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How to Keep it Cool?
Magnetic resonance imaging (MRI) devices use liquid nitrogen to cool the superconducting magnets. Nitrogen is a gas at room temperature and liquefies at $-195.8^\text{o} \text{C}$. Its neighbor on the periodic table (oxygen) boils at $-182.95^\text{o} \text{C}$. The interactions between nitrogen molecules $\left( \ce{N_2} \right)$ are weaker, so the boiling point is lower. Interactions between nonpolar molecules depend on the degree of electron fluctuation within the molecule.
Van der Waals Forces
The first type of intermolecular forces that we consider are van der Waals forces, after Dutch chemist Johannes van der Waals (1837-1923). Van der Waals forces are the weakest intermolecular force, and consist of dipole-dipole forces and dispersion forces.
Dipole-Dipole Forces
Dipole-dipole forces are the attractive forces that occur between polar molecules. A molecule of hydrogen chloride has a partially positive hydrogen atom and a partially negative chlorine atom. In a collection of many hydrogen chloride molecules, the molecules will align themselves so that the oppositely charged regions of neighboring molecules are near each other.
Dipole-dipole forces are similar in nature to ionic bonds, but much weaker.
London Dispersion Forces
Dispersion forces are also considered a type of van der Waals force and are the weakest of all intermolecular forces. They are often called London dispersion forces after Fritz London (1900-1954), who first proposed their existence in 1930. London dispersion forces are the intermolecular forces that occur between atoms, and between nonpolar molecules as a result of the motion of electrons.
The electron cloud of a helium atom contains two electrons, which can normally be expected to be equally distributed spatially around the nucleus. However, at any given moment the electron distribution may be uneven, resulting in an instantaneous dipole. This weak and temporary dipole subsequently influences neighboring helium atoms through electrostatic attraction and repulsion. It induces a dipole on nearby helium atoms (see figure below).
The instantaneous and induced dipoles are weakly attracted to one another. The strength of dispersion forces increases as the number of electrons in the atoms or nonpolar molecules increases.
The halogen group consists of four elements that all take the form of nonpolar diatomic molecules. The table below shows a comparison of the melting and boiling points for each.
Melting and Boiling Points of Halogens
Table $1$: Melting and Boiling Points of Halogens
Molecule Total Number of Electrons Melting Point $\left( ^\text{o} \text{C} \right)$ Boiling Point $\left( ^\text{o} \text{C} \right)$ Physical State at Room Temperature
$\ce{F_2}$ 18 -220 -188 gas
$\ce{Cl_2}$ 34 -102 -34 gas
$\ce{Br_2}$ 70 -7 59 liquid
$\ce{I_2}$ 106 114 184 solid
The dispersion forces are strongest for iodine molecules because they have the greatest number of electrons. The relatively stronger forces result in melting and boiling points that are the highest of the halogen group. These forces are strong enough to hold iodine molecules close together in the solid state at room temperature. The dispersion forces are progressively weaker for bromine, chloride, and fluorine; this is illustrated in their steadily lower melting and boiling points. Bromine is a liquid at room temperature, while chlorine and fluorine are gases whose molecules are much further apart from one another. Intermolecular forces are nearly nonexistent in the gas state, and so the dispersion forces in chlorine and fluorine only become measurable as the temperature decreases and they condense into the liquid state.
Simulation
Why does water form droplets? Hint: it has to do with the interactions between water molecules. Try out this simulation to learn more.
Summary
• Van der Waals forces are weak interactions between molecules that involve dipoles.
• Polar molecules have permanent dipole-dipole interactions.
• Nonpolar molecules can interact by way of London dispersion forces.
Review
1. What attractive forces develop between polar molecules?
2. What creates London dispersion forces?
3. Are London dispersion forces permanent or temporary?
4. Are the dispersion forces for Cl2 stronger or weaker than the ones for Br2? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/09%3A_Covalent_Bonding/9.18%3A_Van_der_Waals_Forces.txt |
What's the difference between these two molecules?
A rough rule of thumb is that higher molecular-weight materials have higher boiling points than their lower molecular-weight counterparts. More energy is needed to move the larger molecule from the liquid state to the vapor state. However, ammonia has a boiling point of $-33.34^\text{o} \text{C}$ and a molecular weight of 17, while nitrogen (molecular weight 28) has a boiling point of $-195.8^\text{o} \text{C}$. The lighter ammonia molecule must have other factors that influence its physical properties.
Hydrogen Bonding
The attractive force between water molecules is a dipole interaction. The hydrogen atoms are bound to the highly electronegative oxygen atom (which also possesses two lone pair sets of electrons, making for a very polar bond). The partially positive hydrogen atom of one molecule is then attracted to the oxygen atom of a nearby water molecule (see figure below).
A hydrogen bond is an intermolecular attractive force in which a hydrogen atom that is covalently bonded to a small, highly electronegative atom is attracted to a lone pair of electrons on an atom in a neighboring molecule. Hydrogen bonds are very strong compared to other dipole interactions. The strength of a typical hydrogen bond is about $5\%$ of that of a covalent bond.
Hydrogen bonding occurs only in molecules where hydrogen is covalently bonded to one of three elements: fluorine, oxygen, or nitrogen. These three elements are so electronegative that they withdraw the majority of the electron density in the covalent bond with hydrogen, leaving the $\ce{H}$ atom very electron-deficient. The $\ce{H}$ atom nearly acts as a bare proton, leaving it very attracted to lone pair electrons on a nearby atom.
The hydrogen bonding that occurs in water leads to some unusual, but very important, properties. Most molecular compounds that have a mass similar to water are gases at room temperature. Because of the strong hydrogen bonds, water molecules are able to stay condensed in the liquid state. The figure below shows how the bent shape, and two hydrogen atoms per molecule, allows each water molecule to be able to hydrogen bond to two other molecules.
In the liquid state, the hydrogen bonds of water can break and reform as the molecules flow from one place to another. When water is cooled, the molecules begin to slow down. Eventually, when water is frozen to ice, the hydrogen bonds become permanent and form a very specific network (see figure below).
The bent shape of the molecules leads to gaps in the hydrogen bonding network of ice. Ice has a very unusual property—its solid state is less dense than its liquid state. Ice floats in water. Virtually all other substances are denser in the solid state than in the liquid state. Hydrogen bonds play a very important biological role in the physical structures of proteins and nucleic acids.
Simulation
Why does water form droplets? Hint: it has to do with the attraction between water molecules. Learn more (and see how many drops of water you can get a penny can hold) in this simulation.
Summary
• Hydrogen bonds form when a $\ce{H}$ attached to a $\ce{N}$, $\ce{O}$, or $\ce{F}$ atom interacts with another $\ce{N}$, $\ce{O}$, or $\ce{F}$ atom.
Review
1. How strong is a hydrogen bond?
2. What happens when H is covalently bonded to N, O, or F?
3. How does the shape of the water molecule affect the properties of ice? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/09%3A_Covalent_Bonding/9.19%3A_Hydrogen_Bonding.txt |
Can you guess how much this is worth?
Carbon is an interesting and versatile element. There are over twenty million known compounds containing carbon, encompassed in the growing field of organic chemistry. The element itself can exist in two major forms. Diamond is a form of carbon that is extremely hard and is one of the few materials that can scratch glass. The other form of carbon is graphite, a very soft material that we find in “lead” pencils. The two forms differ mainly in how the carbon atoms are connected to one another. The differences in the arrangement of atoms affect the properties of the material.
Physical Properties and Intermolecular Forces
The physical state and properties of a particular compound depend in large part on the type of chemical bonding it displays. Molecular compounds, sometimes called covalent compounds, display a wide range of physical properties due to the different types of intermolecular attractions such as different kinds of polar interactions. The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds. This is because the energy required to disrupt the intermolecular forces between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound. Since molecular compounds are composed of neutral molecules, their electrical conductivity is generally quite poor, whether in the solid or liquid state. Ionic compounds do not conduct electricity in the solid state because of their rigid structure, but conduct well when either molten or dissolved into a solution. The water solubility of molecular compounds is variable and depends primarily on the type of intermolecular forces involved. Substances that exhibit hydrogen bonding or dipole-dipole forces are generally water soluble, whereas those that exhibit only London dispersion forces are generally insoluble. Most, but not all, ionic compounds are quite soluble in water. The table below summarizes some of the differences between ionic and molecular compounds.
Table \(1\): Comparison of Ionic and Molecular Compounds
Property Ionic Compounds Molecular Compounds
Type of elements metal and nonmetal nonmetals only
Bonding ionic – transfer of electron(s) between atoms covalent – sharing of pair(s) of electrons between atoms
Representative unit formula unit molecule
Physical state at room temp. solid gas, liquid, or solid
Water solubility usually high variable
Melting and boiling temps generally high generally low
Electrical conductivity good when molten or in solution poor
One type of molecular compound behaves quite differently than that described so far. A covalent network solid is a compound in which all of the atoms are connected to one another by covalent bonds. Diamond is composed entirely of carbon atoms, each bonded to four other carbon atoms in a tetrahedral geometry. Melting a covalent network solid is not accomplished by overcoming the relatively weak intermolecular forces. Rather, all of the covalent bonds must be broken, a process that requires extremely high temperatures. Diamond, in fact, does not melt at all. Instead, it vaporizes to a gas at temperatures above 3500°C.
Summary
• The physical properties of a material are affected by the intermolecular forces holding the molecules together.
Review
1. Are melting points of molecular compounds generally higher or lower than those of ionic compounds?
2. Do ionic compounds conduct electricity in the solid state?
3. What types of substances are generally water-soluble?
9.21: Valence Bond Theory
What happens next?
We have seen that the old fish-hook idea of atoms connecting that Democritus liked so much just doesn’t work. Electrons don’t have little hooks on them, but they are the basis for connecting atoms to form molecules. You have learned how to write Lewis electron-dot structures for molecules and predict their shape using VSEPR theory. Now it is time to apply these abilities to understand how the electrons behave in their atomic orbitals when a covalent bond forms.
Valence Bond Theory
You have learned that a covalent bond forms when the electron clouds of two atoms overlap one another. In a simple $\ce{H_2}$ molecule, the single electron in each atom becomes attracted to the nucleus of the other atom in the molecule as the atoms come closer together. An optimum distance, equal to the bond length, is eventually attained, and the potential energy reaches a minimum. A stable, single covalent bond has formed between the two hydrogen atoms. Other covalent bonds form in the same way as unpaired electrons, when two atoms "match up" to form the bond. In a fluorine atom, there is an unpaired electron in one of the $2p$ orbitals. When a $\ce{F_2}$ molecule forms, the $2p$ orbitals from each of the two atoms overlap to produce the $\ce{F-F}$ covalent bond. The overlapping orbitals do not have to be of the same type. In a molecule of $\ce{HF}$, the $1s$ orbital of the hydrogen atom overlaps with the $2p$ orbital of the $2p$ orbital of the fluorine atom (see figure below).
In essence, any covalent bond results from the overlap of atomic orbitals. This idea forms the basis for a quantum mechanical theory called valence bond (VB) theory. In valence bond theory, the electrons in a molecule are assumed to occupy atomic orbitals of the individual atoms, and a bond results from overlap of those orbitals.
Summary
• Electrons occupy atomic orbitals.
• Covalent bonds result from the overlap of atomic orbitals.
Review
1. Where are electrons according to valence bond theory?
2. How do covalent bonds form?
3. Do the orbitals of the two electrons involved in the bond need to be the same? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/09%3A_Covalent_Bonding/9.20%3A_Physical_Properties_and_Intermolecular_Forces.txt |
Do you recognize this plant?
If we were walking on the beach, the plants shown above would look very different. They would be short and sticking out of the sand. When we see them this way, we may not immediately recognize them as beach plants. Often, we need to look at the world around us in different ways to understand things better.
Hybrid Orbitals - sp3
The bonding scheme described by valence bond theory must account for molecular geometries as predicted by VSEPR theory. To do that, we must introduce the concept of hybrid orbitals.
$sp^3$ Hybridization
Unfortunately, overlap of existing atomic orbitals ($s$, $p$, etc.) is not sufficient to explain some of the bonding and molecular geometries that are observed. Consider the element carbon and the methane $\left( \ce{CH_4} \right)$ molecule. A carbon atom has the electron configuration of $1s^2 \: 2s^2 \: 2p^2$, meaning that it has two unpaired electrons in its $2p$ orbitals, as shown in the figure below.
According to the description of valence bond theory so far, carbon would be expected to form only two bonds, corresponding to its two unpaired electrons. However, methane is a common and stable molecule, with four equivalent $\ce{C-H}$ bonds. To account for this, one of the $2s$ electrons is promoted to the empty $2p$ orbital (see figure below).
Now, four bonds are possible. The promotion of the electron "costs" a small amount of energy, but recall that the process of bond formation is accompanied by a decrease in energy. The two extra bonds that can now be formed results in a lower overall energy, and thus greater stability to the $\ce{CH_4}$ molecule. Carbon normally forms four bonds in most of its compounds.
The number of bonds is now correct, but the geometry is wrong. The three $p$ orbitals, ($p_x$, $p_y$, and $p_z$), are oriented at $90^\text{o}$ relative to one another. However, as seen in VSEPR theory, the observed $\ce{H-C-H}$ bond angle in the tetrahedral $\ce{CH_4}$ molecule is actually $109.5^\text{o}$. Therefore, the methane molecule cannot be adequately represented by simple overlap of the $2s$ and $2p$ orbitals of carbon with the $1s$ orbitals of each hydrogen atom.
To explain the bonding in methane, it is necessary to introduce the concept of hybridization and hybrid atomic orbitals. Hybridization is the mixing of the atomic orbitals in an atom to produce a set of hybrid orbitals. When hybridization occurs, it must do so as a result of the mixing of nonequivalent orbitals. In other words, $s$ and $p$ orbitals can hybridize, but $p$ orbitals cannot hybridize with other $p$ orbitals. Hybrid orbitals are the atomic orbitals obtained when two or more nonequivalent orbitals from the same atom combine in preparation for bond formation. In the current case of carbon, the single $2s$ orbital hybridizes with the three $2p$ orbitals to form a set of four hybrid orbitals, called $sp^3$ hybrids (see figure below).
The $sp^3$ hybrids are all equivalent to one another. Spatially, the hybrid orbitals point towards the four corners of a tetrahedron (see figure below).
Summary
• Electrons hybridize in order to form covalent bonds.
• Nonequivalent orbitals mix to form hybrid orbitals.
Review
• Why is carbon expected to form only two covalent bonds?
• How many covalent bonds does carbon actually form?
• What needs to happen to allow carbon to form four bonds?
9.23: Hybrid Orbitals - sp and sp
How do you open the closed circle?
Romeo and Juliet were two of the greatest-known fictional lovers of all time. Their embrace allowed no other person to be a part of it—they only wanted to be with each other. It took outside intervention to get them away from one another. Paired electrons are similar to Romeo and Juliet. They do not bond covalently until they are unpaired; then, they can become a part of a larger chemical structure.
sp Hybridization
A beryllium hydride $\left( \ce{BeH_2} \right)$ molecule is predicted to be linear by VSEPR. The beryllium atom contains all paired electrons and so must also undergo hybridization. One of the $2s$ electrons is first promoted to the empty $2p_x$ orbital (see figure below).
Now the hybridization takes place only with occupied orbitals, and the result is a pair of $sp$ hybrid orbitals. The two remaining $p$ orbitals ($p_y$ and $p_z$) do not hybridize and remain unoccupied (see figure below).
The geometry of the $sp$ hybrid orbitals is linear, with the lobes of the orbitals pointing in opposite directions along one axis, arbitrarily defined as the $x$-axis (see figure below). Each can bond with a $1s$ orbital from a hydrogen atom to form the linear $\ce{BeH_2}$ molecule.
Other molecules whose electron domain geometry is linear and for whom hybridization is necessary also form $sp$ hybrid orbitals. Examples include $\ce{CO_2}$ and $\ce{C_2H_2}$, which will be discussed in further detail later.
sp² Hybridization
Boron trifluoride $\left( \ce{BF_3} \right)$ is predicted to have a trigonal planar geometry by VSEPR. First, a paired $2s$ electron is promoted to the empty $2p_y$ orbital (see figure below).
This is followed by hybridization of the three occupied orbitals to form a set of three $sp^2$ hybrids, leaving the $2p_z$ orbital unhybridized (see figure below).
The geometry of the $sp^2$ hybrid orbitals is trigonal planar, with the lobes of the orbitals pointing towards the corners of a triangle (see figure below). The angle between any two of the hybrid orbital lobes is $120^\text{o}$. Each can bond with a $2p$ orbital from a fluorine atom to form the trigonal planar $\ce{BF_3}$ molecule.
Other molecules with a trigonal planar electron domain geometry form $sp^2$ hybrid orbitals. Ozone $\left( \ce{O_3} \right)$ is an example of a molecule whose electron domain geometry is trigonal planar, though the presence of a lone pair on the central oxygen makes the molecular geometry bent. The hybridization of the central $\ce{O}$ atom of ozone is $sp^2$.
Summary
• Paired electrons can be hybridized and then participate in covalent bonding.
Review
1. Does the ground state beryllium atom contain any unpaired electrons?
2. Why does one 2s electron in Be get promoted to a 2p orbital?
3. What is the geometry of the two sp orbitals? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/09%3A_Covalent_Bonding/9.22%3A_Hybrid_Orbitals_-_sp.txt |
How many people do you think are squeezed on this street?
Our minds can handle two electrons interacting with one another in a sphere of space. But then we start putting in double bonds and triple bonds. The way we draw these bonds on paper suggests we are squeezing more electrons into the same space, and that doesn’t work. Electrons don’t like to be pushed together (especially since they all have negative charges that repel one another). So we need a more complex picture that works for all these electrons.
Sigma and Pi Bonds
The hybridization model helps explain molecules with double or triple bonds (see figure below). Ethene $\left( \ce{C_2H_4} \right)$ contains a double covalent bond between the two carbon atoms, and single bonds between the carbon atoms and the hydrogen atoms. The entire molecule is planar.
As can be seen in the figure below, the electron domain geometry around each carbon independently is trigonal planar. This corresponds to $sp^2$ hybridization. Previously, we saw carbon undergo $sp^3$ hybridization in a $\ce{CH_4}$ molecule, so the electron promotion is the same for ethene, but the hybridization occurs only between the single $s$ orbital and two of the three $p$ orbitals. This generates a set of three $sp^2$ hybrids, along with an unhybridized $2p_z$ orbital. Each contains one electron and so is capable of forming a covalent bond.
The three $sp^2$ hybrid orbitals lie in one plane, while the unhybridized $2p_z$ orbital is oriented perpendicular to that plane. The bonding in $\ce{C_2H_4}$ is explained as follows: one of the three $sp^2$ hybrids forms a bond by overlapping with the identical hybrid orbital on the other carbon atom. The remaining two hybrid orbitals form bonds by overlapping with the $1s$ orbital of a hydrogen atom. Finally, the $2p_z$ orbitals on each carbon atom form another bond by overlapping with one another sideways.
It is necessary to distinguish between the two types of covalent bonds in a $\ce{C_2H_4}$ molecule. A sigma bond ($\sigma$ bond) is a bond formed by the overlap of orbitals in an end-to-end fashion, with the electron density concentrated between the nuclei of the bonding atoms. A pi bond ($\pi$ bond) is a bond formed by the overlap of orbitals in a side-by-side fashion with the electron density concentrated above and below the plane of the nuclei of the bonding atoms. The figure below shows the two types of bonding in $\ce{C_2H_4}$. The $sp^2$ hybrid orbitals are purple and the $p_z$ orbital is blue. Three sigma bonds are formed from each carbon atom for a total of six sigma bonds in the molecule. The pi bond is the "second" bond of the double bonds between the carbon atoms, and is shown as an elongated green lobe that extends both above and below the plane of the molecule. This plane contains the six atoms and all of the sigma bonds.
In a conventional Lewis electron-dot structure, a double bond is shown as a double dash between the atoms, as in $\ce{C=C}$. It is important to realize, however, that the two bonds are different: one is a sigma bond, while the other is a pi bond.
Ethyne $\left( \ce{C_2H_2} \right)$ is a linear molecule with a triple bond between the two carbon atoms (see figure below). The hybridization is therefore $sp$.
The promotion of an electron in the carbon atom occurs in the same way. However, the hybridization now involves only the $2s$ orbital and the $2p_x$ orbital, leaving the $2p_y$ and the $2p_z$ orbitals unhybridized.
The $sp$ hybrid orbitals form a sigma bond between each other as well as sigma bonds to the hydrogen atoms. Both the $p_y$ and the $p_z$ orbitals on each carbon atom form pi bonds between each other. As with ethene, these side-to-side overlaps are above and below the plane of the molecule. The orientation of the two pi bonds is that they are perpendicular to one another (see figure below). One pi bond is above and below the line of the molecule as shown, while the other is in front of and behind the page.
In general, single bonds between atoms are always sigma bonds. Double bonds are comprised of one sigma and one pi bond. Triple bonds are comprised of one sigma bond and two pi bonds.
Summary
• Sigma bonds form between two atoms.
• Pi bonds form from p orbital overlap.
Review
1. What is the hybridization around each carbon in ethene?
2. What are the two types of bonds in C=C?
3. What is the shape of the ethene molecule?
4. How are the ethyne pi bonds oriented in relation to each other? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/09%3A_Covalent_Bonding/9.24%3A_Sigma_and_Pi_Bonds.txt |
The mole is the unit of measurement in the International System of Units (SI) for amount of substance. It is defined as the amount of a chemical substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons, or photons). This number is expressed by the Avogadro constant, which has a value of $6.022140857 \times 10^{23} \;mol^{-1}$. The mole is one of the base units of the SI, and has the unit symbol mol.
10: The Mole
Is there an easier way to load this truck?
When the weather is nice, many people begin to work on their yards and homes. For many projects, sand is needed as a foundation for a walk or to add to other materials. You could order up twenty million grains of sand and have people really stare at you. You could order by the pound, but that takes a lot of time weighing out. The best bet is to order by the yard, meaning a cubic yard. The loader can easily scoop up what you need and put it directly in your truck.
Avogadro's Number
It certainly is easy to count objects such as bananas, or something as large as elephants (as long as you stay out of their way). However, counting grains of sugar from a sugar canister would take a long, long time. Atoms and molecules are extremely small—far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about $10^{22}$ molecules of water. If you counted 10 molecules every second for 50 years, without stopping, you would have counted only $1.6 \times 10^{10}$ molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.
Chemists of the past needed a name that could stand for a very large number of items. Amadeo Avogadro (1776-1856), an Italian scientist, provided such a number. He is responsible for the counting unit of measure called the mole. A mole $\left( \text{mol} \right)$ is the amount of a substance that contains $6.02 \times 10^{23}$ representative particles of that substance. The mole is the SI unit for amount of a substance. Just like the dozen and the gross, it is a name that stands for a number. There are therefore $6.02 \times 10^{23}$ water molecules in a mole of water molecules. There also would be $6.02 \times 10^{23}$ bananas in a mole of bananas, if such a huge number of bananas ever existed.
The number $6.02 \times 10^{23}$ is called Avogadro's number, the number of representative particles in a mole. It is an experimentally determined number. A representative particle is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are $\ce{H_2}$, $\ce{N_2}$, $\ce{O_2}$, $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$. The representative particle for these elements is the molecule. Likewise, all molecular compounds such as $\ce{H_2O}$ and $\ce{CO_2}$ exist as molecules and so the molecule is their representative particle. For ionic compounds such as $\ce{NaCl}$ and $\ce{Ca(NO_3)_2}$, the representative particle is the formula unit. A mole of any substance contains Avogadro's number $\left( 6.02 \times 10^{23} \right)$ of representative particles.
Summary
• A mole of any substance contains Avogadro's number $\left( 6.02 \times 10^{23} \right)$ of representative particles.
Review
1. What is the SI unit for amount of a substance?
2. What is the representative particle for an element?
3. The formula unit is the representative particle for what? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.01%3A_Avogadro%27s_Number.txt |
Big numbers or little numbers?
Do you hate to type subscripts and superscripts? Even with a good word-processing program, having to click on an icon to get a superscript and then remembering to click off after you type the number can be a real hassle. If we did not know about moles and just knew about numbers of atoms or molecules (those big numbers that require lots of superscripts), life would be much more complicated and we would make many more typing errors.
Conversions Between Moles and Atoms
Conversions Between Moles and Number of Particles
Using our unit conversion techniques, we can use the mole label to convert back and forth between the number of particles and moles.
Example $1$: Converting Number of Particles to Moles
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Known
• number of $\ce{C}$ atoms $= 4.72 \times 10^{24}$
• $1$ mole $= 6.02 \times 10^{23}$ atoms
Unknown
• 4.72 x 1024 = ? mol C
One conversion factor will allow us to convert from the number of $\ce{C}$ atoms to moles of $\ce{C}$ atoms.
Step 2: Calculate.
$4.72 \times 10^{24} \: \text{atoms} \: \ce{C} \times \frac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \text{atoms} \: \ce{C}} = 7.84 \: \text{mol} \: \ce{C}\nonumber$
Step 3: Think about your result.
The given number of carbon atoms was greater than Avogadro's number, so the number of moles of $\ce{C}$ atoms is greater than 1 mole. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.
Suppose that you want to know how many hydrogen atoms are in a mole of water molecules. First, you need to know the chemical formula for water, which is $\ce{H_2O}$. There are two atoms of hydrogen in each molecule of water. How many atoms of hydrogen are in two water molecules? There are $2 \times 2 = 4$ hydrogen atoms. How about in a dozen? In that case, a dozen is 12; so $12 \times 2 = 24$ hydrogen atoms in a dozen water molecules. To get the answers (4 and 24), you multiply the given number of molecules by two atoms of hydrogen per molecule. So, to find the number of hydrogen atoms in a mole of water molecules, the problem can be solved using conversion factors:
$1 \: \text{mol} \: \ce{H_2O} \times \frac{6.02 \times 10^{23} \: \text{molecules} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} \times \frac{2 \: \text{atoms} \: \ce{H}}{1 \: \text{molecule} \: \ce{H_2O}} = 1.20 \times 10^{24} \: \text{atoms} \: \ce{H}\nonumber$
The first conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.
Example $2$: Atoms, Molecules, and Moles
Sulfuric acid has the chemical formula $\ce{H_2SO_4}$. A certain quantity of sulfuric acid contains $4.89 \times 10^{25}$ atoms of oxygen. How many moles of sulfuric acid is the sample?
Known
• $4.89 \times 10^{25} = \ce{O}$ atoms
• $1$ mole $= 6.02 \times 10^{23}$ molecules $\ce{H_2SO_4}$
Unknown
• mol of H2SO4 molecules
Two conversion factors will be used. First, convert atoms of oxygen to molecules of sulfuric acid. Then, convert molecules of sulfuric acid to moles of sulfuric acid.
Step 2: Calculate.
$4.89 \times 10^{25} \: \text{atoms} \: \ce{O} \times \frac{1 \: \text{molecule} \: \ce{H_2SO_4}}{4 \: \text{atoms} \: \ce{O}} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{6.02 \times 10^{23} \: \text{molecules} \: \ce{H_2SO_4}} = 20.3 \: \text{mol} \: \ce{H_2SO_4}\nonumber$
Step 3: Think about your result.
The original number of oxygen atoms was about 80 times larger than Avogadro's number. Since each sulfuric acid molecule contains 4 oxygen atoms, there are about 20 moles of sulfuric acid molecules.
Summary
• Methods are described for conversions between moles, atoms, and molecules.
Review
1. What conversion factor would we need to convert moles of helium to atoms of helium?
2. I want to convert atoms to moles. My friend tells me to multiply the number of atoms by 6.02 × 1023 atoms/mole. Is my friend correct?
3. Why do you need to know the formula for a molecule in order to calculate the number of moles of one of the atoms?
4. How many atoms of fluorine are in 5.6×1022 molecules of MgF2? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.02%3A_Conversions_Between_Moles_and_Atoms.txt |
When creating a solution, how do I know how much of each substance to put in?
I want to make a solution that contains 1.8 moles of potassium dichromate. You do not have a balance calibrated in moles, but you do have one calibrated in grams. If you know the relationship between moles and the number of grams in a mole, you can use your balance to measure out the needed amount of material.
Molar Mass
Molar mass is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is $6.94 \: \text{g}$, the molar mass of zinc is $65.38 \: \text{g}$, and the molar mass of gold is $196.97 \: \text{g}$. Each of these quantities contains $6.02 \times 10^{23}$ atoms of that particular element. The units for molar mass are grams per mole, or $\text{g/mol}$.
Molar Masses of Compounds
The molecular formula of the compound carbon dioxide is $\ce{CO_2}$. One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen. We can calculate the mass of one molecule of carbon dioxide by adding together the masses of 1 atom of carbon and 2 atoms of oxygen:
$12.01 \: \text{amu} + 2 \left( 16.00 \: \text{amu} \right) = 44.01 \: \text{amu}\nonumber$
The molecular mass of a compound is the mass of one molecule of that compound. The molecular mass of carbon dioxide is $44.01 \: \text{amu}$.
The molar mass of any compound is the mass in grams of one mole of that compound. One mole of carbon dioxide molecules has a mass of $44.01 \: \text{g}$, while one mole of sodium sulfide formula units has a mass of $78.04 \: \text{g}$. The molar masses are $44.01 \: \text{g/mol}$ and $78.04 \: \text{g/mol}$ respectively. In both cases, that is the mass of $6.02 \times 10^{23}$ representative particles. The representative particle of $\ce{CO_2}$ is the molecule, while for $\ce{Na_2S}$ it is the formula unit.
Example $1$: Molar Mass of a Compound
Calcium nitrate, $\ce{Ca(NO_3)_2}$, is used as a component in fertilizer. Determine the molar mass of calcium nitrate.
Solution
Step 1: List the known and unknown quantities and plan the problem.
Known
• Formula $= \ce{Ca(NO_3)_2}$
• Molar mass $\ce{Ca} = 40.08 \: \text{g/mol}$
• Molar mass $\ce{N} = 14.01 \: \text{g/mol}$
• Molar mass $\ce{O} = 16.00 \: \text{g/mol}$
Unknown
• molar mass Ca(NO3)2
First we need to analyze the formula. Since the $\ce{Ca}$ lacks a subscript, there is one $\ce{Ca}$ atom per formula unit. The 2 outside the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consists of one nitrogen atom and three oxygen atoms per formula unit. Thus, $1 \: \text{mol}$ of calcium nitrate contains $1 \: \text{mol}$ of $\ce{Ca}$ atoms, $2 \: \text{mol}$ of $\ce{N}$ atoms, and $6 \: \text{mol}$ of $\ce{O}$ atoms.
Step 2: Calculate
Use the molar masses of each atom together with the number of atoms in the formula and add together.
$1 \: \text{mol} \: \ce{Ca} \times \frac{40.08 \: \text{g} \: \ce{Ca}}{1 \: \text{mol} \: \ce{Ca}} = 40.08 \: \text{g} \: \ce{Ca}\nonumber$
$2 \: \text{mol} \: \ce{N} \times \frac{14.01 \: \text{g} \: \ce{N}}{1 \: \text{mol} \: \ce{N}} = 28.02 \: \text{g} \: \ce{N}\nonumber$
$6 \: \text{mol} \: \ce{O} \times \frac{16.00 \: \text{g} \: \ce{O}}{1 \: \text{mol} \: \ce{O}} = 96.00 \: \text{g} \: \ce{O}\nonumber$
Molar mass of $\ce{Ca(NO_3)_2} = 40.08 \: \text{g} + 28.02 \: \text{g} + 96.00 \: \text{g} = 164.10 \: \text{g/mol}$
Summary
• Calculations are described for the determination of molar mass of an atom or a compound.
Review
1. What is the molar mass of Pb?
2. Where do you find the molar mass of an element?
3. How many moles of Cl are in one mole of the CaCl2?
4. How many moles of H are in one mole of the compound (NH4)3PO4?
5. Calculate the molar mass of CaCl2. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.03%3A_Molar_Mass.txt |
How can we get more product?
Chemical manufacturing plants are always seeking to improve their processes. One way that improvement comes about is through measuring the amount of material produced in a reaction. By knowing how much is made, the scientists and engineers can try different ways of getting more product at less cost.
Conversions Between Moles and Mass
The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment, you need 3.00 moles of calcium chloride $\left( \ce{CaCl_2} \right)$. Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. The molar mass of $\ce{CaCl_2}$ is $110.98 \: \text{g/mol}$. The conversion factor that can be used is then based on the equality that $1 \: \text{mol} = 110.98 \: \text{g} \: \ce{CaCl_2}$. Dimensional analysis will allow you to calculate the mass of $\ce{CaCl_2}$ that you should measure.
$3.00 \: \text{mol} \: \ce{CaCl_2} \times \frac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \text{mol} \: \ce{CaCl_2}} = 333 \: \text{g} \: \ce{CaCl_2}\nonumber$
When you measure the mass of $333 \: \text{g}$ of $\ce{CaCl_2}$, you are measuring 3.00 moles of $\ce{CaCl_2}$.
Example $1$: Converting Moles to Mass
Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.
Known
• Molar mass of $\ce{Cr} = 52.00 \: \text{g/mol}$
• $0.560 \: \text{mol} \: \ce{Cr}$
Unknown
• $0.560 \: \text{mol} \: \ce{Cr}$ = ? g
One conversion factor will allow us to convert from the moles of $\ce{Cr}$ to mass.
Step 2: Calculate.
$0.560 \: \text{mol} \: \ce{Cr} \times \frac{52.00 \: \text{g} \: \ce{Cr}}{1 \: \text{mol} \: \ce{Cr}} = 29.1 \: \text{g} \: \ce{Cr}\nonumber$
Step 3: Think about your result.
Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass. The answer has three significant figures because of the $0.560 \: \text{mol}$.
A similar conversion factor utilizing molar mass can be used to convert from the mass of a substance to moles. In a laboratory situation, you may perform a reaction and produce a certain amount of a product which can be massed. It will often then be necessary to determine the number of moles of the product that was formed. The next problem illustrates this situation.
Example $2$: Converting Mass to Moles
A certain reaction produces $2.81 \: \text{g}$ of copper (II) hydroxide, $\ce{Cu(OH)_2}$. Determine the number of moles produced in the reaction.
• mass = 2.81g
Unknown
• mol Cu(OH)2
One conversion factor will allow us to convert from mass to moles.
Step 2: Calculate.
First, it is necessary to calculate the molar mass of $\ce{Cu(OH)_2}$ from the molar masses of $\ce{Cu}$, $\ce{O}$, and $\ce{H}$. The molar mass is $97.57 \: \text{g/mol}$.
$2.81 \: \text{g} \: \ce{Cu(OH)_2} \times \frac{1 \: \text{mol} \: \ce{Cu(OH)_2}}{97.57 \: \text{g} \: \ce{Cu(OH)_2}} = 0.0288 \: \text{mol} \: \ce{Cu(OH)_2}\nonumber$
Step 3: Think about your result.
The relatively small mass of product formed results in a small number of moles.
Simulation
Practice converting mass to moles in everyday objects in this simulation:
How many atoms are in a gold ring, a grain of sand, or the human body?
Summary
• Calculations involving conversions between moles of a material and the mass of that material are described.
Review
1. You have 19.7 grams of a material and wonder how many moles were formed. Your friend tells you to multiply the mass by grams/mole. Is your friend correct?
2. How many grams of MgO are in 3.500 moles?
3. How many moles of H2O are in 15.2 grams of pure ice? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.04%3A_Conversions_Between_Moles_and_Mass.txt |
How much gas is there?
Avogadro was interested in studying gases. He theorized that equal volumes of gases under the same conditions contained the same number of particles. Other researchers studied how many gas particles were in specific volumes of gases. Eventually, scientists were able to develop the relationship between number of particles and mass, using the idea of moles.
Conversions Between Mass and Number of Particles
In "Conversions Between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and the mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to grams. In order to convert from mass to number of particles or vice-versa, a conversion to moles is required.
Example $1$: Converting Mass to Particles
How many molecules is $20.0 \: \text{g}$ of chlorine gas, $\ce{Cl_2}$?
