes for coordination isomerism, there is no ligand that can bond to the atom in more than one way for it to exhibit linkage isomerism, and there are no ions outside the coordination sphere for ionization isomerism.
This is a tetrahedral structure which immediately rules out any geometric isomers since they require 90° and/or 180° bond angles. Tetrahedral structures have 109.5° angles.
To confirm that the structure has no optical isomer, we must determine if there is a plane of symmetry. Structures that have no plane of symmetry are considered chiral and would have optical isomers.
(Fig 2.) We can rotate the structure and find that there is indeed a plane of symmetry through the two chloro ligands and central atom and between the carbonyl ligands.
Since there is a plane of symmetry, we can conclude that there are no optical isomers.
Overall, there are no isomers that exist for this compound.
2. trigonal byprimidal $\mathrm{[Mn(CO)_4(NO)]}$
(Fig 3.) The central purple atom is manganese, the carbonyl ligands are the pink atoms, and the nitrosyl ligand is the fuschia atom.
There are no ions, other coordination complex, and ambidentate ligands. Therefore, no structural isomers exist for this structure.
Geometric isomers do not exist for this compound because there is only one nitrosyl ligand.
(Fig 4.) Dashed line bisects molecule and shows plane of symmetry. The molecule is rotated in this image.
In the image above, after the structure has been rotated, we can see that there is a plane of symmetry. Thus, there are no optical isomers.
No isomers (the ones mentioned above) exist for this compound.
3. $\mathrm{[Pt(en)_{2}Cl_2]Cl_2}$
(Fig 5.) The green atoms are the chloro ligands, the the central atom is platinum, and the grey/blue atoms are ethyldiamine ligands.
Coordination isomerism cannot exist for this complex because there are no other complexes. There are no linkage isomers because there are no ambidentate ligands. Ionization isomers cannot exist in this complex either, even though there is a neutral molecule outside the coordination sphere. If we exchange $\mathrm{Cl_2}$ with one ethyldiamine molecule, There would be 5 ligands in the coordination sphere instead of 4. This difference in the ratio of metal atom to ligands means that an ionization isomer cannot exist.
(Fig 6.) Here, one chloro ligand exchanged places with the ethyldiamine so that it can be at a 90° angle with the other chloro ligand.
The image above, shows the chloro and ethyldiamine ligands at a 90° angle with its other identical ligand. This is the cis isomer, while Fig. 5 shows the trans isomer.
Fig 5. shows that there is a plane of symmetry in the trans isomer. Therefore, that structure does not have an optical isomer. On the other hand, the cis isomer does not have a plane of symmetry and therefore has an optical isomer.
A19.2.8
none; none; The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer.
Q19.2.9
Predict whether the carbonate ligand $\ce{CO3^2-}$ will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.
S19.2.9
$\ce{CO3^2−}$ can be either monodentate or bidentate, since two of its oxygen atoms have lone pairs as shown above and can form covalent bonds with a transition metal ion. In most cases carbonate is monodentate because of its trigonal planar geometry (there is 120 degrees between the oxygens so it's hard for both to bind to the same metal). However, in some cases it will bind to two different metals, making it bidentate.
A19.2.9
CO3-2 will coordinate to a metal center as a monodentate ligand.
Q19.2.10
Draw the geometric, linkage, and ionization isomers for [CoCl5CN][CN].
S19.2.10
Isomers are compounds with same formula but different atom arrangement. There are two subcategories: structural isomers, which are isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another, and stereoisomers, isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space.
There are three subcategories under structural isomers: ionization isomers, which are isomers that are identical except for a ligand has exchanging places with an anion or neutral molecule that was originally outside the coordination complex; coordination isomers, isomers that have an interchange of some ligands from the cationic part to the anionic part; and linkage isomers, in two or more coordination compounds in which the donor atom of at least one of the ligands is different.
There are also two main kinds of stereoisomers: geometric isomers, metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal, and optical isomers, which occurs when the mirror image of an object is non-superimposable on the original object.
Some of the isomers look almost identical, but that is because the CN ligand can be attached by both (but not at the same time) the C or N.
19.3: Spectroscopic and Magnetic Properties of Coordination Compounds
Q19.3.1
Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.
S19.3.1
• The crystal field theory is is a model that describes the breaking of degeneracies of electron orbital states, usually d or f orbitals, due to a static electric field produced by a surrounding charge distribution.
• The degenerate d-orbitals split into two levels, e$_{g}$ and t$_{2g}$, in the presence of ligands.
• The energy difference between the two levels is called the crystal-field splitting energy, $\Delta_{\circ}$.
• After 1 electron each has been filled in the three t$_{2g}$ orbitals, the filling of the fourth electron takes place either in the e$_{g}$ orbital or in the t$_{2g}$, where the electrons pair up. depending on whether the complex is high spin or low spin.
• If the $\Delta_{\circ}$ value of a ligand is less than the pairing energy (P), then the electrons enter the e$_{g}$ orbital, but if the $\Delta_{\circ}$ value of a ligand is more than the pairing energy (P), then the electrons enter the t$_{2g}$ orbital.
• when the $\Delta_o$ is less than the pairing energy, the electrons prefer then eg orbitals because there is not enough energy to pair the electrons together. It will be high spin
• when the $\Delta_o$ is more then the pairing energy, the electrons prefer the t2g because there is enough energy to pair the electrons. It will be low spin.
Step 1: Determine the oxidation state of the Fe
For $[Fe(NO_{2}){_{6}}]^{3-}$ and $[FeF_{6}]^{3-}$, both $NO_{2}$ and $F_{6}$ have a charge of -1. Since there is 6 of them then that means the charge is -6 and in order for there to be an overall charge of -3, Fe has to have a +3 charge.
Step 2: Determine type of ligand
Based on the spectrochemical series we can see that $NO_{2}^{-}$ is a stronger field ligand than F$^{-}$, and therefore is a low spin complex because it has a high $\Delta_{\circ}$ unlike F$^{-}$ which is a high spin.
Step 3: Draw the crystal field
$[Fe(NO_{2}){_{6}}]^{3-}$
$[FeF_{6}]^{3-}$
There is 1 unpaired electron for $[Fe(NO_{2}){_{6}}]^{3-}$, and 5 for $[FeF_{6}]^{3-}$ based on the crystal field theory.
A19.3.1
[Fe(NO2)6]3−:1 electron
[FeF6]3−:5 electrons
Q19.3.2
Draw the crystal field diagrams for [Fe(NO2)6]4− and [FeF6]2−. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δoct to P for each complex.
S19.3.2
a) $[Fe(NO_2)_6]^{_{-4}}$
• NO2- has a -1 charge. The overall ion has a -4 charge, therefore Fe must be +2 charge. (The math: $x+(6)(-1)=-4, x+-6=-4, x=+2$ or 2+(6*-1)=-4)
• Fe2+ has 6 valence electrons.
• Next we look at the ligand bonded to Fe, which is NO2- . Based on the spectrochemical Series, NO2- is a strong field ligand meaning that it has a large DELTAo large splitting energy in comparison to the pairing energy, P. So the electrons would rather pair up, as it takes the least amount of energy.
• So [Fe(NO2)6]4− is low spin.
• All 6 electrons are paired up, so it is diamagnetic.
b) $[FeF_6]^{_{-3}}$
• F- has a -1 charge. The overall ion has a -3 charge, therefore Fe must be +3 charge. (The math: $x+(-1)(6)=-3, x+-6=-3, x=+3$ or 3+(6*-1)=-3)
• F3+ has 5 valence electrons.
• Next we look at the ligand bonded to Fe, which is F- . Based on the Spectrochemical Series, F- is a weak field ligand, meaning that it has a small DELTAo or small splitting energy in comparison to the pairing energy, P. So the electrons would rather split up and move up to the higher energy level, rather than pairing up, as it takes the least amount of energy.
• So [Fe(NO2)6]4− is high spin.
• [FeF6]3− is paramagnetic because it has unpaired electrons.
Q19.3.3
Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH3)6]Cl3.
S19.3.3
The oxidation state of the metal can be found by identifying the charge of one of each molecule in the coordinate compound, multiplying each molecule's charge by the respective number of molecules present, and adding the products. This final sum represents the charge of the overall coordination compound. You can then solve for the oxidation state of the metal algebraically. In this case, one chloride anion Cl- has a charge of -1. So three chloride anions have a total charge of -3. One ammine ligand NH3 has no charge so six ammine ligands have a total charge of zero. Finally, we are trying to solve for the oxidation state of a cobalt ion. Now we can write the equation that adds the total charges of each molecule or ion and is equal to the total charge of the overall coordinate compound.
$(\text{oxidation}\text{ state}\text{ of}\text{ Co})+(-3)+0=0$
$\text{oxidation}\text{ state}\text{ of}\text{ Co}=+3$
So the oxidation state of Co is +3. Now we need to identify the number of d-electrons in the Co3+ ion. The electron configuration for cobalt that has no charge is
$[\text{Ar}]4\text{s}^23\text{d}^7$
However, a Co3+ ion has 3 less electrons than its neutral counterpart and has an electron configuration of
$[\text{Ar}]3\text{d}^6$
For transition metals, the $\text{s}$ electrons are lost first. So cobalt loses its two $4\text{s}$ electrons first and then loses a single $3\text{d}$ electron meaning Co3+ ion has 6 $\text{d}$ electrons. To predict the number of unpaired electrons, we must first determine if the complex is high spin or low spin. Whether the complex is high spin or low spin is determined by the ligand in the coordinate complex. Specifically, the ligand must be identified as either a weak-field ligand or a strong-field ligand based on the spectrochemical series. Weak-field ligands induce high spin while strong-field ligands induce low spin. We can then construct the energy diagram or crystal field diagram of the designated spin that has the proper electron placings. The geometric shape of the compound must also be identified to construct the correct diagram. Finally, from this crystal field diagram we can determine the number of unpaired electrons. The number of electrons in the diagram is equal to the number of $\text{d}$ electrons of the metal. The ligand in this case is NH3, which is a strong field ligand according to the spectrochemical series. This means that the complex is low spin. Additionally, six monodentate ligands means the ligand field is octahedral. The number of electrons that will be in the diagram is 6 since the metal ion Co3+ has 6 $\text{d}$ electrons. Now the proper crystal field diagram can be constructed.
From the crystal field diagram, we can tell that the complex has no unpaired electrons.
A19.3.3
a) 3+
b) 6 d electrons
c) No unpaired electrons
Q19.3.4
The solid anhydrous solid CoCl2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.
S19.3.4
From our knowledge of ligands and coordination compounds (or if you need a refresher Coordination Compounds), we can assume the product of CoCl2 in water. H2O is a common weak field ligand that forms six ligand bonds around the central Cobalt atom while the Chloride stays on the outer sphere. We can use this to determine the complex:
$[Co(H_2O)_6]Cl_2$
From this formation, we can use the Crystal Field Theory (CFT)(Crystal Field Theory) to determine the number of unpaired electrons. This coordination compound has six ligand bonds attached to the central atom which means the CFT model will follow the octahedral splitting. Keep in mind that we know H2O is a weak field ligand and will produce a high spin. High spin is when the electrons pairing energy (P) is greater than the octahedral splitting energy. Thus, the electrons spread out and maximize spin.
In order to fill out our crystal field diagram, we need to determine the charge of cobalt. Because the H2O ligand is neutral, and there are two chlorine ions, we can deduce the charge of cobalt is plus two in order to make the coordination complex neutral. From here, we can use the electron configuration of Co2+ is [Ar]4s23d7. The electrons that are taken away from the cobalt atom in order to form the plus two charge will from the 4s orbital and leave the 3d orbital untouched. Thus, there will be 7 electrons in the crystal field diagram and appear as:
We can see here that there are 3 unpaired electrons.
A19.3.4
[Co(H2O)6]Cl2 with three unpaired electrons.
Q19.3.5
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
A19.3.5
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
It is not possible for a metal in the transition series to have six unpaired electrons. This is because transition metals have a general electron configuration of (n-1)d1-10 ns1-2 where n is the quantum number. The last electron will go into the d orbital which has 5 orbitals that can each contain 2 electrons, yielding 10 electrons total. According to Hund's Rule, electrons prefer to fill each orbital singly before they pair up. This is more energetically favorable. Since there are only 5 orbitals and due to Hund's Rule, the maximum number of unpaired electrons a transition metal can have is 5. Therefore, there cannot be a complex of a transition metal that has 6 unpaired electrons.