Known
• Molar mass $\ce{Cl_2} = 70.90 \: \text{g/mol}$
• $20.0 \: \text{g} \: \ce{Cl_2}$
Unknown
• number of molecules of Cl2
Use two conversion factors. The first converts grams of $\ce{Cl_2}$ to moles. The second converts moles of $\ce{Cl_2}$ to the number of molecules.
Step 2: Calculate.
$20.0 \: \text{g} \: \ce{Cl_2} \times \frac{1 \: \text{mol} \: \ce{Cl_2}}{70.90 \: \text{g} \: \ce{Cl_2}} \times \frac{6.02 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}}{1 \: \text{mol} \: \ce{Cl_2}} = 1.70 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}\nonumber$
The problem is done using two consecutive conversion factors. There is no need to explicitly calculate the moles of $\ce{Cl_2}$.
Step 3: Think about your result.
Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro's number.
Summary
• Calculations are illustrated for conversions between mass and number of particles.
Review
1. Why can’t we convert directly from number of particles to grams?
2. How many atoms of chlorine are present in 1.70×1023 molecules Cl2?
3. How many molecules of BH3 are in 14.32 grams BH3?
10.06: Avogadro's Hypothesis and Molar Volume
How do scuba divers know if they will run out of gas?
Knowing how much gas is available for a dive is crucial to a diver's survival. The tank on the diver's back is equipped with gauges to indicate how much gas is present and what the pressure is. A basic knowledge of gas behavior allows the diver to assess how long they can stay underwater without developing problems.
Avogadro's Hypothesis and Molar Volume
Volume is a third way to measure the amount of matter, after item count and mass. With liquids and solids, volume varies greatly depending on the density of the substance. This is because solid and liquid particles are packed close together with very little space in between the particles. However, gases are largely composed of empty spaces between the actual gas particles (see figure below).
In 1811, Amadeo Avogadro explained that the volumes of all gases can be easily determined. Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles. Since the total volume that a gas occupies is made up primarily of the empty space between the particles, the actual size of the particles themselves is nearly negligible. A given volume of a gas with small light particles, such as hydrogen $\left( \ce{H_2} \right)$, contains the same number of particles as the same volume of a heavy gas with large particles, such as sulfur hexafluoride, $\ce{SF_6}$.
Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another. This decreases the amount of empty space and reduces the volume of the gas. Gas volume is also affected by temperature. When a gas is heated, its molecules move faster and the gas expands. Because of the variation in gas volume due to pressure and temperature changes, the comparison of gas volumes must be done at standard temperature and pressure. Standard temperature and pressure (STP) is defined as $0^\text{o} \text{C}$ $\left( 273.15 \: \text{K} \right)$ and $1 \: \text{atm}$ pressure. The molar volume of a gas is the volume of one mole of a gas at STP. At STP, one mole ($6.02 \times 10^{23}$ representative particles) of any gas occupies a volume of $22.4 \: \text{L}$ (figure below).
The figure below illustrates how molar volume can be seen when comparing different gases. Samples of helium $\left( \ce{He} \right)$, nitrogen $\left( \ce{N_2} \right)$, and methane $\left( \ce{CH_4} \right)$ are at STP. Each contains 1 mole or $6.02 \times 10^{23}$ particles. However, the mass of each gas is different and corresponds to the molar mass of that gas: $4.00 \: \text{g/mol}$ for $\ce{He}$, $28.0 \: \text{g/mol}$ for $\ce{N_2}$, and $16.0 \: \text{g/mol}$ for $\ce{CH_4}$.
Summary
• Equal volumes of gases at the same conditions contain the same number of particles.
• Standard temperature and pressure is abbreviated (STP).
• Standard temperature is 0°C (273.15 K) and standard pressure is 1 atm.
• At STP, one mole of any gas occupies a volume of 22.4 L
Review
• A container is filled with gas, what do we know about the space actually taken up by a gas?
• Why do we need to do all our comparisons at the same temperature and pressure?
• At standard temperature and pressure, 1 mole of gas is always equal to how many liters? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.05%3A_Conversions_Between_Mass_and_Number_of_Particles.txt |
How can you tell how much gas is in these containers?
Small gas tanks are often used to supply gases for chemistry reactions. A gas gauge will give some information about how much is in the tank, but quantitative estimates are needed so that the reaction will be able to proceed to completion. Knowing how to calculate the necessary parameters for gases is very helpful to avoid running out earlier than desired.
Conversions Between Moles and Gas Volume
Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles. The equality of $1 \: \text{mol} = 22.4 \: \text{L}$ is the basis for the conversion factor.
Example $1$: Converting Gas Volume to Moles
Many metals react with acids to produce hydrogen gas. A certain reaction produces $86.5 \: \text{L}$ of hydrogen gas at STP. How many moles of hydrogen were produced?
Known
• $86.5 \: \text{L} \: \ce{H_2}$
• $1 \: \text{mol} = 22.4 \: \text{L}$
Unknown
• moles of H2
Apply a conversion factor to convert from liters to moles.
Step 2: Calculate.
$86.5 \: \text{L} \: \ce{H_2} \times \frac{1 \: \text{mol} \: \ce{H_2}}{22.4 \: \text{L} \: \ce{H_2}} = 3.86 \: \text{mol} \: \ce{H_2}\nonumber$
Step 3: Think about your result.
The volume of gas produced is nearly four times larger than the molar volume. The fact that the gas is hydrogen plays no role in the calculation.
Example $2$: Convertig Moles to Gas Volume
What volume does $4.96 \: \text{mol}$ of $\ce{O_2}$ occupy at STP?
Solution
Step 1: List the known quantities and plan the problem.
Known
• $4.96 \: \text{mol} \: \ce{O_2}$
• $1 \: \text{mol} = 22.4 \: \text{L}$
• volume of O2
Step 2: Calculate.
$4.96 \: \text{mol} \times 22.4 \: \text{L/mol} = 111.1 \: \text{L}\nonumber$
Step 3: Think about your result.
The volume seems correct given the number of moles.
Example $3$: Converting Volume to Mass
If we know the volume of a gas sample at STP, we can determine how much mass is present. Assume we have $867 \: \text{L}$ of $\ce{N_2}$ at STP. What is the mass of the nitrogen gas?
Known
• $867 \: \text{L} \: \ce{N_2}$
• $1 \: \text{mol} = 22.4 \: \text{L}$
• Molar mass $\ce{N_2} = 28.02 \: \text{g/mol}$
• mass of N2
Step 2: Calculate.
We start by determining the number of moles of gas present. We know that 22.4 liters of a gas at STP equals one mole, so:
$867 \: \text{L} \times \frac{1 \: \text{mol}}{22.4 \: \text{L}} = 38.7 \: \text{mol}\nonumber$
We also know the molecular weight of $\ce{N_2}$ $\left( 28.0 \: \text{g/mol} \right)$, so we can then calculate the weight of nitrogen gas in 867 liters:
$38.7 \: \text{mol} \times \frac{28 \: \text{g}}{1 \: \text{mol}} = 1083.6 \: \text{g} \: \ce{N_2}\nonumber$
Step 3: Think about your result.
In a multi-step problem, be sure that the units check out.
Summary
• Conversions between moles and volume of a gas are shown.
Review
1. In the problems above, why was the gas always at standard temperature and pressure?
2. A container contains 45.2 L of N2 gas at STP. How many moles of N2 gas are in the container?
3. If the gas in the previous problem was CH4 gas at STP instead of N2 gas, then how many moles of CH4 gas would there be? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.07%3A_Conversions_Between_Moles_and_Gas_Volume.txt |
Why does carbon dioxide sink in air?
When we run a reaction to produce a gas, we expect it to rise into the air. Many students have done experiments where gases such as hydrogen are formed. The gas can be trapped in a test tube held upside-down over the reaction. Carbon dioxide, on the other hand, sinks when it is released. Carbon dioxide has a density greater than air, so it will not rise like the hydrogen gas.
Gas Density
As you know, density is defined as the mass per unit volume of a substance. Since gases all occupy the same volume on a per mole basis, the density of a particular gas is dependent on its molar mass. A gas with a small molar mass will have a lower density than a gas with a large molar mass. Gas densities are typically reported in $\text{g/L}$. Gas density can be calculated from molar mass and molar volume.
Example $1$: Gas Density
What is the density of nitrogen gas at STP?
Known
• $\ce{N_2} = 28.02 \: \text{g/mol}$
• $1 \: \text{mol} = 22.4 \: \text{L}$
Unknown
• density = ? g/L
Molar mass divided by molar volume yields the gas density at STP.
Step 2: Calculate.
$\frac{28.02 \: \text{g}}{1 \: \text{mol}} \times \frac{1 \: \text{mol}}{22.4 \: \text{L}} = 1.25 \: \text{g/L}\nonumber$
When set up with a conversion factor, the $\text{mol}$ unit cancels, leaving $\text{g/L}$ as the unit in the result.
Step 3: Think about your result.
The molar mass of nitrogen is slightly larger than molar volume, so the density is slightly greater than $1 \: \text{g/L}$.
Alternatively, the molar mass of a gas can be determined if the density of the gas at STP is known.
Example $2$: Molar Mass from Gas Density
What is the molar mass of a gas whose density is $0.761 \: \text{g/L}$ at STP?
Solution
Step 1: List the known quantities and plan the problem.
Known
• $\ce{N_2} = 28.02 \: \text{g/mol}$
• $1 \: \text{mol} = 22.4 \: \text{L}$
Unknown
• molar mass = ? g/mol
Molar mass is equal to density multiplied by molar volume.
Step 2: Calculate.
$\frac{0.761 \: \text{g}}{1 \: \text{L}} \times \frac{22.4 \: \text{L}}{1 \: \text{mol}} = 17.0 \: \text{g/mol}\nonumber$
Step 3: Think about your result.
Because the density of the gas is less than $1 \: \text{g/L}$, the molar mass is less than 22.4.
Summary
• Calculations are described showing conversions between molar mass and density for gases.
Review
1. How is density calculated?
2. How is molar mass calculated?
3. What would be the volume of 3.5 moles of a gas?
10.08: Mole Road Map
How do I get from here to there?
If I want to visit the town of Manteo, North Carolina, out on the coast, I will need a map of how to get there. I may have a printed map or I may download directions from the internet, but I need something to get me going in the right direction. Chemistry road maps serve the same purpose. How do I handle a certain type of calculation? There is a process and a set of directions to help.
Mole Road Map
Previously, we saw how the conversions between mass and number of particles required two steps, with moles as the intermediate. This concept can now be extended to also include gas volume at STP. The resulting diagram is referred to as a mole road map (see figure below).
The mole is at the center of any calculation involving amount of a substance. The sample problem below is one of many different problems that can be solved using the mole road map.
Example $1$: Mole Road Map
What is the volume of $79.3 \: \text{g}$ of neon gas at STP?
Known
• $\ce{Ne} = 20.18 \: \text{g/mol}$
• $1 \: \text{mol} = 22.4 \: \text{L}$
Unknown
• volume = ? L
The conversion factors will be grams $\rightarrow$ moles $\rightarrow$ gas volume.
Step 2: Calculate.
$79.3 \: \text{g} \: \ce{Ne} \times \frac{1 \: \text{mol} \: \ce{Ne}}{20.18 \: \text{g} \: \ce{Ne}} \times \frac{22.4 \: \text{L} \: \ce{Ne}}{1 \: \text{mol} \: \ce{Ne}} = 88.0 \: \text{L} \: \ce{Ne}\nonumber$
Step 3: Think about your result.
The given mass of neon is equal to about 4 moles, resulting in a volume that is about 4 times larger than molar volume.
Summary
• An overall process is given for calculations involving moles, grams, and gas volume.
Review
1. In the problem above, what is the formula weight of neon?
2. What value is at the center of all the calculations?
3. If we had 79.3 grams of Xe, would we expect a volume that is greater than or less than that obtained with neon? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.08%3A_Gas_Density.txt |
Is there anything healthy in this jar?
Packaged foods that you eat typically have nutritional information provided on the label. The label on a jar of peanut butter (shown above) reveals that one serving size is considered to be $32 \: \text{g}$. The label also gives the masses of various types of compounds that are present in each serving. One serving contains $7 \: \text{g}$ of protein, $15 \: \text{g}$ of fat, and $3 \: \text{g}$ of sugar. By calculating the fraction of protein, fat, or sugar in one serving size of peanut butter and converting to percent values, we can determine the composition of peanut butter on a percent by mass basis.
Percent Composition
Chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound. It is calculated in a similar way to what was just indicated for the peanut butter.
$\% \: \text{by mass} = \frac{\text{mass of element}}{\text{mass of compound}} \times 100\%\nonumber$
Percent Composition from Mass Data
The sample problem below shows the calculation of the percent composition of a compound based on mass data.
Example $1$: Percent Composition from Mass
A certain newly synthesized compound is known to contain the elements zinc and oxygen. When a $20.00 \: \text{g}$ sample of the compound is decomposed, $16.07 \: \text{g}$ of zinc remains. Determine the percent composition of the compound.
Known
• Mass of compound $= 20.00 \: \text{g}$
• Mass of $\ce{Zn} = 16.07 \: \text{g}$
Unknown
• Percent $\ce{Zn} = ? \%$
• Percent $\ce{O} = ? \%$
Subtract to find the mass of oxygen in the compound. Divide each element's mass by the mass of the compound to find the percent by mass.
Step 2: Calculate.
$\text{Mass of oxygen} = 20.00 \: \text{g} - 16.07 \: \text{g} = 3.93 \: \text{g} \: \ce{O}\nonumber$
$\% \: \ce{Zn} = \frac{16.07 \: \text{g} \: \ce{Zn}}{20.00 \: \text{g}} \times 100\% = 80.35\% \: \ce{Zn}\nonumber$
$\% \: \ce{O} = \frac{3.93 \: \text{g} \: \ce{O}}{20.00 \: \text{g}} \times 100\% = 19.65\% \: \ce{O}\nonumber$
Step 3: Think about your result.
The calculations make sense because the sum of the two percentages adds up to $100\%$. By mass, the compound is mostly zinc.
Percent Composition from a Chemical Formula
The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. That is divided by the molar mass of the compound and multiplied by $100\%$.
$\% \: \text{by mass} = \frac{\text{mass of element in} \: 1 \: \text{mol}}{\text{molar mass of compound}} \times 100\%\nonumber$
The percent composition of a given compound is always the same as long as the compound is pure.
Example $2$: Percent Composition from Chemical Formula
Dichlorineheptoxide $\left( \ce{Cl_2O_7} \right)$ is a highly reactive compound used in some organic synthesis reactions. Calculate the percent composition of dichlorineheptoxide.
Known
• Mass of $\ce{Cl}$ in $1 \: \text{mol} \: \ce{Cl_2O_7} = 70.90 \: \text{g}$
• Mass of $\ce{O}$ in $1 \: \text{mol} \: \ce{Cl_2O_7} = 112.00 \: \text{g}$
• Molar mass of $\ce{Cl_2O_7} = 182.90 \: \text{g/mol}$
Unknown
• Percent $\ce{Cl} = ? \%$
• Percent $\ce{O} = ? \%$
Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by $100\%$.
Step 2: Calculate.
$\% \ce{Cl} = \frac{70.90 \: \text{g} \: \ce{Cl}}{182.90 \: \text{g}} \times 100\% = 38.76\% \: \ce{Cl}\nonumber$
$\% \: \ce{O} = \frac{112.00 \: \text{g} \: \ce{O}}{182.90 \: \text{g}} \times 100\% = 61.24\% \: \ce{O}\nonumber$
Step 3: Think about your result.
The percentages add up to $100\%$.
Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorine heptoxide is $38.76\% \: \ce{Cl}$ and $61.24\% \: \ce{O}$. Suppose that you need to know the masses of chlorine and oxygen present in a $12.50 \: \text{g}$ sample of dichlorine heptoxide. You can set up a conversion factor based on the percent by mass of each element:
$12.50 \: \text{g} \: \ce{Cl_2O_7} \times \frac{38.76 \: \text{g} \: \ce{Cl}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 4.845 \: \text{g} \: \ce{Cl}\nonumber$
$12.50 \: \text{g} \: \ce{Cl_2O_7} \times \frac{61.24 \: \text{g} \: \ce{O}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 7.655 \: \text{g} \: \ce{O}\nonumber$
The sum of the two masses is $12.50 \: \text{g}$, the mass of the sample size.
Science Friday: Stained Glass Conservation
Stained glass from the Middle Ages is often hundreds of years old. Unfortunately, many of these relics are in need of cleaning and maintenance. In this video by Science Friday, conservator Mary Higgins discusses the methods used to protect the stained glass.
Summary
• Processes are described for calculating the percent composition of a material based on mass or on chemical composition.
Review
1. What is the formula for calculating percent composition?
2. What information do you need to calculate percent composition by mass?
3. What do subscripts in a chemical formula tell you? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.10%3A_Percent_Composition.txt |
Why does the color change?
If you look at a typical bottle of copper sulfate, it will be a bluish-green. If someone were to tell you that copper sulfate is white, you likely would not believe them. However, it turns out that you are both right; it just depends on the copper sulfate. Your blue-green copper sulfate has several water molecules attached to it, while your friend's copper sulfate is anhydrous (no water attached). Why the difference? The water molecules interact with some of the $d$ electrons in the copper ion and produce the color. When the water is removed, the electron configuration changes and the color disappears.
Percent of Water in a Hydrate
Many ionic compounds naturally contain water as part of the crystal lattice structure. A hydrate is a compound that has one or more water molecules bound to each formula unit. Ionic compounds that contain a transition metal are often highly colored. Interestingly, it is common for the hydrated form of a compound to be of a different color than the anhydrous form, which has no water in its structure. A hydrate can usually be converted to the anhydrous compound by heating. For example, the anhydrous compound cobalt (II) chloride is blue, while the hydrate is a distinctive magenta color.
The hydrated form of cobalt (II) chloride contains six water molecules in each formula unit. The name of the compound is cobalt (II) chloride hexahydrate and its formula is $\ce{CoCl_2} \cdot 6 \ce{H_2O}$. The formula for water is set apart at the end of the formula with a dot, followed by a coefficient that represents the number of water molecules per formula unit.
It is useful to know the percent of water contained within a hydrate. The sample problem below demonstrates the procedure.
Example $1$: Percent of Water in a Hydrate
Find the percent water in cobalt (II) chloride hexahydrate, $\ce{CoCl_2} \cdot 6 \ce{H_2O}$.
Solution
Step 1: List the known quantities and plan the problem.
The mass of water in the hydrate is the coefficient (6) multiplied by the molar mass of $\ce{H_2O}$. The molar mass of the hydrate is the molar mass of the $\ce{CoCl_2}$ plus the mass of water.
Known
• Mass of $\ce{H_2O}$ in $1 \: \text{mol}$ hydrate $= 108.12 \: \text{g}$
• Molar mass of hydrate $= 237.95 \: \text{g/mol}$
Unknown
• percent H2O = ?%
Calculate the percent by mass of water by dividing the mass of $\ce{H_2O}$ in 1 mole of the hydrate by the molar mass of the hydrate and multiplying by $100\%$.
Step 2: Calculate.
$\% \: \ce{H_2O} = \frac{108.12 \: \text{g} \: \ce{H_2O}}{237.95 \: \text{g}} \times 100\% = 45.44\% \: \ce{H_2O}\nonumber$
Step 3: Think about your result.
Nearly half of the mass of the hydrate is composed of water molecules within the crystal.
Summary
• The process of calculating the percent water in a hydrate is described.
Review
1. What is a hydrate?
2. How can you convert a hydrate to an anhydrous compound?
3. What does hexahydrate mean?
10.12: Determining Empirical Formulas
What is occuring in this picture?
In the early days of chemistry, there were few tools available for the detailed study of compounds. Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials. The "new" field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely. The relative amounts of elements could be determined, but many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios. Chemists had no way to determine the exact amounts of these atoms that were contained in specific molecules.
Determining Empirical Formulas
An empirical formula is one that shows the lowest whole-number ratio of the elements in a compound. Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical. However, we can also consider the empirical formula of a molecular compound. Ethene is a small hydrocarbon compound with the formula $\ce{C_2H_4}$ (see figure below). While $\ce{C_2H_4}$ is its molecular formula and represents its true molecular structure, it has an empirical formula of $\ce{CH_2}$. The simplest ratio of carbon to hydrogen in ethene is 1:2. There are two ways to view that ratio. Considering one molecule of ethene, the ratio is 1 carbon atom for every 2 atoms of hydrogen. Considering one mole of ethene, the ratio is 1 mole of carbon for every 2 moles of hydrogen. So, the subscripts in a formula represent the mole ratio of the elements in that formula.
In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it. These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula. The steps to be taken are outlined below.
1. Assume a $100 \: \text{g}$ sample of the compound, so that the given percentages can be directly converted into grams.
2. Use each element's molar mass to convert the grams of each element to moles.
3. In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest.
4. If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.
5. In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply each of the moles by the smallest whole number that will convert each into a whole number. Write the empirical formula.
Example $1$: Determining the Empirical Formula of a Compound
A compound of iron and oxygen is analyzed and found to contain $69.94\%$ iron and $30.06\%$ oxygen. Find the empirical formula of the compound.
Known
• $\%$ of $\ce{Fe} = 69.94\%$
• $\%$ of $\ce{O} = 30.06\%$
Unknown
• Empirial formula = Fe?O?
Steps to follow are outlined in the text.
Step 2: Calculate.
1. Assume a $100 \: \text{g}$ sample.
$69.94 \: \text{g} \: \ce{Fe}\nonumber$
$30.06 \: \text{g} \: \ce{O}\nonumber$
2. Convert to moles.
$69.94 \: \text{g} \: \ce{Fe} \times \frac{1 \: \text{mol} \: \ce{Fe}}{55.85 \: \text{g} \: \ce{Fe}} = 1.252 \: \text{mol} \: \ce{Fe}\nonumber$
$30.06 \: \text{g} \: \ce{O} \times \frac{1 \: \text{mol} \: \ce{O}}{16.00 \: \text{g} \: \ce{O}} = 1.879 \: \text{mol} \: \ce{O}\nonumber$
3. Divide both moles by the smallest of the results.
$\frac{1.252 \: \text{mol} \: \ce{Fe}}{1.252} = 1 \: \text{mol} \: \ce{Fe} \: \: \: \: \: \frac{1.879 \: \text{mol} \: \ce{O}}{1.252} = 1.501 \: \text{mol} \ce{O}\nonumber$
4/5. Since the moles of $\ce{O}$ is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.
$1 \: \text{mol} \: \ce{Fe} \times 2 = 2 \: \text{mol} \: \ce{Fe} \: \: \: \: \: 1.501 \: \text{mol} \: \ce{O} \times 2 = 3 \: \text{mol} \: \ce{O}\nonumber$
The empirical formula of the compound is $\ce{Fe_2O_3}$.
Step 3: Think about your result.
The subscripts are whole numbers and represent the mole ratio of the elements in the compound. The compound is the ionic compound iron (III) oxide.
Summary
• A process is described for the calculation of the empirical formula for a compound based on the percent composition of that compound.
Review
1. What is an empirical formula?
2. What does an empirical formula tell you?
3. What does it not tell you?
10.13: Determining Molecular Formulas
How can you determine the differences between these two molecules?
Above we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar. Some people could distinguish them on the basis of taste, but it's not a good idea to go around tasting chemicals. The best way is to determine the molecular weights—this approach allows you to easily tell which compound is which.
Molecular Formulas
Molecular formulas give the kind and number of atoms of each element present in a molecular compound. In many cases, the molecular formula is the same as the empirical formula. The molecular formula of methane is $\ce{CH_4}$ and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is $\ce{C_2H_4O_2}$. Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is $\ce{C_6H_{12}O_6}$. The structures of both molecules are shown in the figure below. They are very different compounds, yet both have the same empirical formula of $\ce{CH_2O}$.
Empirical formulas can be determined from the percent composition of a compound. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:
1. Calculate the empirical formula mass (EFM), which is simply the molar mass represented by the empirical formula.
2. Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.
3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.
Example $1$: Determining the Molecular Formula of a Compound
The empirical formula of a compound of boron and hydrogen is $\ce{BH_3}$. Its molar mass is $27.7 \: \text{g/mol}$. Determine the molecular formula of the compound.
Known
• Empirical formula $= \ce{BH_3}$
• Molar mass $= 27.7 \: \text{g/mol}$
Unknown
• molecular formula = ?
Steps to follow are outlined in the text.
Step 2: Calculate.
$\text{Empirical formula mass (EFM)} = 13.84 \: \text{g/mol}\nonumber$
$\frac{\text{molar mass}}{\text{EFM}} = \frac{17.7}{13.84} = 2\nonumber$
$\ce{BH_3} \times 2 = \ce{B_2H_6}\nonumber$
The molecular formula of the compound is $\ce{B_2H_6}$.
Step 3: Think about your result.
The molar mass of the molecular formula matches the molar mass of the compound.
Summary
• A procedure is described for the calculation of the exact molecular formula of a compound.
Review
1. What is the difference between an empirical formula and a molecular formula?
2. In addition to the elemental analysis, what do you need to know to calculate the molecular formula?
3. What does the empirical formula mass tell you? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/10%3A_The_Mole/10.11%3A_Percent_of_Water_in_a_Hydrate.txt |
What's for Dinner?
Various ways of recording recipes have developed over the centuries. The cookbook shown above was written by a woman who probably collected all her own recipes. Later, printed cookbooks became available (even guys had no excuse for not being able to cook). Today we can find recipes on a number of internet sites and can quickly search for information on how to cook anything we want. Reading a recipe sometimes requires we understand a few codes and symbols (what’s the difference between a tsp and a Tsp?), but the information on what we start with and what we end up with is there.
Writing Chemical Equations
Chemical reactions are occurring all around us. Plants use sunlight to drive their photosynthetic process and produce energy. Cars and other vehicles burn gasoline in order to power their engines. Batteries use electrochemical reactions to produce energy and power many everyday devices. Many chemical reactions are going on within the human body as well, particularly during the digestion of food.
In math class, you have written and solved many mathematical equations. Chemists keep track of chemical reactions by writing equations as well. In any chemical reaction, one or more substances—called reactants—are converted into one or more new substances—called products. The general form of the equation for such a process looks like this:
$\text{Reactants} \rightarrow \text{Products}\nonumber$
Unlike a math equation, a chemical equation does not use an equal sign. Instead, the arrow is called a yield sign and so the equation is described as "reactants yield products".
Word Equations
You can describe a chemical reaction by writing a word equation. When silver metal is exposed to sulfur, it reacts to form silver sulfide. Silver sulfide is commonly known as tarnish and turns the surface of silver objects dark and streaky black (see figure below). The sulfur that contributes to tarnish can come from traces of sulfur in the air, or from foods such as eggs. The word equation for the process is:
$\text{Silver} + \text{sulfur} \rightarrow \text{Silver sulfide}\nonumber$
The silver and the sulfur are the reactants in the equation, while the silver sulfide is the product.
Another common chemical reaction is the burning of methane gas. Methane is the major component of natural gas and is commonly burned on a gas stove or in a Bunsen burner (figure below). Burning is a chemical reaction in which some type of fuel is reacted with oxygen gas. The products of the reaction in the burning of methane, as well as other fuels, are carbon dioxide and water. The word equation for this reaction is:
$\text{Methane} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water}\nonumber$
Word equations can be very useful, but do have one major drawback—they cannot be used for any quantitative work. A word equation does not tell how many moles of each material are needed, or how many moles of product are formed.
Summary
• Word equations are used to describe the conversion of reactants to products.
Review
1. Write the generic form of a chemical reaction.
2. What are reactants?
3. What are products?
11.02: Chemical Equations
How do you make shrimp gumbo?
Shrimp gumbo is one of many dishes that are part of the Cajun culture in Louisiana. It's a spicy dish that needs careful control of all the ingredients so that it has a "kick", but is not overwhelming. Recipes convey not only what the preparation entails, but also describe how much of each ingredient is needed, and the details of how to cook the meal. Similarly, we need this type of information in order to carry out chemical reactions successfully and safely.
Chemical Equations
Word equations are time-consuming to write and do not prove to be convenient for many of the things that chemists need to do with equations. A chemical equation is a representation of a chemical reaction that displays the reactants and products with chemical formulas. The chemical equation for the reaction of methane with oxygen is shown:
$\ce{CH_4} + \ce{O_2} \rightarrow \ce{CO_2} + \ce{H_2O}\nonumber$
The equation above, called a skeleton equation, is an equation that shows only the formulas of the reactants and products with nothing to indicate the relative amounts. The first step in writing an accurate chemical equation is to write the skeleton equation, making sure that the formulas of all substances involved are written correctly. All reactants are written to the left of the yield arrow, separated from one another by a plus sign. Likewise, products are written to the right of the yield arrow, also separated with a plus sign.
It is often important to know the physical states of the reactants and products taking part in a reaction. To do this, put the appropriate symbol in parentheses after each formula: $\left( s \right)$ for solid, $\left( l \right)$ for liquid, $\left( g \right)$ for gas, and $\left( aq \right)$ for an aqueous (water-based) solution. The previous reaction becomes:
$\ce{CH_4} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber$
The table below shows a listing of symbols used in chemical equations. Some, such as the double arrow which represents equilibrium, and the use of a catalyst in a reaction, will be detailed in other concepts.
Symbols Description
Table $1$: Symbols Used in Chemical Equations
$+$ Used to separate multiple reactants or products.
$\rightarrow$ Yield sign; separates reactants from products.
$\rightleftharpoons$ Replaces the yield sign for reversible reactions that reach equilibrium.
$\left( s \right)$ Reactant or product in the solid state.
$\left( l \right)$ Reactant or product in the liquid state.
$\left( g \right)$ Reactant or product in the gaseous state.
$\left( aq \right)$ Reactant or product in an aqueous solution (dissolved in water).
$\overset{\ce{Pt}}{\rightarrow}$ Formula written above the arrow is used as a catalyst in the reaction.
$\overset{\Delta}{\rightarrow}$ Triangle indicates that the reaction is being heated.
Summary
• A chemical equation is representation a chemical reaction using chemical formulas reactants and products, as well as symbols.
• Symbols used in chemical equations are described and explained.
Review
1. What does a skeleton equation tell you?
2. Why would you want to know the physical state of materials in a chemical reaction?
3. In a chemical reaction what does symbol → mean?
4. If I see Δ over the arrow, what does this indicate? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.01%3A_Word_Equations.txt |
Any leftovers?
When you cook a meal, quite often there are leftovers because you prepared more than people would eat at one sitting. Sometimes when you repair a piece of equipment, you end up with what are called "pocket parts"—small pieces you put in your pocket because you're not sure where they belong. Chemistry tries to avoid leftovers and pocket parts. In normal chemical processes, we cannot create or destroy matter (law of conservation of mass). If we start out with ten carbon atoms, we need to end up with ten carbon atoms. John Dalton's atomic theory said that chemical reactions basically involve the rearrangement of atoms. Chemical equations need to follow these principles in order to be correct.
Balancing Chemical Equations
A balanced equation is a chemical equation in which mass is conserved and there are equal numbers of atoms of each element on both sides of the equation. We can write a chemical equation for the reaction of carbon with hydrogen gas to form methane $\left( \ce{CH_4} \right)$:
$\begin{array}{ccccc} \ce{C} \left( s \right) & + & \ce{H_2} \left( g \right) & \rightarrow & \ce{CH_4} \left( g \right) \ 2 \: \ce{C} \: \text{atoms} & & 2 \: \ce{H} \: \text{atoms} & & 1 \: \ce{C} \: \text{atom,} \: 4 \: \ce{H} \: \text{atoms} \end{array}\nonumber$
In order to write a correct equation, you must first write the correct skeleton equation with the correct chemical formulas. Recall that hydrogen is a diatomic molecule and so is written as $\ce{H_2}$.
When we count the number of atoms of both elements, shown under the equation, we see that the equation is not balanced. There are only 2 atoms of hydrogen on the reactant side of the equation, while there are 4 atoms of hydrogen on the product side. We can balance the above equation by adding a coefficient of 2 in front of the formula for hydrogen.
$\ce{C} \left( s \right) + 2 \ce{H_2} \left( g \right) \rightarrow \ce{CH_4} \left( g \right)\nonumber$
A coefficient is a small whole number placed in front of a formula in an equation in order to balance it. The 2 in front of the $\ce{H_2}$ means that there are a total of $2 \times 2 = 4$ atoms of hydrogen as reactants. Visually, the reaction looks like the figure below.
In the balanced equation, there is one atom of carbon and four atoms of hydrogen on both sides of the arrow. Below are guidelines for writing and balancing chemical equations.
1. Determine the correct chemical formulas for each reactant and product.
2. Write the skeleton equation.
3. Count the number of atoms of each element that appears as a reactant and as a product. If a polyatomic ion is unchanged on both sides of the equation, count it as a unit.
4. Balance each element one at a time by placing coefficients in front of the formulas.
1. It is best to begin by balancing elements that only appear in one chemical formula on each side of the equation.
2. No coefficient is written for a 1.
3. NEVER change the subscripts in a chemical formula—you can only balance equations by using coefficients.
5. Check each atom or polyatomic ion to be sure that they are equal on both sides of the equation.
6. Make sure that all coefficients are in the lowest possible ratio. If necessary, reduce to the lowest ratio.
Example $1$: Balancing Chemical Equations
Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.
Step 1: Plan the problem.
Follow the steps for writing and balancing a chemical equation listed in the text.
Step 2: Solve.
Write the skeleton equation with the correct formulas.
$\ce{Pb(NO_3)_2} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)\nonumber$
Count the number of each atom or polyatomic ion on both sides of the equation.
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \ 2 \: \ce{NO_3^-} \: \text{ions} & 1 \: \ce{NO_3^-} \: \text{ions} \ 1 \: \ce{Na} \: \text{atom} & 1 \: \ce{Na} \: \text{atom} \ 1 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}\nonumber$
The nitrate ions and the chlorine atoms are unbalanced. Start by placing a 2 in front of the $\ce{NaCl}$. This increases the reactant counts to 2 $\ce{Na}$ atoms and 2 $\ce{Cl}$ atoms. Then place a 2 in front of the $\ce{NaNO_3}$. The result is:
$\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{NaCl} \left( aq \right) \rightarrow 2 \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)\nonumber$
The new count for each atom and polyatomic ion becomes:
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \ 2 \: \ce{NO_3^-} \: \text{ions} & 2 \: \ce{NO_3^-} \: \text{ions} \ 2 \: \ce{Na} \: \text{atom} & 2 \: \ce{Na} \: \text{atom} \ 2 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}\nonumber$
Step 3: Think about your result.
The equation is now balanced since there are equal numbers of atoms of each element on both sides of the equation.
Simulation
Practice balancing chemical equations with this simulation:
Can you balance a chemical equation?
Summary
• The process of balancing chemical equations is described.
Review
1. What is the law of conservation of mass?
2. How did Dalton describe the process of a chemical reaction?
3. Why don’t we change the subscripts in order to balance an equation? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.03%3A_Balancing_Equations.txt |
How useful is a wheel rim?
A wheel rim is not very useful by itself. Driving on the rim can damage it and make for a very rough ride. When the rim is combined with a tire, the product can be put on a car and used for a safe and comfortable ride. The two separate items combine to make something that improves the car ride.
Combination Reactions
A combination reaction is a reaction in which two or more substances combine to form a single new substance. Combination reactions can also be called synthesis reactions. The general form of a combination reaction is:
$\ce{A} + \ce{B} \rightarrow \ce{AB}\nonumber$
One combination reaction is two elements combining to form a compound. Solid sodium metal reacts with chlorine gas to produce solid sodium chloride.
$2 \ce{Na} \left( s \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{NaCl} \left( s \right)\nonumber$
In order to write and balance the equation correctly, it is important to remember the seven elements that exist in nature as diatomic molecules: $\ce{H_2}$, $\ce{N_2}$, $\ce{O_2}$, $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$.
One sort of combination reaction that occurs frequently is the reaction of an element with oxygen to form an oxide. Metals and nonmetals both react readily with oxygen under most conditions. Magnesium reacts rapidly and dramatically when ignited, combining with oxygen from the air to produce a fine powder of magnesium oxide.
$2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right)\nonumber$
Sulfur reacts with oxygen to form sulfur dioxide.
$\ce{S} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{SO_2} \left( g \right)\nonumber$
When nonmetals react with one another, the product is a molecular compound. Often, the nonmetal reactants can combine in different ratios and produce different products. Sulfur can also combine with oxygen to form sulfur trioxide.
$2 \ce{S} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right)\nonumber$
Transition metals are capable of adopting multiple positive charges within their ionic compounds. Therefore, most transition metals are capable of forming different products in a combination reaction. Iron reacts with oxygen to form both iron (II) oxide and iron (III) oxide:
$2 \ce{Fe} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{FeO} \left( s \right)\nonumber$
$4 \ce{Fe} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{Fe_2O_3} \left( s \right)\nonumber$
Example $1$: Combination Reactions
Potassium is a very reactive alkali metal that must be stored under oil in order to prevent it from reacting with air. Write the balanced chemical equation for the combination reaction of potassium with oxygen.
Step 1: Plan the problem.
Make sure formulas of all reactants and products are correct before balancing the equation. Oxygen gas is a diatomic molecule. Potassium oxide is an ionic compound and so its formula is constructed by the crisscross method. Potassium as an ion becomes $\ce{K^+}$, while the oxide ion is $\ce{O^{2-}}$.
Step 2: Solve.
The skeleton (unbalanced) equation:
$\ce{K} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{K_2O} \left( s \right) \nonumber\nonumber$
The equation is then easily balanced with coefficients.
$4 \ce{K} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{K_2O} \left( s \right) \nonumber\nonumber$
Step 3: Think about your result.
Formulas are correct and the resulting combination reaction is balanced.
Exercise $1$
Can you build a balanced, combination reaction for the formation of aluminum oxide (Al2O3)?
Answer
4Al + 3O2 → 2Al2O3
Combination reactions can also take place when an element reacts with a compound to form a new compound composed of a larger number of atoms. Carbon monoxide reacts with oxygen to form carbon dioxide, according to the equation:
$2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right)\nonumber$
Two compounds may also react to form a more complex compound. A very common example is the reaction of an oxide with water. Calcium oxide reacts readily with water to produce an aqueous solution of calcium hydroxide.
$\ce{CaO} \left( s \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{Ca(OH)_2} \left( aq \right)\nonumber$
Sulfur trioxide gas reacts with water to form sulfuric acid. This is an unfortunately common reaction that occurs in the atmosphere in some places where oxides of sulfur are present as pollutants. The acid formed in the reaction falls to the ground as acid rain.
$\ce{SO_3} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_2SO_4} \left( aq \right)\nonumber$
Summary
• Combination reactions occur when two or more substances combine to form a new substance.
Review
1. What are combination reactions?
2. Write the product of the following reaction: Mg+H2O→
3. Is CH4+2O2→CO2+2H2O a combination reaction? Explain your answer. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.04%3A_Combination_Reactions.txt |
How does a decomposition reaction work?
Antoine Lavoisier is widely known as the “father of modern chemistry.” He was one of the first to study chemical reactions in detail. Lavoisier reacted mercury with oxygen to form mercuric oxide as part of his studies on the composition of the atmosphere. He was then able to show that the decomposition of mercuric oxide produced mercury and oxygen. The diagram above shows the apparatus used by Lavoisier to study the formation and decomposition of mercuric oxide.
Decomposition Reactions
A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances. The general form of a decomposition reaction is:
$\ce{AB} \rightarrow \ce{A} + \ce{B}\nonumber$
Most decomposition reactions require an input of energy in the form of heat, light, or electricity.
Binary compounds are compounds composed of just two elements. The simplest kind of decomposition reaction is when a binary compound decomposes into its elements. Mercury (II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas.
$2 \ce{HgO} \left( s \right) \rightarrow 2 \ce{Hg} \left( l \right) + \ce{O_2} \left( g \right)\nonumber$
A reaction is also considered to be a decomposition reaction even when one or more of the products are still compounds. A metal carbonate decomposes into a metal oxide and carbon dioxide gas. For example, calcium carbonate decomposes into calcium oxide and carbon dioxide:
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber$
Metal hydroxides decompose on heating to yield metal oxides and water. Sodium hydroxide decomposes to produce sodium oxide and water:
$2 \ce{NaOH} \left( s \right) \rightarrow \ce{Na_2O} \left( s \right) + \ce{H_2O} \left( g \right)\nonumber$
Some unstable acids decompose to produce nonmetal oxides and water. Carbonic acid decomposes easily at room temperature into carbon dioxide and water:
$\ce{H_2CO_3} \left( aq \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber$
Exercise $1$
Can you build a balanced Chemical reaction for the decomposition of Potassium chlorate (KClO3)?
Answer
2KClO3 → 2KCl + 3O2
Example $1$: Decomposition Reactions
When an electric current is passed through pure water, it decomposes into its elements. Write a balanced equation for the decomposition of water.
Solution
Step 1: Plan the problem.
Water is a binary compound composed of hydrogen and oxygen. The hydrogen and oxygen gases produced in the reaction are both diatomic molecules.
Step 2: Solve.
The skeleton (unbalanced) equation:
$\ce{H_2O} \left( l \right) \overset{\text{elec}}{\rightarrow} \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right)\nonumber$
Note the abbreviation "$\text{elec}$" above the arrow to indicate the passage of an electric current to initiate the reaction. Balance the equation.
$2 \ce{H_2O} \left( l \right) \overset{\text{elec}}{\rightarrow} 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right)\nonumber$
Step 3: Think about your result.
The products are elements and the equation is balanced.
Simulations
How does an airbag explode at just the right time in the event of a car accident? Learn more in this simulation:
How does a chemical reaction inflate and airbag
Summary
• A definition of decomposition reaction and example reactions are given.
Review
1. What is a decomposition reaction?
2. What is usually needed for a decomposition reaction to take place?
3. Are elements always the product of a decomposition reaction?
11.06: Combustion Reactions
How do you cook the perfect marshmallow?
Roasting marshmallows over an open fire is a favorite past-time for campers, outdoor cook-outs, and just gathering around a fire in the back yard. The trick is to get the marshmallow a nice golden brown without catching it on fire. Too often we are not successful and we see the marshmallow burning on the stick – a combustion reaction taking place right in front of us.
Combustion Reactions
A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve $\ce{O_2}$ as one reactant. The combustion of hydrogen gas produces water vapor:
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)\nonumber$
Notice that this reaction also qualifies as a combination reaction.
The Hindenberg was a hydrogen-filled airship that suffered an accident upon its attempted landing in New Jersey in 1937. The hydrogen immediately combusted in a huge fireball, destroying the airship and killing 36 people. The chemical reaction was a simple one: hydrogen combining with oxygen to produce water.
Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy. Propane $\left( \ce{C_3H_8} \right)$ is a gaseous hydrocarbon that is commonly used as the fuel source in gas grills.
$\ce{C_3H_8} \left( g \right) + 5 \ce{O_2} \left( g \right) \rightarrow 3 \ce{CO_2} \left( g \right) + 4 \ce{H_2O} \left( g \right)\nonumber$
Example $1$: Combustion Reactions
Ethanol can be used as a fuel source in an alcohol lamp. The formula for ethanol is $\ce{C_2H_5OH}$. Write the balanced equation for the combustion of ethanol.
Solution
Step 1: Plan the problem.
Ethanol and oxygen are the reactants. As with a hydrocarbon, the products of the combustion of an alcohol are carbon dioxide and water.
Step 2: Solve.
Write the skeleton equation:
$\ce{C_2H_5OH} \left( l \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( g \right)\nonumber$
Balance the equation.
$\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( g \right)\nonumber$
Step 3: Think about your result.
Combustion reactions must have oxygen as a reactant. Note that the water produced is in the gas state, rather than the liquid state, because of the high temperatures that accompany a combustion reaction.
Summary
• Combustion reaction is defined and examples are given.
Review
1. What is needed for a combustion reaction to take place?
2. What is formed in any combustion reaction?
3. Mercury reacts with oxygen to form mercuric oxide. Is this a combustion reaction?
4. What are the products of any combustion reaction involving a hydrocarbon? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.05%3A_Decomposition_Reactions.txt |
Why is the silver dark?
The platter and pitcher shown above provides an example of tarnish, a chemical reaction caued when silver metal reacts with hydrogen sulfide gas produced by some industrial processes or as a result of decaying animal or plant materials:
$2 \ce{Ag} + \ce{H_2S} \rightarrow \ce{Ag_2S} + \ce{H_2}\nonumber$
The tarnish can be removed using a number of polishes, but the process also removes a small amount of silver along with the tarnish.
Single-Replacement Reactions
A single-replacement reaction is a reaction in which one element replaces a similar element in a compound. The general form of a single-replacement (also called single-displacement) reaction is:
$\ce{A} + \ce{BC} \rightarrow \ce{AC} + \ce{B}\nonumber$
In this general reaction, element $\ce{A}$ is a metal and replaces element $\ce{B}$ (also a metal) in the compound. When the element that is doing the replacing is a nonmetal, it must replace another nonmetal in a compound, and the general equation becomes:
$\ce{Y} + \ce{XZ} \rightarrow \ce{XY} + \ce{Z}\nonumber$
(Where $\ce{Y}$ is a nonmetal and replaces the nonmetal $\ce{Z}$ in the compound with $\ce{X}$.)
Metal Replacement
Magnesium is a more reactive metal than copper. When a strip of magnesium metal is placed in an aqueous solution of copper (II) nitrate, it replaces the copper. The products of the reaction are aqueous magnesium nitrate and solid copper metal.
$\ce{Mg} \left( s \right) + \ce{Cu(NO_3)_2} \left( aq \right) \rightarrow \ce{Mg(NO_3)_2} \left( aq \right) + \ce{Cu} \left( s \right)\nonumber$
This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (copper).
Hydrogen Replacement
Many metals react easily with acids and when they do so, one of the products of the reaction is hydrogen gas. Zinc reacts with hydrochloric acid to produce aqueous zinc chloride and hydrogen (figure below).
$\ce{Zn} \left( s \right) + 2 \ce{HCl} \left( aq \right) \rightarrow \ce{ZnCl_2} \left( aq \right) + \ce{H_2} \left( g \right)\nonumber$
In a hydrogen replacement reaction, the hydrogen in the acid is replaced by an active metal.
Some metals are so reactive that they are capable of replacing the hydrogen in water. The products of such a reaction are the metal hydroxide and hydrogen gas. All group 1 metals undergo this type of reaction. Sodium reacts vigorously with water to produce aqueous sodium hydroxide and hydrogen (see figure below).
$2 \ce{Na} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow 2 \ce{NaOH} \left( aq \right) + \ce{H_2} \left( g \right)\nonumber$
Halogen Replacement
The element chlorine reacts with an aqueous solution of sodium bromide to produce aqueous sodium chloride and elemental bromine:
$\ce{Cl_2} \left( g \right) + 2 \ce{NaBr} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{Br_2} \left( l \right)\nonumber$
The reactivity of the halogen group (group 17) decreases from top to bottom within the group. Fluorine is the most reactive halogen, while iodine is the least. Since chlorine is above bromine, it is more reactive than bromine and can replace it in a halogen replacement reaction.
Summary
• The activity series describes the relative reactivities of metals and halogens.
Review
1. What is a metal replacement reaction?
2. Will a non-metal replace a metal?
3. What is the most reactive halogen?
4. What products will I get if I add potassium metal to water? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.07%3A_Single_Replacement_Reactions.txt |
What's the difference between the two pictures above?
We see above two metals that can be exposed to water. The picture on the left is of sodium, which gives a violent reaction when it comes in contact with water. The picture on the right is of silver, a metal so unreactive with water that it can be made into drinking vessels. Both metals have a single s electron in their outer shell, so you would predict similar reactivities. However, we have a better tool that allows us to make better prediction about what will react with what.
The Activity Series
Single-replacement reactions only occur when the element that is doing the replacing is more reactive than the element that is being replaced. Therefore, it is useful to have a list of elements in order of their relative reactivities. The activity series is a list of elements in decreasing order of reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series. The table below is an activity series of most common metals, and of the halogens.
Activity of Metals Activity of Halogens
Table $1$: Activity Series
Elements Reaction Occurring Elements
$\ce{Li}$
$\ce{K}$
$\ce{Ba}$
$\ce{Sr}$
$\ce{Ca}$
$\ce{Na}$
React with cold water, replacing hydrogen.
$\ce{F_2}$
$\ce{Cl_2}$
$\ce{Br_2}$
$\ce{I_2}$
$\ce{Mg}$
$\ce{Al}$
$\ce{Zn}$
$\ce{Cr}$
$\ce{Fe}$
$\ce{Cd}$
React with steam, but not cold water, replacing hydrogen.
$\ce{Co}$
$\ce{Ni}$
$\ce{Sn}$
$\ce{Pb}$
Do not react with water. React with acids, replacing hydrogen.
$\ce{H_2}$
$\ce{Cu}$
$\ce{Hg}$
$\ce{Ag}$
$\ce{Pt}$
$\ce{Au}$
Unreactive with water or acids.
For a single-replacement reaction, a given element is capable of replacing an element that is below it in the activity series. This can be used to predict if a reaction will occur. Suppose that small pieces of the metal nickel were placed into two separate aqueous solutions: one of iron (III) nitrate and one of lead (II) nitrate. Looking at the activity series, we see that nickel is below iron, but above lead. Therefore, the nickel metal will be capable of replacing the lead in a reaction, but will not be capable of replacing iron.
$\ce{Ni} \left( s \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow \ce{Ni(NO_3)_2} \left( aq \right) + \ce{Pb} \left( s \right)\nonumber$
$\ce{Ni} \left( s \right) + \ce{Fe(NO_3)_3} \left( aq \right) \rightarrow \text{NR (no reaction)}\nonumber$
In the descriptions that accompany the activity series of metals, a given metal is also capable of undergoing the reactions described below that section. For example, lithium will react with cold water, replacing hydrogen. It will also react with steam and with acids, since that requires a lower degree of reactivity.
Example $1$: Single-Replacement Reactions
Use the activity series to predict if the following reactions will occur. If not, write $\text{NR}$. If the reaction does occur, write the products of the reaction and balance the equation.
1. $\ce{Al} \left( s \right) + \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow$
2. $\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow$
Step 1: Plan the problem.
For A, compare the placements of aluminum and zinc on the activity series. For B, compare the placements of silver and hydrogen.
Step 2: Solve.
Since aluminum is above zinc, it is capable of replacing it and a reaction will occur. The products of the reaction will be aqueous aluminum nitrate and solid zinc. Take care to write the correct formulas for the products before balancing the equation. Aluminum adopts a $+3$ charge in an ionic compound, so the formula for aluminum nitrate is $\ce{Al(NO_3)_3}$. The balanced equation is:
$2 \ce{Al} \left( s \right) + 3 \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{Al(NO_3)_3} \left( aq \right) + 3 \ce{Zn} \left( s \right)\nonumber$
Since silver is below hydrogen, it is not capable of replacing hydrogen in a reaction with an acid.
$\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow \text{NR}\nonumber$
Summary
• Metals and halogens are ranked according to their ability to displace other metals or halogens below them in the series.
Review
1. What does the activity series tell us?
2. Can a metal undergo any of the reactions listed below it in the series?
3. List two metals that cobalt will displace and two that will displace it. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.08%3A_Activity_Series.txt |
Want to trade?
The practice of barter (trading one thing for another) has been in existence for a very long time. In the illustration above, items like chickens were bartered for newspapers. You have something I want, and I have something you want. So we trade and we each have something new. Some chemical reactions are like that. Compounds swap parts and you have new materials.
Double-Replacement Reactions
A double-replacement reaction is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement (also called double-displacement) reaction is:
$\ce{AB} + \ce{CD} \rightarrow \ce{AD} + \ce{CB}\nonumber$
In this reaction, $\ce{A}$ and $\ce{C}$ are positively-charged cations, while $\ce{B}$ and $\ce{D}$ are negatively-charged anions. Double-replacement reactions generally occur between substances in aqueous solution. In order for a reaction to occur, one of the products is usually a solid precipitate, a gas, or a molecular compound such as water.
Formation of a Precipitate
A precipitate forms in a double-replacement reaction when the cations from one of the reactants combine with the anions from the other reactant to form an insoluble ionic compound. When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, the following reaction occurs:
$2 \ce{KI} \left( aq \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{KNO_3} \left( aq \right) + \ce{PbI_2} \left( s \right)\nonumber$
There are very strong attractive forces that occur between $\ce{Pb^{2+}}$ and $\ce{I^-}$ ions and the result is a brilliant yellow precipitate (see figure below). The other product of the reaction, potassium nitrate, remains soluble.
Formation of a Gas
Some double-replacement reactions produce a gaseous product which then bubbles out of the solution and escapes into the air. When solutions of sodium sulfide and hydrochloric acid are mixed, the products of the reaction are aqueous sodium chloride and hydrogen sulfide gas.
$\ce{Na_2S} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{H_2S} \left( g \right)\nonumber$
Formation of a Molecular Compound
Another kind of double-replacement reaction is one that produces a molecular compound as one of its products. Many examples in this category are reactions that produce water. When aqueous hydrochloric acid is reacted with aqueous sodium hydroxide, the products are aqueous sodium chloride and water:
$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
Example $1$: Double-Replacement Reactions
Write a complete and balanced chemical equation for the following double-replacement reactions. One product is indicated as a guide.
1. $\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow$ (hydrogen cyanide gas is formed)
2. $\ce{(NH_4)_2SO_4} \left( aq \right) + \ce{Ba(NO_3)_2} \left( aq \right) \rightarrow$ (a precipitate of barium sulfate forms)
Solution
Step 1: Plan the problem.
In A, the production of a gas drives the reaction. In B, the production of a precipitate drives the reaction. In both cases, use the ionic charges of both reactants to construct the correct formulas of the products.
Step 2: Solve.
A. The cations of both reactants are $+1$ charged ions, while the anions are $-1$ charged ions. After exchanging partners, the balanced equation is:
$\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow \ce{NaBr} \left( aq \right) + \ce{HCN} \left( g \right)\nonumber$
B. Ammonium ion and nitrate ion are $1+$ and $1-$ respectively, while barium and sulfate are $2+$ and $2-$. This must be taken into account when exchanging partners and writing the new formulas. Then, the equation is balanced.
$\ce{(NH_4)_2SO_4} \left( aq \right) + \ce{Ba(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{NH_4NO_3} \left( aq \right) + \ce{BaSO_4} \left( s \right)\nonumber$
Step 3: Think about your result.
Both are double replacement reactions. All formulas are correct and the equations are balanced.
Occasionally, a reaction will produce both a gas and a molecular compound. The reaction of a sodium carbonate solution with hydrochloric acid produces aqueous sodium chloride, carbon dioxide gas, and water.
$\ce{Na_2CO_3} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber$
Summary
• The double-replacement reaction generally takes the form of AB + CDAD + CB where A and C are positively-charged cations, while B and D are negatively-charged anions.
• In a double replacement reactions, typically one of the products is a precipitate, a gas, or a molecular compound.
Review
1. In a double-replacement reaction, what type of compounds are usually the reactants? A double-replacement reaction occurs between sodium sulfide and hydrogen chloride. Write the rest of the chemical equation and balance it.
2. Na2S + HCl →
3. In double-replacement reaction, one of three possible types of products usually form. What are the three types? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/11%3A_Chemical_Reactions/11.09%3A_Double_Replacement_Reactions.txt |
How much equipment do you need for an experiment?
You are in charge of setting out the lab equipment for a chemistry experiment. If you have twenty students in the lab (and they will be working in teams of two) and the experiment calls for three beakers and two test tubes, how much glassware do you need to set out? Figuring this out involves a type of balanced equation and the sort of calculations that you would do for a chemical reaction.
Everyday Stoichiometry
You have learned about chemical equations and the techniques used in order to balance them. Chemists use balanced equations to allow them to manipulate chemical reactions in a quantitative manner. Before we look at a chemical reaction, let's consider an equation for the ideal ham sandwich.
Our ham sandwich is composed of 2 slices of ham $\left( \ce{H} \right)$, a slice of cheese $\left( \ce{C} \right)$, a slice of tomato $\left( \ce{T} \right)$, 5 pickles $\left( \ce{P} \right)$, and 2 slices of bread $\left( \ce{B} \right)$. The equation for our sandwich is:
$2 \ce{H} + \ce{C} + \ce{T} + 5 \ce{P} + 2 \ce{B} \rightarrow \ce{H_2CTP_5B_2}\nonumber$
Now let us suppose that you are having some friends over and need to make five ham sandwiches. How much of each sandwich ingredient do you need? You take the number of each ingredient required for one sandwich (its coefficient in the above equation) and multiply by five. Using ham and cheese as examples, and using a conversion factor, you can calculate:
$5 \ce{H_2CTP_5B_2} \times \frac{2 \: \ce{H}}{1 \ce{H_2CTP_5B_2}} = 10 \: \ce{H}\nonumber$
$5 \ce{H_2CTP_5B_2} \times \frac{1 \ce{C}}{1 \ce{H_2CTP_5B_2}} = 5 \: \ce{C}\nonumber$
The conversion factors contain the coefficient of each specific ingredient as the numerator and the formula of one sandwich as the denominator. The result is what you would expect. In order to make five ham sandwiches, you need 10 slices of ham and 5 slices of cheese.
This type of calculation demonstrates the use of stoichiometry. Stoichiometry is the calculation of the amount of substances in a chemical reaction from the balanced equation. The sample problem below is another stoichiometry problem involving ingredients of the ideal ham sandwich.
Example $1$: Ham Sandwich Stoichiometry
Kim looks in the refrigerator and finds that she has 8 slices of ham. In order to make as many sandwiches as possible, how many pickles does she need? Use the equation above.
Step 1: List the known quantities and plan the problem.
• Have 8 ham slices $\left( \ce{H} \right)$
• $2 \: \ce{H} = 5 \: \ce{P}$ (conversion factor)
Unknown
• How many pickles (P) needed?
The coefficients for the two reactants (ingredients) are used to make a conversion factor between ham slices and pickles.
Step 2: Solve.
$8 \: \ce{H} \times \frac{5 \: \ce{P}}{2 \: \ce{H}} = 20 \: \ce{P}\nonumber$
Since 5 pickles combine with 2 ham slices in each sandwich, 20 pickles are needed to fully combine with 8 ham slices.
Step 3: Think about your result.
The 8 ham slices will make 4 ham sandwiches. With 5 pickles per sandwich, the 20 pickles are used in the 4 sandwiches.
Summary
• An example of everyday stoichiometry is given.
Review
1. A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equation to describe the relationship.
2. Write a conversion factor to show the relationship between the number of ice cubes and the number of smoothies produced.
3. How many strawberries would you need to make 12 smoothies? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.01%3A_Everyday_Stoichiometry.txt |
What does this porch need?
You want to add some sections to the porch seen above. Before you go to the hardware store to buy lumber, you need to determine the unit composition (the material between two large uprights). You count how many posts, how many boards, how many rails – then you decide how many sections you want to add before you calculate the amount of building material needed for your porch expansion.
Mole Ratios
Stoichiometry problems can be characterized by two things: (1) the information given in the problem, and (2) the information that is to be solved for, referred to as the unknown. The given and the unknown may both be reactants, both be products, or one may be a reactant while the other is a product. The amounts of the substances can be expressed in moles. However, in a laboratory situation, it is common to determine the amount of a substance by finding its mass in grams. The amount of a gaseous substance may be expressed by its volume. In this concept, we will focus on the type of problem where both the given and the unknown quantities are expressed in moles.
Chemical equations express the amounts of reactants and products in a reaction. The coefficients of a balanced equation can represent either the number of molecules or the number of moles of each substance. The production of ammonia $\left( \ce{NH_3} \right)$ from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\nonumber$
The balanced equation can be analyzed in several ways, as shown in the figure below.
We see that 1 molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia. This is the smallest possible relative amount of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia.
The most useful quantity for counting particles is the mole. So if each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation.
Finally, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. $1 \: \ce{mol}$ of nitrogen has a mass of $28.02 \: \text{g}$, while $3 \: \text{mol}$ of hydrogen has a mass of $6.06 \: \text{g}$, and $2 \: \text{mol}$ of ammonia has a mass of $34.08 \: \text{g}$.
$28.02 \: \text{g} \: \ce{N_2} + 6.06 \: \text{g} \: \ce{H_2} \rightarrow 34.08 \: \text{g} \: \ce{NH_3}\nonumber$
Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved.
A mole ratio is a conversion factor that relates the amounts in moles of any two substances in a chemical reaction. The numbers in a conversion factor come from the coefficients of the balanced chemical equation. The following six mole ratios can be written for the ammonia forming reaction above.
$\begin{array}{ccc} \dfrac{1 \: \text{mol} \: \ce{N_2}}{3 \: \text{mol} \: \ce{H_2}} & or & \dfrac{3 \: \text{mol} \: \ce{H_2}}{1 \: \text{mol} \: \ce{N_2}} \ \dfrac{1 \: \text{mol} \: \ce{N_2}}{2 \: \text{mol} \: \ce{NH_3}} & or & \dfrac{2 \: \text{mol} \: \ce{NH_3}}{1 \: \text{mol} \: \ce{N_2}} \ \dfrac{3 \: \text{mol} \: \ce{H_2}}{2 \: \text{mol} \: \ce{NH_3}} & or & \dfrac{2 \: \text{mol} \: \ce{NH_3}}{3 \: \text{mol} \: \ce{H_2}} \end{array}\nonumber$
In a mole ratio problem, the given substance, expressed in moles, is written first. The appropriate conversion factor is chosen in order to convert from moles of the given substance to moles of the unknown.
Example $1$: Mole Ratio
How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?
Known
• given: H2 = 4.20 mol
Unknown
• mol of NH3
The conversion is from $\text{mol} \: \ce{H_2}$ to $\text{mol} \: \ce{NH_3}$. The problem states that there is an excess of nitrogen, so we do not need to be concerned with any mole ratio involving $\ce{N_2}$. Choose the conversion factor that has the $\ce{NH_3}$ in the numerator and the $\ce{H_2}$ in the denominator.
Step 2: Solve.
$4.20 \: \text{mol} \: \ce{H_2} \times \dfrac{2 \: \text{mol} \: \ce{NH_3}}{3 \: \text{mol} \: \ce{H_2}} = 2.80 \: \text{mol} \: \ce{NH_3}\nonumber$
The reaction of $4.20 \: \text{mol}$ of hydrogen with excess nitrogen produces $2.80 \: \text{mol}$ of ammonia.
Step 3: Think about your result.
The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.
Summary
• Mole ratios allow comparison of the amounts of any two materials in a balanced equation.
• Calculations can be made to predict how much product can be obtained from a given number of moles of reactant.
Review
• If a reactant is in excess, why do we not worry about the mole ratios involving that reactant?
• What is the mole ratio of H to N in the ammonia molecule?
• The formula for ethanol is CH3CH2OH. What is the mole ratio of H to C in this molecule? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.02%3A_Mole_Ratios.txt |
Need nails?
When you are doing a large construction project, you have a good idea of how many nails you will need (lots!). When you go to the hardware store, you don’t want to sit there and count out several hundred nails. You can buy nails by weight, so you determine how many nails are in a pound, calculate how many pounds you need, and you’re on your way to begin building.
While the mole ratio is ever-present in all stoichiometry calculations, amounts of substances in the laboratory are most often measured by mass. Therefore, we need to use mole-mass calculations in combination with mole ratios to solve several different types of mass-based stoichiometry problems.
Mass to Moles Problems
In this type of problem, the mass of one substance is given, usually in grams. From this, you are to determine the amount in moles of another substance that will either react with or be produced from the given substance.
$\text{mass of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown}\nonumber$
The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation.
Example $1$: Mass-Mole Stoichiometry
Tin metal reacts with hydrogen fluoride to produce tin (II) fluoride and hydrogen gas, according to the following balanced equation.
$\ce{Sn} \left( s \right) + 2 \ce{HF} \left( g \right) \rightarrow \ce{SnF_2} \left( s \right) + \ce{H_2} \left( g \right)\nonumber$
How many moles of hydrogen fluoride are required to react completely with $75.0 \: \text{g}$ of tin?
Known
• Given: $75.0 \: \text{g} \: \ce{Sn}$
• Molar mass of $\ce{Sn} = 118.69 \: \text{g/mol}$
• $1 \: \text{mol} \: \ce{Sn} = 2 \: \ce{mol} \: \ce{HF}$ (mole ratio)
Unknown
• mol HF
Use the molar mass of $\ce{Sn}$ to convert the grams of $\ce{Sn}$ to moles. Then use the mole ratio to convert from $\text{mol} \: \ce{Sn}$ to $\text{mol} \: \ce{HF}$. This will be done in a single two-step calculation.
$\text{g} \: \ce{Sn} \rightarrow \text{mol} \: \ce{Sm} \rightarrow \text{mol} \: \ce{HF}$
Step 2: Solve.
$75.0 \: \text{g} \: \ce{Sn} \times \frac{1 \: \text{mol} \: \ce{Sn}}{118.69 \: \text{g} \: \ce{Sn}} \times \frac{2 \: \text{mol} \: \ce{HF}}{1 \: \text{mol} \: \ce{Sn}} = 1.26 \: \text{mol} \: \ce{HF}\nonumber$
Step 3: Think about your result.
The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of $\ce{HF}$ is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
Moles to Mass Problems
In this type of problem, the amount of one substance is given in moles. From this, you are to determine the mass of another substance that will either react with or be produced from the given substance.
$\text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{mass of unknown}\nonumber$
The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into mass in grams by use of the molar mass of that substance from the periodic table.
Example $2$: Mole-Mass Stoichiometry
Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor:
$2 \ce{H_2S} \left( g \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\nonumber$
What mass of oxygen gas is consumed in a reaction that produces $4.60 \: \text{mol} \: \ce{SO_2}$?
Known
• Given: $4.60 \: \text{mol} \: \ce{SO_2}$
• $2 \: \text{mol} \: \ce{SO_2} = 3 \: \text{mol} \: \ce{O_2}$ (mole ratio)
• Molar mass of $\ce{O_2} = 32.00 \: \text{g/mol}$
Unknown
• mass O2 = ? g
Use the mole ratio to convert from $\text{mol} \: \ce{SO_2}$ to $\text{mol} \: \ce{O_2}$. Then convert $\text{mol} \: \ce{O_2}$ to grams. This will be done in a single two-step calculation.
$\text{mol} \: \ce{SO_2} \rightarrow \text{mol} \: \ce{O_2} \rightarrow \text{g} \: \ce{O_2}$
Step 2: Solve.
$4.60 \: \text{mol} \: \ce{SO_2} \times \frac{3 \: \text{mol} \: \ce{O_2}}{2 \: \text{mol} \: \ce{SO_2}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \text{mol} \: \ce{O_2}} = 221 \: \text{g} \: \ce{O_2}\nonumber$
Step 3: Think about your result.
According to the mole ratio, $6.90 \: \text{mol} \: \ce{O_2}$ is produced with a mass of $221 \: \text{g}$. The answer has three significant figures because the given number of moles has three significant figures.
Summary
• Calculations involving conversions of mass to moles and moles to mass are described.
Review
1. In the first problem, what would happen if you multiply grams Sn by 118.69 grams/mole Sn?
2. Why is a balanced equation needed?
3. Does the physical form of the material matter for these calculations? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.03%3A_Mass-Mole_Stoichiometry.txt |
How many walnuts are needed to equal 250 grams?
I want to send 250 grams of shelled walnuts to a friend (don't ask why- just go with the question). How many walnuts in shells do I need to buy? To figure this out, I need to know how much the shell of a walnut weighs (about $40\%$ of the total weight of the unshelled walnut). I can then calculate the mass of walnuts that will yield 250 grams of shelled walnuts, and then determine how many walnuts I need to buy.
Mass to Mass Problems
Mass-mass calculations are the most practical of all mass-based stoichiometry problems. Moles cannot be measured directly, while the mass of any substance can generally be easily measured in the lab. This type of problem is three steps, and is a combination of the two previous types.
$\text{mass of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{mass of unknown}\nonumber$
The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Finally, the moles of the unknown are converted to mass by use of its molar mass.
Example $1$: Mass-Mass Stoichiometry
Ammonium nitrate decomposes to dinitrogen monoxide and water, according to the following equation.
$\ce{NH_4NO_3} \left( s \right) \rightarrow \ce{N_2O} \left( g \right) + 2 \ce{H_2O} \left( l \right)\nonumber$
In a certain experiment, $45.7 \: \text{g}$ of ammonium nitrate is decomposed. Find the mass of each of the products formed.
Solution
Step 1: List the known quantities and plan the problem.
Known
• Given: $45.7 \: \text{g} \: \ce{NH_4NO_3}$
• $1 \: \text{mol} \: \ce{NH_4NO_3} = 1 \: \text{mol} \: \ce{N_2O} = 2 \: \text{mol} \: \ce{H_2O}$
• Molar mass of $\ce{NH_4NO_3} = 80.06 \: \text{g/mol}$
• Molar mass of $\ce{N_2O} = 44.02 \: \text{g/mol}$
• Molar mass of $\ce{H_2O} = 18.02 \: \text{g/mol}$
Unknown
• Mass $\ce{N_2O} = ? \: \text{g}$
• Mass $\ce{H_2O} = ? \: \text{g}$
Perform two separate three-step mass-mass calculations, as shown below.
$\text{g} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{N_2O} \rightarrow \text{g} \: \ce{N_2O}\nonumber$
$\text{g} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{NH_4NO_3} \rightarrow \text{mol} \: \ce{H_2O} \rightarrow \text{g} \: \ce{H_2O}\nonumber$
Step 2: Solve.
$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \frac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \frac{1 \: \text{mol} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \frac{44.02 \: \text{g} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{N_2O}} = 25.1 \: \text{g} \: \ce{N_2O}\nonumber$
$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \frac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \frac{2 \: \ce{H_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \frac{18.02 \: \text{g} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} = 20.6 \: \text{g} \: \ce{H_2O}\nonumber$
Step 3: Think about your result.
The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.
Summary
• Mass-mass calculations involve converting the mass of a reactant to moles of reactant, then using mole ratios to determine moles of product which can then be converted to mass of product.
Review
1. If matter is neither created nor destroyed, why can’t we just go directly from grams of reactant to grams of product?
2. Why is it important to get the subscripts correct in the formulas?
3. Why do the coefficients need to be correct? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.04%3A_Mass-Mass_Stoichiometry.txt |
How much propane is left in the tank?
As the weather gets warmer, more and more people want to cook out on the back deck or back yard. Many folks still use charcoal for grilling because of the added flavor. But increasing numbers of back yard cooks like to use a propane grill. The gas burns clean, the grill is ready to go as soon as the flame is lit – but how do you know how much propane is left in the tank? You can buy gauges at hardware stores that measure gas pressure and tell you how much is left in the tank.
Volume-Volume Stoichiometry
Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of gas particles. Further, one mole of any gas at standard temperature and pressure ($0^\text{o} \text{C}$ and $1 \: \text{atm}$) occupies a volume of $22.4 \: \text{L}$. These characteristics make stoichiometry problems involving gases at STP very straightforward. Consider the reaction of nitrogen and oxygen gases to form nitrogen dioxide:
$\begin{array}{lllll} \ce{N_2} \left( g \right) & + & 2 \ce{O_2} \left( g \right) & \rightarrow & 2 \ce{NO_2} \left( g \right) \ 1 \: \text{molecule} & & 2 \: \text{molecules} & & 2 \: \text{molecules} \ 1 \: \text{mol} & & 2 \: \text{mol} & & 2 \: \text{mol} \ 1 \: \text{volume} & & 2 \: \text{volumes} & & 2 \: \text{volumes} \end{array}\nonumber$
Because of Avogadro's work, we know that the mole ratios between substances in a gas-phase reaction are also volume ratios. The six possible volume ratios for the above equation are:
\begin{align*} \frac{1 \: \text{volume} \: \ce{N_2}}{2 \: \text{volumes} \: \ce{O_2}} &\text{or} \frac{2 \: \text{volumes} \: \ce{O_2}}{1 \: \text{volume} \: \ce{N_2}} \ \frac{1 \: \text{volume} \: \ce{N_2}}{2 \: \text{volumes} \: \ce{NO_2}} &\text{or} \frac{2 \: \text{volumes} \: \ce{NO_2}}{1 \: \text{volume} \: \ce{N_2}} \ \frac{2 \: \text{volumes} \: \ce{O_2}}{2 \: \text{volumes} \: \ce{NO_2}} &\text{or} \frac{2 \: \text{volumes} \: \ce{NO_2}}{2 \: \text{volumes} \: \ce{O_2}} \end{align*}\nonumber
The volume ratios above can easily be used when the volume of one gas in a reaction is known, and you need to determine the volume of another gas that will either react with or be produced from the first gas. The pressure and temperature conditions of both gases need to be the same.