For example, lets look at iron's electron configuration. Iron has an electron configuration of 1s22s22p63s23p64s23d6. Now the most important orbital to look at is the d orbital which has 6 electrons in it, but there are only 4 unpaired electrons as you can see by this diagram:
3d: [↿⇂] [↿] [↿][↿][↿]
Each [ ] represents an orbital within the d orbital. This diagram follows Hund's rule and shows why no transition metal can have 6 unpaired electrons.
Q19.3.6
How many unpaired electrons are present in each of the following?
1. [CoF6]3− (high spin)
2. [Mn(CN)6]3− (low spin)
3. [Mn(CN)6]4− (low spin)
4. [MnCl6]4− (high spin)
5. [RhCl6]3− (low spin)
S19.3.6
1. For [CoF6]3-, we first found the oxidation state of Co, which is 3+ since F has a 1- charge and since there is 6 F, Co's charge has to be 3+ for the overall charge to be 3-.
$\textrm{charge of Co} + \textrm{-6} = \textrm{-6}$ $\textrm{charge of Co} = \textrm{+3}$
After finding the oxidation state, I then go to the periodic table to find its electron configuration: [Ar]3d6
We distribute the 6 d-orbital electrons along the complex and since it is high spin, the electrons is distributed once in each energy level before it is paired. There is only one pair and the other 4 electrons are unpaired, making the answer 4.
2. The same process is repeated. We find the charge of Mn, which is 3+, making the electron configuration: [Ar]3d4
$\textrm{charge of Mn} + \textrm{-6} = \textrm{-3}$ $\textrm{charge of Mn} = \textrm{+3}$
There is a difference between this and number 1. This is low spin so instead of distributing one electron in each level before pairing it, I must distribute one electron on the bottom and then pair them all up before I'm able to move to the top portion. So since there is 4, there is only a pair at dyz and the other two electrons are unpaired. Making the number of unpaired electrons 2.
3.The same process as number 2 is applied. The only difference is that the charge of Mn is now 2+ so the electron configuration: [Ar]3d5.
$\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}$ $\textrm{charge of Mn} = \textrm{+2}$
Since is it low spin like number 2, I only need to add an extra electron to the next level, making that 2 pairs of electron and only 1 electron unpaired.
4. Since Cl has a -1 charge like CN, Mn's charge is also 2+ with the same electron configuration as number 3, which is 5.
$\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}$ $\textrm{charge of Mn} = \textrm{+2}$
With 5 electrons, this is high spin instead of low. So as stated in number 1, we pair distribute the electrons on all levels first. Since there are 5 electrons and 5 levels and they are al distributed, there are zero pairs, making that 5 unpaired electrons.
5. Using the same process as the problems above, Rh's charge is 3+, with the electron configuration: [Kr]4d6.
$\textrm{charge of Rh} + \textrm{-6} = \textrm{-3}$ $\textrm{charge of Mn} = \textrm{+3}$
With a low spin and 6 electrons, all electrons are paired up, making it 0 electrons that are unpaired.
4; 2; 1; 5; 0
Q19.3.7
Explain how the diphosphate ion, [O3P−O−PO3]4−, can function as a water softener that prevents the precipitation of Fe2+ as an insoluble iron salt.
S19.3.7
The diphosphate ion, [O3P−O−PO3]4− can function as a water softener keeping the iron in a water soluble form because of its more negative electrochemical potential than water's. This is similar to the way plating prevents metals from reacting with oxygen to corrode.
Mineral deposits are formed by ionic reactions. The Fe2+ will form an insoluble iron salt of iron(III) oxide-hydroxide when a salt of ferric iron hydrolyzes water. However, with the addition of [O3P−O−PO3]4−, the Fe2+ cations are more attracted to the PO3 group, forming a Fe(PO3) complex.
The excess minerals in this type of water is considered hard thus its name hard water.
Q19.3.8
For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?
1. [Fe(H2O)6]2+ or [Fe(CN)6]4−
2. [Co(NH3)6]3+ or [CoF6]3−
3. [Mn(CN)6]4− or [MnCl6]4−
S19.3.8
The Spectrochemical Series is as follows $I^{-}<Br^{-}<SCN^{-}\approx Cl^{-}<F^{-}<OH^{-}<ONO^{-}<ox<H_{2}O<SCN^{-}<EDTA<NH_{3}<en<NO_{2}^{-}<CN$
The strong field ligands (on the right) are low spin which fills in more electrons in the t2g orbitals. The weak field ligands (on the left) are high spin so it can fill electrons in the t2g orbitals and eg orbitals. In conclusion, more electrons are filled up from the strong field ligands because the electrons don't move up to the eg orbitals.
a. $[Fe(CN)_{6}]^{4-}$
$CN$ is a stronger ligand than $H_{2}O$ so it is low spin, which fills up the t2g orbitals.
b. $[Co(NH_{3})_{6}]^{3+}$
$NH_{3}$ is a stronger ligand than $F$.
c. $[Mn(CN)_{6}]^{4-}$
$CN$ is a stronger ligand than $Cl^{-}$.
For more information regarding the shape of the complex and d-electron configuration, libretext provides more information on how to classify high and low spin complexes.
A19.3.8
[Fe(CN)6]4−; [Co(NH3)6]3+; [Mn(CN)6]4−
Q19.3.9
Trimethylphosphine, P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?
S19.3.9
1)Find the empirical formula. There is a total of 270 grams. To find out how many grams of each element/compound there are, multiply the percentage by the mass (270).
$(270g)(0.215)=58.05gNi$
$(270g)(0.26)=70.2gCl$
$(270g)(0.525)=141.75gP(CH_3)_3$
Now that we have the grams of each element/compound, we can convert them to moles by using their molar mass.
$(58.055gNi)(\frac{1mol.}{58.69gNi})=0.989mol.Ni$
$(70.2gCl)(\frac{1mol.}{35.45gCl})=1.98mol.Cl$
$(141.75gP(CH_3)_3)(\frac{1mol.}{76.07gP(CH_3)_3})=1.86mol.P(CH_3)_3$
Now that we have the moles of all elements/compounds, we can find the ratio of all them to each other. To do this, we take the element/compound with the least amount of moles and divide all element/compound moles by this amount. In this case, Ni has the least number of moles.
$\frac{0.989mol.Ni}{0.989mol.Ni}=1$
$\frac{1.98mol.Cl}{0.989mol.Ni}=approx.2$
$\frac{1.86mol.P(CH_3)_3}{0.989mol.Ni}=approx.2$
We now know the ratio of all element/compounds in the blue compound.
The empirical formula is: NiCl(P(CH3)3)2
This formula shows us there are 4 ligands. There are 2 chlorine ligands and 2 trimethylphosphine ligands. This means that the blue compound has either a tetrahedral or square planar shape, where tetrahedral shapes are capable of different isomeric forms when all ligands are different (because if not, there is only 1 way for them to be arranged), and square planar shapes are capable of cis/trans forms. In the problem, it states this compound does not have any isomeric forms, therefore this has a tetrahedral shape.
A19.3.9
a) NiCl(P(CH3)3)2
b) Tetrahedral
Q19.3.10
Would you expect the complex [Co(en)3]Cl3 to have any unpaired electrons? Any isomers?
S19.3.10
Assign oxidation states to each element. Cl- has a -1 oxidation state. En is neutral, so 0. The entire complex is also neutral, so in order to balance the charges out, Co must be +3 because there are 3 chlorides, which gives a -3 charge.
STEP 2:
Write the electron configuration for $Co^{3+}$. $[Ar]3d^6$. There are 6 electrons.
STEP 3:
Check where en lies on the spectrochemical series. Does it have a strong field strength? It does, so these electrons will exist at the d-level with high splitting energy because the magnitude of the pairing energy is less than the crystal field splitting energy in the octahedral field.
You will the notice that there aren't any unpaired electrons when you draw the Crystal Field Theory (CFT) diagram.
This complex does not have any geometric isomers because cis-trans structures cannot be formed. The mirror image is nonsuperimpoasable, which means the enantiomers are chiral molecules; if the mirror image is placed on top on the original molecule, then they will never be perfectly aligned to give the same molecule.
A19.3.10
The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer.
Q19.3.11
Would you expect the Mg3[Cr(CN)6]2 to be diamagnetic or paramagnetic? Explain your reasoning.
S19.3.11
The first step to determine the magnetism of the complex is to calculate the oxidation state of the transition metal. In this case, the transition metal is Cr.
Before doing so, we need to find charge of the of the complex ion [Cr(CN)6]2 given that the oxidation state of Mg3 is 2+. Using the subscripts of the $\displaystyle Mg^{2+}$ ion and the [Cr(CN)6]2 complex, we find that the oxidation state of [Cr(CN)6]2 , x, to be:
$\displaystyle 3(+2)+2(x)=0$
$\displaystyle x=3$
Now that we found the charge of the coordination complex, we are able to find the charge of the transition metal Cr given that the charge of CN is -1. Again, using the subscripts we find the oxidation state of Cr, y, to be:
$\displaystyle y + 6(-1)=-3$
$\displaystyle y=3$
Therefore, the oxidation state of the transition metal Cr is $\displaystyle Cr^{3+}$
Next, using the transition metal $\displaystyle Cr^{3+}$ and the periodic table as reference, we can determine the electron configuration of $\displaystyle Cr^{3+}$ to be $\displaystyle [Ar]d^{3}$. This means that $\displaystyle Cr^{3+}$ has 3 unpaired electrons in the 3d sublevel. Therefore, we find that since at least one electron is unpaired(in this case all 3 electrons are unpaired), Mg3[Cr(CN)6]2 is paramagnetic.
A19.3.11
a) Paramagnetic
Q19.3.12
Would you expect salts of the gold ion, Au+, to be colored? Explain.
S19.3.12
No. Colored ions have unpaired electrons in their outmost orbital. A partially filled d orbital, for example, can yield various colors. After completing the noble gas configuration, we see that Au+ has a configuration of [Xe] 4f145d10. Since Au+ has a completely filled d sublevel, we are certain that any salts of the gold ion, Au+ will be colorless.
*An example of a colored ion would be copper(II). Cu2+ has an electron configuration of [Ar]3d9. It has one unpaired electron. Copper(II) appears blue.
A19.3.12
No. Au+ has a complete 5d sublevel.
Q19.3.13
[CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?
S19.3.13
Although a color might appear a certain way, it actual absorbs a different color, opposite of it on the color wheel.
In this case;
[CuCl4]2- appears green but is opposite of red on the color wheel which is absorbed and is characterized by wavelengths 620-800 nanometers.
[Cu(H2O)6]2+ appears blue but is opposite of orange on the color wheel which is absorbed and is characterized by wavelengths 580-620 nanometers.
When determining which absorbs the higher energy photons, one must look at the complex itself. A higher energy indicates a high energy photon absorbed and a lower energy indicates a lower energy photon absorbed. How can we determine this? By looking at the complex and more specifically the ligand attached and its location in the spectrochemical series.
The ligands attached are Water and Chlorine and since Water is a stronger ligand than Chlorine according to the series, it also has larger energy, indicating a higher energy. This means that the complex [Cu(H2O)6]2+ absorbs a higher energy photon because of its a stronger ligand than chlorine.
Part 2 of this question also asks which complex is predicted to have a larger crystal field splitting. To determine this you also use the spectrochemical series and see which ligand is stronger. Since H2O is stronger than Cl- on the spectrochemical series, we can say [Cu(H2O)6]2+ has a higher crystal field splitting.
A19.3.13
a) [Cu(H2O)6]2+
b) [Cu(H2O)6]2+ has a higher crystal field splitting | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.E%3A_Transition_Metals_and_Coordination_Chemistry_%28Exercises%29.txt |
The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions. This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas.
20: Nuclear Chemistry
Learning Objectives
• Describe nuclear structure in terms of protons, neutrons, and electrons
• Calculate mass defect and binding energy for nuclei
• Explain trends in the relative stability of nuclei
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^1_1H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:
$\ce{^{A}_{Z}X} \label{Eq1}$
where
• $X$ is the symbol for the element,
• $A$ is the mass number, and
• $Z$ is the atomic number.
Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger). Example $1$ demonstrates just how great nuclear densities can be in the natural world.