Example $1$: Volume-Volume Stoichiometry
The combustion of propane gas produces carbon dioxide and water vapor.
$\ce{C_3H_8} \left( g \right) + 5 \ce{O_2} \left( g \right) \rightarrow 3 \ce{CO_2} \left( g \right) + 4 \ce{H_2O} \left( g \right)\nonumber$
What volume of oxygen is required to completely combust $0.650 \: \text{L}$ of propane? What volume of carbon dioxide is produced in the reaction?
Known
• Given: $0.650 \: \text{L} \: \ce{C_3H_8}$
• 1 volume $\ce{C_3H_8} = 5$ volumes $\ce{O_2}$
• 1 volume $\ce{C_3H_8} = 3$ volumes $\ce{CO_2}$
Unknown
• Volume $\ce{O_2} = ? \: \text{L}$
• Volume $\ce{CO_2} = ? \: \text{L}$
Two separate calculations can be done using the volume ratios.
Step 2: Solve.
$0.650 \: \text{L} \: \ce{C_3H_8} \times \frac{5 \: \text{L} \: \ce{O_2}}{1 \: \text{L} \: \ce{C_3H_8}} = 3.25 \: \text{L} \: \ce{O_2}\nonumber$
$0.650 \: \text{L} \: \ce{C_3H_8} \times \frac{3 \: \text{L} \: \ce{CO_2}}{1 \: \text{L} \: \ce{C_3H_8}} = 1.95 \: \text{L} \: \ce{CO_2}\nonumber$
Step 3: Think about your result.
Because the coefficients of the $\ce{O_2}$ and the $\ce{CO_2}$ are larger than that of the $\ce{C_3H_8}$, the volumes for those two gases are greater. Note that total volume is not necessarily conserved in a reaction because moles are not necessarily conserved. In this reaction, 6 total volumes of reactants become 7 total volumes of products.
Summary
• Calculations of volume-volume ratios are based on Avogadro's hypothesis.
• Pressures and temperatures of the gases involved need to be the same.
Review
1. What is Avogadro’s hypothesis?
2. How much volume is occupied by one mole of a gas at STP?
3. In the sample problem above, assume we combust 1.3 L of propane. How much CO2 will be produced? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.05%3A_Volume-Volume_Stoichiometry.txt |
How much azide is needed to fill an air bag?
Cars and many other vehicles have air bags in them. In case of a collision, a reaction is triggered so that the rapid decomposition of sodium azide produces nitrogen gas, filling the air bag. If too little sodium azide is used, the air bag will not fill completely and will not protect the person in the vehicle. Too much sodium azide could cause the formation of more gas that the bag can safely handle. If the bag breaks from the excess gas pressure, all protection is lost.
Mass to Volume and Volume to Mass Problems
Chemical reactions frequently involve both solid substances whose masses can be measured, as well as gases, for which volume measurements are more appropriate. Stoichiometry problems of this type are called either mass-volume or volume-mass problems.
$\text{mass of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{volume of unknown}\nonumber$
$\text{volume of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{mass of unknown}\nonumber$
Because both types of problems involve a conversion from either moles of gas to volume or vice-versa, we can use the molar volume of $22.4 \: \text{L/mol}$, provided that the conditions for the reaction are at STP.
Example $1$: Mass-Volume Stoichiometry
Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas:
$2 \: \text{Al} \left( s \right) + 3 \ce{H_2SO_4} \left( aq \right) \rightarrow \ce{Al_2(SO_4)_3} \left( aq \right) + 3 \ce{H_2} \left( g \right)\nonumber$
Determine the volume of hydrogen gas produced at STP when a $2.00 \: \text{g}$ piece of aluminum completely reacts.
Known
• Given: $2.00 \: \text{g} \: \ce{Al}$
• Molar mass $\ce{Al} = 26.98 \: \text{g/mol}$
• $2 \: \text{mol} \: \ce{Al} = 3 \: \text{mol} \: \ce{H_2}$
Unknown
• volume H2 = ?
The grams of aluminum will first be converted to moles. Then the mole ratio will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.
$\text{g} \: \ce{Al} \rightarrow \text{mol} \: \ce{Al} \rightarrow \text{mol} \: \ce{H_2} \rightarrow \text{L} \: \ce{H_2}\nonumber$
Step 2: Solve.
$2.00 \: \text{g} \: \ce{Al} \times \frac{1 \: \text{mol} \: \ce{Al}}{26.98 \: \text{g} \: \ce{Al}} \times \frac{3 \: \text{mol} \: \ce{H_2}}{2 \: \text{mol} \: \ce{Al}} \times \frac{22.4 \: \text{L} \: \ce{H_2}}{1 \: \text{mol} \: \ce{H_2}} = 2.49 \: \text{L} \: \ce{H_2}\nonumber$
Step 3: Think about your result.
The volume result is in liters. For much smaller amounts, it may be convenient to convert to milliliters. The answer here has three significant figures. Because the molar volume is a measured quantity of $22.4 \: \text{L/mol}$, three is the maximum number of significant figures for this type of problem.
Example $2$: Volume-Mass Stoichiometry
Calcium oxide is used to remove sulfur dioxide generated in coal-burning power plants, according to the following reaction.
$2 \ce{CaO} \left( s \right) + 2 \ce{SO_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CaSO_4} \left( s \right)\nonumber$
What mass of calcium oxide is required to react completely with $1.4 \times 10^3 \: \text{L}$ of sulfur dioxide?
Known
• Given: $1.4 \times 10^3 \: \text{L} = \ce{SO_2}$
• $2 \: \text{mol} \: \ce{SO_2} = 2 \: \text{mol} \ce{CaO}$
• Molar mass $\ce{CaO} = 56.08 \: \text{g/mol}$
Unknown
• mass CaO = ? g
The volume of $\ce{SO_2}$ will be converted to moles, followed by the mole ratio, and finally a conversion of moles of $\ce{CaO}$ to grams.
$\text{L} \: \ce{SO_2} \rightarrow \text{mol} \: \ce{SO_2} \rightarrow \text{mol} \: \ce{CaO} \rightarrow \text{g} \: \ce{CaO}\nonumber$
Step 2: Solve.
$1.4 \times 10^3 \: \text{L} \: \ce{SO_2} \times \frac{1 \: \text{mol} \: \ce{SO_2}}{22.4 \: \text{L} \: \ce{SO_2}} \times \frac{2 \: \text{mol} \: \ce{CaO}}{2 \: \text{mol} \: \ce{SO_2}} \times \frac{56.08 \: \text{g} \: \ce{CaO}}{1 \: \text{mol} \: \ce{CaO}} = 3.5 \times 10^3 \: \text{g} \: \ce{CaO}\nonumber$
Step 3: Think about your result.
The resultant mass could be reported as $3.5 \: \text{kg}$, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.
Summary
• Calculations are described for determining the amount of gas formed in a reaction.
• Calculations are described for determining amounts of a material needed to react with a gas.
Review
1. What are the conditions for all gases in these calculations?
2. How can you tell if all the ratios were set up correctly?
3. Why was 2 mol CaO/2mol SO2 included in the second example if it did not affect the final number? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.06%3A_Mass-Volume_Stoichiometry.txt |
Don't you hate running out of cooking ingredients?
Cooking is a great example of everyday chemistry. In order to correctly follow a recipe, a cook needs to make sure that he has plenty of all the necessary ingredients in order to make his dish. Let us suppose that you are deciding to make some pancakes for a large group of people. The recipe on the box indicates that the following ingredients are needed for each batch of pancakes:
$1$ cup of pancake mix
$\frac{3}{4}$ cup milk
$1$ egg
$1$ tablespoon vegetable oil
Now you check the pantry and the refrigerator and see that you have the following ingredients available:
2 boxes of pancake mix (8 cups)
Half a gallon of milk (4 cups)
2 eggs
Full bottle of vegetable oil (about 3 cups)
The question that you must ask is: how many batches of pancakes can I make? The answer is two. Even though you have enough pancake mix, milk, and oil to make many more batches of pancakes, you are limited by the fact that you only have two eggs. As soon as you have made two batches of pancakes, you will be out of eggs and your "reaction" will be complete.
Limiting Reactant
For a chemist, the balanced chemical equation is the recipe that must be followed. As you have seen earlier, the Haber process is a reaction in which nitrogen gas is combined with hydrogen gas to form ammonia. The balanced equation is shown below.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\nonumber$
We know that the coefficients of the balanced equation indicate the mole ratio that is required for this reaction to occur. One mole of $\ce{N_2}$ will react with three moles of $\ce{H_2}$ to form two moles of $\ce{NH_3}$.
Now let us suppose that a chemist were to react three moles of $\ce{N_2}$ with six moles of $\ce{H_2}$ (see figure below).
So what happened in this reaction? The chemist started with 3 moles of $\ce{N_2}$. You may think of this as being 3 times as much as the "recipe" (the balanced equation) requires since the coefficient for the $\ce{N_2}$ is a 1. However, the 6 moles of $\ce{H_2}$ that the chemist started with is only two times as much as the "recipe" requires, since the coefficient for the $\ce{H_2}$ is a 3 and $3 \times 2 = 6$. So, after the reaction is complete, the hydrogen gas will be completely used up; while there will be 1 mole of nitrogen gas left over. Finally, the reaction will produce 4 moles of $\ce{NH_3}$ because that is also two times as much as shown in the balanced equation. The overall reaction that occurred in words:
$2 \: \text{mol} \: \ce{N_2} + 6 \: \text{mol} \: \ce{H_2} \rightarrow 4 \: \text{mol} \: \ce{NH_3}\nonumber$
All the amounts are doubled from the original balanced equation.
The limiting reactant (or limiting reagent) is the reactant that determines the amount of product that can be formed in a chemical reaction. The reaction proceeds until the limiting reactant is completely used up. In our example above, the $\ce{H_2}$ is the limiting reactant. The excess reactant (or excess reagent) is the reactant that is initially present in a greater amount than will eventually be reacted. In other words, there is always excess reactant left over after the reaction is complete. In the above example, the $\ce{N_2}$ is the excess reactant.
Summary
• The amount of limiting reactant determines how much product will be formed in a chemical reaction.
Review
1. In the Haber reaction illustrated above, how do we know that hydrogen is the limiting reactant?
2. What if hydrogen were left over?
3. Which material would be limiting if no hydrogen or nitrogen were left over? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.07%3A_Limiting_Reactant.txt |
Who's coming for dinner?
You have ten people that show up for a dinner party. One of the guest brings twenty brownies for dessert. The decision about serving dessert is easy: two brownies are placed on every plate. If someone wants more brownies, they will have to wait until they go to the store. There are only enough brownies for everyone to have two.
Determining the Limiting Reactant
In the real world, amounts of reactants and products are typically measured by mass or by volume. It is first necessary to convert the given quantities of each reactant to moles in order to identify the limiting reactant.
Example $1$: Determining the Limiting Reactant
Silver metal reacts with sulfur to form silver sulfide according to the following balanced equation:
$2 \ce{Ag} \left( s \right) + \ce{S} \left( s \right) \rightarrow \ce{Ag_2S} \left( s \right)\nonumber$
What is the limiting reactant when $50.0 \: \text{g} \: \ce{Ag}$ is reacted with $10.0 \: \text{g} \: \ce{S}$?
Known
• Given: $50.0 \: \text{g} \: \ce{Ag}$
• Given: $10.0 \: \text{g} \: \ce{S}$
Unknown
• limiting reactant
Use the atomic masses of $\ce{Ag}$ and $\ce{S}$ to determine the number of moles of each present. Then, use the balanced equation to calculate the number of moles of sulfur that would be needed to react with the number of moles of silver present. Compare this result to the actual number of moles of sulfur present.
Step 2: Solve.
First, calculate the number of moles of $\ce{Ag}$ and $\ce{S}$ present:
$50.0 \: \text{g} \: \ce{Ag} \times \frac{1 \: \text{mol} \: \ce{Ag}}{107.87 \: \text{g} \: \ce{Ag}} = 0.464 \: \text{mol} \: \ce{Ag}\nonumber$
$10.0 \: \text{g} \: \ce{S} \times \frac{1 \: \text{mol} \: \ce{S}}{32.07 \: \text{g} \: \ce{S}} = 0.312 \: \text{mol} \: \ce{S}\nonumber$
Second, find the moles of $\ce{S}$ that would be required to react with all of the given $\ce{Ag}$:
$0.464 \: \text{mol} \: \ce{Ag} \times \frac{1 \: \text{mol} \: \ce{S}}{2 \: \text{mol} \: \ce{Ag}} = 0.232 \: \text{mol} \: \ce{S} \: \text{(required)}\nonumber$
The amount of $\ce{S}$ actually present is 0.312 moles. The amount of $\ce{S}$ that is required to fully react with all of the $\ce{Ag}$ is 0.232 moles. Since there is more sulfur present than what is required to react, the sulfur is the excess reactant. Therefore, silver is the limiting reactant.
Step 3: Think about your result.
The balanced equation indicates that the necessary mole ratio of $\ce{Ag}$ to $\ce{S}$ is 2:1. Since there were not twice as many moles of $\ce{Ag}$ present in the original amounts, that makes silver the limiting reactant.
There is a very important point to consider about the preceding problem. Even though the mass of silver present in the reaction $\left( 50.0 \: \text{g} \right)$ was greater than the mass of sulfur $\left( 10.0 \: \text{g} \right)$, silver was the limiting reactant. This is because chemists must always convert to molar quantities and consider the mole ratio from the balanced chemical equation.
There is one other component to be determined in a limiting reactant problem—the quantity of the excess reactant that will be left over after the reaction is complete. We will return to the sample problem, above, to answer this question below.
Example $2$: Determining the Amount of Excess Reactant Left Over
What is the mass of excess reactant remaining when $50.0 \: \text{g} \: \ce{Ag}$ reacts with $10.0 \: \text{g} \: \ce{S}$?
$2 \ce{Ag} \left( s \right) + \ce{S} \left( s \right) \rightarrow \ce{Ag_2S} \left( s \right)\nonumber$
Known
• Excess reactant $= 0.312 \: \text{mol} \: \ce{S}$ (from Example $1$)
• Amount of excess reactant needed $= 0.232 \: \text{mol} \: \ce{S}$ (from Example $1$)
Unknown
• Mass of excess reactant remaining after the reaction = ? g
Subtract the amount (in moles) of the excess reactant that will react from the amount that is originally present. Convert moles to grams.
Step 2: Solve.
$0.312 \: \text{mol} \: \ce{S} - 0.232 \: \text{mol} \: \ce{S} = 0.080 \: \text{mol} \: \ce{S} \: \text{(remaining after reaction)}\nonumber$
$0.080 \: \text{mol} \: \ce{S} \times \frac{32.07 \: \text{g} \: \ce{S}}{1 \: \text{mol} \: \ce{S}} = 2.57 \: \text{g} \: \ce{S}\nonumber$
There are $2.57 \: \text{g}$ of sulfur remaining when the reaction is complete.
Step 3: Think about your result.
There were $10.0 \: \text{g}$ of sulfur present before the reaction began. If $2.57 \: \text{g}$ of sulfur remain after the reaction, then $7.43 \: \text{g} \: \ce{S}$ reacted.
$7.43 \: \text{g} \: \ce{S} \times \frac{1 \: \text{mol} \: \ce{S}}{32.07 \: \text{g} \: \ce{S}} = 0.232 \: \text{mol} \: \ce{S}\nonumber$
This is the amount of sulfur that reacted. The problem is internally consistent.
Summary
• Determining the limiting reactant requires that all mass quantities first be converted to moles to evaluate the equation.
Review
1. Why do all mass values need to be converted to moles before determining the limiting reactant?
2. Silver metal reacts with sulfur to form silver sulfide according to the following balanced equation:
$2 \ce{Ag} \left( s \right) + \ce{S} \left( s \right) \rightarrow \ce{Ag_2S} \left( s \right)\nonumber$
1. If 0.700 moles Ag is reacted with 10.0 g S, is sulfur or aluminum the limiting reactant?
2. How many grams of Ag2S will be produced? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.08%3A_Determining_the_Limiting_Reactant.txt |
Can we save some money?
The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.
Percent Yield
Chemical reactions in the real world do not always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are often losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.
To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage:
$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\nonumber$
Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.
Typically, percent yields are understandably less than $100\%$ because of the reasons previously indicated. However, percent yields greater than $100\%$ are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction.
Example $1$: Calculating the Theoretical Yield and the Percent Yield
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below.
$2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)\nonumber$
In a certain experiment, $40.0 \: \text{g} \: \ce{KClO_3}$ is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed, the oxygen gas is collected, and its mass is found to be $14.9 \: \text{g}$. What is the percent yield for the reaction?
Solution
First, we will calculate the theoretical yield based on the stoichiometry.
Known
• Given: Mass of $\ce{KClO_3} = 40.0 \: \text{g}$
• Molar mass $\ce{KClO_3} = 122.55 \: \text{g/mol}$
• Molar mass $\ce{O_2} = 32.00 \: \text{g/mol}$
Unknown
• theoretical yield O2 = ? g
Apply stoichiometry to convert from the mass of a reactant to the mass of a product:
$\text{g} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{O_2} \rightarrow \text{g} \: \ce{O_2} \nonumber\nonumber$
Step 2: Solve.
$40.0 \: \text{g} \: \ce{KClO_3} \times \frac{1 \: \text{mol} \: \ce{KClO_3}}{122.55 \: \text{g} \: \ce{KClO_3}} \times \frac{3 \: \text{mol} \: \ce{O_2}}{2 \: \text{mol} \: \ce{KClO_3}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \text{mol} \: \ce{O_2}} = 15.7 \: \text{g} \: \ce{O_2} \nonumber\nonumber$
The theoretical yield of $\ce{O_2}$ is $15.7 \: \text{g}$.
Step 3: Think about your result.
The mass of oxygen gas must be less than the $40.0 \: \text{g}$ of potassium chlorate that was decomposed.
Now we will use the actual yield and the theoretical yield to calculate the percent yield.
Known
• Actual yield $= 14.9 \: \text{g}$
• Theoretical yield $= 15.7 \: \text{g}$
Unknown
• Percent yield = ? %
$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \nonumber\nonumber$
Use the percent yield equation above.
Step 2: Solve.
$\text{Percent Yield} = \frac{14.9 \: \text{g}}{15.7 \: \text{g}} \times 100\% = 94.9\% \nonumber\nonumber$
Step 3: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $100\%$.
Summary
• Theoretical yield is calculated based on the stoichiometry of the chemical equation.
• The actual yield is experimentally determined.
• The percent yield is determined by calculating the ratio of actual yield/theoretical yield.
Review
1. What do we need in order to calculate theoretical yield?
2. If I spill some of the product before I weigh it, how will that affect the actual yield?
3. How will spilling some of the product affect the percent yield?
4. I make a product and weigh it before it is dry. How will that affect the actual yield? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.09%3A_Theoretical_Yield_and_Percent_Yield.txt |
How much oxygen is in this container?
Approximately \(20\%\) of the atmosphere is oxygen. This gas is essential for life. In environments where oxygen is in low supply, it can be provided from a tank. Since gases are very compressible, a large amount of oxygen can be stored in a relatively small container. When it is released, the volume expands and the pressure decreases. The gas is then available for ventilation under normal pressure.
Kinetic-Molecular Theory
The kinetic-molecular theory explains the states of matter, and is based on the idea that matter is composed of tiny particles that are always in motion. This theory helps explain observable properties and behaviors of solids, liquids, and gases. However, kinetic-molecular theory is most easily understood as it applies to gases, and it is with gases that we will begin our detailed study. The theory applies specifically to a model of gas called an ideal gas. An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.
The kinetic-molecular theory, as it applies to gases, has five basic assumptions:
1. Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size. The particles of a gas may be either atoms or molecules. The distance between the particles of a gas is much, much greater than the distance between the particles of a liquid or a solid. Most of the volume of a gas, therefore, is composed of the empty space between the particles. In fact, the volume of the particles themselves is considered to be insignificant compared to the volume of the empty space.
2. Gas particles are in constant rapid motion in random directions. The fast motion of gas particles gives them a relatively large amount of kinetic energy. Recall that kinetic energy is the energy that an object possesses because of its motion. The particles of a gas move in straight-line motion until they collide with another particle, or with one of the walls of the gas container.
3. Collisions between gas particles and between particles and the container walls are elastic collisions. An elastic collision is one in which there is no overall loss of kinetic energy. Kinetic energy may be transferred from one particle to another during an elastic collision, but there is no change in the total energy of the colliding particles.
4. There are no forces of attraction or repulsion between gas particles. Attractive forces are responsible for particles of a real gas condensing together to form a liquid. It is assumed that the particles of an ideal gas have no such attractive forces. The motion of each particle is completely independent of the motion of all other particles.
5. The average kinetic energy of gas particles is dependent upon the temperature of the gas. As the temperature of a sample of gas is increased, the speeds of the particles are increased. This results in an increase in the kinetic energy of the particles. Not all particles of gas in a sample have the same speed, and so they do not have the same kinetic energy. The temperature of a gas is proportional to the average kinetic energy of the gas particles.
Summary
• Assumptions of the kinetic-molecular theory:
• Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size.
• Gas particles are in constant rapid motion in random directions.
• Collisions between gas particles and between particles and the container walls are elastic collisions.
• There are no forces of attraction or repulsion between gas particles.
• The average kinetic energy of gas particles is dependent upon the temperature of the gas.
Review
1. Describe the motion of gas particles.
2. What kind of collisions occurs?
3. What is the relationship between the kinetic energy of gas particles and the temperature of the gas? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.01%3A_Kinetic_Molecular_Theory.txt |
How are hot air balloons able to move smoothly in the air?
Many people enjoy riding in hot air balloons. Some use them for romantic picnics and marriage proposals. Others race in competitions. Being above the earth gives a whole new perspective on the world around us. As the beginning of a hot air balloon ride, the balloon is flat because the pressure inside the balloon equals the pressure outside. When the air inside the balloon is heated, the speed of movement of those air molecules increases and the pressure goes up. After a while the balloon is completely expanded and the flight is ready to take off.
Gas Pressure
Pressure is defined as the force per unit area on a surface.
$\text{Pressure} = \frac{\text{force}}{\text{area}}\nonumber$
When a person stands on the floor, his feet exert pressure on the surface. That pressure is related to both the mass of the person and the surface area of his feet. If the person were holding a heavy object, the pressure would increase because of a greater force. Alternatively, if the person stands on his toes, the pressure also increases because of a decrease in the surface area.
Gas molecules also exert pressure. Earth's atmosphere exerts pressure because gravity acts on the huge number of gas particles contained in the atmosphere, holding it in place. Pressure is also exerted by a small sample of gas, such as that which is contained in a balloon. Gas pressure is the pressure that results from collisions of gas particles with an object. Inside the balloon, the gas particles collide with the balloon's inner walls. It is those collisions which keep the balloon inflated. If the gas particles were to suddenly stop moving, the balloon would instantly deflate. The figure below is an illustration of gas particles exerting pressure inside a container.
The pressure inside the hot air balloon is affected by temperature. As the molecules heat up, they move faster and strike the inside wall of the balloon harder. This increased motion of the gas particles increases the force on an area of the balloon, producing a rise in the pressure.
Summary
• Pressure is defined as $\frac{\text{force}}{\text{volume}}$.
• Gas pressure is the result of collisions between gas particles and an object.
• An increase in temperature will produce an increase in pressure of a gas.
Review
1. What is pressure?
2. How does a gas create pressure on an object or container?
3. What would happen to the pressure if gas particles suddenly stopped moving?
4. How does temperature affect pressure?
13.03: Atmospheric Pressure
What do storm reports reveal?
The pressure in the atmosphere is an important factor in determining what the weather will be like. If the barometric pressure is high in an area, this will cause air to move to a region of lower pressure. The greater the difference in pressure between the two areas, the stronger the winds will develop. Under certain conditions, the winds can produce a tornado (a violent rotating column of air that reaches from a thunderstorm down to the ground).
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by gas particles in Earth's atmosphere as those particles collide with objects. A barometer is an instrument used to measure atmospheric pressure. A traditional mercury barometer consists of an evacuated tube immersed in a container of mercury. Air molecules push down on the surface of the mercury. Because the inside of the tube is a vacuum, the mercury rises inside the tube. The height to which the mercury rises is dependent on the external air pressure.
A more convenient barometer, called an aneroid barometer, measures pressure by the expansion and contraction of a small spring within an evacuated metal capsule.
Atmospheric Pressure and Altitude
At sea level, a mercury column will rise a distance of $760 \: \text{mm}$. This atmospheric pressure is reported as $760 \: \text{mm} \: \ce{Hg}$ (millimeters of mercury). At higher altitudes, the atmospheric pressure is decreased and so the column of mercury will not rise as high. On the summit of Mt. Everest (elevation of $8848 \: \text{m}$), the air pressure is $253 \: \text{mm} \: \ce{Hg}$. Atmospheric pressure is slightly dependent on weather conditions. On the graph below, we can see the decrease in atmospheric pressure as the altitude increases. At sea level, the atmospheric pressure would be a little over $100 \: \text{kPa}$ (one atmosphere or $760 \: \text{mm} \: \ce{Hg}$). If we climb to the top of Mount Everest (the highest mountain in the world at 29,029 feet or 8848 meters), the atmospheric pressure will drop to slightly over $30 \: \text{kPa}$ (about 0.30 atmospheres or $228 \: \text{mm} \: \ce{Hg}$). This marked decrease in atmospheric pressure results in much lower levels of oxygen. Teams that climb this mountain must bring supplies of oxygen with them in order to breathe at these high altitudes.
Simulation
Ever try cooking at the top of a mountain or at sea level? Compare how atmospheric pressure, vapor pressure and boiling point relate in this simulation:
Does water boil at 100°C everywhere?
Summary
• Atmospheric pressure is the pressure exerted by gas particles in Earth’s atmosphere as those particles collide with objects.
• A barometer measures atmospheric pressure.
• Atmospheric pressure decreases as the altitude increases.
Review
1. Define atmospheric pressure.
2. What is an aneroid barometer?
3. How does the atmospheric pressure change as the altitude increases? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.02%3A_Gas_Pressure.txt |
Can you guess how old this car is?
There are several benefits to maintaining the proper air pressure in a car tire. The ride is smoother and safer than it is with too low of pressure. The car gets better gas mileage and the tires do not wear as fast. The recommended pressure for that model of car (usually somewhere between $32$-$35 \: \text{psi}$) is generally listed in the owner's manual or stamped somewhere inside the door. The pressure on the tire is the maximum pressure for that tire, not the recommended one. Tire pressure is best measured when the tire is cold, since driving the car will heat up the air in the tire and increase the pressure.
Pressure Units and Conversions
A barometer measures gas pressure by the height of the column of mercury. One unit of gas pressure is the millimeter of mercury $\left( \text{mm} \: \ce{Hg} \right)$. An equivalent unit to the $\text{mm} \: \ce{Hg}$ is called the $\text{torr}$, in honor of the inventor of the barometer, Evangelista Torricelli. The pascal $\left( \text{Pa} \right)$ is the standard unit of pressure. A pascal is a very small amount of pressure, so the most useful unit for everyday gas pressures is the kilopascal $\left( \text{kPa} \right)$. A kilopascal is equal to 1000 pascals. Another commonly used unit of pressure is the atmosphere $\left( \text{atm} \right)$. Standard atmospheric pressure is defined as $1 \: \text{atm}$ of pressure and is equal to $760 \: \text{mm} \: \ce{Hg}$ and $101.3 \: \text{kPa}$. Atmospheric pressure is also often stated as pounds per square inch $\left( \text{psi} \right)$. The atmospheric pressure at sea level is $14.7 \: \text{psi}$.
$1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg} = 760 \: \text{torr} = 101.3 \: \text{kPa} = 14.7 \: \text{psi}\nonumber$
It is important to be able to convert between different units of pressure. To do so, we will use the equivalent standard pressures shown above.
Example $1$: Pressure Unit Conversions
The atmospheric pressure in a mountainous location is measured to be $613 \: \text{mm} \: \ce{Hg}$. What is this pressure in $\text{atm}$ and in $\text{kPa}$?
Known
• Given: $613 \: \text{mm} \: \ce{Hg}$
• $1 \: \text{atm} = 760 \: \text{mm} \: \ce{Hg}$
• $101.3 \: \text{kPa} = 760 \: \text{mm} \: \ce{Hg}$
Unknown
• Pressure $= ? \: \text{atm}$
• Pressure $= ? \: \text{kPa}$
Use conversion factors from the equivalent pressure units to convert from $\text{mm} \: \ce{Hg}$ to $\text{atm}$ and from $\text{mm} \: \ce{Hg}$ to $\text{kPa}$.
Step 2: Solve.
$613 \: \text{mm} \: \ce{Hg} \times \frac{1 \: \text{atm}}{760 \: \text{mm} \: \ce{Hg}} = 0.807 \: \text{atm}\nonumber$
$613 \: \text{mm} \: \ce{Hg} \times \frac{101.3 \: \text{kPa}}{760 \: \text{mm} \: \ce{Hg}} = 81.7 \: \text{kPa}\nonumber$
Step 3: Think about your result.
The air pressure is about $80\%$ of standard atmospheric pressure at sea level. For significant figure purposes, the standard pressure of $760 \: \text{mm} \: \ce{Hg}$ has three significant figures.
Summary
• Calculations are described for converting between different pressure units.
Review
1. What instrument measures gas pressure by the height of the column of mercury?
2. 1 atm = ___ torr
3. 32.02 atm = ___ kPa
4. 542 mmHg = ___ psi
5. The pressure in a car tire is 35 psi. How many atmospheres is that?
13.05: Average Kinetic Energy and Temperature
How much energy does it take to hit a baseball?
Kinetic energy is the energy of motion. Any object that is moving possesses kinetic energy. Baseball involves a great deal of kinetic energy. The pitcher throws a ball, imparting kinetic energy to the ball. When the batter swings, the motion of swinging creates kinetic energy in the bat. The collision of the bat with the ball changes the direction and speed of the ball, with the idea of kinetic energy being involved again.
Kinetic Energy and Temperature
As stated in the kinetic-molecular theory, the temperature of a substance is related to the average kinetic energy of the particles of that substance. When a substance is heated, some of the absorbed energy is stored within the particles, while some of the energy increases the motion of the particles. This is registered as an increase in the temperature of the substance.
Average Kinetic Energy
At any given temperature, not all of the particles of a sample of matter have the same kinetic energy. Instead, the particles display a wide range of kinetic energies. Most of the particles have a kinetic energy near the middle of the range. However, a small number of particles have kinetic energies a great deal lower or a great deal higher than the average (see figure below).
The blue curve in the figure above is for a sample of matter at a relatively low temperature, while the red curve is for a sample at a relatively high temperature. In both cases, most of the particles have intermediate kinetic energies, close to the average. Notice that as the temperature increases, the range of kinetic energies increases and the distribution curve "flattens out". At a given temperature, the particles of any substance have the same average kinetic energy.
Absolute Zero
As a sample of matter is continually cooled, the average kinetic energy of its particles decreases. Eventually, one would expect the particles to stop moving completely. Absolute zero is the temperature at which the motion of particles theoretically ceases. Absolute zero has never been attained in the laboratory, but temperatures on the order of $1 \times 10^{-10} \: \text{K}$ have been achieved. The Kelvin temperature scale is the scale that is based on molecular motion, and so absolute zero is also called $0 \: \text{K}$. The Kelvin temperature of a substance is directly proportional to the average kinetic energy of the particles of the substance. For example, the particles in a sample of hydrogen gas at $200 \: \text{K}$ have twice the average kinetic energy as the particles in a hydrogen sample at $100 \: \text{K}$.
Summary
• Kinetic energy is the energy of motion.
• At a given temperature, individual particles of a substance have a range of kinetic energies.
• The motion of particles theoretically ceases at absolute zero.
Review
1. What is kinetic energy?
2. If the temperature increases, will particles move faster or slower than they would at a lower temperature?
3. What is absolute zero? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.04%3A_Pressure_Units_and_Conversions.txt |
Carbon is an interesting and versatile element. There are over twenty million known compounds containing carbon, encompassed in the growing field of organic chemistry. The element itself can exist in two major forms. Diamond is a form of carbon that is extremely hard, and is one of the few materials that can scratch glass. The other form of carbon is graphite, a very soft material that we find in "lead" pencils. The two forms differ mainly in how the carbon atoms are connected to one another. The differences in the arrangement of atoms affect the properties of the material.
Physical Properties and Intermolecular Forces
The physical state and properties of a particular compound depend in large part on the type of chemical bonding it displays. Molecular compounds, sometimes called covalent compounds, display a wide range of physical properties due to the different types of intermolecular attractions, such as different kinds of polar interactions. The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds. This is because the energy required to disrupt the intermolecular forces between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound. Since molecular compounds are composed of neutral molecules, their electrical conductivity is generally quite poor, whether in the solid or liquid state. Ionic compounds do not conduct electricity in the solid state because of their rigid structure, but conduct well when either molten or dissolved into a solution. The water solubility of molecular compounds is variable and depends primarily on the type of intermolecular forces involved. Substances that exhibit hydrogen bonding or dipole-dipole forces are generally water soluble, whereas those that exhibit only London dispersion forces are generally insoluble. Most, but not all, ionic compounds are quite soluble in water. The table below summarizes some of the differences between ionic and molecular compounds.
Comparison of Ionic and Molecular Compounds
Table $1$: Comparison of Ionic and Molecular Compounds
Property Ionic Compounds Molecular Compounds
Type of elements Metal and nonmetal Nonmetals only
Bonding Ionic - transfer of electron(s) between atoms Covalent - sharing of pair(s) of electrons between atoms
Representative unit Formula unit Molecule
Physical state at room temperature Solid Gas, liquid, or solid
Water solubility Usually high Variable
Melting and boiling temperatures Generally high Generally low
Electrical conductivity Good when molten or in solution Poor
One type of molecular compound behaves quite differently than those described thus far. A covalent network solid is a compound in which all of the atoms are connected to one another by covalent bonds. Diamond is composed entirely of carbon atoms, each bonded to four other carbon atoms in a tetrahedral geometry. Melting a covalent network solid is not accomplished by overcoming the relatively weak intermolecular forces. Rather, all of the covalent bonds must be broken, a process that requires extremely high temperatures. Diamond, in fact, does not melt at all. Instead, it vaporizes to a gas at temperatures above $3500^\text{o} \text{C}$.
Summary
• The physical properties of a material are affected by the intermolecular forces holding the molecules together.
• The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds.
13.06: Surface Tension
How is this insect able to stand on water?
The next time you are by a still body of water, take a close look at the creatures scooting along on the surface. You may see insects seemingly floating on top of the water. These creatures are known by a variety of names including water skaters, water striders, and pond skaters. They take advantage of a property called surface tension to stay above the water and not sink. The force they exert downward is less than the forces exerted among the water molecules on the surface of the pond, so the insect does not penetrate the surface of the water.
Surface Tension
Molecules within a liquid are pulled equally in all directions by intermolecular forces. However, molecules at the surface are pulled downwards and sideways by other liquid molecules, but not upwards away from the surface. The overall effect is that the surface molecules are pulled into the liquid, creating a surface that is tightened like a film (see A in the figure below). The surface tension of a liquid is a measure of the elastic force within the liquid's surface. Liquids that have strong intermolecular forces, like the hydrogen bonding in water, exhibit the greatest surface tension. Surface tension allows objects that are denser than water, such as the paper clip shown in B in the figure below, to nonetheless float on its surface. It is also responsible for the beading up of water droplets on a freshly waxed car because there are no attractions between the polar water molecules and the nonpolar wax.