Example $1$: Neutron Stars
Density of a Neutron Star Neutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star’s outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses (1 solar mass = $M_☉$ = mass of the sun = $\mathrm{1.99 \times 10^{30}\; kg}$) and a diameter of 26 km.
1. What is the density $\rho$ of this neutron star?
2. How does this neutron star’s density compare to the density of a uranium nucleus, which has a diameter of about 15 fm (1 fm = 10–15 m)?
Solution
We can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:
$\rho = \dfrac{m}{V} \nonumber$
with
$V = \dfrac{4}{3} \pi r^3 \nonumber$
(a) The radius of the neutron star is $\mathrm{\dfrac{1}{2}\times 26\; km = \dfrac{1}{2} \times 2.6 \times 10^4\; m = 1.3 \times 10^4\; m}$ so the density of the neutron star is:
\begin{align*} \rho &= \dfrac{m}{V} \[4pt] &=\dfrac{m}{\frac{4}{3}\pi r^3} \[4pt] &= \dfrac{2.4(1.99 \times 10^{30}\;kg)}{\frac{4}{3} \pi (1.3 \times 10^4m)^3} \[4pt] &=5.2 \times 10^{17}\;kg/m^3 \end{align*} \nonumber
(b) The radius of the U-235 nucleus is $\mathrm{\dfrac{1}{2} \times 15 \times 10^{−15}\;m=7.5 \times 10^{−15}\;m}$, so the density of the U-235 nucleus is:
\begin{align*} \rho &=\dfrac{m}{V} \[4pt] &=\dfrac{m}{\frac{4}{3}\pi r^3} \[4pt] &= \dfrac{235\;amu \left(\frac{1.66 \times 10^{-27}\;kg}{1\;amu}\right)}{ \frac{4}{3} \pi (7.5 \times 10^{-15}m)^3} \[4pt] &=2.2 \times 10^{17} \; kg/m^3 \end{align*} \nonumber
These values are fairly similar (same order of magnitude), but the nucleus is more than twice as dense as the neutron star.
Exercise $1$
Find the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km, and compare it to the density of a hydrogen nucleus, which has a diameter of 1.75 fm ($\mathrm{1\; fm = 1 \times 10^{–15}\; m}$).
Answer
The density of the neutron star is $\mathrm{3.4 \times 10^{18}\; kg/m^3}$. The density of a hydrogen nucleus is $\mathrm{6.0 \times 10^{17}\; kg/m^3}$. The neutron star is 5.7 times denser than the hydrogen nucleus.
To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than 10−15 meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent.
Nuclear Binding Energy
As a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as:
$\underset{\Large\text{protons}}{(2 \times 1.0073\; \text{amu})} + \underset{\Large\text{neutrons}}{(2 \times 1.0087\; \text{amu})} + \underset{\Large\text{electrons}}{(2 \times 0.00055\; \text{amu})}= 4.0331\; \text{amu }\label{Eq2}$
However, mass spectrometric measurements reveal that the mass of an $\ce{_2^4 He}$ atom is 4.0026 amu, less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium, the mass defect indicates a “loss” in mass of 4.0331 amu – 4.0026 amu = 0.0305 amu. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms’ nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are vastly greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions.
The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:
$E=mc^2 \label{Eq3}$
where E is energy, m is mass of the matter being converted, and c is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this mass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its mass defect, as demonstrated in Example $2$. A variety of units are commonly used for nuclear binding energies, including electron volts (eV), with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making $\mathrm{1\; eV = 1.602 \times 10^{-19}\; J}$.
Example $2$: Calculation of Nuclear Binding Energy
Determine the binding energy for the nuclide $\ce{^4_2 He}$ in:
1. joules per mole of nuclei
2. joules per nucleus
3. MeV per nucleus
Solution
The mass defect for a $\ce{^4_2He}$ nucleus is 0.0305 amu, as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that 1 J = 1 kg m2/s2).
(a) First, express the mass defect in g/mol. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass (g/mol) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore 0.0305 g/mol. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg, since 1 J = 1 kg m2/s2. Converting grams into kilograms yields a mass defect of $\mathrm{3.05 \times 10^{–5}\; kg/mol}$. Substituting this quantity into the mass-energy equivalence equation yields:
\begin{align*} E &=mc^2 \[4pt] &= \dfrac{3.05 \times 10^{-5}\;kg}{mol} \times \left(\dfrac{2.998 \times 10^8\;m}{s}\right)^2 \[4pt] &= 2.74×10^{12}\:kg\:m^2s^{-2}mol^{-1} \[4pt] &=2.74 \times 10^{12}\;J/mol=2.74\: TJ /mol \end{align*} \nonumber
Note that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg, roughly the mass of typical drop of water).
(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro’s number:
\begin{align*} E &= 2.74×10^{12}\:J\:mol^{-1}×\dfrac{1\: mol}{6.022×10^{23}\:nuclei} \[4pt] &=4.55×10^{-12} \: J =4.55\: pJ \end{align*} \nonumber
(c) Recall that $\mathrm{1\; eV = 1.602 \times 10^{-19}\; J}$. Using the binding energy computed in part (b):
\begin{align*} E &= 4.55×10^{-12} \: J× \dfrac{1\: eV}{1.602×10^{-19}\:J} \[4pt] &=2.84×10^7\:eV=28.4\: MeV \end{align*} \nonumber
Exercise $2$
What is the binding energy for the nuclide$\ce{^{19}_9F}$ (atomic mass: 18.9984 amu) in MeV per nucleus?
Answer
148.4 MeV
Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit enthalpies on the order of thousands of kJ/mol, which is equivalent to mass differences in the nanogram range (10–9 g). On the other hand, nuclear binding energies are typically on the order of billions of kJ/mol, corresponding to mass differences in the milligram range (10–3 g).
Nuclear Stability
A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure $1$ represents nuclei that have a 1:1 ratio of protons to neutrons (n:p ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead-207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.
The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.
Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (Table $1$). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of both protons and neutrons, such as $\ce{^4_2He}$, $\ce{^{16}_8O}$, $\ce{^{40}_{20}Ca}$, and $\ce{^{208}_{82}Pb}$ and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.
Table $1$: Stable Nuclear Isotopes
Number of Stable Isotopes Proton Number Neutron Number
157 even even
53 even odd
50 odd even
5 odd odd
The relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, the binding energy for a $\ce{^4_2He}$ nucleus is therefore:
$\mathrm{\dfrac{28.4\; MeV}{4\; nucleons}=7.10\; MeV/nucleon} \label{Eq3a}$
The binding energy per nucleon of a nuclide on the curve shown in Figure $2$
Example $3$: Calculation of Binding Energy per Nucleon
The iron nuclide $\ce{^{56}_{26}Fe}$ lies near the top of the binding energy curve (Figure $2$) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide $\ce{^{56}_{26}Fe}$ (atomic mass of 55.9349 amu)?
Solution
As in Example, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an $\ce{^{56}_{26}Fe}$ atom:
\begin{align*} \mathrm{Mass\: defect}&=\mathrm{[(26×1.0073\: amu)+(30×1.0087\: amu)+(26×0.00055\: amu)]−55.9349\: amu}\ &=\mathrm{56.4651\: amu−55.9349\: amu}\ &=\mathrm{0.5302\: amu} \end{align*} \nonumber
We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:
\begin{align*} E&=mc^2=\mathrm{0.5302\: amu×\dfrac{1.6605×10^{-27}\:kg}{1\: amu}×(2.998×10^8\:m/s)^2}\ &=\mathrm{7.913×10^{−11}\:\textrm{kg⋅m}/s^2}\ &=\mathrm{7.913×10^{−11}\:J} \end{align*} \nonumber
We then convert the binding energy in joules per nucleus into units of MeV per nuclide:
$\mathrm{7.913×10^{−11}\:J×\dfrac{1\: MeV}{1.602×10^{−13}\:J}=493.9\: MeV} \nonumber$
Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:
$\textrm{Binding energy per nucleon}=\mathrm{\dfrac{493.9\: MeV}{56}=8.820\: MeV/nucleon} \nonumber$
Note that this is almost 25% larger than the binding energy per nucleon for $\ce{^4_2He}$.(Note also that this is the same process as in Example $2\, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.) Exercise \(3$
What is the binding energy per nucleon in $\ce{^{19}_9F}$ (atomic mass, 18.9984 amu)?
Answer
7.810 MeV/nucleon
Summary
An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc2. Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56; these are the most stable nuclei.
• E = mc2
Glossary
band of stability
(also, belt of stability, zone of stability, or valley of stability) region of graph of number of protons versus number of neutrons containing stable (nonradioactive) nuclides
binding energy per nucleon
total binding energy for the nucleus divided by the number of nucleons in the nucleus
electron volt (eV)
measurement unit of nuclear binding energies, with 1 eV equaling the amount energy due to the moving an electron across an electric potential difference of 1 volt
magic number
nuclei with specific numbers of nucleons that are within the band of stability
mass defect
difference between the mass of an atom and the summed mass of its constituent subatomic particles (or the mass “lost” when nucleons are brought together to form a nucleus)
mass-energy equivalence equation
Albert Einstein’s relationship showing that mass and energy are equivalent
nuclear binding energy
energy lost when an atom’s nucleons are bound together (or the energy needed to break a nucleus into its constituent protons and neutrons)
nuclear chemistry
study of the structure of atomic nuclei and processes that change nuclear structure
nucleon
collective term for protons and neutrons in a nucleus
nuclide
nucleus of a particular isotope
radioactivity
phenomenon exhibited by an unstable nucleon that spontaneously undergoes change into a nucleon that is more stable; an unstable nucleon is said to be radioactive
radioisotope
isotope that is unstable and undergoes conversion into a different, more stable isotope
strong nuclear force
force of attraction between nucleons that holds a nucleus together | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.1%3A_Nuclear_Structure_and_Stability.txt |
Learning Objectives
• Identify common particles and energies involved in nuclear reactions
• Write and balance nuclear equations
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Types of Particles in Nuclear Reactions
Many entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure $1$. Protons $(\ce{^{1}_{1}p}$, also represented by the symbol $\ce{^1_1H})$ and neutrons $(\ce{^1_0n})$ are the constituents of atomic nuclei, and have been described previously. Alpha particles $(\ce{^4_2He}$, also represented by the symbol $\ce{^{4}_{2}\alpha})$ are high-energy helium nuclei. Beta particles $(\ce{^{0}_{−1}\beta}$, also represented by the symbol $\ce{^0_{-1}e})$ are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons $(\ce{^0_{+1}e}$, also represented by the symbol $\ce{^0_{+1}β})$ are positively charged electrons (“anti-electrons”). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4, so it is symbolized $\ce{^4_2He}$. This works because, in general, the ion charge is not important in the balancing of nuclear equations.
Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter, particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays (γ)—and other much smaller subnuclear particles, which are beyond the scope of this chapter—according to the mass-energy equivalence equation $E = mc^2$, seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created:
$\ce{^0_{−1}e + ^0_{+1}e } \rightarrow \gamma + \gamma \label{21.3.1}$
Gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays. Gamma rays are produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions.
Balancing Nuclear Reactions
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, was one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an $α$ particle with magnesium-25 $(\ce{^{25}_{12}Mg})$ produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber$
where
• $\ce A$ is the mass number and
• $\ce Z$ is the atomic number of the new nuclide, $\ce X$.
Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$\mathrm{25+4=A+1} \nonumber$
so
$\mathrm{A=28} \nonumber$
Similarly, the charges must balance, so:
$\mathrm{12+2=Z+1} \nonumber$
so
$\mathrm{Z=13} \nonumber$
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{^{28}_{13}Al}$.
Exercise $1$
The nuclide $\ce{^{125}_{53}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber$
Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry:
• The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting α particles: $\ce{^{212}_{84}Po⟶ ^{208}_{82}Pb + ^4_2He}\nonumber$
• The first nuclide to be prepared by artificial means was an isotope of oxygen, 17O. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with α particles: $\ce{^{14}_7N + ^4_2α⟶ ^{17}_8O + ^1_1H} \nonumber$
• James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with 12C by the nuclear reaction between 9Be and 4He: $\ce{^9_4Be + ^4_2He⟶ ^{12}_6C + ^1_0n} \nonumber$
• The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, $\ce{^2_1H}$), by Emilio Segre and Carlo Perrier in 1937: $\ce{^2_1H + ^{97}_{42}Mo⟶2^1_0n + ^{97}_{43}Tc}\nonumber$
• The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was: $\ce{^{235}_{92}U + ^1_0n⟶ ^{87}_{35}Br + ^{146}_{57}La + 3^1_0n} \nonumber$
Summary
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.