Other liquids, such as diethyl ether, do not demonstrate strong surface tension interactions. The intermolecular forces for the ether are relatively weak dipole-dipole interactions that do not draw the molecules together as tightly as hydrogen bonds would.
Science Friday: Candy Corn in Space
Candy corn is a very tasty treat. In this video by Science Friday, astronaut Don Pettit uses Candy Corn to demonstrate the effects of hydrophobic and hydrophilic interactions.
Summary
• The surface tension of a liquid is a measure of the elastic force in the liquid’s surface.
• Liquids with strong intermolecular forces have higher surface tensions than liquids with weaker forces.
Review
1. Define surface tension.
2. What is responsible for the strong surface tension in water?
3. Does diethyl ether have a stronger or weaker surface tension than water? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.06%3A_Physical_Properties_and_Intermolecular_Forces.txt |
What is the box on the house's roof?
On the roof of the house pictured below is a device known as a "swamp cooler". This piece of equipment traces its origin back to the ancient Egyptians, who hung wet blankets across the doors of their homes. As the warm air passed through the blankets, water would evaporate and cool the air. The royalty went one step further, and had servants fan wet cloths over jugs of water to get more evaporation and cooling.
The origin of the term "swamp cooler" is not known; they certainly don't work in a swamp. The best conditions for cooling include a high temperature (over $80^\text{o} \text{F}$) and a low humidity (preferably less than $30\%$). These coolers work well in desert areas, but don't provide any cooling in the humid areas of the country.
Evaporation
A puddle of water left undisturbed eventually disappears. The liquid molecules escape into the gas phase, becoming water vapor. Vaporization is the process in which a liquid is converted to a gas. Evaporation is the conversion of a liquid to its vapor below the boiling temperature of the liquid. If the water is instead kept in a closed container, the water vapor molecules do not have a chance to escape into the surroundings, and so the water level does not change. As some water molecules become vapor, an equal number of water vapor molecules condense back into the liquid state. Condensation is the change of state from a gas to a liquid.
In order for a liquid molecule to escape into the gas state, the molecule must have enough kinetic energy to overcome the intermolecular attractive forces in the liquid. Recall that a given liquid sample will have molecules with a wide range of kinetic energies. Liquid molecules that have this certain threshold kinetic energy escape the surface and become vapor. As a result, the liquid molecules that remain now have lower kinetic energy. As evaporation occurs, the temperature of the remaining liquid decreases. You have observed the effects of evaporative cooling. On a hot day, the water molecules in your perspiration absorb body heat and evaporate from the surface of your skin. The evaporation process leaves the remaining perspiration cooler, which in turn absorbs more heat from your body.
A given liquid will evaporate more quickly when it is heated. This is because the heating process results in a greater fraction of the liquid's molecules having the necessary kinetic energy to escape the surface of the liquid. The figure below shows the kinetic energy distribution of liquid molecules at two temperatures. The numbers of molecules that have the required kinetic energy to evaporate are shown in the shaded area under the curve at the right. The higher temperature liquid $\left( T_2 \right)$ has more molecules that are capable of escaping into the vapor phase than the lower temperature liquid $\left( T_1 \right)$.
Interactive Element
Explore how heat and temperature relate to phase change in this simulation:
Does the temperature of water rise while it is boiling?
Summary
• Evaporation is the conversion of a liquid to its vapor below the boiling temperature of the liquid.
• Condensation is the change of state from a gas to a liquid.
• As the temperature increases, the rate of evaporation increases.
Review
1. Define vaporization.
2. Define evaporation.
3. Define condensation.
4. How does temperature affect the rate of evaporation? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.07%3A_Evaporation.txt |
What causes this toy to move?
The drinking duck is a toy that many kids (and adults) enjoy playing with. You can see the drinking duck in action in the video below:
The motion of the duck illustrates a physical principle called vapor pressure. As the vapor pressure changes, the liquid in the duck moves up and down, causing the duck to move.
Vapor Pressure
When a partially filled container of liquid is sealed with a stopper, some liquid molecules at the surface evaporate into the vapor phase. However, the vapor molecules cannot escape from the container. So, after a certain amount of time, the space above the liquid reaches a point where it cannot hold any more vapor molecules. Now, some of the vapor molecules condense back into a liquid. The system reaches the point where the rate of evaporation is equal to the rate of condensation (see figure below). This is considered a dynamic equilibrium between the liquid and vapor phase.
A dynamic equilibrium can be illustrated by an equation with a double arrow, meaning that the reaction is occurring in both directions and at the same rate.
$\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_2O} \left( g \right)\nonumber$
The forward direction represents the evaporation process, while the reverse direction represents the condensation process.
Because they cannot escape the container, the vapor molecules above the surface of the liquid exert a pressure on the walls of the container. The vapor pressure is a measure of the pressure (force per unit area) exerted by a gas above a liquid in a sealed container. Vapor pressure is a property of a liquid based on the strength of its intermolecular forces. A liquid with weak intermolecular forces evaporates more easily and has a higher vapor pressure. A liquid with stronger intermolecular forces does not evaporate easily, and thus has a lower vapor pressure. For example, diethyl ether is a nonpolar liquid with weak dispersion forces. Its vapor pressure at $20^\text{o} \text{C}$ is $58.96 \: \text{kPa}$. Water is a polar liquid whose molecules are attracted to one another by relatively strong hydrogen bonding. The vapor pressure of water at $20^\text{o} \text{C}$ is only $2.33 \: \text{kPa}$, far less than that of diethyl ether.
Vapor Pressure and Temperature
Vapor pressure is dependent upon temperature. When the liquid in a closed container is heated, more molecules escape the liquid phase and evaporate. The greater number of vapor molecules strike the container walls more frequently, resulting in an increase in pressure. The table below shows the temperature dependence of the vapor pressure of three liquids.
Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures
Table $1$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures
$0^\text{o} \text{C}$ $20^\text{o} \text{C}$ $40^\text{o} \text{C}$ $60^\text{o} \text{C}$ $80^\text{o} \text{C}$ $100^\text{o} \text{C}$
Water 0.61 2.33 7.37 19.92 47.34 101.33
Ethanol 1.63 5.85 18.04 47.02 108.34 225.75
Diethyl Ether 24.70 58.96 122.80 230.65 399.11 647.87
Notice that the temperature dependence of the vapor pressure is not linear. From $0^\text{o} \text{C}$ to $80^\text{o} \text{C}$, the vapor pressure of water increases by $46.73 \: \text{kPa}$, while it increases by $53.99 \: \text{kPa}$ in only a span of twenty degrees from $80^\text{o} \text{C}$ to $100^\text{o} \text{C}$.
Simulation
Ever try to boil water on the top of a mountain? Explore the relationship between altitude, vapor pressure and boiling point in this simulation:
Does water boil at 100°C everywhere?
Summary
• Vapor pressure is a measure of the pressure exerted by a gas above a liquid in a sealed container.
• Strong intermolecular forces produce a lower rate of evaporation and a lower vapor pressure.
• Weak intermolecular forces produce a higher rate of evaporation and a higher vapor pressure.
• As the temperature increases, the vapor pressure increases.
Review
1. Define vapor pressure.
2. How do intermolecular forces affect vapor pressure?
3. How does temperature affect vapor pressure? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.08%3A_Vapor_Pressure.txt |
Why is it harder to breathe at the top of Mount Everest than at the bottom?
At 29,029 feet $\left( 8848 \: \text{m} \right)$, Mount Everest in the Himalayan range, on the border between China and Nepal, is the highest point on the earth. Its altitude presents many practical problems to climbers. The oxygen content of the air is much lower than at sea level, making it necessary to bring oxygen tanks along (although a few climbers have reached the peak without oxygen). One other problem is that of boiling water for cooking food. Although water boils at $100^\text{o} \text{C}$ at sea level, the boiling point on top of Mount Everest is only about $70^\text{o} \text{C}$. This difference makes it very difficult to get a decent cup of tea (which definitely frustrates some British climbers).
Boiling Point
As a liquid is heated, the average kinetic energy of its particles increases. The rate of evaporation increases as more and more molecules are able to escape the liquid's surface into the vapor phase. Eventually a point is reached when the molecules all throughout the liquid have enough kinetic energy to vaporize. At this point, the liquid begins to boil. The boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure. The figure below illustrates the boiling of liquid.
In the picture on the left, the liquid is below its boiling point, yet some of the liquid evaporates. On the right, the temperature has been increased until bubbles begin to form in the body of the liquid. When the vapor pressure inside the bubbles is equal to the external atmospheric pressure, the bubbles rise to the surface of the liquid and burst. The temperature at which this process occurs is the boiling point of the liquid.
The normal boiling point is the temperature at which the vapor pressure of the liquid is equal to standard pressure. Because atmospheric pressure can change based on location, the boiling point of a liquid changes with the external pressure. The normal boiling point is a constant because it is defined relative to the standard atmospheric pressure of $760 \: \text{mm} \: \ce{Hg}$ (or $1 \: \text{atm}$ or $101.3 \: \text{kPa}$).
Summary
• The boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure.
• As the altitude increases, the boiling point decreases.
Review
1. What happens when a liquid is heated?
2. What is the boiling point?
3. What is the pressure at which the normal boiling point is determined?
4. Are the boiling point and normal boiling point always the same value, sometimes the same value, or never the same value? Explain your answer.
Explore More
Use the video below to answer the following questions:
1. The boiling point of water is 100°C. What is an example of liquid that boils a temperature higher than the boiling point of water?
2. What is an example of liquid that boiling point is below room temperature?
3. What phase change is the opposite of boiling? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.09%3A_Boiling.txt |
What is the fastest way to boil water?
One of the first lessons in cooking is how to boil water. Yes, it sounds simple, but there are a couple of hints that speed things up. One hint is to put a lid on the pot. The picture above has water boiling uncovered with the steam escaping to the atmosphere. If the lid is on the pot, less water will be boiled off and the water will boil faster. The buildup of pressure inside the pot helps speed up the boiling process.
Vapor Pressure Curves
The boiling points of various liquids can be illustrated in a vapor pressure curve (figure below). A vapor pressure curve is a graph of vapor pressure as a function of temperature. To find the normal boiling point of liquid, a horizontal line is drawn from the y-axis at a pressure equal to standard pressure. A vertical line starting at the x-axis can be drawn connected to the point at which the standard pressure and the vapor pressure curve of a liquid intersect—the corresponding temperature is the boiling point of that liquid.
The boiling point of a liquid also correlates to the strength of its intermolecular forces. Recall that diethyl ether has weak dispersion forces, which means that the liquid has a high vapor pressure. The weak forces also mean that it does not require a large input of energy to make diethyl ether boil, and so it has a relatively low normal boiling point of $34.6^\text{o} \text{C}$. Water, with its much stronger hydrogen bonding, has a low vapor pressure and a higher normal boiling point of $100^\text{o} \text{C}$.
As stated earlier, boiling points are affected by external pressure. At higher altitudes, the atmospheric pressure is lower. With less pressure pushing down on the surface of the liquid, it boils at a lower temperature. This can also be seen from the vapor pressure curves. If one draws a horizontal line at a lower vapor pressure, it intersects each curve at a lower temperature. The boiling point of water is $100^\text{o} \text{C}$ at sea level, where the atmospheric pressure is standard. In Denver, Colorado at $1600 \: \text{m}$ above sea level, the atmospheric pressure is about $640 \: \text{mm} \: \ce{Hg}$ and water boils at about $95^\text{o} \text{C}$. On the summit of Mt. Everest, the atmospheric pressure is about $255 \: \text{mm} \: \ce{Hg}$, and water boils at only $70^\text{o} \text{C}$. On the other hand, water boils at greater than $100^\text{o} \text{C}$ if the external pressure is higher than standard. Pressure cookers do not allow the vapor to escape and the vapor pressure increases. Since water in a pressure cooker boils at a temperature above $100^\text{o} \text{C}$, the food cooks more quickly.
The effect of decreased air pressure can be demonstrated by placing a beaker of water in a vacuum chamber. At a low enough pressure, about $20 \: \text{mm} \: \ce{Hg}$, water will boil at room temperature.
Summary
• A vapor pressure curve is a graph of vapor pressure as a function of temperature.
• Boiling points are affected by external pressure.
Review
1. What does a vapor pressure curve show?
2. Why does diethyl ether have a low boiling point?
3. What intermolecular forces hold water molecules together?
4. Why does water boil at a lower temperature when at a high altitude?
Explore More
Use the resource below to answer the questions that follow.
1. What does a vacuum pump do?
2. What is the role of the O-ring?
3. What happens when the vacuum pump is turned on?
4. What happened when the vacuum pump was turned on?
5. What was the temperature of this boiling water? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.10%3A_Vapor_Pressure_Curves.txt |
Have you ever gone ice skating?
In the winter, many people find the snow and ice beautiful. They enjoy getting out to ski or ice-skate. Others don’t find that time of year to be so much fun. When the snow melts, the roads get very sloppy and messy. Those people look forward to spring when all the ice and snow are gone and the weather is warmer.
Melting Point
Solids are similar to liquids in that both are condensed states, with particles that are far closer together than those of a gas. However, while liquids are fluid, solids are not. The particles of most solids are packed tightly together in an orderly arrangement. The motion of individual atoms, ions, or molecules in a solid is restricted to vibrational motion about a fixed point. Solids are almost completely incompressible and are the densest of the three states of matter.
As a solid is heated, its particles vibrate more rapidly as it absorbs kinetic energy. Eventually, the organization of the particles within the solid structure begins to break down, and the solid starts to melt. The melting point is the temperature at which a solid changes into a liquid. At its melting point, the disruptive vibrations of the particles of the solid overcome the attractive forces operating within the solid. As with boiling points, the melting point of a solid is dependent on the strength of those attractive forces. Sodium chloride $\left( \ce{NaCl} \right)$ is an ionic compound that consists of a multitude of strong ionic bonds. Sodium chloride melts at $801^\text{o} \text{C}$. Ice (solid $\ce{H_2O}$) is a molecular compound whose molecules are held together by hydrogen bonds. Though hydrogen bonds are the strongest of the intermolecular forces, the strength of hydrogen bonds is much less than that of ionic bonds. The melting point of ice is 0 °C.
The melting point of a solid is the same as the freezing point of the liquid. At that temperature, the solid and liquid states of the substance are in equilibrium. For water, this equilibrium occurs at $0^\text{o} \text{C}$.
$\ce{H_2O} \left( s \right) \rightleftharpoons \ce{H_2O} \left( l \right)\nonumber$
We tend to think of solids as those materials that are solid at room temperature. However, all materials have melting points of some sort. Gases become solids at extremely low temperatures, and liquids will also become solid if the temperature is low enough.
The table below gives the melting points of some common materials.
Materials Melting Point (°C)
Table $1$: Melting Points of Common Materials
Hydrogen -259
Oxygen -219
Diethyl ether -116
Ethanol -114
Water 0
Pure silver 961
Pure gold 1063
Iron 1538
Simulation
Explore how heat and temperature relate to phase change in this simulation:
Does the temperature of water rise while it is boiling?
Summary
• The melting point is the temperature at which a solid changes into a liquid.
• Intermolecular forces have a strong influence on melting point.
Review
1. Define melting point.
2. What happens when a material melts?
3. Would you expect ethane (C2H6) to have a higher or lower melting point than water? Explain your answer. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.11%3A_Melting.txt |
How did early settlers do their laundry in the winter?
Of course, they could wash the clothes inside, but where to dry them? There were no dryers available, so the clothes had to be hung up to dry. In the winter, you would expect that ice would form on the clothes, but that didn’t happen. Even in freezing weather, the water might freeze initially, but would eventually go off as a vapor – from solid ice directly to the gas stage. There are still families today who have to rely on this cumbersome process to get their laundry done.
Sublimation
Vapor Pressure of a Solid
"Sublimation" surveys the vapor pressure of a liquid and its dependence upon temperature. Solids also have a vapor pressure, though it is generally much less than that of a liquid. A snow bank will gradually disappear even if the temperature stays below $0^\text{o} \text{C}$. The snow does not melt, but instead passes directly from the solid state to the vapor state. Sublimation is the change of state from a solid to a gas, without passing through the liquid state.
Examples of Sublimation
Sublimation of Iodine
When iodine is heated sublimation can be readily observed. Iodine's vapor is a distinctive purple color and has a very strong scent, making it easy to detect. The video below shows the sublimation and deposition of iodine. The beaker containing iodine is covered with a round-bottom flask that contains ice. When the iodine is heated it sublimes. When iodine vapors cool, deposition occurs. Deposition is the change of state from a gas to a solid.
Dry Ice
"Dry ice", or solid carbon dioxide, is a substance that sublimes at atmospheric pressures. Dry ice is very cold $\left( -78^\text{o} \text{C} \right)$ and so is used as a coolant for goods, such as ice cream, that must remain frozen during shipment. Because the dry ice sublimes rather than melts, there is no liquid mess associated with its change of state as it warms. As you may have seen in demonstrations, dry ice merely transforms from a solid to a cold, steamy looking gas.
Ferrocene
Ferrocene (an iron-containing compound) is usually purified by sublimation. A sample of crude ferrocene is gently heated causing it to sublime. When the ferrocene is cooled, reddish ferrocene crystals deposited on the outside of that tube as shown in the figure below.
Summary
• Sublimation is the change of state from a solid to a gas, without passing through the liquid state.
• Deposition is the change of state from a gas to a solid.
• Carbon dioxide is an example of a material that easily undergoes sublimation.
Review
1. Define sublimation.
2. What is another name for solid carbon dioxide?
3. How is ferrocene purified?
4. Name a substance which undergoes sublimation.
13.13: Crystal Systems
What are the different uses of lasers?
The development of modern lasers has opened many doors to both research and applications. A laser beam was used to measure the distance from the Earth to the moon. Lasers are important components of CD players. As the image above illustrates, lasers can provide precise focusing of beams to selectively destroy cancer cells in patients. The ability of a laser to focus precisely is due to high-quality crystals that help give rise to the laser beam. A variety of techniques are used to manufacture pure crystals for use in lasers.
Crystalline Solids
The majority of solids are crystalline in nature. A crystal is a substance in which the particles are arranged in an orderly, repeating, three-dimensional pattern. Particles of a solid crystal may be ions, atoms, or molecules, depending on the type of substance. The three-dimensional arrangement of a solid crystal is referred to as the crystal lattice. Different arrangements of the particles within a crystal cause them to adopt several different shapes.
Crystal Systems
Crystals are classified into general categories based on their shapes. A crystal is defined by its faces, which intersect one another at specific angles; these intersections are characteristic of the given substance. The seven crystal systems are shown below, along with an example of each. The edge lengths of a crystal are represented by the letters $a$, $b$, and $c$. The angles at which the faces intersect are represented by the Greek letters $\alpha$, $\beta$, and $\gamma$. Each of the seven crystal systems differ in terms of the angles between the faces, and in the number of edges of equal length on each face.
Table $1$: Seven Basic Crystal Systems and Examples of Each
Seven Basic Crystal Systems and Examples of Each
Crystal System Diagram Example
Cubic
$a = b = c$; $\alpha = \beta = \gamma = 90^\text{o}$
Tetragonal
$a = b \neq c$; $\alpha = \beta = \gamma = 90^\text{o}$
Orthorhombic
$a \neq b \neq c$; $\alpha = \beta = \gamma = 90^\text{o}$
Monoclinic
$a \neq b \neq c$; $\alpha \neq 90^\text{o} = \beta = \gamma$
Rhombohedral
$a = b = c$; $\alpha = \beta = \gamma \neq 90^\text{o}$
Triclinic
$a \neq b \neq c$; $\alpha \neq \beta \neq \gamma \neq 90^\text{o}$
Hexagonal
$a = b \neq c$; $\alpha = \beta = 90^\text{o}$, $\gamma = 120^\text{o}$
Summary
• A crystal is a substance in which the particles are arranged in an orderly, repeating, three-dimensional pattern.
• The crystal lattice is the three-dimensional arrangement of a solid crystal.
Review
1. What is a crystal?
2. List the seven crystal systems. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.12%3A_Sublimation.txt |
How can we measure size of atoms?
We have many data tables available to us in chemistry. If we wanted to know the size of the lithium atom, we can easily look it up and find that this atom is 134 picometers across. If we remove the outer electron, the lithium ion is now only 90 picometers in size. How do we know this? We don’t have a ruler small enough to measure these tiny distances. But we can use a technique called x-ray diffraction to shine beams of x-rays through a crystal of a lithium compound. By measuring how much the beams are bent after they come through the crystal, we can calculate the size of the molecule. This technique works both for small materials like lithium compounds and equally well for very large protein molecules.
Unit Cells
A unit cell is the smallest portion of a crystal lattice that shows the three-dimensional pattern of the entire crystal. A crystal can be thought of as the same unit cell repeated over and over in three dimensions. The figure below illustrates the relationship of a unit cell to the entire crystal lattice.
Unit cells occur in many different varieties. As one example, the cubic crystal system is composed of three different types of unit cells: (1) simple cubic, (2) face-centered cubic, and (3) body-centered cubic. These are shown in three different ways in the figure below.
Pay special attention to the last diagram for each type of cell. You will notice that the atoms or ions at the edges of each face, or at the corners, are not complete spheres. In the simple cubic cell, each corner atom is shared by 8 different unit cells. The same situation exists for the edge or corner particles in the face-centered and body-centered cubic forms. In addition, each particle in the center of the face-centered cubic cell is shared by 2 unit cells. Body-centered cells have an additional atom in the middle of the cell which is contained entirely in that cell.
Note that we have only considered the unit cells of a cubic crystal. Other crystal forms also have unit cells. These unit cells are:
• Rhombohedral, hexagonal, triclinic - one unique form each.
• Tetragonal - simple and body-centered.
• Monoclinic - simple and base-centered.
• Orthorhombic - simple, face-centered, body-centered, and base-centered.
Summary
• A unit cell is the smallest portion of a crystal lattice that shows the three-dimensional pattern of the entire crystal.
• There are three different types of unit cells in the cubic crystal system.
Review
1. What is a unit cell?
2. List the three cubic unit cells.
3. Does each unit cell exist as an entity by itself?
13.15: Classes of Crystalline Solids
What are common things that we connect to wires?
We often take a lot of things for granted. We just assume that we will get electric power when we connect a plug to an electrical outlet. The wire that comprises that outlet is almost always copper, a material that conducts electricity well. The unique properties of the solid copper allow electrons to flow freely through the wire and into whatever device we connect it to. Then we can enjoy music, television, work on the computer, or whatever other activity we want to undertake.
Classes of Crystalline Solids
Crystalline substances can be described by the types of particles in them, and the types of chemical bonding that take place between the particles. There are four types of crystals: (1) ionic, (2) metallic, (3) covalent network, and (4) molecular. Properties and examples of each type are described in the table below.
Type of Crystalline Solid Examples (formulas) Melting Point $\left( ^\text{o} \text{C} \right)$ Normal Boiling Point $\left( ^\text{o} \text{C} \right)$
Table $1$: Crystalline Solids - Melting and Boiling Points
Ionic $\ce{NaCl}$ 801 1413
$\ce{CaF_2}$ 1418 1533
Metallic $\ce{Hg}$ -39 630
$\ce{Na}$ 371 883
$\ce{Au}$ 1064 2856
$\ce{W}$ 3410 5660
Covalent Network $\ce{B}$ 2076 3927
$\ce{C}$ (diamond) 3500 3930
$\ce{SiO_2}$ 1600 2230
Molecular $\ce{H_2}$ -259 -253
$\ce{I_2}$ 114 184
$\ce{NH_3}$ -78 -33
$\ce{H_2O}$ 0 100
1. Ionic crystals: The ionic crystal structure consists of alternating positively-charged cations and negatively-charged anions (see figure below). The ions may either be monatomic or polyatomic. Generally, ionic crystals form from a combination of Group 1 or 2 metals, and Group 16 or 17 nonmetals or nonmetallic polyatomic ions. Ionic crystals are hard, brittle, and have high melting points. Ionic compounds do not conduct electricity as solids, but do conduct when molten or in aqueous solution.
2. Metallic crystal: Metallic crystals consist of metal cations surrounded by a "sea" of mobile valence electrons (see figure below). These electrons (also referred to as delocalized electrons) do not belong to any one atom, but are capable of moving through the entire crystal. As a result, metals are good conductors of electricity. As seen in the table above, the melting points of metallic crystals span a wide range.
3. Covalent network crystals: A covalent network crystal consists of atoms at the lattice points of the crystal, with each atom covalently bonded to its nearest neighbor atoms (see figure below). The covalently bonded network is three-dimensional and contains a very large number of atoms. Network solids include diamond, quartz, many metalloids, and oxides of transition metals and metalloids. Network solids are hard and brittle, with extremely high melting and boiling points. Being composed of atoms rather than ions, they do not conduct electricity in any state.
4. Molecular crystals: Molecular crystals typically consist of molecules at the lattice points of the crystal, held together by relatively weak intermolecular forces (see figure below). The intermolecular forces may be dispersion forces in the case of nonpolar crystals, or dipole-dipole forces in the case of polar crystals. Some molecular crystals, such as ice, have molecules held together by hydrogen bonds. When one of the noble gases is cooled and solidified, the lattice points are individual atoms rather than molecules. In all cases, the intermolecular forces holding the particles together are far weaker than either ionic or covalent bonds. As a result, the melting and boiling points of molecular crystals are much lower. Lacking ions or free electrons, molecular crystals are poor electrical conductors.
Summary
• Ionic crystals are composed of alternating positive and negative ions.
• Metallic crystals consist of metal cations surrounded by a "sea" of mobile valence electrons.
• Covalent crystals are composed of atoms which are covalently bonded to one another.
• Molecular crystals are held together by weak intermolecular forces.
Review
1. What is an ionic crystal?
2. What type of crystal is a diamond?
3. What forces hold molecular crystals together?
4. Which type of crystal is a good conductor of electricity? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.14%3A_Unit_Cells.txt |
When a tire goes flat, its shape changes. A tire may be flat because of a slow leak in the tire valve. It could be flat because it ran over a nail or screw, and ended up with a small hole where the air leaks out over a period of time. Or, it could go flat after hitting a large rock or other object while traveling at high speeds (this one is for those readers who enjoy detective movies or TV shows!). What if a crystalline solid like \(\ce{LiBr}\) were made into a tire? When it encountered a blow, the crystal would break into small pieces. Since rubber is an amorphous solid, it has a very different set of physical properties.
Amorphous Solids
Unlike a crystalline solid, an amorphous solid is a solid that lacks an ordered internal structure. Some examples of amorphous solids include rubber, plastic, and gels. Glass is a very important amorphous solid that is made by cooling a mixture of materials in such a way that it does not crystallize. Glass is sometimes referred to as a supercooled liquid, rather than a solid. If you were to watch a glassblower in action, you may notice that they take advantage of the fact that amorphous solids do not have a distinct melting point like crystalline solids do. Instead, as glass is heated, it slowly softens and can be shaped into all sorts of interesting forms. When a glass object shatters, it does so in a very irregular way—unlike crystalline solids, which always break into fragments of the same shape as dictated by their crystal system.
Properties of amorphous solids are different in many ways from those of crystalline solids. The intermolecular forces in amorphous solids do not have a regular external structure, and they do not have sharp melting points. Unlike crystalline solids that have regular planes of cleavage, the physical properties of amorphous solids are the same in all directions.
Plastics are used for many purposes because they are inexpensive to produce, and do not shatter like glass or ceramic materials. Since they are easily disposed of, the accumulation of plastic garbage has become a serious problem in many parts of the world. Recycling programs that help reuse plastics are growing in popularity.
Summary
• An amorphous solid is a solid that lacks an ordered internal structure.
• Examples of amorphous solids include glass, rubber, and plastics.
• The physical properties of amorphous solids differ from those of crystalline solids.
13.18: Heating and Cooling Curves
During the lifetime of Mark Twain (real name: Samuel Langhorne Clemens, 1835-1910), the steamboat was a major means of transportation on the rivers and lakes of the United States. Twain himself was a steamboat pilot on the Mississippi River for a period of time, and took his pen name from the measurement of water depth (twelve feet, which was a safe depth for the boats). The boats got their power from steam—liquid water converted to a gas at high temperatures. The steam would push the pistons of the engine, causing the paddle wheels to turn and propel the boat.
Heating Curves
Imagine that you have a block of ice that is at a temperature of $-30^\text{o} \text{C}$, well below its melting point. The ice is in a closed container. As heat is steadily added to the ice block, the water molecules will begin to vibrate faster and faster as they absorb kinetic energy. Eventually, when the ice has warmed to $0^\text{o} \text{C}$, the added energy will start to break apart the hydrogen bonding that keeps the water molecules in place when it is in the solid form. As the ice melts, its temperature does not rise. All of the energy that is being put into the ice goes into the melting process and not into any increase in temperature. During the melting process, the two states—solid and liquid—are in equilibrium with one another. If the system was isolated at that point and no energy was allowed to enter or leave, the ice-water mixture at $0^\text{o} \text{C}$ would remain. Temperature is always constant during a change of state.
Continued heating of the water after the ice has completely melted will now increase the kinetic energy of the liquid molecules and the temperature will rise. Assuming that the atmospheric pressure is standard, the temperature will rise steadily until it reaches $100^\text{o} \text{C}$. At this point, the added energy from the heat will cause the liquid to begin to vaporize. As with the previous state change, the temperature will remain at $100^\text{o} \text{C}$ while the water molecules are going from the liquid to the gas or vapor state. Once all the liquid has completely boiled away, continued heating of the steam (since the container is closed) will increase its temperature above $100^\text{o} \text{C}$.
The experiment described above can be summarized in a graph called a heating curve (figure below).
The change of state behavior of all substances can be represented with a heating curve of this type. The melting and boiling points of the substance can be determined by the horizontal lines or plateaus on the curve. Other substances have melting and boiling points that are different from those of water. An exception to this blueprint heating curve is for a substance such as carbon dioxide, which sublimes rather than melts at standard pressure. The heating curve for carbon dioxide would have only one plateau, at the sublimation temperature of $\ce{CO_2}$.
The entire experiment could be run in reverse. Steam above $100^\text{o} \text{C}$ could be steadily cooled down to $100^\text{o} \text{C}$, at which point it would condense to liquid water. The water could then be cooled to $0^\text{o} \text{C}$, at which point continued cooling would freeze the water to ice. The ice could then be cooled to a point below $0^\text{o} \text{C}$. This could be diagrammed in a cooling curve that would be the reverse of the heating curve.
Summary of State Changes
All of the changes of state that occur between solid, liquid, and gas are summarized in the diagram in the figure below. Freezing is the opposite of melting, and both represent the equilibrium between the solid and liquid states. Evaporation occurs when a liquid turns to a gas. Condensation is the opposite of vaporization, and both represent the equilibrium between the liquid and gas states. Deposition is the opposite of sublimation, and both represent the equilibrium between the solid and gas states.
Summary
• A change of state can be brought about by putting heat into a system or removing it from the system.
• The temperature of a system will not change as long as the substance is undergoing a change from solid to liquid to gas, as well as the reverse.
• Freezing is the opposite of melting.
• Evaporation occurs when a liquid turns to a gas.
• Condensation is the opposite of vaporization.
• Deposition is the opposite of sublimation. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.17%3A_Amorphous_Solids.txt |
Many rockets use a combination of kerosene and liquid oxygen for their fuel. Oxygen can be reduced to the liquid state either by cooling or by using high pressure. In the case of a rocket, since the oxygen is in a container essentially out in the open, maintaining a temperature of $-183^\text{o} \text{C}$ (the boiling point of oxygen) is not very practical. However, high pressure can be used to force the oxygen into tanks, causing it to liquefy so that it can mix with the kerosene, and provide a powerful ignition to move the rocket.
Phase Diagrams
The relationship among the solid, liquid, and vapor (gas) states of a substance can be shown as a function of temperature and pressure in a single diagram. A phase diagram is graph showing the conditions of temperature and pressure under which a substance exists in the solid, liquid, and gas phases. Examine the general phase diagram shown in the figure below. In each of the three colored regions of the diagram, the substance is in a single state (or phase). The dark lines that act as the boundary between those regions represent the conditions under which the two phases are in equilibrium.
Find the $X$ on the pressure axis and presume that the value of $X$ is standard pressure of $1 \: \text{atm}$. As one moves left to right across the red line, the temperature of the solid substance is being increased while the pressure remains constant. When point $A$ is reached, the substance melts and the temperature $B$ on the horizontal axis represents the normal melting point of the substance. Moving further to the right, the substance boils at point $Y$ and so point $C$ on the horizontal axis represents the normal boiling point of the substance. As the temperature increases at a constant pressure, the substance changes from solid to liquid to gas.
Start right above point $B$ on the temperature axis and follow the red line vertically. At very low pressure, the particles of the substance are far apart from one another and the substance is in the gas state. As the pressure is increased, the particles of the substance are forced closer and closer together. Eventually the particles are pushed so close together that attractive forces cause the substance to condense into the liquid state. Continually increasing the pressure on the liquid will eventually cause the substance to solidify. For the majority of substances, the solid state is denser than the liquid state and so putting a liquid under great pressure will cause it to turn into a solid. The line segment $R$-$S$ represents the process of sublimation, where the substance changes directly from a solid to a gas. At a sufficiently low pressure, the liquid phase does not exist. The point labeled $TP$ is called the triple point. The triple point is the one condition of temperature and pressure where the solid, liquid, and vapor states of a substance can all coexist at equilibrium.
Summary
• A phase diagram graphs the conditions of temperature and pressure under which a substance exists in the solid, liquid, and gas states.
• The triple point is the one condition of temperature and pressure where the solid, liquid, and vapor states of a substance can all coexist at equilibrium.
13.20: Phase Diagram for Water
A specific consistency of snow is required to make the best snowballs. Dry snow can be tightly pressed, and will form snowballs because the higher pressure causes the snowflakes to melt somewhat. However, when you release the pressure, the snow goes back to a more solid form and the flakes no longer stick together. Ideally, instead, the snow needs to be a little bit wet so that the particles will stick together.
Phase Diagram for Water
Water is a unique substance in many ways. One of these special properties is the fact that solid water (ice) is less dense than liquid water just above the freezing point. The phase diagram for water is shown in the figure below.
Notice one key difference between last section's general phase diagram, and the above phase diagram for water: in water's diagram, the slope of the line between the solid and liquid states is negative rather than positive. The reason is that water is an unusual substance, in that its solid state is less dense than the liquid state. Ice floats in liquid water. Therefore, a pressure change has the opposite effect on those two phases. If ice is relatively near its melting point, it can be changed into liquid water by the application of pressure. The water molecules are actually closer together in the liquid phase than they are in the solid phase.
Refer again to water's phase diagram (figure above). Notice point $E$, labeled the critical point. What does that mean? At $373.99^\text{o} \text{C}$, particles of water in the gas phase are moving very, very rapidly. At any temperature higher than that, the gas phase cannot be made to liquefy, no matter how much pressure is applied to the gas. The critical pressure $\left( P_\text{C} \right)$ is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid. For water, the critical pressure is very high, $217.75 \: \text{atm}$. The critical point is the intersection point of the critical temperature and the critical pressure.
Summary
• Solid water is less dense than liquid water just above the freezing point.
• The critical temperature $\left( T_\text{C} \right)$ of a substance is the highest temperature at which the substance can possibly exist as a liquid.
• The critical pressure $\left( P_\text{C} \right)$ is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid.
• The critical point is the intersection point of the critical temperature and the critical pressure. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.19%3A_General_Phase_Diagram.txt |
When we pack to go on vacation, there seems to always be "one more" thing that we need to get in our suitcase. Maybe it's another bathing suit, pair of shoes, book—whatever the item, we need to get it in. Fortunately, we can usually squeeze things together somehow. Perhaps there is a little space between folds of clothing, or we can rearrange the shoes; somehow we can get that last item in and close the suitcase.
Compressibility
Scuba diving is a form of underwater diving in which a diver carries his own breathing gas, usually in the form of a tank of compressed air. The pressure in most commonly used scuba tanks ranges from 200 to 300 atmospheres. Gases are unlike other states of matter in that a gas expands to fill the shape and volume of its container. For this reason, gases can also be compressed so that a relatively large amount of gas can be forced into a small container. If the air in a typical scuba tank were transferred to a container at the standard pressure of $1 \: \text{atm}$, the volume of that container would need to be about 2500 liters.