Glossary
alpha particle
(α or $\ce{^4_2He}$ or $\ce{^4_2α}$) high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons
antimatter
particles with the same mass but opposite properties (such as charge) of ordinary particles
beta particle
($β$ or $\ce{^0_{-1}e}$ or $\ce{^0_{-1}β}$) high-energy electron
gamma ray
(γ or $\ce{^0_0γ}$) short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality
nuclear reaction
change to a nucleus resulting in changes in the atomic number, mass number, or energy state
positron ($\ce{^0_{+1}β}$ or $\ce{^0_{+1}e}$)
antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.2%3A_Nuclear_Equations.txt |
Learning Objectives
• Recognize common modes of radioactive decay
• Identify common particles and energies involved in nuclear decay reactions
• Write and balance nuclear decay equations
• Calculate kinetic parameters for decay processes, including half-life
• Describe common radiometric dating techniques
Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.
The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure $1$).
Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.
Types of Radioactive Decay
Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure $2$) helped him determine that one type of radiation consisted of positively charged and relatively massive $α$ particles; a second type was made up of negatively charged and much less massive $β$ particles; and a third was uncharged electromagnetic waves, $γ$ rays. We now know that $α$ particles are high-energy helium nuclei, $β$ particles are high-energy electrons, and $γ$ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.
Alpha ($α$) decay is the emission of an α particle from the nucleus. For example, polonium-210 undergoes α decay:
$\ce{^{210}_{84}Po⟶ ^4_2He + ^{206}_{82}Pb} \hspace{40px}\ce{or}\hspace{40px} \ce{^{210}_{84}Po ⟶ ^4_2α + ^{206}_{82}Pb}\nonumber$
Alpha decay occurs primarily in heavy nuclei (A > 200, Z > 83). Because the loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing α decay lies below the band of stability, the daughter nuclide will lie closer to the band.
Beta (β) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes β decay:
$\ce{^{131}_{53}I ⟶ ^0_{-1}e + ^{131}_{54}X} \hspace{40px}\ce{or}\hspace{40px} \ce{^{131}_{53}I ⟶ ^0_{-1}β + ^{131}_{54}Xe}\nonumber$
Beta decay, which can be thought of as the conversion of a neutron into a proton and a β particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Gamma emission (γ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a γ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits γ radiation and is used in many applications including cancer treatment:
$\mathrm{^{60}_{27}Co^* ⟶\, ^0_0γ +\, ^{60}_{27}Co}\nonumber$
There is no change in mass number or atomic number during the emission of a γ ray unless the γ emission accompanies one of the other modes of decay.
Positron emission ($β^+$ decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:
$\ce{^{15}_8O ⟶ ^0_{+1}e + ^{15}_7N} \hspace{40px}\ce{or}\hspace{40px} \ce{^{15}_8O ⟶ ^0_{+1}β + ^{15}_7N}\nonumber$
Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture:
$\ce{^{40}_{19}K + ^0_{-1}e ⟶ ^{40}_{18}Ar}\nonumber$
Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur. Figure $3$ summarizes these types of decay, along with their equations and changes in atomic and mass numbers.
PET Scan
Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure $4$). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.
For example, F-18 is produced by proton bombardment of 18O $(\ce{^{18}_8O + ^1_1p⟶ ^{18}_9F + ^1_0n})$ and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The 18F emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.
Radioactive Decay Series
The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure $5$). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.
Radioactive Half-Lives
Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life (t1/2), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.
For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure $6$). In a given cobalt-60 source, since half of the $\ce{^{60}_{27}Co}$ nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.
Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, N, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:
$\text{decay rate} = \lambda N\nonumber$
with $\lambda$ is the decay constant for the particular radioisotope.
The decay constant, $\lambda$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, t1/2:
$λ=\dfrac{\ln 2}{t_{1/2}}=\dfrac{0.693}{t_{1/2}} \hspace{40px}\ce{or}\hspace{40px} t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}\nonumber$
The first-order equations relating amount, N, and time are:
$N_t=N_0e^{−kt} \hspace{40px}\ce{or}\hspace{40px} t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber$
where N0 is the initial number of nuclei or moles of the isotope, and Nt is the number of nuclei/moles remaining at time t. Example $1$ applies these calculations to find the rates of radioactive decay for specific nuclides.
Example $1$: Rates of Radioactive Decay
$\ce{^{60}_{27}Co}$ decays with a half-life of 5.27 years to produce $\ce{^{60}_{28}Ni}$.
1. What is the decay constant for the radioactive disintegration of cobalt-60?
2. Calculate the fraction of a sample of the $\ce{^{60}_{27}Co}$ isotope that will remain after 15 years.
3. How long does it take for a sample of $\ce{^{60}_{27}Co}$ to disintegrate to the extent that only 2.0% of the original amount remains?
Solution
(a) The value of the rate constant is given by:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5.27\:y}=0.132\:y^{−1}} \nonumber$
(b) The fraction of $\ce{^{60}_{27}Co}$ that is left after time t is given by $\dfrac{N_t}{N_0}$. Rearranging the first-order relationship Nt = N0eλt to solve for this ratio yields:
$\dfrac{N_t}{N_0}=e^{-λt}=e^\mathrm{-(0.132/y)(15.0/y)}=0.138 \nonumber$
The fraction of $\ce{^{60}_{27}Co}$ that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the $\ce{^{60}_{27}Co}$ originally present will remain after 15 years.
(c) 2.00% of the original amount of $\ce{^{60}_{27}Co}$ is equal to 0.0200 × N0. Substituting this into the equation for time for first-order kinetics, we have:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)=−\dfrac{1}{0.132\:\ce y^{−1}}\ln\left(\dfrac{0.0200×N_0}{N_0}\right)=29.6\:\ce y \nonumber$
Exercise $1$
Radon-222, $\ce{^{222}_{86}Rn}$, has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?
Answer
11.1 days
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of $\ce{^{209}_{83}Bi}$ is 1.9 × 1019 years; $\ce{^{239}_{94}Ra}$ is 24,000 years; $\ce{^{222}_{86}Rn}$ is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 × 10–3 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table $1$, and others are listed in Appendix N1.
Table $1$: Half-lives of Radioactive Isotopes Important to Medicine
Type Decay Mode Half-Life Uses
F-18 β+ decay 110. minutes PET scans
Co-60 β decay, γ decay 5.27 years cancer treatment
Tc-99m1 γ decay 8.01 hours scans of brain, lung, heart, bone
I-131 β decay 8.02 days thyroid scans and treatment
Tl-201 electron capture 73 hours heart and arteries scans; cardiac stress tests
The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit $γ$ radiation to rid themselves of excess energy and become (more) stable.
Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
Radioactive Dating Using Carbon-14
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: $\ce{^{12}_6C}$, which constitutes about 99% of the carbon on earth; $\ce{^{13}_6C}$, about 1% of the total; and trace amounts of $\ce{^{14}_6C}$. Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
$\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C + ^1_1H}\nonumber$
All isotopes of carbon react with oxygen to produce CO2 molecules. The ratio of $\ce{^{14}_6CO2}$ to $\ce{^{12}_6CO2}$ depends on the ratio of $\ce{^{14}_6CO}$ to $\ce{^{12}_6CO}$ in the atmosphere. The natural abundance of $\ce{^{14}_6CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of $\ce{^{14}_6C ^{14}_6CO2}$ and $\ce{^{12}_6CO2}$ into plants is a regular part of the photosynthesis process, which means that the $\ce{^{14}_6C: ^{12}_6C}$ ratio found in a living plant is the same as the $\ce{^{14}_6C: ^{12}_6C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because $\ce{^{12}_6C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years:
$\ce{^{14}_6C⟶ ^{14}_7N + ^0_{-1}e}\nonumber$
Thus, the $\ce{^{14}_6C: ^{12}_6C}$ ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure $7$ visually depicts this process.
For example, with the half-life of $\ce{^{14}_6C}$ being 5730 years, if the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of $\ce{^{14}_6C : ^{12}_6C}$ ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
Example $2$: Radiocarbon Dating
A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls.
Solution
The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, N, in the relationship:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)⟶t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right) \nonumber$
where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript t represents the current time.
The decay constant can be determined from the half-life of C-14, 5730 years:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5730\: y}=1.21×10^{−4}\:y^{−1}} \nonumber$
Substituting and solving, we have:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right)=\mathrm{−\dfrac{1}{1.21×10^{−4}\:y^{−1}}\ln\left(\dfrac{10.8\:dis/min/g\: C}{13.6\:dis/min/g\: C}\right)=1910\: y}\nonumber$
Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure $8$).
Exercise $2$
More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end?
Answer
about 3350 years ago, or approximately 1340 BC
There have been some significant, well-documented changes to the $\ce{^{14}_6C : ^{12}_6C}$ ratio. The accuracy of a straightforward application of this technique depends on the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of CO2 molecules (largely $\ce{^{12}_6CO2}$) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the $\ce{^{14}_6C}$ has decayed), the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the atmosphere may be changing. This manmade increase in $\ce{^{12}_6CO2}$ in the atmosphere causes the $\ce{^{14}_6C : ^{12}_6C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radioactive Dating Using Nuclides Other than Carbon-14
Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Example $3$: Radioactive Dating of Rocks
An igneous rock contains 9.58 × 10–5 g of U-238 and 2.51 × 10–5 g of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed.
Solution
The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.
The amount of U-238 currently in the rock is:
$\mathrm{9.58×10^{−5}\cancel{g\: U}×\left( \dfrac{1\: mol\: U}{238\cancel{g\: U}}\right )=4.03×10^{−7}\:mol\: U}\nonumber$
Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is:
$\mathrm{2.51×10^{-5}\cancel{g\: Pb}×\left( \dfrac{1\cancel{mol\: Pb}}{206\cancel{g\: Pb}}\right )×\left(\dfrac{1\: mol\: U}{1\cancel{mol\: Pb}}\right)=1.22×10^{-7}\:mol\: U}\nonumber$
The total amount of U-238 originally present in the rock is therefore:
$\mathrm{4.03×10^{−7}\:mol+1.22×10^{−7}\:mol=5.25×10^{−7}\:mol\: U}\nonumber$
The amount of time that has passed since the formation of the rock is given by:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber$
with N0 representing the original amount of U-238 and Nt representing the present amount of U-238.
U-238 decays into Pb-206 with a half-life of 4.5 × 109 y, so the decay constant λ is:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{4.5×10^9\:y}=1.54×10^{−10}\:y^{−1}}\nonumber$
Substituting and solving, we have:
$t=\mathrm{−\dfrac{1}{1.54×10^{−10}\:y^{−1}}\ln\left(\dfrac{4.03×10^{−7}\cancel{mol\: U}}{5.25×10^{−7}\cancel{mol\: U}}\right)=1.7×10^9\:y}\nonumber$
Therefore, the rock is approximately 1.7 billion years old.
Exercise $3$
A sample of rock contains 6.14 × 10–4 g of Rb-87 and 3.51 × 10–5 g of Sr-87. Calculate the age of the rock. (The half-life of the β decay of Rb-87 is 4.7 × 1010 y.)
Answer
3.7 × 109 y
Summary
Nuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some substances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.
Key Equations
• decay rate = λN
• $t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}$
Glossary
alpha (α) decay
loss of an alpha particle during radioactive decay
beta (β) decay
breakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle
daughter nuclide
nuclide produced by the radioactive decay of another nuclide; may be stable or may decay further
electron capture
combination of a core electron with a proton to yield a neutron within the nucleus
gamma (γ) emission
decay of an excited-state nuclide accompanied by emission of a gamma ray
half-life (t1/2)
time required for half of the atoms in a radioactive sample to decay
parent nuclide
unstable nuclide that changes spontaneously into another (daughter) nuclide
positron emission
(also, β+ decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted
radioactive decay
spontaneous decay of an unstable nuclide into another nuclide
radioactive decay series
chains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product
radiocarbon dating
highly accurate means of dating objects 30,000–50,000 years old that were derived from once-living matter; achieved by calculating the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the object vs. the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the present-day atmosphere
radiometric dating
use of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.3%3A_Radioactive_Decay.txt |
Learning Objectives
• Describe the synthesis of transuranium nuclides
• Explain nuclear fission and fusion processes
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission and fusion reactors
After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Science learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace.