Compressibility is the measure of how much a given volume of matter decreases when placed under pressure. If we put pressure on a solid or a liquid, there is essentially no change in volume. The atoms, ions, or molecules that make up the solid or liquid are very close together. There is no space between the individual particles, so they cannot pack together.
The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles. At room temperature and standard pressure, the average distance between gas molecules is about ten times the diameter of the molecules themselves. When a gas is compressed, as when the scuba tank is being filled, the gas particles are forced closer together.
Compressed gases are used in many situations. In hospitals, oxygen is often used for patients who have damaged lungs to help them breathe better. If a patient is having a major operation, the anesthesia that is administered will frequently be a compressed gas. Welding requires very hot flames produced by compressed acetylene and oxygen mixtures. Many summer barbeque grills are fueled by compressed propane.
14.02: Factors Affecting Gas Pressure
The pressure of the air in a basketball has to be adjusted so that the ball bounces to the correct height. Before a game, officials check the ball by dropping it from shoulder height and seeing how far back up it bounces. What would an official do if the ball did not bounce up high enough, or if it bounced too high?
The pressure inside a container is dependent on the amount of gas inside the container. If a basketball does not bounce high enough, an official could remedy the situation by using a hand pump and adding more air to the ball. Conversely, if it bounces too high, the official could let some air out of the ball.
Factors Affecting Gas Pressure
Recall from the kinetic-molecular theory that gas particles move randomly and in straight lines until they elastically collide with either other gas particles, or with one of the walls of the container. It is these collisions with the walls of the container that define the pressure of the gas. Four variables are used to describe the condition of a gas: pressure $\left( P \right)$, volume $\left( V \right)$, temperature $\left( T \right)$, and the amount of the gas as measured by the number of moles (\left( n \right)\). We will examine separately how the volume, temperature, and amount of gas each affect the pressure of an enclosed gas sample.
Amount of Gas
The figure below shows what happens when air is added to a rigid container. A rigid container is one that is incapable of expanding or contracting. A steel canister is an example of a rigid container.
The canister on the left contains a gas at a certain pressure. The attached air pump is then used to double the amount of gas in the canister. Since the canister cannot expand, the increased number of air molecules will strike the inside walls of the canister twice as frequently as they did before. The result is that the pressure inside the canister doubles. As you might imagine, if more and more air is continually added to a rigid container, it may eventually burst. Reducing the number of molecules in a rigid container has the opposite effect, and the pressure decreases.
Volume
Pressure is also affected by the volume of the container. If the volume of a container is decreased, the gas molecules have less space in which to move around. As a result, they will strike the walls of the container more often, and the pressure increases.
The figure below shows a cylinder of gas whose volume is controlled by an adjustable piston. On the left, the piston is pulled mostly out and the gauge reads a certain pressure. On the right, the piston has been pushed so that the volume of the enclosed portion of the container where the gas is located has been cut in half. The pressure of the gas doubles. Increasing the volume of the container would have the opposite effect, and the pressure of the gas would decrease.
Temperature
It would be very inadvisable to place a can of soup over a campfire without venting the can. As the can heats up, it may explode. The kinetic-molecular theory explains why. The air inside the rigid can of soup is given more kinetic energy by the heat coming from the campfire. The kinetic energy causes the air molecules to move faster; they impact the container walls more frequently and with more force. The increase in pressure inside may eventually exceed the strength of the can and it will explode. An additional factor is that the soup may begin boiling, which will then aid even more gas and more pressure to the inside of the can.
Shown in the figure below is a cylinder of gas (left) that is at room temperature $\left( 300 \: \text{K} \right)$. On the right, the cylinder has been heated until the Kelvin temperature has doubled to $600 \: \text{K}$. The kinetic energy of the gas molecules increases, so collisions with the walls of the container are now more forceful than they were before. As a result, the pressure of the gas doubles. Decreasing the temperature would have the opposite effect, and the pressure of an enclosed gas would decrease.
Summary
• An increase in the number of gas molecules, while container volume stays constant, increases pressure.
• A decrease in container volume increases gas pressure.
• An increase in temperature of a gas in a rigid container increases the pressure. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.01%3A_Compressibility.txt |
Each day, hundreds of weather balloons are launched. Made of a synthetic rubber and carrying a box of instruments, each helium-filled balloon rises up into the sky. As a balloon gains altitude, the atmospheric pressure becomes less and the balloon expands. At some point the balloon bursts due to the expansion; the instruments drop (aided by a parachute) to be retrieved and studied for information about the weather.
Boyle's Law
Robert Boyle (1627-1691), an English chemist, is widely considered to be one of the founders of the modern experimental science of chemistry. He discovered that doubling the pressure of an enclosed sample of gas, while keeping its temperature constant, caused the volume of the gas to be reduced by half. Boyle's law states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases.
Physically, what is happening? The gas molecules are moving and are a certain distance apart from one another. An increase in pressure pushes the molecules closer together, reducing the volume. If the pressure is decreased, the gases are free to move about in a larger volume.
Mathematically, Boyle's law can be expressed by the equation:
$P \times V = k\nonumber$
The $k$ is a constant for a given sample of gas and depends only on the mass of the gas and the temperature. The table below shows pressure and volume data for a set amount of gas at a constant temperature. The third column represents the value of the constant $\left( k \right)$ for this data and is always equal to the pressure multiplied by the volume. As one of the variables changes, the other changes in such a way that the product of $P \times V$ always remains the same. In this particular case, that constant is $500 \: \text{atm} \cdot \text{mL}$.
Pressure $\left( \text{atm} \right)$ Volume $\left( \text{mL} \right)$ $P \times V = k$ $\left( \text{atm} \cdot \text{mL} \right)$
Table $1$: Pressure-Volume Data
0.5 1000 500
0.625 800 500
1.0 500 500
2.0 250 500
5.0 100 500
8.0 62.5 500
10.0 50 500
A graph of the data in the table further illustrates the inverse relationship nature of Boyle's Law (see figure below). Volume is plotted on the $x$-axis, with the corresponding pressure on the $y$-axis.
Boyle's Law can be used to compare changing conditions for a gas. We use $P_1$ and $V_1$ to stand for the initial pressure and initial volume of a gas. After a change has been made, $P_2$ and $V_2$ stand for the final pressure and volume. The mathematical relationship of Boyle's Law becomes:
$P_1 \times V_1 = P_2 \times V_2\nonumber$
This equation can be used to calculate any one of the four quantities if the other three are known.
Example $1$
A sample of oxygen gas has a volume of $425 \: \text{mL}$ when the pressure is equal to $387 \: \text{kPa}$. The gas is allowed to expand into a $1.75 \: \text{L}$ container. Calculate the new pressure of the gas.
Known
• $P_1 = 387 \: \text{kPa}$
• $V_1 = 425 \: \text{mL}$
• $V_2 = 1.75 \: \text{L} = 1750 \: \text{mL}$
Unknown
Use Boyle's Law to solve for the unknown pressure $\left( P_2 \right)$. It is important that the two volumes ($V_1$ and $V_2$) are expressed in the same units, so $V_2$ has been converted to $\text{mL}$.
Step 2: Solve.
First, rearrange the equation algebraically to solve for $P_2$.
$P_2 = \frac{P_1 \times V_1}{V_2}\nonumber$
Now substitute the known quantities into the equation and solve.
$P_2 = \frac{387 \: \text{kPa} \times 425 \: \text{mL}}{1750 \: \text{mL}} = 94.0 \: \text{kPa}\nonumber$
Step 3: Think about your result.
The volume has increased to slightly over 4 times its original value and so the pressure is decreased by about one fourth. The pressure is in $\text{kPa}$ and the value has three significant figures. Note that any pressure or volume units can be used as long as they are consistent throughout the problem. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.03%3A_Boyle%27s_Law.txt |
Freshly-baked bread is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end result is an enjoyable treat, especially when covered with melted butter.
Charles's Law
French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.
Mathematically, the direct relationship of Charles's Law can be represented by the following equation:
$\frac{V}{T} = k\nonumber$
As with Boyle's Law, $k$ is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.
Temperature $\left( \text{K} \right)$ Volume $\left( \text{mL} \right)$ $\frac{V}{T} = k$ $\left( \frac{\text{mL}}{\text{K}} \right)$
Table $1$: Temperature-Volume Data
50 20 0.40
100 40 0.40
150 60 0.40
200 80 0.40
300 120 0.40
500 200 0.40
1000 400 0.40
When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below.
Figure $2$: The volume of a gas increases as the Kelvin temperature increases.
Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.
Charles's Law can also be used to compare changing conditions for a gas. Now we use $V_1$ and $T_1$ to stand for the initial volume and temperature of a gas, while $V_2$ and $T_2$ stand for the final volume and temperature. The mathematical relationship of Charles's Law becomes:
$\frac{V_1}{T_1} = \frac{V_2}{T_2}\nonumber$
This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that $\text{K} = \: ^\text{o} \text{C} + 273$.
Example $1$
A balloon is filled to a volume of $2.20 \: \text{L}$ at a temperature of $22^\text{o} \text{C}$. The balloon is then heated to a temperature of $71^\text{o} \text{C}$. Find the new volume of the balloon.
Known
• $V_1 = 2.20 \: \text{L}$
• $T_1 = 22^\text{o} \text{C} = 295 \: \text{K}$
• $T_2 = 71^\text{o} \text{C} = 344 \: \text{K}$
Unknown
Use Charles's Law to solve for the unknown volume $\left( V_2 \right)$. The temperatures have first been converted to Kelvin.
Step 2: Solve.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \frac{V_1 \times T_2}{T_1}\nonumber$
Now substitute the known quantities into the equation and solve.
$V_2 = \frac{2.20 \: \text{L} \times 344 \: \text{K}}{295 \: \text{K}} = 2.57 \: \text{L}\nonumber$
Step 3: Think about your result.
The volume increases as the temperature increases. The result has three significant figures.
14.05: Gay-Lussac's Law
Propane tanks are widely used with barbeque grills. However, it's not fun to find out half-way through grilling that you've run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So, you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.
Gay-Lussac's Law
When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment.
The mathematical expressions for Gay-Lussac's Law are likewise similar to those of Charles's Law:
$\frac{P}{T} \: \: \: \text{and} \: \: \: \frac{P_1}{T_1} = \frac{P_2}{T_2}\nonumber$
A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume, its pressure continually decreases until the gas condenses to a liquid.
Example $1$
The gas in an aerosol can is under a pressure of $3.00 \: \text{atm}$ at a temperature of $25^\text{o} \text{C}$. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of $845^\text{o} \text{C}$?
Known
• $P_1 = 3.00 \: \text{atm}$
• $T_1 = 25^\text{o} \text{C} = 298 \: \text{K}$
• $T_2 = 845^\text{o} \text{C} = 1118 \: \text{K}$
Unknown
Use Gay-Lussac's Law to solve for the unknown pressure $\left( P_2 \right)$. The temperatures have first been converted to Kelvin.
Step 2: Solve.
First, rearrange the equation algebraically to solve for $P_2$.
$P_2 = \frac{P_1 \times T_2}{T_1}\nonumber$
Now substitute the known quantities into the equation and solve.
$P_2 = \frac{3.00 \: \text{atm} \times 1118 \: \text{K}}{298 \: \text{K}} = 11.3 \: \text{atm}\nonumber$
Step 3: Think about your result.
The pressure increases dramatically due to large increase in temperature. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.04%3A_Charles%27s_Law.txt |
The modern refrigerator takes advantage of the gas laws to remove heat from a system. Compressed gas in the coils is allowed to expand. This expansion lowers the temperature of the gas and transfers heat energy from the material in the refrigerator to the gas. As the gas is pumped through the coils, the pressure on the gas compresses it and raises the gas temperature. This heat is then dissipated through the coils into the outside air. As the compressed gas is pumped through the system again, the process repeats itself.
Combined Gas Law
To this point, we have examined the relationships between any two of the variables of $P$, $V$, and $T$, while the third variable is held constant. However, situations do arise where all three variables change. The combined gas law expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant.
$\frac{P \times V}{T} = k \: \: \: \text{and} \: \: \: \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\nonumber$
Example $1$
$2.00 \: \text{L}$ of a gas at $35^\text{o} \text{C}$ and $0.833 \: \text{atm}$ is brought to standard temperature and pressure (STP). What will be the new gas volume?
Known
• $P_1 = 0.833 \: \text{atm}$
• $V_1 = 2.00 \: \text{L}$
• $T_1 = 35^\text{o} \text{C} = 308 \: \text{K}$
• $P_2 = 1.00 \: \text{atm}$
• $T_2 = 0^\text{o} \text{C} = 273 \: \text{K}$
Unknown
Use the combined gas law to solve for the unknown volume $\left( V_2 \right)$. STP is $273 \: \text{K}$ and $1 \: \text{atm}$. The temperatures have been converted to Kelvin.
Step 2: Solve.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1}\nonumber$
Now substitute the known quantities into the equation and solve.
$V_2 = \frac{0.833 \: \text{atm} \times 2.00 \: \text{L} \times 273 \: \text{K}}{1.00 \: \text{atm} \times 308 \: \text{K}} = 1.48 \: \text{L}\nonumber$
Step 3: Think about your result.
Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease. Since both changes are relatively small, the volume does not decrease dramatically.
It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle's, Charles's, and Gay-Lussac's laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that $T_1 = T_2$. Look at the combined gas law and cancel the $T$ variable out from both sides of the equation. What is left over is Boyle's Law: $P_1 \times V_1 = P_2 \times V_2$. Likewise, if the pressure is constant, then $P_1 = P_2$ and cancelling $P$ out of the equation leaves Charles's Law. If the volume is constant, then $V_1 = V_2$ and cancelling $V$ out of the equation leaves Gay-Lussac's Law.
Summary
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
14.07: Avogadro's Law
A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. The amount of air that should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst.
Avogadro's Law
You have learned about Avogadro's hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles of gas, when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is:
$V = k \times n \: \: \: \text{and} \: \: \: \frac{V_1}{n_1} = \frac{V_2}{n_2}\nonumber$
(Where $n$ is the number of moles of gas and $k$ is a constant). Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.
If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro's Law. Adding gas to a rigid container makes the pressure increase.
Example $1$
A balloon has been filled to a volume of $1.90 \: \text{L}$ with $0.0920 \: \text{mol}$ of helium gas. If $0.0210 \: \text{mol}$ of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon?
Solution
Step 1: List the known quantities and plan the problem.
Known
• $V_1 = 1.90 \: \text{L}$
• $n_1 = 0.0920 \: \text{mol}$
• $n_2 = 0.0920 + 0.0210 = 0.1130 \: \text{mol}$
Unknown
Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. Use Avogadro's Law to solve for the final volume.
Step 2: Solve.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \frac{V_1 \times n_2}{n_1}\nonumber$
Now substitute the known quantities into the equation and solve.
$V_2 = \frac{1.90 \: \text{L} \times 0.1130 \: \text{mol}}{0.0920 \: \text{mol}} = 2.33 \: \text{L}\nonumber$
Step 3: Think about your result.
Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.
Summary
• Avogadro's Law states that the volume of a gas is directly proportional to the number of moles of gas, when the temperature and pressure are held constant.
• Calculations are shown for relationships between volume and number of moles of a gas. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.06%3A_Combined_Gas_Law.txt |
There are a number of chemical reactions that require ammonia. In order to carry out a reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.
Ideal Gas Law
The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these laws together gives us the following equation:
$\frac{P_1 \times V_1}{T_1 \times n_1} = \frac{P_2 \times V_2}{T_2 \times n_2}\nonumber$
As with the other gas laws, we can also say that $\frac{\left( P \times V \right)}{\left( T \times n \right)}$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable $R$ for the constant, the equation becomes:
$\frac{P \times V}{T \times n} = R\nonumber$
The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted:
$PV = nRT\nonumber$
The variable $R$ in the equation is called the ideal gas constant.
Evaluating the Ideal Gas Constant
The value of $R$, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: $\text{kPa}$, $\text{atm}$, or $\text{mm} \: \ce{Hg}$. Therefore, $R$ can have three different values.
We will demonstrate how $R$ is calculated when the pressure is measured in $\text{kPa}$. Recall that the volume of $1.00 \: \text{mol}$ of any gas at STP is measured to be $22.414 \: \text{L}$. We can substitute $101.325 \: \text{kPa}$ for pressure, $22.414 \: \text{L}$ for volume, and $273.15 \: \text{K}$ for temperature into the ideal gas equation and solve for $R$.
$R = \frac{PV}{nT} = \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} = 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol}\nonumber$
This is the value of $R$ that is to be used in the ideal gas equation when the pressure is given in $\text{kPa}$. The table below shows a summary of this and the other possible values of $R$. It is important to choose the correct value of $R$ to use for a given problem.
Unit of $P$ Unit of $V$ Unit of $n$ Unit of $T$ Value and Unit of $R$
Table $1$: Values of the Ideal Gas Constant
$\text{kPa}$ $\text{L}$ $\text{mol}$ $\text{K}$ $8.314 \: \text{J/K} \cdot \text{mol}$
$\text{atm}$ $\text{L}$ $\text{mol}$ $\text{K}$ $0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$
$\text{mm} \: \ce{Hg}$ $\text{L}$ $\text{mol}$ $\text{K}$ $62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}$
Notice that the unit for $R$ when the pressure is in $\text{kPa}$ has been changed to $\text{J/K} \cdot \text{mol}$. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule $\left( \text{J} \right)$.
Example $1$
What volume is occupied by $3.760 \: \text{g}$ of oxygen gas at a pressure of $88.4 \: \text{kPa}$ and a temperature of $19^\text{o} \text{C}$? Assume the oxygen is an ideal gas.
Known
• $P = 88.4 \: \text{kPa}$
• $T = 19^\text{o} \text{C} = 292 \: \text{K}$
• Mass $\ce{O_2} = 3.760 \: \text{g}$
• $\ce{O_2} = 32.00 \: \text{g/mol}$
• $R = 8.314 \: \text{J/K} \cdot \text{mol}$
Unknown
In order to use the ideal gas law, the number of moles of $\ce{O_2}$ $\left( n \right)$ must be found from the given mass and the molar mass. Then, use $PV = nRT$ to solve for the volume of oxygen.
Step 2: Solve.
$3.760 \: \text{g} \times \frac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \text{g} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2}\nonumber$
Rearrange the ideal gas law and solve for $V$.
$V = \frac{nRT}{P} = \frac{0.1175 \: \text{mol} \times 8.314 \: \text{J/K} \cdot \text{mol} \times 292 \: \text{K}}{88.4 \: \text{kPa}} = 3.23 \: \text{L} \: \ce{O_2}\nonumber$
Step 3: Think about your result.
The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume $\left( 22.4 \: \text{L/mol} \right)$ since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for $T$ and $P$. Since a joule $\left( \text{J} \right) = \text{kPa} \cdot \text{L}$, the units cancel out correctly, leaving a volume in liters.
Summary
• The ideal gas constant is calculated.
• An example of calculations using the ideal gas law is shown. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.08%3A_Ideal_Gas_Law.txt |
Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. Small balloons filled with helium are often affordable and available at stores, but large ones are much more expensive (and require a lot more helium).
Calculating Molar Mass and Density of a Gas
A chemical reaction, which produces a gas, is performed. The produced gas is then collected and its mass and volume are determined. The molar mass and volume are determined. The molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known.
Example $1$
A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of $1.211 \: \text{g}$ and occupies a volume of $677 \: \text{mL}$. The temperature in the laboratory is $23^\text{o} \text{C}$ and the air pressure is $0.987 \: \text{atm}$. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.
Known
• Mass $= 1.211 \: \text{g}$
• $V = 677 \: \text{mL} = 0.677 \: \text{L}$
• $T = 23^\text{o} \text{C} = 296 \: \text{K}$
• $P = 0.987 \: \text{atm}$
• $R = 0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$
Unknown
• $n = ? \: \text{mol}$
• Molar Mass $= ? \: \text{g/mol}$
First the ideal gas law will be used to solve for the moles of unknown gas $\left( n \right)$. Then the mass of the gas divided by the moles will give the molar mass.
Step 2: Solve.
$n = \frac{PV}{RT} = \frac{0.987 \: \text{atm} \times 0.677 \: \text{L}}{0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol} \times 296 \: \text{K}} = 0.0275 \: \text{mol}\nonumber$
Now divide $\text{g}$ by $\text{mol}$ to get the molar mass.
$\text{molar mass} = \frac{1.211 \: \text{g}}{0.0275 \: \text{mol}} = 44.0 \: \text{g/mol}\nonumber$
Since $\ce{N}$ has a molar mass of $14 \: \text{g/mol}$ and $\ce{O}$ has a molar mass of $16 \: \text{g/mol}$, the formula $\ce{N_2O}$ would produce the correct molar mass.
Step 3: Think about your result.
The $R$ value that corresponds to a pressure in $\text{atm}$ was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide.
Calculating Density of a Gas
The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas $\left( \ce{NH_3} \right)$ at $0.913 \: \text{atm}$ and $20^\text{o} \text{C}$, assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be $17.04 \: \text{g/mol}$. Next, assume exactly $1 \: \text{mol}$ of ammonia $\left( n = 1 \right)$ and calculate the volume that such an amount would occupy at the given temperature and pressure.
$V = \frac{nRT}{P} = \frac{1.00 \: \text{mol} \times 0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol} \times 293 \: \text{K}}{0.913 \: \text{atm}} = 26.3 \: \text{L}\nonumber$
Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above.
$\text{Density} = \frac{17.04 \: \text{g}}{26.3 \: \text{L}} = 0.648 \: \text{g/L}\nonumber$
As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to $\frac{\left( 170.4 \: \text{g/mol} \right)}{\left( 22.4 \: \text{L/mol} \right)} = 0.761 \: \text{g/L}$. It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from $0^\text{o} \text{C}$ to $20^\text{o} \text{C}$) and the decrease in pressure (from $1 \: \text{atm}$ to $0.913 \: \text{atm}$) would cause the $\ce{NH_3}$ molecules to spread out a bit further from one another.
14.10: Gas Stoichiometry
The Haber cycle reaction of gaseous nitrogen and hydrogen to form ammonia is a critical step in the production of fertilizer from ammonia. It is important to have an excess of the starting materials so that a maximum yield of ammonia can be achieved. By knowing how much ammonia is needed for the manufacture of a batch of fertilizer, the proper amounts of nitrogen and hydrogen gases can be incorporated into the process.
Gas Stoichiometry
You have learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.
Example $1$
What volume of carbon dioxide is produced by the combustion of $25.21 \: \text{g}$ of ethanol $\left( \ce{C_2H_5OH} \right)$ at $54^\text{o} \text{C}$ and $728 \: \text{mm} \: \ce{Hg}$?
Solution
Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that in most combustion reactions, the given substance reacts with $\ce{O_2}$ to form $\ce{CO_2}$ and $\ce{H_2O}$. Here is the balanced equation for the combustion of ethanol:
$\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( l \right)\nonumber$
Known
• Mass $\ce{C_2H_5OH} = 25.21 \: \text{g}$
• Molar mass $\ce{C_2H_5OH} = 46.08 \: \text{g/mol}$
• $P = 728 \: \text{mm} \: \ce{Hg}$
• $T = 54^\text{o} \text{C} = 327 \: \text{K}$
Unknown
The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then, the ideal gas law is used to calculate the volume of $\ce{CO_2}$ produced.
Step 2: Solve.
$25.21 \: \text{g} \: \ce{C_2H_5OH} \times \frac{1 \: \text{mol} \: \ce{C_2H_5OH}}{46.08 \: \text{g} \: \ce{C_2H_5OH}} \times \frac{2 \: \text{mol} \: \ce{CO_2}}{1 \: \text{mol} \: \ce{C_2H_5OH}} = 1.094 \: \text{mol} \: \ce{CO_2}\nonumber$
The moles of carbon dioxide $\left( n \right)$ is now substituted into $PV = nRT$ to solve for the volume.
$V = \frac{nRT}{P} = \frac{1.094 \: \text{mol} \times 62.36 \: \text{L} \cdot \text{mm} \: \text{Hg}/\text{K} \cdot \text{mol} \times 327 \: \text{K}}{728 \: \text{mm} \: \ce{Hg}} = 30.6 \: \text{L}\nonumber$
Step 3: Think about your result.
The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than $22.4 \: \text{L}$. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.09%3A_Calculating_the_Molar_Mass_of_a_Gas.txt |
The behavior of a molecule depends a lot on its structure. Two compounds with the same number of atoms can act very differently. Ethanol $\left( \ce{C_2H_5OH} \right)$ is a clear liquid that has a boiling point of about $79^\text{o} \text{C}$. Dimethylether $\left( \ce{CH_3OCH_3} \right)$ has the same number of carbons, hydrogens, and oxygens, but boils at a much lower temperature $\left( -25^\text{o} \text{C} \right)$. The difference lies in the amount of intermolecular interaction (strong $\ce{H}$-bonds for ethanol, weak van der Waals force for the ether).
Real and Ideal Gases
An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas needs to completely abide by the kinetic-molecular theory. The gas particles need to occupy zero volume and they need to exhibit no attractive forces whatsoever toward each other. Since neither of those conditions can be true, there is no such thing as an ideal gas. A real gas is a gas that does not behave according to the assumptions of the kinetic-molecular theory. Fortunately, at the conditions of temperature and pressure that are normally encountered in a laboratory, real gases tend to behave very much like ideal gases.
Under what conditions then, do gases behave least ideally? When a gas is put under high pressure, its molecules are forced closer together as the empty space between the particles is diminished. A decrease in the empty space means that the assumption that the volume of the particles themselves is negligible is less valid. When a gas is cooled, the decrease in kinetic energy of the particles causes them to slow down. If the particles are moving at slower speeds, the attractive forces between them are more prominent. Another way to view it is that continued cooling of the gas will eventually turn it into a liquid and a liquid is certainly not an ideal gas anymore (see liquid nitrogen in the figure below). In summary, a real gas deviates most from an ideal gas at low temperatures and high pressures. Gases are most ideal at high temperature and low pressure.
The figure below shows a graph of $\frac{PV}{RT}$ plotted against pressure for $1 \: \text{mol}$ of a gas at three different temperatures—$200 \: \text{K}$, $500 \: \text{K}$, and 1000 \: \text{K}\). An ideal gas would have a value of 1 for that ratio at all temperatures and pressures, and the graph would simply be a horizontal line. As can be seen, deviations from an ideal gas occur. As the pressure begins to rise, the attractive forces cause the volume of the gas to be less than expected and the value of $\frac{PV}{RT}$ drops under 1. Continued pressure increase results in the volume of the particles to become significant and the value of $\frac{PV}{RT}$ rises to greater than 1. Notice that the magnitude of the deviations from ideality is greatest for the gas at $200 \: \text{K}$ and least for the gas at $1000 \: \text{K}$.
The ideality of a gas also depends on the strength and type of intermolecular attractive forces that exist between the particles. Gases whose attractive forces are weak are more ideal than those with strong attractive forces. At the same temperature and pressure, neon is more ideal than water vapor because neon's atoms are only attracted by weak dispersion forces, while water vapor's molecules are attracted by relatively strong hydrogen bonds. Helium is a more ideal gas than neon because its smaller number of electrons means that helium's dispersion forces are even weaker than those of neon.
Summary
• A real gas is a gas that does not behave according to the assumptions of the kinetic-molecular theory.
• The properties of real gases and their deviations from ideality are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.11%3A_Real_and_Ideal_Gases.txt |
Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions. Humans produce sulfur dioxide by burning coal. When in the atmosphere, the gas has a cooling effect by reflecting sunlight away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce $\ce{SO_2}$ levels to lower acid rain production. However, $\ce{SO_2}$ reduction efforts have an unforeseen complication: as we lower the concentration of this gas in the atmosphere, we lower its ability to cool, and consequently have global warming concerns.
Mole Fraction
One way to express relative amounts of substances in a mixture is with the mole fraction. Mole fraction $X$ is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances, $\ce{A}$ and $\ce{B}$, the mole fractions of each would be written as follows:
$X_A = \frac{\text{mol} \: \ce{A}}{\text{mol} \: \ce{A} + \text{mol} \: \ce{B}} \: \: \: \text{and} \: \: \: X_B = \frac{\text{mol} \: \ce{B}}{\text{mol} \: \ce{A} + \text{mol} \: \ce{B}}\nonumber$
If a mixture consists of $0.50 \: \text{mol} \: \ce{A}$ and $1.00 \: \text{mol} \: \ce{B}$, then the mole fraction of $\ce{A}$ would be $X_A = \frac{0.5}{1.5} + 0.33$. Similarly, the mole fraction of $\ce{B}$ would be $X_B = \frac{1.0}{1.5} = 0.67$. Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton's law of partial pressures. Consider the following situation... A 20.0 liter vessel contains $1.0 \: \text{mol}$ of hydrogen gas at a pressure of $600 \: \text{mm} \: \ce{Hg}$. Another 20.0 liter vessels contains $3.0 \: \text{mol}$ of helium at a pressure of $1800 \: \text{mm} \: \ce{Hg}$. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton's law, we can express the partial pressures as follows:
$P_{H_2} = X_{H_2} \times P_\text{Total} \: \: \: \text{and} \: \: \: P_{He} = X_{He} \times P_\text{Total}\nonumber$
The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:
$X_{H_2} = \frac{1.0 \: \text{mol}}{1.0 \: \text{mol} + 3.0 \: \text{mol}} = 0.25 \: \: \: \text{and} \: \: \: X_{He} = \frac{3.0 \: \text{mol}}{1.0 \: \text{mol} + 3.0 \: \text{mol}} = 0.75\nonumber$
The total pressure according to Dalton's law is $600 \: \text{mm} \: \ce{Hg} + 1800 \: \text{mm} \: \ce{Hg} = 2400 \: \text{mm} \: \ce{Hg}$. So, each partial pressure will be:
$P_{H_2} = 0.25 \times 2400 \: \text{mm} \: \ce{Hg} = 600 \: \text{mm} \: \ce{Hg}\nonumber$
$P_{He} = 0.75 \times 2400 \: \text{mm} \: \ce{Hg} = 1800 \: \text{mm} \: \ce{Hg}\nonumber$
The partial pressures of each gas in the mixture do not change, since they were mixed into the same size vessel and the temperature was not changed.
Example $1$
A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is $104 \: \text{kPa}$, what is the partial pressure of each gas?
Known
• $1.24 \: \text{mol} \: \ce{H_2}$
• $2.91 \: \text{mol} \: \ce{O_2}$
• $P_\text{Total} = 104 \: \text{kPa}$
Unknown
• $P_{H_2} = ? \: \text{kPa}$
• $P_{O_2} = ? \: \text{kPa}$
First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.
Step 2: Solve.
$\begin{array}{ll} X_{H_2} = \frac{1.24 \: \text{mol}}{1.24 \: \text{mol} + 2.91 \: \text{mol}} = 0.299 & X_{O_2} = \frac{2.91 \: \text{mol}}{1.24 \: \text{mol} + 2.91 \: \text{mol}} = 0.701 \ P_{H_2} = 0.299 \times 104 \: \text{kPa} = 31.1 \: \text{kPa} & P_{O_2} = 0.701 \times 104 \: \text{kPa} = 72.9 \: \text{kPa} \end{array}\nonumber$
Step 3: Think about your result.
The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.12%3A_Mole_Fraction.txt |
Imagine that you need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple—you don't. All you need is the atmospheric pressure in the room. As the gas pushes out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside.
Gas Collection by Water Displacement
Gases that are produced in laboratory experiments are often collected by a technique called water displacement (see figure below). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid.
Because the gas is collected over water, it is not pure, but is mixed with vapor from the evaporation of the water. Dalton's law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor:
$\begin{array}{ll} P_\text{Total} = P_g + P_{H_2O} & P_g \: \text{is the pressure of the desired gas} \ P_g = P_{Total} - P_{H_2O} & \end{array}\nonumber$
In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see table below). The sample problem illustrates the use of Dalton's law when a gas is collected over water.
Vapor Pressure of Water $\left( \text{mm} \: \ce{Hg} \right)$ at Selected Temperatures $\left( ^\text{o} \text{C} \right)$
Table $1$: Vapor Pressure of Water $\left( \text{mm} \: \ce{Hg} \right)$ at Selected Temperatures $\left( ^\text{o} \text{C} \right)$
0 5 10 15 20 25 30 35 40 45 50 55 60
4.58 6.54 9.21 12.79 17.54 23.76 31.82 42.18 55.32 71.88 92.51 118.04 149.38
Example $1$
A certain experiment generates $2.58 \: \text{L}$ of hydrogen gas, which is collected over water. The temperature is $20^\text{o} \text{C}$ and the atmospheric pressure is $98.60 \: \text{kPa}$. Find the volume that the dry hydrogen would occupy at STP.
Known
• $V_\text{Total} = 2.58 \: \text{L}$
• $T = 20^\text{o} \text{C} = 293 \: \text{K}$
• $P_\text{Total} = 98.60 \: \text{kPa} = 739.7 \: \text{mm} \: \ce{Hg}$
Unknown
The atmospheric pressure is converted from $\text{kPa}$ to $\text{mm} \: \ce{Hg}$ in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law.
Step 2: Solve.
$P_{H_2} = P_\text{Total} - P_{H_2O} = 739.7 \: \text{mm} \: \ce{Hg} - 17.54 \: \text{mm} \: \ce{Hg} = 722.2 \: \text{mm} \: \ce{Hg}\nonumber$
Now the combined gas law is used, solving for $V_2$, the volume of hydrogen at STP.
$V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1} = \frac{722.2 \: \text{mm} \: \ce{Hg} \times 2.58 \: \text{L} \times 273 \: \text{K}}{760 \: \text{mm} \: \ce{Hg} \times 293 \: \text{K}} = 2.28 \: \text{L} \: \ce{H_2}\nonumber$
Step 3: Think about your result.
If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be $2.28 \: \text{L}$. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes.
Summary
• Gases that are produced in laboratory experiments are often collected by a technique called water displacement.
• The vapor pressure due to water in a sample can be corrected for in order to get the true value of the pressure of the gas.
14.14: Dalton's Law of Partial Pressures
The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are $96.5\%$ carbon dioxide and $3\%$ nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus contributes pressure well over $2700 \: \text{mm} \: \ce{Hg}$. And there is no oxygen present, so humans could not breathe there. Not that anyone would want to go to Venus—the surface temperature is usually over $460^\text{o} \text{C}$.
Dalton's Law of Partial Pressures
Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about $78\%$ nitrogen and $21\%$ oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up $78\%$ of the gas particles in a given sample of air, it exerts $78\%$ of the pressure. If the overall atmospheric pressure is $1.00 \: \text{atm}$, then the pressure of just the nitrogen in the air is $0.78 \: \text{atm}$. The pressure of the oxygen in the air is $0.21 \: \text{atm}$.
The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of nitrogen is represented by $P_{N_2}$. Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton's law can be expressed with the following equation:
$P_\text{total} = P_1 + P_2 + P_3 + \cdots\nonumber$
The figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure, $P_1$ and $P_2$, reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If $P_1 = 300 \: \text{mm} \: \ce{Hg}$ and $P_2 = 500 \: \text{mm} \: \ce{Hg}$, then $P_\text{total} = 800 \: \text{mm} \: \ce{Hg}$.
Summary
• The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture.
• The total pressure in a system is equal to the sums of the partial pressures of the gases present, according to Dalton's Law of Partial Pressures. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.13%3A_Gas_Collection_by_Water_Displacement.txt |
We usually cannot see gases, so we need methods to detect their movements indirectly. The relative rates of diffusion of ammonia to hydrogen chloride can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride $\left( \ce{NH_4Cl} \right)$, a white solid that appears in the glass tube as a ring.
Graham's Law
When a person opens a bottle of perfume in one corner of a large room, it doesn't take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. Diffusion is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly. Solids essentially do not diffuse.
A related process to diffusion is effusion. Effusion is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass.