Synthesis of Nuclides
Nuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:
$\ce{^{14}_7N + ^4_2He ⟶ ^{17}_8O + ^1_1H} \nonumber$
The $\ce{^{17}_8O}$ and $\ce{^1_1H}$ nuclei that are produced are stable, so no further (nuclear) changes occur.
To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news.
CERN Particle Accelerator
Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure $1$). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.
In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2103 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.
Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:
\begin{align*} \ce{^{238}_{92}U + ^1_0n &⟶ ^{239}_{92}U} && \[4pt] &\ce{^{239}_{92}U &⟶ ^{239}_{93}Np + ^0_{−1}e \,\,\,\mathit{t}_{1/2}} &&\textrm{half-life}=\mathrm{23.5\: min} \[4pt] &\ce{^{239}_{93}Np &⟶ ^{239}_{94}Pu + ^0_{−1}e\,\,\, \mathit{t}_{1/2}} &&\textrm{half-life}=\mathrm{2.36\: days} \end{align*} \nonumber
Plutonium is now mostly formed in nuclear reactors as a byproduct during the decay of uranium. Some of the neutrons that are released during U-235 decay combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. It is possible to summarize these equations as:
$\mathrm{\ce{^{238}_{92}U} + {^1_0n}⟶ \ce{^{239}_{92}U} \xrightarrow{β^-} \ce{^{239}_{93}Np} \xrightarrow{β^-} \ce{^{239}_{94}Pu}} \nonumber$
Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.
Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.
The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table $1$.
Table $1$: Preparation of Some of the Transuranium Elements
Name Symbol Atomic Number Reaction
americium Am 95 $\ce{^{239}_{94}Pu + ^1_0n ⟶ ^{240}_{95}Am + ^0_{−1}e}$
curium Cm 96 $\ce{^{239}_{94}Pu + ^4_2He ⟶ ^{242}_{96}Cm + ^1_0n}$
californium Cf 98 $\ce{^{242}_{96}Cm + ^4_2He⟶ ^{245}_{98}Cf + ^1_0n}$
einsteinium Es 99 $\ce{^{238}_{92}U + 15^1_0n⟶ ^{253}_{99}Es + 7^0_{−1}e}$
mendelevium Md 101 $\ce{^{253}_{99}Es + ^4_2He ⟶ ^{256}_{101}Md + ^1_0n}$
nobelium No 102 $\ce{^{246}_{96}Cm + ^{12}_6C ⟶ ^{254}_{102}No + 4 ^1_0n}$
rutherfordium Rf 104 $\ce{^{249}_{98}Cf + ^{12}_6C⟶ ^{257}_{104}Rf + 4 ^1_0n}$
seaborgium
Sg
106
$\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}$
$\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}$
meitnerium Mt 107 $\ce{^{209}_{83}Bi + ^{58}_{26}Fe ⟶ ^{266}_{109}Mt + ^1_0n}$
Nuclear Fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure $2$.
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure $3$. Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure $4$). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure $5$).
An atomic bomb (Figure $6$) contains several pounds of fissionable material, $\ce{^{235}_{92}U}$ or $\ce{^{239}_{94}Pu}$, a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure $7$). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (Figure 21.5.5). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water $\ce{( ^2_1H2O)}$ or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure $8$) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
$\ce{^{10}_5B + ^1_0n⟶ ^7_3Li + ^4_2He} \nonumber$
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
$\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g) \nonumber$
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure $9$).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure $10$).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed.
Nuclear Fusion and Fusion Reactors
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
$\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}} \nonumber$
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1$ and a triton, $^3_1$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
$\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n} \nonumber$
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $11$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.
Summary
It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way.
Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). Because the neutrons may induce additional fission reactions when they combine with other heavy nuclei, a chain reaction can result. Useful power is obtained if the fission process is carried out in a nuclear reactor. The conversion of light nuclei into heavier nuclei (fusion) also produces energy. At present, this energy has not been contained adequately and is too expensive to be feasible for commercial energy production.
Glossary
chain reaction
repeated fission caused when the neutrons released in fission bombard other atoms
containment system
(also, shield) a three-part structure of materials that protects the exterior of a nuclear fission reactor and operating personnel from the high temperatures, pressures, and radiation levels inside the reactor
control rod
material inserted into the fuel assembly that absorbs neutrons and can be raised or lowered to adjust the rate of a fission reaction
critical mass
amount of fissionable material that will support a self-sustaining (nuclear fission) chain reaction
fissile (or fissionable)
when a material is capable of sustaining a nuclear fission reaction
fission
splitting of a heavier nucleus into two or more lighter nuclei, usually accompanied by the conversion of mass into large amounts of energy
fusion
combination of very light nuclei into heavier nuclei, accompanied by the conversion of mass into large amounts of energy
fusion reactor
nuclear reactor in which fusion reactions of light nuclei are controlled
nuclear fuel
fissionable isotope present in sufficient quantities to provide a self-sustaining chain reaction in a nuclear reactor
nuclear moderator
substance that slows neutrons to a speed low enough to cause fission
nuclear reactor
environment that produces energy via nuclear fission in which the chain reaction is controlled and sustained without explosion
nuclear transmutation
conversion of one nuclide into another nuclide
particle accelerator
device that uses electric and magnetic fields to increase the kinetic energy of nuclei used in transmutation reactions
reactor coolant
assembly used to carry the heat produced by fission in a reactor to an external boiler and turbine where it is transformed into electricity
subcritical mass
amount of fissionable material that cannot sustain a chain reaction; less than a critical mass
supercritical mass
amount of material in which there is an increasing rate of fission
transmutation reaction
bombardment of one type of nuclei with other nuclei or neutrons
transuranium element
element with an atomic number greater than 92; these elements do not occur in nature | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.4%3A_Transmutation_and_Nuclear_Energy.txt |
Learning Objectives
• List common applications of radioactive isotopes
Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.
Radioisotopes have revolutionized medical practice, where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 $\ce{(^{99}_{43}Tc)}$, thallium-201 $\ce{(^{201}_{81}Tl)}$, iodine-131 $\ce{(^{131}_{53}I)}$, and sodium-24 $\ce{(^{24}_{11}Na)}$. Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure $1$) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.
Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure $2$). The parent nuclide Mo-99 is part of a molybdate ion, $\ce{MoO4^2-}$; when it decays, it forms the pertechnetate ion, $\ce{TcO4-}$. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.
Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure $3$). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that has been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.
Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is:
$\ce{^{59}_{27}Co + ^1_0n⟶ ^{60}_{27}Co⟶ ^{60}_{28}Ni + ^0_{−1}β + 2^0_0γ} \nonumber$
The overall decay scheme for this is shown graphically in Figure $4$.
Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.
For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:
$\ce{6CO2}(g)+\ce{6H2O}(l)⟶\ce{C6H12O6}(s)+\ce{6O2}(g), \nonumber$
but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO2 containing a high concentration of $\ce{^{14}_6C}$. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.
Commercial applications of radioactive materials are equally diverse (Figure $5$). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.
Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure $6$). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm.
Summary
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.
Glossary
chemotherapy
similar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells
external beam radiation therapy
radiation delivered by a machine outside the body
internal radiation therapy
(also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells
radiation therapy
use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing
radioactive tracer
(also, radioactive label) radioisotope used to track or follow a substance by monitoring its radioactive emissions | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.5%3A_Uses_of_Radioisotopes.txt |
Learning Objectives
• Describe the biological impact of ionizing radiation.
• Define units for measuring radiation exposure.
• Explain the operation of common tools for detecting radioactivity.
• List common sources of radiation exposure in the US.
The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure $1$).
Ionizing vs. Nonionizing Radiation
There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure $2$).
Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H2O (the most abundant molecule in living organisms), which forms a H2O+ ion that reacts with water, forming a hydronium ion and a hydroxyl radical:
Biological Effects of Exposure to Radiation
Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.
Different types of radiation have differing abilities to pass through material (Figure $4$). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays.
For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure $5$).
Radon is found in buildings across the country, with amounts dependent on location. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the level found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.
Measuring Radiation Exposure
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure $6$). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.
A variety of units are used to measure various aspects of radiation (Table $1$). The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = $3.7 \times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose.
Table $1$: Units Used for Measuring Radiation
Measurement Purpose Unit Quantity Measured Description
activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second
curie (Ci) amount of sample that undergoes $\mathrm{3.7 \times 10^{10}\; decays/second}$
absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue
radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue
biologically effective dose sievert (Sv) tissue damage Sv = RBE × Gy
roentgen equivalent for man (rem) Rem = RBE × rad
The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (1 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy), along with a biological factor referred to as the RBE (for relative biological effectiveness), that is an approximate measure of the relative damage done by the radiation. These are related by:
$\text{number of rems}=\text{RBE} \times \text{number of rads} \label{Eq2}$
with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation.
Example $1$: Amount of Radiation
Cobalt-60 (t1/2 = 5.26 y) is used in cancer therapy since the $\gamma$ rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment.
1. What is its activity in Bq?
2. What is its activity in Ci?
Solution
The activity is given by:
$\textrm{Activity}=λN=\left( \dfrac{\ln 2}{t_{1/2} } \right) N=\mathrm{\left( \dfrac{\ln 2}{5.26\ y} \right) \times 5.00 \ g=0.659\ \dfrac{g}{y} \ of\ \ce{^{60}Co} \text{ that decay}} \nonumber$
And to convert this to decays per second:
$\mathrm{0.659\; \frac{g}{y} \times \dfrac{y}{365 \;day} \times \dfrac{1\; day}{ 24\; hours} \times \dfrac{1\; h}{3,600 \;s} \times \dfrac{1\; mol}{59.9\; g} \times \dfrac{6.02 \times 10^{23} \;atoms}{1 \;mol} \times \dfrac{1\; decay}{1\; atom}} \nonumber$
$\mathrm{=2.10 \times 10^{14} \; \frac{decay}{s}} \nonumber$
(a) Since $\mathrm{1\; Bq = 1\; \frac{ decay}{s}}$, the activity in Becquerel (Bq) is:
$\mathrm{2.10 \times 10^{14} \dfrac{decay}{s} \times \left(\dfrac{1\ Bq}{1 \; \frac{decay}{s}} \right)=2.10 \times 10^{14} \; Bq} \nonumber$
(b) Since $\mathrm{1\ Ci = 3.7 \times 10^{11}\; \frac{decay}{s}}$, the activity in curie (Ci) is:
$\mathrm{2.10 \times 10^{14} \frac{decay}{s} \times \left( \dfrac{1\ Ci}{3.7 \times 10^{11} \frac{decay}{s}} \right) =5.7 \times 10^2\;Ci} \nonumber$
Exercise $1$
Tritium is a radioactive isotope of hydrogen ($t_{1/2} = \mathrm{12.32\; years}$) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci?
Answer a
$\mathrm{3.56 \times 10^{11} Bq}$
Answer b
$\mathrm{0.962\; Ci}$
Effects of Long-term Radiation Exposure on the Human Body
The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure $8$, the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).
A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table $2$.
Table $2$: Health Effects of Radiation
Exposure (rem) Health Effect Time to Onset (Without Treatment)
5–10 changes in blood chemistry
50 nausea hours
55 fatigue
70 vomiting
75 hair loss 2–3 weeks
90 diarrhea
100 hemorrhage
400 possible death within 2 months
1000 destruction of intestinal lining
internal bleeding
death 1–2 weeks
2000 damage to central nervous system
loss of consciousness minutes
death hours to days
It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.
Summary
We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating.
Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source of radiation, and limiting time of exposure.
Footnotes
1. 1 Source: US Environmental Protection Agency
Glossary
becquerel (Bq)
SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s.
curie (Ci)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
Geiger counter
Instrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube.
gray (Gy)
SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue.
ionizing radiation
Radiation that can cause a molecule to lose an electron and form an ion.
millicurie (mCi)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
nonionizing radiation
Radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules.
radiation absorbed dose (rad)
SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy.
radiation dosimeter
Device that measures ionizing radiation and is used to determine personal radiation exposure.
relative biological effectiveness (RBE)
Measure of the relative damage done by radiation.
roentgen equivalent man (rem)
Unit for radiation damage, frequently used in medicine; 1 rem = 1 Sv.
scintillation counter
Instrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation.
sievert (Sv)
SI unit measuring tissue damage caused by radiation; takes energy and biological effects of radiation into account. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.6%3A_Biological_Effects_of_Radiation.txt |
21.2: Nuclear Structure and Stability
Q21.2.1
Write the following isotopes in hyphenated form (e.g., “carbon-14”)
1. $\ce{^{24}_{11}Na}$
2. $\ce{^{29}_{13}Al}$
3. $\ce{^{73}_{36}Kr}$
4. $\ce{^{194}_{77}Ir}$
Q21.2.2
Write the following isotopes in nuclide notation (e.g., " $\ce{^{14}_6C}$ ")
1. oxygen-14
2. copper-70
3. tantalum-175
4. francium-217
Q21.2.3
For the following isotopes that have missing information, fill in the missing information to complete the notation
1. $\ce{^{34}_{14}X}$
2. $\ce{^{36}_P}$
3. $\ce{^{57}_{X}Mn}$
4. $\ce{^{121}_{56}X}$
Q21.2.4
For each of the isotopes in Question 21.2.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.
Q21.2.5
Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:
1. 25 protons, 20 neutrons, 24 electrons
2. 45 protons, 24 neutrons, 43 electrons
3. 53 protons, 89 neutrons, 54 electrons
4. 97 protons, 146 neutrons, 97 electrons
Q21.2.6
Calculate the density of the $\ce{^{24}_{12}Mg}$ nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 × 10–13 cm and is spherical in shape.
Q21.2.7
What are the two principal differences between nuclear reactions and ordinary chemical changes?
Q21.2.8
The mass of the atom $\ce{^{23}_{11}Na}$ is 22.9898 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
Q21.2.9
Which of the following nuclei lie within the band of stability?
1. chlorine-37
2. calcium-40
3. 204Bi
4. 56Fe
5. 206Pb
6. 211Pb
7. 222Rn
8. carbon-14
Q21.2.10
Which of the following nuclei lie within the band of stability?
1. argon-40
2. oxygen-16
3. 122Ba
4. 58Ni
5. 205Tl
6. 210Tl
7. 226Ra
8. magnesium-24
21.3: Nuclear Equations
Q21.3.1
Write a brief description or definition of each of the following:
1. nucleon
2. α particle
3. β particle
4. positron
5. γ ray
6. nuclide
7. mass number
8. atomic number
Q21.3.2
Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei?
Q21.3.3
Complete each of the following equations by adding the missing species:
1. $\ce{^{27}_{13}Al + ^4_2He⟶\:? + ^1_0n}$
2. $\ce{^{239}_{94}Pu +\, ? ⟶ ^{242}_{96}Cm + ^1_0n}$
3. $\ce{^{14}_7N + ^4_2He⟶\:? + ^1_1H}$
4. $\ce{^{235}_{92}U⟶\:? + ^{135}_{55}Cs + 4^1_0n}$
Q21.3.4
Complete each of the following equations:
1. $\ce{^7_3Li +\, ?⟶2^4_2He}$
2. $\ce{^{14}_6C⟶ ^{14}_7N +\, ?}$
3. $\ce{^{27}_{13}Al + ^4_2He⟶\,? + ^1_0n}$
4. $\ce{^{250}_{96}Cm ⟶\, ? + ^{98}_{38}Sr + 4^1_0n}$
Q21.3.5
Write a balanced equation for each of the following nuclear reactions:
1. the production of 17O from 14N by α particle bombardment
2. the production of 14C from 14N by neutron bombardment
3. the production of 233Th from 232Th by neutron bombardment
4. the production of 239U from 238U by $\ce{^2_1H}$ bombardment
Q21.3.6
Technetium-99 is prepared from 98Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as 99Tc*. This excited nucleus relaxes to the ground state, represented as 99Tc, by emitting a γ ray. The ground state of 99Tc then emits a β particle. Write the equations for each of these nuclear reactions.
Q21.3.7
The mass of the atom $\ce{^{19}_9F}$ is 18.99840 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
Q21.3.8
For the reaction $\ce{^{14}_6C ⟶ ^{14}_7N +\, ?}$, if 100.0 g of carbon reacts, what volume of nitrogen gas (N2) is produced at 273 K and 1 atm?
21.4: Radioactive Decay
Q21.4.1
What are the types of radiation emitted by the nuclei of radioactive elements?
Q21.4.2
What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?
1. an α particle is emitted
2. a β particle is emitted
3. γ radiation is emitted
4. a positron is emitted
5. an electron is captured
Q21.4.3
What is the change in the nucleus that results from the following decay scenarios?
1. emission of a β particle
2. emission of a β+ particle
3. capture of an electron
Q21.4.4
Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include α particles.
Q21.4.5
Why is electron capture accompanied by the emission of an X-ray?
Q21.4.6
Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability.
Q21.4.7
Which of the following nuclei is most likely to decay by positron emission? Explain your choice.
1. chromium-53
2. manganese-51
3. iron-59
Q21.4.8
The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.
1. $\ce{^{34}_{15}P}$
2. $\ce{^{239}_{92}U}$
3. $\ce{^{38}_{20}Ca}$
4. $\ce{^3_1H}$
5. $\ce{^{245}_{94}Pu}$
Q21.4.9
The following nuclei do not lie in the band of stability. How would they be expected to decay?
1. $\ce{^{28}_{15}P}$
2. $\ce{^{235}_{92}U}$
3. $\ce{^{37}{20}Ca}$
4. $\ce{^9_3Li}$
5. $\ce{^{245}_{96}Cm}$
Q21.4.10
Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:
1. $\ce{^6_2He}$
2. $\ce{^{60}_{30}Zn}$
3. $\ce{^{235}_{91}Pa}$
4. $\ce{^{241}_{94}Np}$
5. 18F
6. 129Ba
7. 237Pu
Q21.4.11
Write a nuclear reaction for each step in the formation of $\ce{^{218}_{84}Po}$ from $\ce{^{238}_{92}U}$, which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α, α particles, in that order.
Q21.4.12
Write a nuclear reaction for each step in the formation of $\ce{^{208}_{82}Pb}$ from $\ce{^{228}_{90}Th}$, which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.
Q21.4.13
Define the term half-life and illustrate it with an example.
Q21.4.14
A 1.00 × 10–6-g sample of nobelium, $\ce{^{254}_{102}No}$, has a half-life of 55 seconds after it is formed. What is the percentage of $\ce{^{254}_{102}No}$ remaining at the following times?
1. 5.0 min after it forms
2. 1.0 h after it forms
Q21.4.15
239Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y?
Q21.4.16
The isotope 208Tl undergoes β decay with a half-life of 3.1 min.
1. What isotope is produced by the decay?
2. How long will it take for 99.0% of a sample of pure 208Tl to decay?
3. What percentage of a sample of pure 208Tl remains un-decayed after 1.0 h?
Q21.4.17
If 1.000 g of $\ce{^{226}_{88}Ra}$ produces 0.0001 mL of the gas $\ce{^{222}_{86}Rn}$ at STP (standard temperature and pressure) in 24 h, what is the half-life of 226Ra in years?
Q21.4.18
The isotope $\ce{^{90}_{38}Sr}$ is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life.
Q21.4.19
Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of $\ce{^{99}_{43}Tc}$.
Q21.4.20
What is the age of mummified primate skin that contains 8.25% of the original quantity of 14C?
Q21.4.21
A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87.
1. Calculate the age of the rock if the half-life of the decay of rubidium by β emission is 4.7 × 1010 y.
2. If some $\ce{^{87}_{38}Sr}$ was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.
Q21.4.22
A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of $\ce{^{238}_{92}U}$ and 2.52 mg of $\ce{^{206}_{82}Pb}$. Calculate the age of the ore. The half-life of $\ce{^{238}_{92}U}$ is 4.5 × 109 yr.
Q21.4.23
Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this 239Pu was the capture of neutrons by 238U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 109 years ago?
Q21.4.24
A $\ce{^7_4Be}$ atom (mass = 7.0169 amu) decays into a $\ce{^7_3Li}$ atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?
Q21.4.25
A $\ce{^8_5B}$ atom (mass = 8.0246 amu) decays into a $\ce{^8_4Be}$ atom (mass = 8.0053 amu) by loss of a β+ particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?
Q21.4.26
Isotopes such as 26Al (half-life: 7.2 × 105 years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.
1. 26Al decays by β+ emission or electron capture. Write the equations for these two nuclear transformations.
2. The earth was formed about 4.7 × 109 (4.7 billion) years ago. How old was the earth when 99.999999% of the 26Al originally present had decayed?
Q21.4.27
Write a balanced equation for each of the following nuclear reactions:
1. bismuth-212 decays into polonium-212
2. beryllium-8 and a positron are produced by the decay of an unstable nucleus
3. neptunium-239 forms from the reaction of uranium-238 with a neutron and then spontaneously converts into plutonium-239
4. strontium-90 decays into yttrium-90
Q21.4.28
Write a balanced equation for each of the following nuclear reactions:
1. mercury-180 decays into platinum-176
2. zirconium-90 and an electron are produced by the decay of an unstable nucleus
3. thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium-228 by beta decay
4. neon-19 decays into fluorine-19
21.5: Transmutation and Nuclear Energy
Q21.5.1
Write the balanced nuclear equation for the production of the following transuranium elements:
1. berkelium-244, made by the reaction of Am-241 and He-4
2. fermium-254, made by the reaction of Pu-239 with a large number of neutrons
3. lawrencium-257, made by the reaction of Cf-250 and B-11
4. dubnium-260, made by the reaction of Cf-249 and N-15
Q21.5.2
How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic?
Q21.5.3
Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?
Q21.5.4
Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.
Q21.5.5
Describe the components of a nuclear reactor.
Q21.5.6
In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.
Q21.5.7
Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.
Q21.5.8
The mass of a hydrogen atom $\ce{(^1_1H)}$ is 1.007825 amu; that of a tritium atom $\ce{(^3_1H)}$ is 3.01605 amu; and that of an α particle is 4.00150 amu. How much energy in kilojoules per mole of $\ce{^4_2He}$ produced is released by the following fusion reaction: $\ce{^1_1H + ^3_1H ⟶ ^4_2He}$.
21.6: Uses of Radioisotopes
Q21.6.1
How can a radioactive nuclide be used to show that the equilibrium:
$\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)$
is a dynamic equilibrium?
Q21.6.2
Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?
Q21.6.3
Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-131 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β emission.
1. Write a nuclear equation for the decay.
2. How long will it take for 95.0% of a dose of I-131 to decay?
21.7: Biological Effects of Radiation
Q21.7.1
If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:
1. cobalt-60 (a strong γ emitter used for irradiation)
2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)
Q21.7.2
Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous?
Q21.7.3
Given specimens uranium-232 ($t_{1/2} = \mathrm{68.9 \;y}$) and uranium-233 ($t_{1/2} = \mathrm{159,200\; y}$) of equal mass, which one would have greater activity and why?
Q21.7.4
A scientist is studying a 2.234 g sample of thorium-229 (t1/2 = 7340 y) in a laboratory.
1. What is its activity in Bq?
2. What is its activity in Ci?
Q21.7.5
Given specimens neon-24 ($t_{1/2} = \mathrm{3.38\; min}$) and bismuth-211 ($t_{1/2} = \mathrm{2.14\; min}$) of equal mass, which one would have greater activity and why? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/20%3A_Nuclear_Chemistry/20.E%3A_Nuclear_Chemistry_%28Exercises%29.txt |
Organic chemistry involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter in its various forms that contain carbon atoms. Study of structure includes many physical and chemical methods to determine the chemical composition and the chemical constitution of organic compounds and materials. Study of properties includes both physical properties and chemical properties, and uses similar methods as well as methods to evaluate chemical reactivity, with the aim to understand the behavior of the organic matter.
21: Organic Chemistry
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and CO2. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
Figure 20.0.1: All organic compounds contain carbon and most are formed by living things, although they are also formed by geological and artificial processes. (credit left: modification of work by Jon Sullivan; credit left middle: modification of work by Deb Tremper; credit right middle: modification of work by “annszyp”/Wikimedia Commons; credit right: modification of work by George Shuklin)
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings.