Scottish chemist Thomas Graham (1805-1869) studied the rates of effusion and diffusion of gases. Graham's law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham's law can be understood by comparing two gases ($\ce{A}$ and $\ce{B}$) at the same temperature, meaning the gases have the same kinetic energy. The kinetic energy of a moving object is given by the equation $KE = \frac{1}{2} mv^2$ where $m$ is mass and $v$ is velocity. Setting the kinetic energies of the two gases equal to one another gives:
$\frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2\nonumber$
The equation can be rearranged to solve for the ratio of the velocity of gas $\ce{A}$ to the velocity of gas $\ce{B}$ $\left( \frac{v_A}{v_B} \right)$.
$\frac{v_A^2}{v_B^2} = \frac{m_B}{m_A} \: \: \: \text{which becomes} \: \: \: \frac{v_A}{v_B} = \sqrt{\frac{m_B}{m_A}}\nonumber$
For the purposes of comparing the rates of effusion or diffusion of two gases at the same temperature, the molar masses of each gas can be used in the equation for $m$.
Example $1$
Calculate the ratio of diffusion rates of ammonia gas $\left( \ce{NH_3} \right)$ to hydrogen chloride $\left( \ce{HCl} \right)$ at the same temperature and pressure.
Solution:
Known
• Molar mass $\ce{NH_3} = 17.04 \: \text{g/mol}$
• Molar mass $\ce{HCl} = 36.46 \: \text{g/mol}$
Unknown
Substitute the molar masses of the gases into Graham's law and solve for the ratio.
Step 2: Solve.
$\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{36.46 \: \text{g/mol}}{17.04 \: \text{g/mol}}} = 1.46\nonumber$
The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride.
Step 3: Think about your result.
Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1.
Summary
• Diffusion is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform.
• Effusion is the process of a confined gas escaping through a tiny hole in its container.
• Graham's law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.15%3A_Diffusion_and_Effusion_and_Graham%27s_Law.txt |
In his well-known poem "The Rime of the Ancient Mariner", Samuel Coleridge wrote "water, water everywhere, nor any drop to drink." Coleridge was talking about being out on the ocean, but not having any water because he had killed an albatross (apparently bringing bad luck to everyone on the ship). About $75\%$ of the Earth's surface is water. The major constituent of the human body (over $60\%$) is water. This simple molecule plays important roles in all kinds of processes.
Structure of Water
Water is a simple molecule consisting of one oxygen atom bonded to two different hydrogen atoms. Because of the higher electronegativity of the oxygen atom, the bonds are polar covalent (polar bonds). The oxygen atom attracts the shared electrons of the covalent bonds to a significantly greater extent than the hydrogen atoms. As a result, the oxygen atom acquires a partial negative charge $\left( \delta - \right)$, while the hydrogen atoms each acquire a partial positive charge $\left( \delta + \right)$. The molecule adopts a bent structure because of the two lone pairs of electrons on the oxygen atom. The $\ce{H-O-H}$ bond angle is about $105^\text{o}$, slightly smaller than the ideal $109.5^\text{o}$ of an $sp^3$ hybridized atomic orbital.
The bent shape of the water molecule is critical because the polar $\ce{O-H}$ bonds do not cancel one another and the molecule as a whole is polar. The figure below illustrates the net polarity of the water molecule. The oxygen is the negative end of the molecule, while the area between the hydrogen atoms is the positive end of the molecule.
Polar molecules attract one another by dipole-dipole forces, as the positive end of one molecule is attracted to the negative end of the nearby molecule. In the case of water, the highly polar $\ce{O-H}$ bonds results in very little electron density around the hydrogen atoms. Each hydrogen atom is strongly attracted to the lone-pair electrons on an adjacent oxygen atom. These are called hydrogen bonds and are stronger than conventional dipole-dipole forces.
Because each oxygen atom has two lone pairs, it can make hydrogen bonds to the hydrogen atoms of two separate other molecules. The figure below shows the result—an approximately tetrahedral geometry around each oxygen atom, consisting of two covalent bonds and two hydrogen bonds.
Summary
• Water is a molecular compound consisting of polar molecules that have a bent shape.
• The oxygen atom acquires a partial negative charge, while the hydrogen atom acquires a partial positive charge.
15.02: Structure of Ice
Ice is an interesting and useful material. It can be used to cool food and keep it fresh. Ice can provide recreation, such as in the case of ice-skating. It can do great damage when it freezes—roads can buckle, houses can be damaged, water pipes can burst. All of this happens because of a unique property of water: when water freezes, it expands in volume as ice is formed.
Structure of Ice
Liquid water is a fluid. The hydrogen bonds in liquid water constantly break and reform as the water molecules tumble past one another. As water cools, its molecular motion slows and the molecules move gradually closer to one another. The density of any liquid increases as its temperature decreases. For most liquids, this continues as the liquid freezes; the solid state is denser than the liquid state. However, water behaves differently. It actually reaches its highest density at about $4^\text{o} \text{C}$.
Temperature $\left( ^\text{o} \text{C} \right)$ Density $\left( \text{g/cm}^3 \right)$
Table $1$: Density of Water and Ice
100 (liquid) 0.9584
50 0.9881
25 0.9971
10 0.9997
4 1.0000
0 (liquid) 0.9998
0 (solid) 0.9168
Between $4^\text{o} \text{C}$ and $0^\text{o} \text{C}$, the density gradually decreases as the hydrogen bonds begin to form a network characterized by a generally hexagonal structure with open spaces in the middle of the hexagons (see figure below).
Ice is less dense than liquid water and so it floats. Ponds or lakes begin to freeze at the surface, closer to the cold air. A layer of ice forms, but does not sink as it would if water did not have this unique structure dictated by its shape, polarity, and hydrogen bonding. If the ice were to sink as it froze, entire lakes would freeze solid. Since the ice does not sink, liquid water remains under the ice all winter long. This is important, as fish and other organisms are capable of surviving through winter. Ice is one of few solids that is less dense than its liquid form.
Summary
• The density of any liquid increases as its temperature decreases. As an exception, ice is less dense than liquid water.
• The intermolecular structure of ice has spaces that are not present in liquid water.
15.03: Physical Properties of Water
Water loss to the atmosphere is a significant problem in many parts of the world. When water supplies are low, anything that can be done to decrease water loss is important for farmers. An evaporation pan can be used to measure how fast water evaporates in a given location. This information can be used as part of project development to cut down on evaporation, and increase the amount of usable water in a region.
Properties of Water
Compared to other molecular compounds of relatively low molar mass, ice melts at a very high temperature. A great deal of energy is required to break apart the hydrogen-bonded network of ice and return it to the liquid state. Likewise, the boiling point of water is very high. Most molecular compounds of similar molar mass are gases at room temperature.
Surface Tension
Water has a high surface tension (attraction between molecules at the surface of a liquid) because of its hydrogen bonding. Liquids that cannot hydrogen bond do not exhibit nearly as much surface tension. Surface tension can be seen by the curved meniscus that forms when water is in a thin column, such as in a graduated cylinder or a buret.
Vapor Pressure
The vapor pressure of a liquid is the pressure of the vapor produced by evaporation of a liquid or solid above the liquid or solid in a closed container. The hydrogen bonding between liquid water molecules explains why water has an unusually low vapor pressure. Relatively few molecules of water are capable of escaping the surface of the liquid and entering the vapor phase. Evaporation is slow, and thus the vapor exerts a low pressure in a closed container. Low vapor pressure is an important physical property of water, since lakes, oceans, and other large bodies of water would otherwise evaporate much more quickly.
Vapor pressure is influenced by temperature. As the temperature increases, more molecules are released from the surface of the liquid. This increases movement above the liquid surface, increasing the pressure in the vapor stage. The image below illustrates the effect of temperature on vapor pressure.
Summary
• Water has high surface tension because of extensive hydrogen bonding.
• The vapor pressure of water is low due to hydrogen bonding.
• Vapor pressure increases as temperature increases. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/15%3A_Water/15.01%3A_Structure_of_Water.txt |
In the winter, the temperature often gets well below the freezing point of water. This condition can create problems in car radiators. If the water freezes, water hoses will break, the engine block can crack, and significant damage can be done to the car.
Solute and Solvent
When one substance dissolves into another, a solution is formed. A solution is a homogeneous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with many different types and forms of solutes and solvents.
We know of many types of solutions. Check out a few examples in the table below.
Type Solvent Solute Example
Table $1$: Types of Solutions
gas/gas nitrogen oxygen air
gas/liquid water carbon dioxide soda pop
liquid/liquid water ethylene glycol antifreeze
solid/liquid water salts seawater
This section's focus is on solutions where the solvent is water. An aqueous solution is water that contains one or more dissolved substances. The dissolved substances in an aqueous solution many be solids, gases, or other liquids. Some examples are listed in the table above. Other examples include vinegar (acetic acid in water), alcoholic beverages (ethanol in water), and liquid cough medicines (various drugs in water).
In order to be a true solution, a mixture must be stable. When sugar is fully dissolved into water, it can stand for an indefinite amount of time and the sugar will not settle out of the solution. Further, if the sugar-water solution is passed through a filter, it will be unchanged. The dissolved sugar particles will pass through the filter along with the water. This is because the dissolved particles in a solution are very small, usually less than $1 \: \text{nm}$ in diameter. Solute particles can be atoms, ions, or molecules, depending on the type of substance that has been dissolved.
Summary
• A solution is a homogeneous mixture of a solute in a solvent.
• A solute is the material present in the smaller amount in the solution.
• A solvent is the material present in the larger amount in the solution.
15.05: Dissolving Process
Many people start their day with a cup of coffee (others drink coffee all day long). The coffee typically consumed is a type of solution—sometimes a very complex one. The coffee itself has been brewed so that material from the coffee bean will dissolve in hot water. Some people add sugar, while others add milk or cream. In some areas of the country, lattés are popular; one or more special flavors might be added along with the milk. Whatever the situation, the end result is an enjoyable solution to drink.
The Process of Dissolution
Water typically dissolves many ionic compounds and polar molecules. Nonpolar molecules such as those found in grease or oil do not dissolve in water. We will first examine the process that occurs when an ionic compound such as table salt (sodium chloride) dissolves in water.
Water molecules move about continuously due to their kinetic energy. When a crystal of sodium chloride is placed into water, the water's molecules collide with the crystal lattice. Recall that the crystal lattice is composed of alternating positive and negative ions. Water is attracted to the sodium chloride crystal because water is polar and has both a positive and a negative end. The positively charged sodium ions in the crystal attract the oxygen end of the water molecules because they are partially negative. The negatively charged chloride ions in the crystal attract the hydrogen end of the water molecules because they are partially positive. The action of the polar water molecules takes the crystal lattice apart (see image below).
After coming apart from the crystal, the individual ions are then surrounded by solvent particles in a process called solvation. Note that the individual $\ce{Na^+}$ ions are surrounded by water molecules with the oxygen atom oriented near the positive ion. Likewise, the chloride ions are surrounded by water molecules with the opposite orientation. Hydration is the process of solute particles being surrounded by water molecules arranged in a specific manner. Hydration helps to stabilize aqueous solutions by preventing the positive and negative ions from coming back together and forming a precipitate.
Table sugar is sucrose $\left( \ce{C_{12}H_{22}O_{11}} \right)$, and is an example of a molecular compound. Solid sugar consists of individual sugar molecules held together by intermolecular attractive forces. When water dissolves sugar, it separates the individual sugar molecules by disrupting the attractive forces, but does not break the covalent bonds between the carbon, hydrogen, and oxygen atoms. Dissolved sugar molecules are also hydrated, but without as distinct an orientation to the water molecules as in the case of the ions. The sugar molecules contain many $\ce{-OH}$ groups that can form hydrogen bonds with the water molecules, helping to form the sucrose solution.
Summary
• Motion of water molecules helps break up interactions between solid ions or molecules.
• Solvation involves surrounding ions with solvent particles.
• Ionic solute molecules are hydrated (surrounded by water molecules in a specific orientation). | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/15%3A_Water/15.04%3A_Solute_and_Solvent.txt |
In 2010, a major oil spill occurred when an explosion on a drilling rig in the Gulf of Mexico released millions of gallons of crude oil into the Gulf. Oil is primarily a mixture of hydrocarbons (organic compounds composed of only carbon and hydrogen atoms). Because of its composition, oil does not dissolve in water. As a result, much of the Gulf of Mexico was contaminated, as was a great deal of shoreline in the affected area.
Liquid-Liquid Solutions
Nonpolar compounds do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. However, the nonpolar molecules are more attracted to themselves than they are to the polar water molecules. When a nonpolar liquid such as oil is mixed with water, two separate layers form because the liquids will not dissolve into each other (figure below). When another polar liquid such as ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are deemed immiscible. The general rule for deciding if one substance is capable of dissolving another is "like dissolves like". A nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but will not dissolve in polar water.
For molecular compounds, the major factor that contributes to the material dissolving in water is the ability to form hydrogen bonds with the water solvent. Small compounds such as methanol, ethanol, acetic acid, and acetone have polar groups that can interact with the polar \(\ce{H}\) of water. However, as the nonpolar portion of the molecule gets larger, solubility with water drops off. The nonpolar portion of the molecule increasingly repels the water and eventually overrides the interaction of the polar component with water.
Summary
• Liquids that dissolve in one another in all proportions are miscible; liquids that do not dissolve in one another are immiscible.
• The general rule for deciding if one substance is capable of dissolving another is "like dissolves like".
• Nonpolar molecules are usually insoluble in water.
• A non-ionized molecule must be relatively polar to interact with water molecules.
15.07: Electrolytes and Nonelectrolytes
People around the world jog for exercise. For the most part, jogging can be a healthy way to stay fit. However, problems can develop for those who jog in the heat. Excessive sweating can lead to electrolyte loss, which can be life-threatening. Early symptoms of electrolyte deficiency can include nausea, fatigue, and dizziness. If not treated, individuals can experience muscle weakness and increased heart rate (which could lead to a heart attack). Sports drinks can be consumed to restore electrolytes quickly in the body.
Electrolytes and Nonelectrolytes
An electrolyte is a compound that conducts an electric current when it is in an aqueous solution or melted. In order to conduct a current, a substance must contain mobile ions that can move from one electrode to the other. All ionic compounds are electrolytes. When ionic compounds dissolve, they break apart into ions which are then able to conduct a current (conductivity). Even insoluble ionic compounds such as \(\ce{CaCO_3}\) are electrolytes because they can conduct a current in the molten (melted) state.
A nonelectrolyte is a compound that does not conduct an electric current in either aqueous solution or in the molten state. Many molecular compounds, such as sugar or ethanol, are nonelectrolytes. When these compounds dissolve in water, they do not produce ions. The figure below illustrates the difference between an electrolyte and a nonelectrolyte.
Roles of Electrolytes in the Body
Several electrolytes play important roles in the body. Here are a few significant electrolytes:
1. Calcium - in bones and teeth. Also important for muscle contraction, blood clotting, and nerve function.
2. Sodium - found outside the cell. Mainly involved in water balance and nerve signaling.
3. Potassium - major cation inside the cell. Important for proper functioning of heart, muscles, kidneys, and nerves.
4. Magnesium - in bone and cells. Involved in muscle, bone, nervous system, and takes part in many biochemical reactions.
Summary
• Electrolytes conduct electric current when in solution or melted.
• Nonelectrolytes do not conduct electric current when in solution or melted.
• Some electrolytes play important roles in the body.
15.08: Dissociation
In many areas, winter ice on the streets and sidewalks represents a serious walking and driving hazard. One common approach to melting the ice is to put some form of deicing salt on the surface. Materials such as sodium chloride or calcium chloride are frequently employed for this purpose. In order to be effective, the solid material must first dissolve and break up into the ions that make up the compound.
Dissociation
An ionic crystal lattice breaks apart when it is dissolved in water. Dissociation is the separation of ions that occurs when a solid ionic compound dissolves. It is important to be able to write dissociation equations. Simply undo the crisscross method that you learned when writing chemical formulas of ionic compounds. The subscripts for the ions in the chemical formulas become the coefficients of the respective ions on the product side of the equation. Shown below are dissociation equations for $\ce{NaCl}$, $\ce{Ca(NO_3)_2}$, and $\ce{(NH_4)_3PO_4}$.
$\ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
$\ce{Ca(NO_3)_2} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right)\nonumber$
$\ce{(NH_4)_3PO_4} \left( s \right) \rightarrow 3 \ce{NH_4} \left( aq \right) + \ce{PO_4} \left( aq \right)\nonumber$
The formula unit of sodium chloride dissociates into one sodium ion and one chloride ion. The calcium nitrate formula unit dissociates into one calcium ion and two nitrate ions. This is because of the $2+$ charge of the calcium ion. Two nitrate ions, each with a $1-$ charge are required to make the equation balance electrically. The ammonium phosphate formula unit dissociates into three ammonium ions and one phosphate ion. Note that the polyatomic ions themselves do not dissociate further, but remain intact.
Do not confuse the subscripts of the atoms within the polyatomic ion for the subscripts that result from the crisscrossing of the charges that make up the original compound neutral. The 3 subscript of the nitrate ion and the 4 subscript of the ammonium ion are part of the polyatomic ion and simply remain as part of its formula after the compound dissociates. Notice that the compounds are solids $\left( s \right)$ which then become ions in aqueous solution $\left( aq \right)$.
Nonelectrolytes do not dissociate when forming an aqueous solution. An equation can still be written that simply shows the solid going into solution. For the dissolution of sucrose:
$\ce{C_{12}H_{22}O_{11}} \left( s \right) \rightarrow \ce{C_{12}H_{22}O_{11}} \left( aq \right)\nonumber$
Summary
• Dissociation is the separation of ions that occurs when a solid ionic compound dissolves.
• Nonionic compounds do not dissociate in water. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/15%3A_Water/15.06%3A_Liquid-Liquid_Solutions.txt |
Car batteries are used around the world to provide the power to start car engines. One essential component of car batteries is the strong electrolyte sulfuric acid. In the battery, this material ionizes into hydrogen ions and sulfate ions. As the battery is used, the concentrations of these ions decreases. Older batteries had openings in the top where new sulfuric acid could be added to replenish the supply. Today, batteries are sealed to prevent leakage of hazardous sulfuric acid.
Strong and Weak Electrolytes
Some polar molecular compounds are nonelectrolytes when they are in their pure state, but become electrolytes when they are dissolved in water. Hydrogen chloride $\left( \ce{HCl} \right)$ is a gas in its pure molecular state and is a nonelectrolyte. However, when $\ce{HCl}$ is dissolved in water, it conducts a current well because the $\ce{HCl}$ molecule ionizes into hydrogen and chloride ions.
$\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
$\ce{HCl}$ dissolved into water is called hydrochloric acid. A strong electrolyte is a solution in which a large fraction of the dissolved solute exists as ions. Ionic compounds, and some polar compounds, are completely broken apart into ions and thus conduct a current very well—this makes them strong electrolytes.
Some other polar molecular compounds become electrolytes upon being dissolved into water, but do not ionize to very great extent. Gaseous nitrous acid ionizes in solution to hydrogen ions and nitrite ions, but does so very weakly. Aqueous nitrous acid is composed of only about $5\%$ ions and $95\%$ intact nitrous acid molecules. A weak electrolyte is a solution in which only a small fraction of the dissolved solute exists as ions. The equation showing the ionization of a weak electrolyte utilizes a double arrow indicating an equilibrium between the reactants and products.
$\ce{HNO_2} \left( g \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)\nonumber$
Summary
• A strong electrolyte exists mainly as ions in solution.
• A weak electrolyte has only a small amount of ionization in solution.
15.10: Suspensions
On a calm day, the Caribbean Sea is clear and smooth. You can see deep into the water—deep enough to see fish and underwater plant life without obstruction. The situation changes markedly when a storm blows up. The sand on the bottom of the sea is stirred up, the water becomes turbid, and you can't see anything below the surface. The sand forms a suspension in the water that slowly clears up again after the storm blows over.
Suspensions
Take a glass of water and throw in a handful of sand or dirt. Stir it and stir it and stir it. The water may become turbid, or unclear. Have you made a solution? Sand and dirt do not dissolve in water and though it may look homogenous for a few moments, the sand or dirt gradually sinks to the bottom of the glass.
A suspension is a heterogeneous mixture in which some of the particles settle out of the mixture upon standing. The particles in a suspension are far larger than those of a solution and thus gravity is able to pull them down out of the dispersion medium (water). The typical diameter for the dispersed particles (the sand) of a suspension is about 1000 times greater than those of a solution (less than approximately two nanometers for particles in solution, compared to greater than 1000 nanometers for particles in suspension). Unlike in a solution, the dispersed particles can be separated from the dispersion medium by filtering. Suspensions are heterogeneous because at least two different substances in the mixture can be identified.
Some Typical Aqueous Suspensions
Milk
Milk is a complex mixture of water, proteins, fats, carbohydrates, and minerals. While the minerals and carbohydrates are water-soluble, the fats and some of the proteins do not dissolve, but are held in suspension.
Paint
Paint can either be water-based or use an organic solvent. We will limit this discussion to water-based paints. The materials will contain binders (organic polymers that help hold the paint to the surface of the material being painted). In addition, pigments are included to provide the desired color. Both the binders and the pigments are not water-soluble, so the paint must be stirred every time it is used. Paint that is allowed to sit for a long period of time will slowly begin to separate as the suspension of binders and pigments settles out.
Summary
• Suspensions are heterogeneous mixtures.
• Some of the material in a suspension will settle out on standing.
• Solid material in a suspension can be removed by filtration. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/15%3A_Water/15.09%3A_Strong_and_Weak_Electrolytes.txt |
Imagine that you are sailing on a yacht. The engine suddenly breaks down and you are stranded in the middle of the ocean. You call the Coast Guard on your radio, but cannot give them an exact location, because your GPS isn't working. Fortunately, you have a smoke flare, which you fire off. The dense colored smoke shows the Coast Guard where you are so that they can rescue you. In using the flare, you are taking advantage of a mixture called a colloid.
Colloids
A colloid is a heterogeneous mixture whose particle size is intermediate between those of a solution and a suspension. The dispersed particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas. Because the dispersed particles of a colloid are not as large as those of a suspension, they do not settle out upon standing. The table below summarizes the properties and distinctions between solutions, colloids, and suspensions.
Properties of Solutions, Colloids, and Suspensions
Table $1$: Properties of Solutions, Colloids, and Suspensions
Solutions Colloids Suspensions
Homogeneous. Heterogeneous. Heterogeneous.
Particle size: $0.01$-$1 \: \text{nm}$; atoms, ions, or molecules. Particle size: $1$-$1000 \: \text{nm}$, dispersed; large molecules or aggregates. Particle size: over $1000 \: \text{nm}$, suspended; large particles or aggregates.
Do not separate on standing. Do not separate on standing. Particles settle out.
Cannot be separated by filtration. Cannot be separated by filtration. Can be separated by filtration.
Do not scatter light. Scatter light (Tyndall effect). May either scatter light or be opaque.
Colloids are unlike solutions because their dispersed particles are much larger than those of a solution. The dispersed particles of a colloid cannot be separated by filtration, but they scatter light—a phenomenon called the Tyndall effect.
The Tyndall Effect
When light is passed through a true solution, the dissolved particles are too small to deflect the light. However, the dispersed particles of a colloid, being larger, do deflect light. The Tyndall effect is the scattering of visible light by colloidal particles. You have undoubtedly "seen" a light beam as it passes through fog, smoke, or a scattering of dust particles suspended in air. All three are examples of colloids. Suspensions may scatter light, but if the number of suspended particles is sufficiently large, the suspension may simply be opaque, and the light scattering will not occur.
Examples of Colloids
The table below lists examples of colloidal systems, most of which are very familiar. The dispersed phase describes the particles, while the dispersion medium is the material in which the particles are distributed.
Classes of Colloids
Table $2$: Classes of Colloids
Class of Colloid Dispersed Phase Dispersion Medium Examples
Solid gel solid liquid paint, jellies, blood, gelatin, mud
Solid aerosol solid gas smoke, dust in air
Solid emulsion liquid solid cheese, butter
Liquid emulsion liquid liquid milk, mayonnaise
Liquid aerosol liquid gas fog, mist, clouds, aerosol spray
Foam gas solid marshmallow
Foam gas liquid whipped cream, shaving cream
Another property of a colloidal system is observed when the colloids are studied under a light microscope. The colloids scintillate, reflecting brief flashes of light because the colloidal particles move in a rapid and random fashion. This phenomenon, called Brownian motion, is caused by collisions between the small colloidal particles and the molecules of the dispersion medium.
Emulsions
Butter and mayonnaise are examples of a class of colloids called emulsions. An emulsion is a colloidal dispersion of a liquid in either a liquid or a solid. A stable emulsion requires an emulsifying agent to be present. Mayonnaise is made in part of oil and vinegar. Since oil is nonpolar and vinegar is an aqueous solution and polar, the two do not mix, and quickly separate into layers. However, the addition of egg yolk causes the mixture to become stable and not separate. Egg yolk is capable of interacting with both the polar vinegar and the nonpolar oil. The egg yolk is called the emulsifying agent. Soap acts as an emulsifying agent between grease and water. Grease cannot be simply rinsed off your hands, or another surface, because it is insoluble. However, the soap stabilizes a grease-water mixture because one end of a soap molecule is polar, and the other end is nonpolar. This allows the grease to be removed from your hands or your clothing by washing with soapy water.
Summary
• A colloid is a heterogeneous mixture whose particle size is intermediate between those of a solution and a suspension.
• The Tyndall effect is the scattering of visible light by colloidal particles.
• An emulsion is a colloidal dispersion of a liquid in either a liquid or a solid. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/15%3A_Water/15.11%3A_Colloids.txt |
Solute-Solvent Combinations
The focus of "Chapter 15: Water" was water's role in the formation of aqueous solutions. We examined the primary characteristics of a solution and how water is able to dissolve solid solutes; we differentiated between a solution, a suspension, and a colloid. There are many examples of solutions that do not involve water at all, or that involve solutes that are not solids. The table below summarizes the possible combinations of solute-solvent states, along with examples of each.
Table $1$: Solute-Solvent Combinations
Solute State Solvent State Example
liquid gas water in air
gas gas oxygen in nitrogen (gas mixture)
solid liquid salt in water
liquid liquid alcohol in water
gas liquid carbon dioxide in water
solid solid zinc in copper (brass alloy)
liquid solid mercury in silver and tin (dental amalgam)
Gas-Gas Solutions
Our air is a homogenous mixture of many different gases and therefore qualifies as a solution. Approximately $78\%$ of the atmosphere is nitrogen, making it the solvent for this solution. The next major constituent is oxygen (about $21\%$), followed by the inert gas argon $\left( 0.9\% \right)$, carbon dioxide $\left( 0.03\% \right)$, and trace amounts of neon, methane, helium, and other gases.
Solid-Solid Solutions
Solid-solid solutions such as brass, bronze, and sterling silver are called alloys. Bronze (composed mainly of copper with added tin) was widely used in making weapons in times past, dating back to at least 2400 B.C. This metal alloy was hard and tough, but was eventually replaced by iron.
Liquid-Liquid Solutions
Perhaps the most familiar liquid-solid solution is dental amalgam, used to fill teeth when there is a cavity. Approximately $50\%$ of the amalgam material is liquid mercury to which a powdered alloy of silver, tin, and copper is added. Mercury is used because it binds well with the solid metal alloy. However, the use of mercury-based dental amalgam has gone under question in recent years, because of concerns regarding the toxicity of mercury.
16.02: Rate of Dissolving
Many people enjoy a cold glass of iced tea on a hot summer day. Some like it unsweetened, while others like to put sugar in it. How sugar dissolves in the tea depends on two factors: how much sugar was put into the tea, and how cold it is. Tea usually has to be stirred for a while to get all the sugar dissolved.
Rate of Dissolution
Dissolution is the process by which a solute dissolves into a solvent and forms a solution. We know that the dissolution of a solid by water depends upon the collisions that occur between the solvent molecules and the particles in the solid crystal. Anything that can be done to increase the frequency of those collisions and/or to give those collisions more energy will increase the rate of dissolution. Imagine that you were trying to dissolve some sugar in a glassful of tea. A packet of granulated sugar would dissolve faster than a cube of sugar. The rate of dissolution would be increased by stirring, or agitating the solution. Finally, the sugar would dissolve faster in hot tea than it would in cold tea.
Surface Area
The rate at which a solute dissolves depends upon the size of the solute particles. Dissolution is a surface phenomenon, since it depends on solvent molecules colliding with the outer surface of the solute. A given quantity of solute dissolves faster when it is ground into small particles, rather than in the form of large pieces, because more surface area is exposed. A packet of granulated sugar exposes far more surface area to the solvent and dissolves more quickly than a sugar cube.
Agitation of the Solution
Dissolving sugar in water will occur more quickly if the water is stirred. The stirring allows fresh solvent molecules to continually be in contact with the solute. If it is not stirred, then the water right at the surface of the solute becomes saturated with dissolved sugar molecules, meaning that it is more difficult for additional solute to dissolve. The sugar cube would eventually dissolve because random motions of the water molecules would bring enough fresh solvent into contact with the sugar, but the process would take much longer. It is important to realize that neither stirring nor breaking up a solute affect the overall amount of solute that dissolves—these actions only affect the rate of dissolution.
Temperature
Heating up a solvent gives the molecules more kinetic energy. The increased rapid motion means that the solvent molecules collide with the solute with greater frequency, and that the collisions occur with more force. Both factors increase the rate at which the solute dissolves. As we will see in the next section, a temperature change not only affects the rate of dissolution, but also affects the amount of solute that dissolves. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.01%3A_Solute-Solvent_Combinations.txt |
When compounds are synthesized, contaminating materials are often mixed in with them. The process of recrystallization can be used to remove these impurities. The crystals are dissolved in a hot solvent, forming a solution. When the solvent is cooled, the compound is no longer as soluble and will precipitate out of solution, leaving contaminating materials still dissolved.
Saturated and Unsaturated Solutions
Table salt $\left( \ce{NaCl} \right)$ readily dissolves in water. Suppose that you have a beaker of water to which you add some salt, stirring until it dissolves. You add more and that dissolves. You keep adding more and more salt, eventually reaching a point at which no more of the salt will dissolve—no matter how long or how vigorously you stir it. Why? On the molecular level, we know that action of the water causes the individual ions to break apart from the salt crystal and enter the solution, where they remain hydrated by water molecules. What also happens is that some of the dissolved ions collide back again with the crystal and remain there. Recrystallization is the process of dissolved solute returning to the solid state. At some point, the rate at which the solid salt is dissolving becomes equal to the rate at which the dissolved solute is recrystallizing. When that point is reached, the total amount of dissolved salt remains unchanged. Solution equilibrium is the physical state described by the opposing processes of dissolution and recrystallization occurring at the same rate. The solution equilibrium for the dissolution of sodium chloride can be represented by one of two equations:
$\ce{NaCl} \left( s \right) \rightleftharpoons \ce{NaCl} \left( aq \right)\nonumber$
While the above equation shows the change of state back and forth between solid and aqueous solution, the preferred equation also shows the dissociation that occurs as an ionic solid dissolves:
$\ce{NaCl} \left( s \right) \rightleftharpoons \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
When the solution equilibrium point is reached and no more solute will dissolve, the solution is said to be saturated. A saturated solution is a solution that contains the maximum amount of solute that is capable of dissolving. At $20^\text{o} \text{C}$, the maximum amount of $\ce{NaCl}$ that will dissolve in \100. \: \text{g}\) of water is $36.0 \: \text{g}$. If any more $\ce{NaCl}$ is added past this point, it will not dissolve, because the solution is saturated. What if more water is added to the solution instead? In that case, more $\ce{NaCl}$ would be capable of dissolving in the additional solvent. An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved. The figure below illustrates the above process and shows the distinction between unsaturated and saturated.
How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium, but which has extra undissolved solute at the bottom of the container, must be saturated.
Summary
• Recrystallization is the process of dissolved solute returning to the solid state.
• A saturated solution is a solution that contains the maximum amount of solute that is capable of dissolving.
• An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved.
• Solution equilibrium exists when the rate of dissolution equals the rate of recrystallization. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.03%3A_Saturated_and_Unsaturated_Solutions.txt |
Nuclear power plants require large amounts of water to generate steam for turbines and to cool equipment. They are usually situated near bodies of water to use that water as a coolant, returning the warmer water back to the lake or river. This increases the overall temperature of the water, which lowers the quantity of dissolved oxygen, affecting the survival of fish and other organisms.
How Temperature Influences Solubility
The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured as the grams of solute per $100 \: \text{g}$ of solvent. The solubility of sodium chloride in water is $36.0 \: \text{g}$ per $100 \: \text{g}$ water at $20^\text{o} \text{C}$. The temperature must be specified because solubility varies with temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. In this section, we will consider solubility of material in water as solvent.
The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve, a graph of the solubility vs. temperature (see figure below).
Notice how the temperature dependence of $\ce{NaCl}$ is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of $\ce{NaCl}$. The curve for $\ce{KNO_3}$, on the other hand, is very steep, and so an increase in temperature dramatically increases the solubility of $\ce{KNO_3}$.
Several substances—$\ce{HCl}$, $\ce{NH_3}$, and $\ce{SO_2}$—have solubility that decreases as temperature increases. They are all gases at standard pressure. When a solvent with a gas dissolved in it is heated, the kinetic energy of both the solvent and solute increase. As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases.
Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that $80 \: \text{g}$ of $\ce{KNO_3}$ is added to $100 \: \text{g}$ of water at $30^\text{o} \text{C}$. According to the solubility curve, approximately $48 \: \text{g}$ of $\ce{KNO_3}$ will dissolve at $30^\text{o} \text{C}$. This means that the solution will be saturated since $48 \: \text{g}$ is less than $80 \: \text{g}$. We can also determine that there will be $80 - 48 = 32 \: \text{g}$ of undissolved $\ce{KNO_3}$ remaining at the bottom of the container. In a second scenario, suppose that this saturated solution is heated to $60^\text{o} \text{C}$. According to the curve, the solubility of $\ce{KNO_3}$ at $60^\text{o} \text{C}$ is about $107 \: \text{g}$. The solution, in this case, is unsaturated since it contains only the original $80 \: \text{g}$ of dissolved solute. Suppose in a third case, that the solution is cooled all the way down to $0^\text{o} \text{C}$. The solubility at $0^\text{o} \text{C}$ is about $14 \: \text{g}$, meaning that $80 - 14 = 66 \: \text{g}$ of the $\ce{KNO_3}$ will recrystallize.
Summary
• The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature.
• A solubility curve is a graph of the solubility vs. temperature
• The solubility of a solid in water increases with an increase in temperature.
• Gas solubility decreases as the temperature increases. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.04%3A_How_Temperature_Influences_Solubility.txt |
In the picture below, a thermal pack is applied to the back. Small packs can be used either for heating or cooling, depending on the material used. A heat pack contains a supersaturated solution of material such as sodium acetate. The solution is clear until a small metal trigger is activated. The sodium acetate then crystallizes out of solution and generates heat in the process.
Supersaturated Solutions
Some solutes, such as sodium acetate, do not recrystallize easily. Suppose an exactly saturated solution of sodium acetate is prepared at $50^\text{o} \text{C}$. As it cools back to room temperature, no crystals appear in the solution, even though the solubility of sodium acetate is lower at room temperature. A supersaturated solution is a solution that contains more than the maximum amount of solute that is capable of being dissolved at a given temperature. The recrystallization of the excess dissolved solute in a supersaturated solution can be initiated by the addition of a tiny crystal of solute, called a seed crystal. The seed crystal provides a nucleation site on which the excess dissolved crystals can begin to grow. Recrystallization from a supersaturated solution is typically very fast.
Summary
• A supersaturated solution is a solution that contains more than the maximum amount of solute that is capable of being dissolved at a given temperature.
• When a seed crystal is added to the solution, a supersaturated solution can recrystallize.
16.06: Henry's Law
Having a soft drink in outer space poses some special problems. Under microgravity, the carbonation can quickly dissipate if not kept under pressure. You can't open the can, or you will lose carbonation. So, a special pressurized container has been developed to get around the problem of gas loss at low gravity.