21.1: Hydrocarbons
Learning Objectives
• Explain the importance of hydrocarbons and the reason for their diversity
• Name saturated and unsaturated hydrocarbons, and molecules derived from them
• Describe the reactions characteristic of saturated and unsaturated hydrocarbons
• Identify structural and geometric isomers of hydrocarbons
The largest database1 of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated2 at 1060—an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities.
The simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. This leads to differences in geometries and in the hybridization of the carbon orbitals.
Alkanes
Alkanes, or saturated hydrocarbons, contain only single covalent bonds between carbon atoms. Each of the carbon atoms in an alkane has sp3 hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. The Lewis structures and models of methane, ethane, and pentane are illustrated in Figure $1$. Carbon chains are usually drawn as straight lines in Lewis structures, but one has to remember that Lewis structures are not intended to indicate the geometry of molecules. Notice that the carbon atoms in the structural models (the ball-and-stick and space-filling models) of the pentane molecule do not lie in a straight line. Because of the sp3 hybridization, the bond angles in carbon chains are close to 109.5°, giving such chains in an alkane a zigzag shape.
The structures of alkanes and other organic molecules may also be represented in a less detailed manner by condensed structural formulas (or simply, condensed formulas). Instead of the usual format for chemical formulas in which each element symbol appears just once, a condensed formula is written to suggest the bonding in the molecule. These formulas have the appearance of a Lewis structure from which most or all of the bond symbols have been removed. Condensed structural formulas for ethane and pentane are shown at the bottom of Figure $1$, and several additional examples are provided in the exercises at the end of this chapter.
A common method used by organic chemists to simplify the drawings of larger molecules is to use a skeletal structure (also called a line-angle structure). In this type of structure, carbon atoms are not symbolized with a C, but represented by each end of a line or bend in a line. Hydrogen atoms are not drawn if they are attached to a carbon. Other atoms besides carbon and hydrogen are represented by their elemental symbols. Figure $2$ shows three different ways to draw the same structure.
Example $1$
Drawing Skeletal Structures Draw the skeletal structures for these two molecules:
Solution
Each carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):
Exercise $1$
Draw the skeletal structures for these two molecules:
Answer
Example $2$
Interpreting Skeletal Structures Identify the chemical formula of the molecule represented here:
Solution
There are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of C8H16.
Location of the hydrogen atoms:
Exercise $2$
Identify the chemical formula of the molecule represented here:
Answer
C9H20
All alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of CnH2n+2. The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table $1$) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change.
Table $1$: Properties of Some Alkanes
Alkane Molecular Formula Melting Point (°C) Boiling Point (°C) Phase at STP4 Number of Structural Isomers
methane CH4 –182.5 –161.5 gas 1
ethane C2H6 –183.3 –88.6 gas 1
propane C3H8 –187.7 –42.1 gas 1
butane C4H10 –138.3 –0.5 gas 2
pentane C5H12 –129.7 36.1 liquid 3
hexane C6H14 –95.3 68.7 liquid 5
heptane C7H16 –90.6 98.4 liquid 9
octane C8H18 –56.8 125.7 liquid 18
nonane C9H20 –53.6 150.8 liquid 35
decane C10H22 –29.7 174.0 liquid 75
tetradecane C14H30 5.9 253.5 solid 1858
octadecane C18H38 28.2 316.1 solid 60,523
Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula C4H10: They are called n-butane and 2-methylpropane (or isobutane), and have the following Lewis structures:
The compounds n-butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The n-butane molecule contains an unbranched chain, meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term normal, or the prefix n, to refer to a chain of carbon atoms without branching. The compound 2–methylpropane has a branched chain (the carbon atom in the center of the Lewis structure is bonded to three other carbon atoms)
Identifying isomers from Lewis structures is not as easy as it looks. Lewis structures that look different may actually represent the same isomers. For example, the three structures in Figure $3$ all represent the same molecule, n-butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms.
The Basics of Organic Nomenclature: Naming Alkanes
The International Union of Pure and Applied Chemistry (IUPAC) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. The nomenclature for alkanes is based on two rules:
1. To name an alkane, first identify the longest chain of carbon atoms in its structure. A two-carbon chain is called ethane; a three-carbon chain, propane; and a four-carbon chain, butane. Longer chains are named as follows: pentane (five-carbon chain), hexane (6), heptane (7), octane (8), nonane (9), and decane (10). These prefixes can be seen in the names of the alkanes described in Table $1$.
2. Add prefixes to the name of the longest chain to indicate the positions and names of substituents. Substituents are branches or functional groups that replace hydrogen atoms on a chain. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. We number the carbon atoms in the chain by counting from the end of the chain nearest the substituents. Multiple substituents are named individually and placed in alphabetical order at the front of the name.
3.
When more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending -o replaces -ide at the end of the name of an electronegative substituent (in ionic compounds, the negatively charged ion ends with -ide like chloride; in organic compounds, such atoms are treated as substituents and the -o ending is used). The number of substituents of the same type is indicated by the prefixes di- (two), tri- (three), tetra- (four), and so on (for example, difluoro- indicates two fluoride substituents).
Example $3$: Naming Halogen-substituted Alkanes
Name the molecule whose structure is shown here:
Solution
The four-carbon chain is numbered from the end with the chlorine atom. This puts the substituents on positions 1 and 2 (numbering from the other end would put the substituents on positions 3 and 4). Four carbon atoms means that the base name of this compound will be butane. The bromine at position 2 will be described by adding 2-bromo-; this will come at the beginning of the name, since bromo- comes before chloro- alphabetically. The chlorine at position 1 will be described by adding 1-chloro-, resulting in the name of the molecule being 2-bromo-1-chlorobutane.
Exercise $3$
Name the following molecule:
Answer
3,3-dibromo-2-iodopentane
We call a substituent that contains one less hydrogen than the corresponding alkane an alkyl group. The name of an alkyl group is obtained by dropping the suffix -ane of the alkane name and adding -yl:
The open bonds in the methyl and ethyl groups indicate that these alkyl groups are bonded to another atom.
Example $4$ Naming Substituted Alkanes
Name the molecule whose structure is shown here:
Solution
The longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right—this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)—so we take our name for two carbons eth- and attach -yl at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane.
Exercise $4$
Name the following molecule:
Answer
4-propyloctane
Some hydrocarbons can form more than one type of alkyl group when the hydrogen atoms that would be removed have different “environments” in the molecule. This diversity of possible alkyl groups can be identified in the following way: The four hydrogen atoms in a methane molecule are equivalent; they all have the same environment. They are equivalent because each is bonded to a carbon atom (the same carbon atom) that is bonded to three hydrogen atoms. (It may be easier to see the equivalency in the ball and stick models in Figure $3$. Removal of any one of the four hydrogen atoms from methane forms a methyl group. Likewise, the six hydrogen atoms in ethane are equivalent and removing any one of these hydrogen atoms produces an ethyl group. Each of the six hydrogen atoms is bonded to a carbon atom that is bonded to two other hydrogen atoms and a carbon atom. However, in both propane and 2–methylpropane, there are hydrogen atoms in two different environments, distinguished by the adjacent atoms or groups of atoms:
Note that alkyl groups do not exist as stable independent entities. They are always a part of some larger molecule. The location of an alkyl group on a hydrocarbon chain is indicated in the same way as any other substituent:
Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:
$\ce{CH4}(g)+\ce{2O2}(g)⟶\ce{CO2}(g)+\ce{2H2O}(g) \nonumber$
Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH4, is the principal component of natural gas. Butane, C4H10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (Figure $5$). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids.
In a substitution reaction, another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction:
The C–Cl portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter.
Alkenes
Organic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. You have likely heard of unsaturated fats. These are complex organic molecules with long chains of carbon atoms, which contain at least one double bond between carbon atoms. Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds, one σ bond and one π bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats.
Ethene, C2H4, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure $6$); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism.
Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules.
Recycling Plastics
Polymers (from Greek words poly meaning “many” and mer meaning “parts”) are large molecules made up of repeating units, referred to as monomers. Polymers can be natural (starch is a polymer of sugar residues and proteins are polymers of amino acids) or synthetic [like polyethylene, polyvinyl chloride (PVC), and polystyrene]. The variety of structures of polymers translates into a broad range of properties and uses that make them integral parts of our everyday lives. Adding functional groups to the structure of a polymer can result in significantly different properties (see the discussion about Kevlar later in this chapter).
An example of a polymerization reaction is shown in Figure $7$. The monomer ethylene (C2H4) is a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of –CH2– units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films).
Polyethylene is a member of one subset of synthetic polymers classified as plastics. Plastics are synthetic organic solids that can be molded; they are typically organic polymers with high molecular masses. Most of the monomers that go into common plastics (ethylene, propylene, vinyl chloride, styrene, and ethylene terephthalate) are derived from petrochemicals and are not very biodegradable, making them candidate materials for recycling. Recycling plastics helps minimize the need for using more of the petrochemical supplies and also minimizes the environmental damage caused by throwing away these nonbiodegradable materials.
Plastic recycling is the process of recovering waste, scrap, or used plastics, and reprocessing the material into useful products. For example, polyethylene terephthalate (soft drink bottles) can be melted down and used for plastic furniture, in carpets, or for other applications. Other plastics, like polyethylene (bags) and polypropylene (cups, plastic food containers), can be recycled or reprocessed to be used again. Many areas of the country have recycling programs that focus on one or more of the commodity plastics that have been assigned a recycling code (Figure $8$). These operations have been in effect since the 1970s and have made the production of some plastics among the most efficient industrial operations today.
The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:
Isomers of Alkenes
Molecules of 1-butene and 2-butene are structural isomers; the arrangement of the atoms in these two molecules differs. As an example of arrangement differences, the first carbon atom in 1-butene is bonded to two hydrogen atoms; the first carbon atom in 2-butene is bonded to three hydrogen atoms.
The compound 2-butene and some other alkenes also form a second type of isomer called a geometric isomer. In a set of geometric isomers, the same types of atoms are attached to each other in the same order, but the geometries of the two molecules differ. Geometric isomers of alkenes differ in the orientation of the groups on either side of a $\mathrm{C=C}$ bond.
Carbon atoms are free to rotate around a single bond but not around a double bond; a double bond is rigid. This makes it possible to have two isomers of 2-butene, one with both methyl groups on the same side of the double bond and one with the methyl groups on opposite sides. When structures of butene are drawn with 120° bond angles around the sp2-hybridized carbon atoms participating in the double bond, the isomers are apparent. The 2-butene isomer in which the two methyl groups are on the same side is called a cis-isomer; the one in which the two methyl groups are on opposite sides is called a trans-isomer (Figure $9$). The different geometries produce different physical properties, such as boiling point, that may make separation of the isomers possible:
Alkenes are much more reactive than alkanes because the $\mathrm{C=C}$ moiety is a reactive functional group. A π bond, being a weaker bond, is disrupted much more easily than a σ bond. Thus, alkenes undergo a characteristic reaction in which the π bond is broken and replaced by two σ bonds. This reaction is called an addition reaction. The hybridization of the carbon atoms in the double bond in an alkene changes from sp2 to sp3 during an addition reaction. For example, halogens add to the double bond in an alkene instead of replacing hydrogen, as occurs in an alkane:
Example $5$: Alkene Reactivity and Naming
Provide the IUPAC names for the reactant and product of the halogenation reaction shown here:
Solution
The reactant is a five-carbon chain that contains a carbon-carbon double bond, so the base name will be pentene. We begin counting at the end of the chain closest to the double bond—in this case, from the left—the double bond spans carbons 2 and 3, so the name becomes 2-pentene. Since there are two carbon-containing groups attached to the two carbon atoms in the double bond—and they are on the same side of the double bond—this molecule is the cis-isomer, making the name of the starting alkene cis-2-pentene. The product of the halogenation reaction will have two chlorine atoms attached to the carbon atoms that were a part of the carbon-carbon double bond:
This molecule is now a substituted alkane and will be named as such. The base of the name will be pentane. We will count from the end that numbers the carbon atoms where the chlorine atoms are attached as 2 and 3, making the name of the product 2,3-dichloropentane.
Exercise $5$
Provide names for the reactant and product of the reaction shown:
Answer
reactant: cis-3-hexene, product: 3,4-dichlorohexane
Alkynes
Hydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The sp-hybridized carbons involved in the triple bond have bond angles of 180°, giving these types of bonds a linear, rod-like shape.
The simplest member of the alkyne series is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:
The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, $\mathrm{CH_3CH_2C≡CH}$ is called 1-butyne.
Example $6$: Structure of Alkynes
Describe the geometry and hybridization of the carbon atoms in the following molecule:
Solution
Carbon atoms 1 and 4 have four single bonds and are thus tetrahedral with sp3 hybridization. Carbon atoms 2 and 3 are involved in the triple bond, so they have linear geometries and would be classified as sp hybrids.
Exercise $6$
Identify the hybridization and bond angles at the carbon atoms in the molecule shown:
Answer
carbon 1: sp, 180°; carbon 2: sp, 180°; carbon 3: sp2, 120°; carbon 4: sp2, 120°; carbon 5: sp3, 109.5°
Chemically, the alkynes are similar to the alkenes. Since the $\mathrm{C≡C}$ functional group has two π bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example:
Acetylene and the other alkynes also burn readily. An acetylene torch takes advantage of the high heat of combustion for acetylene.
Aromatic Hydrocarbons
Benzene, C6H6, is the simplest member of a large family of hydrocarbons, called aromatic hydrocarbons. These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, C6H6, are:
There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:
Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene.
Example $7$: Structure of Aromatic Hydrocarbons
One possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:
Solution
Since the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:
Exercise $7$
Draw three isomers of a six-membered aromatic ring compound substituted with two bromines.
Answer
Summary
Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. The chemistry of these compounds is called organic chemistry. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring structures with delocalized π electron systems.
Footnotes
1. This is the Beilstein database, now available through the Reaxys site (www.elsevier.com/online-tools/reaxys).
2. Peplow, Mark. “Organic Synthesis: The Robo-Chemist,” Nature 512 (2014): 20–2.
3. Physical properties for C4H10 and heavier molecules are those of the normal isomer, n-butane, n-pentane, etc.
4. STP indicates a temperature of 0 °C and a pressure of 1 atm.
Glossary
addition reaction
reaction in which a double carbon-carbon bond forms a single carbon-carbon bond by the addition of a reactant. Typical reaction for an alkene.
alkane
molecule consisting of only carbon and hydrogen atoms connected by single (σ) bonds
alkene
molecule consisting of carbon and hydrogen containing at least one carbon-carbon double bond
alkyl group
substituent, consisting of an alkane missing one hydrogen atom, attached to a larger structure
alkyne
molecule consisting of carbon and hydrogen containing at least one carbon-carbon triple bond
aromatic hydrocarbon
cyclic molecule consisting of carbon and hydrogen with delocalized alternating carbon-carbon single and double bonds, resulting in enhanced stability
functional group
part of an organic molecule that imparts a specific chemical reactivity to the molecule
organic compound
natural or synthetic compound that contains carbon
saturated hydrocarbon
molecule containing carbon and hydrogen that has only single bonds between carbon atoms
skeletal structure
shorthand method of drawing organic molecules in which carbon atoms are represented by the ends of lines and bends in between lines, and hydrogen atoms attached to the carbon atoms are not shown (but are understood to be present by the context of the structure)
substituent
branch or functional group that replaces hydrogen atoms in a larger hydrocarbon chain
substitution reaction
reaction in which one atom replaces another in a molecule | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/21%3A_Organic_Chemistry/21.0%3A_Prelude_to_Organic_Chemistry.txt |
Learning Objectives
• Describe the structure and properties of alcohols
• Describe the structure and properties of ethers
• Name and draw structures for alcohols and ethers
In this section, we will learn about alcohols and ethers.
Alcohols
Incorporation of an oxygen atom into carbon- and hydrogen-containing molecules leads to new functional groups and new families of compounds. When the oxygen atom is attached by single bonds, the molecule is either an alcohol or ether.
Alcohols are derivatives of hydrocarbons in which an –OH group has replaced a hydrogen atom. Although all alcohols have one or more hydroxyl (–OH) functional groups, they do not behave like bases such as NaOH and KOH. NaOH and KOH are ionic compounds that contain OH ions. Alcohols are covalent molecules; the –OH group in an alcohol molecule is attached to a carbon atom by a covalent bond.
Ethanol, CH3CH2OH, also called ethyl alcohol, is a particularly important alcohol for human use. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars:
Large quantities of ethanol are synthesized from the addition reaction of water with ethylene using an acid as a catalyst:
Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):
Naming Alcohols
The name of an alcohol comes from the hydrocarbon from which it was derived. The final -e in the name of the hydrocarbon is replaced by -ol, and the carbon atom to which the –OH group is bonded is indicated by a number placed before the name.1
Example \(1\): Naming Alcohols
Consider the following example. How should it be named?
Solution
The carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol.
Exercise \(1\)
Name the following molecule:
Answer
2-methyl-2-pentanol
Ethers
Ethers are compounds that contain the functional group –O–. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named “methoxy.” The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by “ether.” The common name for the compound shown in below is ethylmethyl ether:
Example \(2\): Naming Ethers
Provide the IUPAC and common name for the ether shown here:
Solution
• IUPAC name: The molecule is made up of an ethoxy group attached to an ethane chain, so the IUPAC name would be ethoxyethane.
• Common name: The groups attached to the oxygen atom are both ethyl groups, so the common name would be diethyl ether.
Exercise \(2\)
Provide the IUPAC and common name for the ether shown:
Answer
IUPAC: 2-methoxypropane; common: isopropylmethyl ether
Ethers can be obtained from alcohols by the elimination of a molecule of water from two molecules of the alcohol. For example, when ethanol is treated with a limited amount of sulfuric acid and heated to 140 °C, diethyl ether and water are formed:
In the general formula for ethers, R—O—R, the hydrocarbon groups (R) may be the same or different. Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. Tertiary-butyl methyl ether, C4H9OCH3 (abbreviated MTBE—italicized portions of names are not counted when ranking the groups alphabetically—so butyl comes before methyl in the common name), is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline.
Carbohydrates and Diabetes
Carbohydrates are large biomolecules made up of carbon, hydrogen, and oxygen. The dietary forms of carbohydrates are foods rich in these types of molecules, like pastas, bread, and candy. The name “carbohydrate” comes from the formula of the molecules, which can be described by the general formula Cm(H2O)n, which shows that they are in a sense “carbon and water” or “hydrates of carbon.” In many cases, m and n have the same value, but they can be different. The smaller carbohydrates are generally referred to as “sugars,” the biochemical term for this group of molecules is “saccharide” from the Greek word for sugar (Figure \(1\)). Depending on the number of sugar units joined together, they may be classified as monosaccharides (one sugar unit), disaccharides (two sugar units), oligosaccharides (a few sugars), or polysaccharides (the polymeric version of sugars—polymers were described in the feature box earlier in this chapter on recycling plastics). The scientific names of sugars can be recognized by the suffix -ose at the end of the name (for instance, fruit sugar is a monosaccharide called “fructose” and milk sugar is a disaccharide called lactose composed of two monosaccharides, glucose and galactose, connected together). Sugars contain some of the functional groups we have discussed: Note the alcohol groups present in the structures and how monosaccharide units are linked to form a disaccharide by formation of an ether.
Organisms use carbohydrates for a variety of functions. Carbohydrates can store energy, such as the polysaccharides glycogen in animals or starch in plants. They also provide structural support, such as the polysaccharide cellulose in plants and the modified polysaccharide chitin in fungi and animals. The sugars ribose and deoxyribose are components of the backbones of RNA and DNA, respectively. Other sugars play key roles in the function of the immune system, in cell-cell recognition, and in many other biological roles.
Diabetes is a group of metabolic diseases in which a person has a high sugar concentration in their blood (Figure \(2\)). Diabetes may be caused by insufficient insulin production by the pancreas or by the body’s cells not responding properly to the insulin that is produced. In a healthy person, insulin is produced when it is needed and functions to transport glucose from the blood into the cells where it can be used for energy. The long-term complications of diabetes can include loss of eyesight, heart disease, and kidney failure.
In 2013, it was estimated that approximately 3.3% of the world’s population (~380 million people) suffered from diabetes, resulting in over a million deaths annually. Prevention involves eating a healthy diet, getting plenty of exercise, and maintaining a normal body weight. Treatment involves all of these lifestyle practices and may require injections of insulin.
Summary
Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.
Footnotes
1. The IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an “infix” rather than a prefix. For example, the new name for 2-propanol would be propan-2-ol. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols.
Glossary
alcohol
organic compound with a hydroxyl group (–OH) bonded to a carbon atom
ether
organic compound with an oxygen atom that is bonded to two carbon atoms | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/21%3A_Organic_Chemistry/21.2%3A_Alcohols_and_Ethers.txt |
Learning Objectives
• Describe the structure and properties of aldehydes, ketones, carboxylic acids and esters
Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section.
Aldehydes and Ketones
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively.
In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms. As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.
In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.
Like the $\mathrm{C=O}$ bond in carbon dioxide, the $\mathrm{C=O}$ bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar $\mathrm{C=O}$ bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure $1$).
The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:
Example $1$: Oxidation and Reduction in Organic Chemistry
Methane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide:
$\ce{CH4⟶CH3OH⟶CH2O⟶HCO2H⟶CO2} \nonumber$
What are the oxidation numbers for the carbon atoms in the molecules shown here?
Solution
In this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing C–H bonds with C–O bonds described).
• For CH4, the carbon atom carries a –4 oxidation number (the hydrogen atoms are assigned oxidation numbers of +1 and the carbon atom balances that by having an oxidation number of –4)
• For the alcohol (in this case, methanol), the carbon atom has an oxidation number of –2 (the oxygen atom is assigned –2, the four hydrogen atoms each are assigned +1, and the carbon atom balances the sum by having an oxidation number of –2; note that compared to the carbon atom in CH4, this carbon atom has lost two electrons so it was oxidized)
• For the aldehyde, the carbon atom’s oxidation number is 0 (–2 for the oxygen atom and +1 for each hydrogen atom already balances to 0, so the oxidation number for the carbon atom is 0)
• For the carboxylic acid, the carbon atom’s oxidation number is +2 (two oxygen atoms each at –2 and two hydrogen atoms at +1)
• For carbon dioxide, the carbon atom’s oxidation number is +4 (here, the carbon atom needs to balance the –4 sum from the two oxygen atoms).
Exercise $1$
Indicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol:
There is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms:
Answer a
reduced (bond to oxygen atom replaced by bond to hydrogen atom);
Answer b
oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom);
Answer c
oxidized (2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom)
Aldehydes are commonly prepared by the oxidation of alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
Alcohols that have their –OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms:
An alcohol with its –OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its –OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the –OH, the molecule will not have a C–H bond to be replaced, so it will not be susceptible to oxidation.
Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.
Dimethyl ketone, CH3COCH3, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals.
Carboxylic Acids and Esters
The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (Figure $2$).
Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples):
The functional groups for an acid and for an ester are shown in red in these formulas.
The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt:
Carboxylic acids are weak acids, meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution.
We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH3CO2CH2CH3, is formed when acetic acid reacts with ethanol:
The simplest carboxylic acid is formic acid, HCO2H, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests.
Acetic acid, CH3CO2H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon.
The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure $3$). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C3H5(OH)3, with large carboxylic acids, such as palmitic acid, CH3(CH2)14CO2H, stearic acid, CH3(CH2)16CO2H, and oleic acid, $\mathrm{CH_3(CH_2)_7CH=CH(CH_2)_7CO_2H}$. Oleic acid is an unsaturated acid; it contains a $\mathrm{C=C}$ double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds.
Summary
Functional groups related to the carbonyl group include the –CHO group of an aldehyde, the –CO– group of a ketone, the –CO2H group of a carboxylic acid, and the –CO2R group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group.
Glossary
aldehyde
organic compound containing a carbonyl group bonded to two hydrogen atoms or a hydrogen atom and a carbon substituent
carbonyl group
carbon atom double bonded to an oxygen atom
carboxylic acid
organic compound containing a carbonyl group with an attached hydroxyl group
ester
organic compound containing a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent
ketone
organic compound containing a carbonyl group with two carbon substituents attached to it | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/21%3A_Organic_Chemistry/21.3%3A_Aldehydes_Ketones_Carboxylic_Acids_and_Esters.txt |
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