Henry's Law
Pressure has very little effect on the solubility of solids or liquids, but has a significant effect on the solubility of gases. Gas solubility increases as the partial pressure of a gas above the liquid increases. Suppose a certain volume of water is in a closed container with the space above it occupied by carbon dioxide gas at standard pressure. Some of the $\ce{CO_2}$ molecules come into contact with the surface of the water and dissolve into the liquid. Now suppose that more $\ce{CO_2}$ is added to the space above the container, causing a pressure increase. In this case, more $\ce{CO_2}$ molecules are in contact with the water and so more of them dissolve. Thus, the solubility increases as the pressure increases. As with a solid, the $\ce{CO_2}$ that is undissolved reaches an equilibrium with the dissolved $\ce{CO_2}$, represented by the equation:
$\ce{CO_2} \left( g \right) \rightleftharpoons \ce{CO_2} \left( aq \right)\nonumber$
At equilibrium, the rate of gaseous $\ce{CO_2}$ dissolution is equal to the rate of dissolved $\ce{CO_2}$ coming out of the solution.
When carbonated beverages are packaged, they are done so under high $\ce{CO_2}$ pressure so that a large amount of carbon dioxide dissolves in the liquid. When the bottle is open, the equilibrium is disrupted because the $\ce{CO_2}$ pressure above the liquid decreases. Immediately, bubbles of $\ce{CO_2}$ rapidly exit the solution and escape out of the top of the open bottle. The amount of dissolved $\ce{CO_2}$ decreases. If the bottle is left open for an extended period of time, the beverage becomes "flat" as more and more $\ce{CO_2}$ comes out of the liquid.
The relationship of gas solubility to pressure is described by Henry's law, named after English chemist William Henry (1774-1836). Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Henry's law can be written as follows:
$\frac{S_1}{P_1} = \frac{S_2}{P_2}\nonumber$
$S_1$ and $P_1$ are the solubility and the pressure at an initial set of conditions; $S_2$ and $P_2$ are the solubility and pressure at another changed set of conditions. The solubility of a gas is typically reported in $\text{g/L}$.
Example $1$
The solubility of a certain gas in water is $0.745 \: \text{g/L}$ at standard pressure. What is its solubility when the pressure above the solution is raised to $4.50 \: \text{atm}$? The temperature is constant at $20^\text{o} \text{C}$.
Known
• $S_1 = 0.745 \: \text{g/L}$
• $P_1 = 1.00 \: \text{atm}$
• $P_2 = 4.50 \: \text{atm}$
Unknown
Substitute into Henry's law and solve for $S_2$.
Step 2: Solve.
$S_2 = \frac{S_1 \times P_2}{P_1} = \frac{0.745 \: \text{g/L} \times 4.50 \: \text{atm}}{1.00 \: \text{atm}} = 3.35 \: \text{g/L}\nonumber$
Step 3: Think about your result.
The solubility is increased to 4.5 times its original value, according to the direct relationship. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.05%3A_Supersaturated_Solutions.txt |
There are human cultures that do not recognize numbers above three. Anything greater than that is simply referred to as "much" or "many". Although this form of calculation may seem very limited, American culture does the same thing to a certain degree. For example, there are several ways to express the amount of solute in a solution in a quantitative manner. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, those terms are vague, and it is often necessary to express concentration with numbers.
Percent Solutions
One way to describe the concentration of a solution is by the percent of a solute in the solvent. The percent can further be determined in one of two ways: (1) the ratio of the mass of the solute divided by the mass of the solution or (2) the ratio of the volume of the solute divided by the volume of the solution.
Mass Percent
When the solute in a solution is a solid, a convenient way to express the concentration is by mass percent $\left( \frac{\text{mass}}{\text{mass}} \right)$, which is the grams of solute per $100 \: \text{g}$ of solution.
$\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%\nonumber$
Suppose that a solution was prepared by dissolving $25.0 \: \text{g}$ of sugar into $100 \: \text{g}$ of water. The percent by mass would be calculated by:
$\text{Percent by mass} = \frac{25 \: \text{g sugar}}{125 \: \text{g solution}} \times 100\% = 20\% \: \text{sugar}\nonumber$
Sometimes you may want to make up a particular mass of solution of a given percent by mass, and need to calculate what mass of the solvent to use. For example, you need to make $3000 \: \text{g}$ of a $5\%$ solution of sodium chloride. You can rearrange and solve for the mass of solute:
$\text{mass of solute} = \frac{\text{percent by mass}}{100\%} \times \text{mass of solution} = \frac{5\%}{100\%} \times 3000 \: \text{g} = 150 \: \text{g} \: \ce{NaCl}\nonumber$
You would need to weigh out $150 \: \text{g}$ of $\ce{NaCl}$ and add it to $2850 \: \text{g}$ of water. Notice that it was necessary to subtract the mass of the $\ce{NaCl}$ $\left( 150 \: \text{g} \right)$ from the mass of solution $\left( 3000 \: \text{g} \right)$ to calculate the mass of the water that would need to be added.
Volume Percent
The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of the solution, expressed as a percent, yields the percent by volume $\left( \frac{\text{volume}}{\text{volume}} \right)$ of the solution. If a solution is made by adding $40 \: \text{mL}$ of ethanol to $20 \: \text{mL}$ of water, the percent by volume is:
\begin{align*} \text{Percent by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \ &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \ &= 16.7\% \: \text{ethanol} \end{align*}\nonumber
Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below).
Summary
• The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution.
• A concentrated solution has a relatively large amount of dissolved solute.
• A dilute solution has a relatively small amount of dissolved solute.
• Techniques for calculation of percent mass and percent volume solution concentrations are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.07%3A_Percent_Solutions.txt |
Chemists deal with amounts of molecules every day. Chemical reactions are described as so many molecules of compound A reacting with so many molecules of compound B to form so many molecules of compound C. When we determine how much reagent to use, we need to know the number of molecules in a given volume of the reagent. Percent solutions only tell us the number of grams, not molecules. A $100 \: \text{mL}$ solution of $2\% \: \ce{NaCl}$ will have a very different number of molecules than a $2\%$ solution of $\ce{CsCl}$. So, we need another way to talk about numbers of molecules.
Molarity
Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles that react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable. The molarity $\left( \text{M} \right)$ of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, divide the moles of solute by the volume of the solution expressed in liters:
$\text{Molarity} \left( \text{M} \right) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{\text{mol}}{\text{L}}\nonumber$
Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol $\text{M}$ and is read as "molar". For example a solution labeled as $1.5 \: \text{M} \: \ce{NH_3}$ is read as "1.5 molar ammonia solution".
Example $1$
A solution is prepared by dissolving $42.23 \: \text{g}$ of $\ce{NH_4Cl}$ into enough water to make $500.0 \: \text{mL}$ of solution. Calculate its molarity.
Known
• Mass $= 42.23 \: \text{g} \: \ce{NH_4Cl}$
• Molar mass $\ce{NH_4Cl} = 53.50 \: \text{g/mol}$
• Volume solution $= 500.0 \: \text{mL} = 0.5000 \: \text{L}$
Unknown
The mass of the ammonium chloride is first converted to moles. Then the molarity is calculated by dividing by liters. Note that the given volume has been converted to liters.
Step 2: Solve.
\begin{align*} 42.23 \ \text{g} \: \ce{NH_4Cl} \times \frac{1 \: \text{mol} \: \ce{NH_4Cl}}{53.50 \: \text{g} \: \ce{NH_4Cl}} &= 0.7893 \: \text{mol} \: \ce{NH_4Cl} \ \frac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L}} &= 1.579 \: \text{M} \end{align*}\nonumber
Step 3: Think about your result.
The molarity is $1.579 \: \text{M}$, meaning that a liter of the solution would contain $1.579 \: \text{mol} \: \ce{NH_4Cl}$. Four significant figures are appropriate.
In a laboratory situation, a chemist must frequently prepare a given volume of solutions of a known molarity. The task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams.
Example $2$
A chemist needs to prepare $3.00 \: \text{L}$ of a $0.250 \: \text{M}$ solution of potassium permanganate $\left( \ce{KMnO_4} \right)$. What mass of $\ce{KMnO_4}$ does she need to make the solution?
Known
• Molarity $= 0.250 \: \text{M}$
• Volume $= 3.00 \: \text{L}$
• Molar mass $\ce{KMnO_4} = 158.04 \: \text{g/mol}$
Unknown
Moles of solute is calculated by multiplying molarity by liters. Then, moles is converted to grams.
Step 2: Solve.
\begin{align*} \text{mol} \: \ce{KMnO_4} = 0.250 \: \text{M} \: \ce{KMnO_4} \times 3.00 \: \text{L} &= 0.750 \: \text{mol} \: \ce{KMnO_4} \ 0.750 \: \text{mol} \: \ce{KMnO_4} \times \frac{158.04 \: \text{g} \: \ce{KMnO_4}}{1 \: \text{mol} \: \ce{KMnO_4}} &= 119 \: \text{g} \: \ce{KMnO_4} \end{align*}\nonumber
Step 3: Think about your result.
When $119 \: \text{g}$ of potassium permanganate is dissolved into water to make $3.00 \: \text{L}$ of solution, the molarity is $0.250 \: \text{M}$.
Summary
• The molarity $\left( \text{M} \right)$ of a solution is the number of moles of solute dissolved in one liter of solution.
• Calculations using the concept of molarity are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.08%3A_Molarity.txt |
Back in the "good ol' days" (a variable time frame, dependent on who you ask), many cooks did not bother with careful measurements. They simply "knew" how much flour to use, or how much water to add. Most of us need somewhat more precise ways to measure when we cook. Chemists are very careful when they prepare solutions, because the results of their experiments need to be quantitative. Simply "knowing" is not accurate enough for scientific purposes.
Preparing Solutions
If you are attempting to prepare $1.00 \: \text{L}$ of a $1.00 \: \text{M}$ solution of $\ce{NaCl}$, you would obtain $58.44 \: \text{g}$ of sodium chloride. However, you cannot simply add the sodium chloride to $1.00 \: \text{L}$ of water. After the solute dissolves, the volume of the solution will be slightly greater than one liter, because the hydrated sodium and chloride ions take up space in the solution. Instead, a volumetric flask needs to be used. Volumetric flasks come in a variety of sizes (see image below) and are designed to allow a chemist to prepare a solution of only one specific volume.
In other words, you cannot use a 1-liter volumetric flask to make $500 \: \text{mL}$ of a solution. It can only be used to prepare 1 liter of a solution. The steps to follow when preparing a solution with a 1-liter volumetric flask are outlined below, and shown in the figure below.
1. The appropriate mass of solute is weighed out and added to a volumetric flask that has been about half-filled with distilled water.
2. The solution is swirled until all of the solute dissolves.
3. More distilled water is carefully added up to the line etched on the neck of the flask.
4. The flask is capped and inverted several times to completely mix.
16.10: Dilution
Muriatic acid (another name for $\ce{HCl}$) is widely used for cleaning concrete and masonry surfaces. The acid must be diluted before use to get it down to a safer strength. Commercially available at concentrations of about $18\%$, muriatic acid can be used to remove scales and deposits (usually composed of basic materials).
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, while the volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).
$\text{mol}_1 = \text{mol}_2\nonumber$
Since the moles of solute in a solution is equal to the molarity multiplied by the liters, we can set those equal.
$M_1 \times L_1 = M_2 \times L_2\nonumber$
Finally, because the two sides of the equation are set equal to one another, the volume can be described in any unit that we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:
$M_1 \times V_1 = M_2 \times V_2\nonumber$
Suppose that you have $100. \: \text{mL}$ of a $2.0 \: \text{M}$ solution of $\ce{HCl}$. You dilute the solution by adding enough water to make the solution volume $500. \: \text{mL}$. The new molarity can easily be calculated by using the above equation and solving for $M_2$:
$M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}\nonumber$
The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value.
Another common dilution problem involves deciding how much of a highly concentrated solution is required to make a desired quantity of solution of lesser concentration. The highly concentrated solution is typically referred to as the stock solution.
Example $1$
Nitric acid $\left( \ce{HNO_3} \right)$ is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is $16 \: \text{M}$. How much of the stock solution of nitric acid needs to be used to make $8.00 \: \text{L}$ of a $0.50 \: \text{M}$ solution?
Known
• Stock $\ce{HNO_3} = 16 \: \text{M}$
• $V_2 = 8.00 \: \text{L}$
Step 2: Solve.
$V_1 = \frac{M_2 \times V_2}{M_1} = \frac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}\nonumber$
Step 3: Think about your result.
$250 \: \text{mL}$ of the stock $\ce{HNO_3}$ needs to be diluted with water to a final volume of $8.00 \: \text{L}$. The dilution is by a factor of 32 to go from $16 \: \text{M}$ to $0.5 \: \text{M}$.
Dilutions can be performed in the laboratory with various tools, depending on the volumes required and the desired accuracy. The images below illustrate the use of two different types of pipettes. In the first figure, a glass pipette is being used to transfer a portion of a solution to a graduated cylinder. Use of a pipette rather than a graduated cylinder for the transfer improves accuracy. The second figure shows a micropipette, which is designed to quickly and accurately dispense small volumes. Micropipettes are adjustable and come in a variety of sizes.
Summary
• A process is described to calculate dilutions.
• Pipettes and micropipettes are used to transfer and dispense solution. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.09%3A_Preparing_Solutions.txt |
For many purposes, the use of molarity is very convenient. However, when we want to know the concentration of solute present in situations where there are temperature changes, molarity will not work. The volume of the solution will change somewhat with temperature, enough to affect the accuracy of data observations and calculations. Another parameter is needed, one not affected by the temperature of the material we are studying.
Molality
A final way to express the concentration of a solution is by its molality. The molality $\left( \textit{m} \right)$ of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains $1.0 \: \text{mol}$ of $\ce{NaCl}$ dissolved into $1.0 \: \text{kg}$ of water is a "one-molal" solution of sodium chloride. The symbol for molality is a lower-case $\textit{m}$ written in italics.
$\text{Molality} \left( \textit{m} \right) = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{\text{mol}}{\text{kg}}\nonumber$
Molality differs from molarity only in the denominator. While molarity is based on the liters of solution, molality is based on the kilograms of solvent. Concentrations expressed in molality are used when studying properties of solutions related to vapor pressure and temperature changes. Molality is used because its value does not change with changes in temperature. The volume of a solution, on the other hand, is slightly dependent upon temperature.
Example $1$
Determine the molality of a solution prepared by dissolving $28.60 \: \text{g}$ of glucose $\left( \ce{C_6H_{12}O_6} \right)$ into $250 \: \text{g}$ of water.
Known
• Mass solute $= 28.60 \: \text{g} \: \ce{C_6H_{12}O_6}$
• Mass solvent $= 250 \: \text{g} = 0.250 \: \text{kg}$
• Molar mass $\ce{C_6H_{12}O_6} = 180.18 \: \text{g/mol}$
Unknown
Convert grams of glucose to moles and divide by the mass of the water in kilograms.
Step 2: Solve.
\begin{align*} 28.60 \: \text{g} \: \ce{C_6H_{12}O_6} \times \frac{1 \: \text{mol} \: \ce{C_6H_{12}O_6}}{180.18 \: \text{g} \: \ce{C_6H_{12}O_6}} &= 0.1587 \: \text{mol} \: \ce{C_6H_{12}O_6} \ \frac{0.1587 \: \text{mol} \: \ce{C_6H_{12}O_6}}{0.250 \: \text{kg} \: \ce{H_2O}} &= 0.635 \: \textit{m} \: \ce{C_6H_{12}O_6} \end{align*}\nonumber
Step 3: Think about your result.
The answer represents the moles of glucose per kilogram of water and has three significant figures.
Molality and molarity are closely related in value for dilute aqueous solutions because the density of those solutions is relatively close to $1.0 \: \text{g/mL}$. This means that $1.0 \: \text{L}$ of solution has nearly a mass of $1.0 \: \text{kg}$. As the solution becomes more concentrated, its density will not be as close to $1.0 \: \text{g/mL}$ and the molality value will be different than the molarity. For solutions with solvents other than water, the molality will be very different than the molarity. Make sure that you are paying attention to which quantity is being used in a given problem.
Summary
• The molality $\left( \textit{m} \right)$ of a solution is the moles of solute divided by the kilograms of solvent.
• The value of molality does not change with temperature.
16.12: The Lowering of Vapor Pressure
A knowledge of Latin can help us to better understand the English language—specifically, scientific terminology. Consider the word "colligative". Where did that come from? If you know a little Latin, you know that it comes from two Latin words meaning "to tie together".
Lowering Vapor Pressure
A colligative property is a property of a solution that depends only on the number of solute particles dissolved in the solution, and not on their identity. Recall that the vapor pressure of a liquid is determined by how easily its molecules are able to escape the surface of the liquid and enter the gaseous phase. When a liquid evaporates easily, it will have a relatively large number of its molecules in the gas phase and thus will have a high vapor pressure. Liquids that do not evaporate easily have a lower vapor pressure. The figure below shows the surface of a pure solvent compared to a solution. In the picture on the left, the surface is entirely occupied by liquid molecules, some of which will evaporate and form a vapor pressure. On the right, a nonvolatile solute has been dissolved into the solvent. Nonvolatile means that the solute itself has little tendency to evaporate. Because some of the surface is now occupied by solute particles, there is less room for solvent molecules. This results in less solvent being able to evaporate. The addition of a nonvolatile solute results in a lowering of the vapor pressure of the solvent.
The lowering of the vapor pressure depends on the number of solute particles that have been dissolved. The chemical nature of the solute is not important because the vapor pressure is merely a physical property of the solvent. The only requirement is that the solute was only dissolved and that it did not undergo a chemical reaction with the solvent.
While the chemical nature of the solute is not a factor, it is necessary to take into account whether the solute is an electrolyte or a nonelectrolyte. Recall that ionic compounds are strong electrolytes, and thus dissociate into ions when they dissolve. This results in a larger number of dissolved particles. For example, consider two different solutions of equal concentration: one is made from the ionic compound sodium chloride, while the other is made from the molecular compound glucose. The following equations show what happens when these solutions dissolve.
$\begin{array}{ll} \ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) & 2 \: \text{dissolved particles} \ \ce{C_6H_{12}O_6} \left( s \right) \rightarrow \ce{C_6H_{12}O_6} \left( aq \right) & 1 \: \text{dissolved particle} \end{array}\nonumber$
The sodium chloride dissociates into two ions, while the glucose does not dissociate. Therefore, equal concentrations of each solution will result in twice as many dissolved particles in the case of the sodium chloride. The vapor pressure of the sodium chloride solution will be lowered twice the amount of the glucose solution.
Summary
• A colligative property is a property of a solution that depends only on the number of solute particles dissolved in the solution, and not on their identity.
• Nonvolatile means that the solute itself has little tendency to evaporate.
• Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.11%3A_Molality.txt |
Colligative properties have practical applications, such as the salting of roads in cold-weather climates. By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly, making driving safer. Sodium chloride $\left( \ce{NaCl} \right)$, and either calcium chloride $\left( \ce{CaCl_2} \right)$ or magnesium chloride $\left( \ce{MgCl_2} \right)$ are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but is less effective because it only dissociates into two ions, instead of three.
Freezing Point Depression
The figure below shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the vapor pressure of the solvent, resulting in a lowering of the freezing point of the solution compared to the solvent. The freezing point depression is the difference in temperature between the freezing point of the pure solvent and that of the solution. On the graph, the freezing point depression is represented by $\Delta T_f$.
When a pure solvent freezes, its particles become more ordered as the intermolecular forces that operate between the molecules become permanent. In the case of water, the hydrogen bonds make the hexagonally-shaped network of molecules that characterizes the structure of ice. By dissolving a solute into the liquid solvent, this ordering process is disrupted. As a result, more energy must be removed from the solution in order to freeze it, and the freezing point of the solution is lower than that of the pure solvent.
The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation is:
$\Delta T_f = K_f \times \textit{m}\nonumber$
The proportionality constant, $K_f$, is called the molal freezing-point depression constant. It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of $K_f$ is $-1.86^\text{o} \text{C}/\textit{m}$. So, the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is $-1.86^\text{o} \text{C}$. Every solvent has a unique molal freezing-point depression constant. These are shown in the table below, along with a related value for the boiling point called $K_b$.
Molal Freezing-Point and Boiling-Point Constants
Table $1$: Molal Freezing-Point and Boiling-Point Constants
Solvent Normal Freezing Point $\left( ^\text{o} \text{C} \right)$ Molal Freezing-Point Depression Constant, $K_f$ $\left( ^\text{o} \text{C}/\textit{m} \right)$ Normal Boiling Point $\left( ^\text{o} \text{C} \right)$ Molal Boiling-Point Elevation Constant, $K_b$ $\left( ^\text{o} \text{C}/\textit{m} \right)$
Acetic acid 16.6 -3.90 117.9 3.07
Camphor 178.8 -39.7 207.4 5.61
Naphthalene 80.2 -6.94 217.7 5.80
Phenol 40.9 -7.40 181.8 3.60
Water 0.00 -1.86 100.00 0.512
Example $1$
Ethylene glycol $\left( \ce{C_2H_6O_2} \right)$ is a molecular compound that is used in many commercial antifreezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point, and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of $400 \: \text{g}$ of ethylene glycol in $500 \: \text{g}$ of water.
Known
• Mass $\ce{C_2H_6O_2} = 400 \: \text{g}$
• Molar mass $\ce{C_2H_6O_2} = 62.08 \: \text{g/mol}$
• Mass $\ce{H_2O} = 500 \: \text{g} = 0.500 \: \text{kg}$
• $K_f \left( \ce{H_2O} \right) = -1.86^\text{o} \text{C}/\textit{m}$
Unknown
This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression.
Step 2: Solve.
\begin{align*} 400. \: \text{g} \: \ce{C_2H_6O_2} \times \frac{1 \: \text{mol} \: \ce{C_2H_6O_2}}{62.08 \: \text{g} \: \ce{C_2H_6O_2}} &= 6.44 \: \text{mol} \: \ce{C_2H_6O_2} \ \frac{6.44 \: \text{mol} \: \ce{C_2H_6O_2}}{0.500 \: \text{kg} \: \ce{H_2O}} &= 12.9 \: \textit{m} \: \ce{C_2H_6O_2} \ \Delta T_f = K_f \times \textit{m} = -1.86^\text{o} \text{C}/\textit{m} &= -24.0^\text{o} \text{C} \ T_f &= -24.0^\text{o} \text{C} \end{align*}\nonumber
The normal freezing point of water is $0.0^\text{o} \text{C}$. Therefore, since the freezing point decreases by $24.0^\text{o} \text{C}$, the freezing point of the solution is $-24.0^\text{o} \text{C}$.
Step 3: Think about your result.
The freezing point of the water decreases by a large amount, protecting the radiator from damage due to the expansion of water when it freezes. There are three significant figures in the result.
Summary
• The freezing point depression is the difference in temperature between the freezing point of the pure solvent and that of the solution.
• The molal freezing-point depression constant is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute.
• Calculations involving freezing point depression are described.
16.14: Boiling Point Elevation
Salt is often added to boiling water when preparing spaghetti or other pasta. One reason is to add flavor to the food. Some people believe that the addition of salt increases the boiling point of the water. Technically, they are correct, but the increase is rather small. You would need to add over 100 grams of $\ce{NaCl}$ to a liter of water to increase the boiling point by a couple of degrees, which is likely unhealthy.
Boiling Point Elevation
The figure below shows the phase diagram of a solution and the effect that the lowered vapor pressure has on the boiling point of the solution compared to the solvent. In this case, the solution has a higher boiling point than the pure solvent. Since the vapor pressure of the solution is lower, more heat must be supplied to the solution to bring its vapor pressure up to the pressure of the external atmosphere. The boiling point elevation is the difference in temperature between the boiling point of the pure solvent and that of the solution. On the graph, the boiling point elevation is represented by $\Delta T_b$.
The magnitude of the boiling point elevation is also directly proportional to the molality of the solution. The equation is:
$\Delta T_b = K_b \times \textit{m}\nonumber$
The proportionality constant, $K_b$, is called the molal boiling-point elevation constant. It is a constant that is equal to the change in the boiling point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of $K_b$ is $0.512^\text{o} \text{C}/\textit{m}$. So, the boiling temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is $100.512^\text{o} \text{C}$.
Summary
• Boiling point elevation is the difference in temperature between the boiling point of the pure solvent and that of the solution.
• The molal boiling-point elevation constant is equal to the change in the boiling point for a 1-molal solution of a nonvolatile molecular solute.
• Calculations involving the molal boiling point elevation constant are outlined. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.13%3A_Freezing_Point_Depression.txt |
The addition of ions creates significant changes in properties of solutions. Water molecules surround the ions and are somewhat tightly bound to them. Colligative properties are affected because the solvent properties are no longer the same as those in the pure solvent.
Electrolytes and Colligative Properties
Ionic compounds are electrolytes and dissociate into two or more ions as they dissolve. This must be taken into account when calculating the freezing and boiling points of electrolyte solutions. The sample problem below demonstrates how to calculate the freezing point and boiling point of a solution of calcium chloride. Calcium chloride dissociates into three ions according to the equation:
$\ce{CaCl_2} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right)\nonumber$
The values of the freezing point depression and the boiling point elevation for a solution of $\ce{CaCl_2}$ will be three times greater than they would be for an equal molality of a nonelectrolyte.
Example $1$
Determine the freezing point and boiling point of a solution prepared by dissolving $82.20 \: \text{g}$ of calcium chloride into $400 \: \text{g}$ of water.
Known
• Mass $\ce{CaCl_2} = 82.20 \: \text{g}$
• Molar mass $\ce{CaCl_2} = 110.98 \: \text{g/mol}$
• Mass $\ce{H_2O} = 400 \: \text{g} = 0.400 \: \text{kg}$
• $K_f \left( \ce{H_2O} \right) = -1.86^\text{o} \text{C}/\textit{m}$
• $K_b \left( \ce{H_2O} \right) = 0.512^\text{o} \text{C}/\textit{m}$
• $\ce{CaCl_2}$ dissociates into 3 ions
Unknown
• $T_f = ? \: ^\text{o} \text{C}$
• $T_b = ? \: ^\text{o} \text{C}$
The moles of $\ce{CaCl_2}$ is first calculated, followed by the molality of the solution. The freezing and boiling points are then determined, including multiplying by 3 for the three ions.
Step 2: Solve.
\begin{align*} 82.20 \: \text{g} \: \ce{CaCl_2} \times \frac{1 \: \text{mol} \: \ce{CaCl_2}}{110.98 \: \text{g} \: \ce{CaCl_2}} &= 0.7407 \: \text{mol} \: \ce{CaCl_2} \ \frac{0.7407 \: \text{mol} \: \ce{CaCl_2}}{0.400 \: \text{kg} \: \ce{H_2O}} &= 1.85 \: \textit{m} \: \ce{CaCl_2} \end{align*}\nonumber
$\begin{array}{ll} \Delta T_f = K_f \times \textit{m} \times 3 = -1.86^\text{o} \text{C}/\textit{m} \times 1.85 \: \textit{m} \times 3 = -10.3^\text{o} \text{C} & T_f = -10.3^\text{o} \text{C} \ \Delta T_b = K_b \times \textit{m} \times 3 = 0.512^\text{o} \text{C}/\textit{m} \times 1.85 \: \textit{m} \times 3 = 2.84^\text{o} \text{C} & T_b = 102.84^\text{o} \text{C} \end{array}\nonumber$
Step 3: Think about your result.
Since the normal boiling point of water is $100.00^\text{o} \text{C}$, the calculated result for $\Delta T_b$ must be added to 100.00 to find the new boiling point.
Summary
• Ionic compounds are electrolytes and dissociate into two or more ions as they dissolve.
• The effect of ionization on colligative properties is described.
16.16: Calculating Molar Mass
Putting antifreeze into a radiator will keep an engine from freezing. By knowing how cold the weather will get and how much water is in the radiator, one can determine how much antifreeze to add to achieve a specific desired freezing point depression. This is made possible by knowing what antifreeze is. Can things be switched around, so as to get some information about the properties of the antifreeze (such as its molecular weight) from the freezing point decrease? As it turns out, this can be done fairly easily and accurately.
Calculating Molar Mass
In the laboratory, freezing point or boiling point data can be used to determine the molar mass of an unknown solute. Since we know the relationship between a decrease in freezing point and the concentration of solute, if we dissolve a known mass of our unknown solute into a known amount of solvent, we can calculate the molar mass of the solute. The $K_f$ or $K_b$ of the solvent must be known. We also need to know if the solute is an electrolyte or a nonelectrolyte. If the solvent is an electrolyte, we would need to know the number of ions produced when it dissociates.
Example $1$
$38.7 \: \text{g}$ of a nonelectrolyte is dissolved into $218 \: \text{g}$ of water. The freezing point of the solution is measured to be $-5.53^\text{o} \text{C}$. Calculate the molar mass of the solute.
Known
• $\Delta T_f = -5.53^\text{o} \text{C}$
• Mass $\ce{H_2O} = 218 \: \text{g} = 0.218 \: \text{kg}$
• Mass solute $= 38.7 \: \text{g}$
• $K_f \left( \ce{H_2O} \right) = -1.86^\text{o} \text{C}/\textit{m}$
Unknown
Use the freezing point depression $\left( \Delta T_f \right)$ to calculate the molality of the solution. Then use the molality equation to calculate the moles of solute. Then divide the grams of solute by the moles to determine the molar mass.
Step 2: Solve.
\begin{align*} \textit{m} = \frac{\Delta T_f}{K_f} &= \frac{-5.53^\text{o} \text{C}}{-1.86^\text{o} \text{C}/\textit{m}} = 2.97 \: \textit{m} \ \text{mol solute} &= \textit{m} \times \text{kg} \: \ce{H_2O} = 2.97 \: \textit{m} \times 0.218 \: \text{kg} = 0.648 \: \text{mol} \ \frac{38.7 \: \text{g}}{0.648 \: \text{mol}} &= 59.7 \: \text{g/mol} \end{align*}\nonumber
Step 3: Think about your result.
The molar mass of the unknown solute is $59.7 \: \text{g/mol}$. Knowing the molar mass is an important step in determining the identity of an unknown. A similar problem could be done with the change in boiling point. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.15%3A_Electrolytes_and_Colligative_Properties.txt |
One of the unfortunate byproducts of industrialized society is acid rain. Sulfur dioxide from burning coal and nitrogen oxides from vehicle emissions both form acids. When these acids react with limestone (calcium carbonate), reactions occur that dissolve the limestone and release water and carbon dioxide. Over a period of time, serious damage is caused to the limestone structure.
Molecular and Ionic Equations
When ionic compounds are dissolved into water, the polar water molecules break apart the solid crystal lattice, resulting in the hydrated ions being evenly distributed through the water. This process is called dissociation and is the reason that all ionic compounds are strong electrolytes. When two different ionic compounds that have been dissolved in water are mixed, a chemical reaction may occur between certain pairs of the hydrated ions. Consider the double-replacement reaction that occurs when a solution of sodium chloride is mixed with a solution of silver nitrate:
$\ce{NaCl} \left( aq \right) + \ce{AgNO_3} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber$
The driving force behind this reaction is the formation of the silver chloride precipitate (see figure below).
This is called a molecular equation. A molecular equation is an equation in which the formulas of the compounds are written as though all substances exist as molecules. However, there is a better way to show what is happening in this reaction. All of the aqueous compounds should be written as ions, because they are present in the water as separated ions—a result of their dissociation.
$\ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber$
This equation is called an ionic equation, an equation in which dissolved ionic compounds are shown as free ions.
Some other double-replacement reactions do not produce a precipitate as one of the products. The production of a gas and/or a molecular compound such as water may also drive the reaction. For example, consider the reaction of a solution of sodium carbonate with a solution of hydrochloric acid $\left( \ce{HCl} \right)$. The products of the reaction are aqueous sodium chloride, carbon dioxide, and water. The balanced molecular equation is:
$\ce{Na_2CO_3} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber$
The ionic equation is:
$2 \ce{Na^+} \left( aq \right) + \ce{CO_3^{2-}} \left( aq \right) + 2 \ce{H^+} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right) \rightarrow 2 \ce{Na^+} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber$
A single-replacement reaction is one in which an element replaces another element in a compound. An element is in either the solid, liquid, or gas state and is not an ion. The example below shows the reaction of solid magnesium metal with aqueous silver nitrate to form aqueous magnesium nitrate and silver metal.
Balanced molecular equation:
$\ce{Mg} \left( s \right) + 2 \ce{AgNO_3} \left( aq \right) \rightarrow \ce{Mg(NO_3)_2} \left( aq \right) + 2 \ce{Ag} \left( s \right)\nonumber$
Ionic equation:
$\ce{Mg} \left( s \right) + 2 \ce{Ag^+} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Mg^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) + \ce{Ag} \left( s \right)\nonumber$
This type of single-replacement reaction is called a metal replacement. Other common categories of single-replacement reactions are: hydrogen replacement and halogen replacement.
Summary
• A molecular equation is an equation in which the formulas of the compounds are written as though all substances exist as molecules.
• An ionic equation is one in which dissolved ionic compounds are shown as free ions.
• Examples of molecular and ionic equations are shown. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.17%3A_Molecular_and_Ionic_Equations.txt |
At sports events around the world, a small number of athletes fiercely compete on fields and in stadiums. They get tired, dirty, and sometimes hurt as they try to win the game. Surrounding them are thousands of spectators watching and cheering. Would the game be different without the spectators? Definitely! Spectators provide encouragement to the team and generate enthusiasm. Although the spectators are not playing the game, they are certainly a part of the process.
Net Ionic Equations
We can write a molecular equation for the formation of silver chloride precipitate:
$\ce{NaCl} + \ce{AgNO_3} \rightarrow \ce{NaNO_3} + \ce{AgCl}\nonumber$
The corresponding ionic equation is:
$\ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber$
If you look carefully at the ionic equation, you will notice that the sodium ion and the nitrate ion appear unchanged on both sides of the equation. When the two solutions are mixed, neither the $\ce{Na^+}$ nor the $\ce{NO_3^-}$ ions participate in the reaction. They can be eliminated from the reaction.
$\cancel{\ce{Na^+} \left( aq \right)} + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \cancel{\ce{NO_3^-} \left( aq \right)} \rightarrow \cancel{\ce{Na^+} \left( aq \right)} + \cancel{\ce{NO_3^-} \left( aq \right)} + \ce{AgCl} \left( s \right)\nonumber$
A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction. In the above reaction, the sodium ion and the nitrate ion are both spectator ions. The equation can now be written without the spectator ions:
$\ce{Ag^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) \rightarrow \ce{AgCl} \left( s \right)\nonumber$
The net ionic equation is the chemical equation that shows only those elements, compounds, and ions that are directly involved in the chemical reaction. Notice that in writing the net ionic equation, the positively-charged silver cation was written first on the reactant side, followed by the negatively-charged chloride anion. This is somewhat customary because that is the order in which the ions must be written in the silver chloride product. However, it is not absolutely necessary to order the reactants in this way.
Net ionic equations must be balanced by both mass and charge. Balancing by mass means ensuring that there are equal masses of each element on the product and reactant sides. Balancing by charge means making sure that the overall charge is the same on both sides of the equation. In the above equation, the overall charge is zero, or neutral, on both sides of the equation. As a general rule, if you balance the molecular equation properly, the net ionic equation will end up being balanced by both mass and charge.
Example $1$
When aqueous solutions of copper (II) chloride and potassium phosphate are mixed, a precipitate of copper (II) phosphate is formed. Write a balanced net ionic equation for this reaction.
Step 1: Plan the problem.
Write and balance the molecular equation first, making sure that all formulas are correct. Then write the ionic equation, showing all aqueous substances as ions. Carry through any coefficients. Finally, eliminate spectator ions and write the net ionic equation.
Step 2: Solve.
Molecular equation:
$3 \ce{CuCl_2} \left( aq \right) + 2 \ce{K_3PO_4} \left( aq \right) \rightarrow 6 \ce{KCl} \left( aq \right) + \ce{Cu_3(PO_4)_2} \left( s \right)\nonumber$
Ionic equation:
$3 \ce{Cu^{2+}} \left( aq \right) + 6 \ce{Cl^-} \left( aq \right) + 6 \ce{K^+} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \rightarrow 6 \ce{K^+} \left( aq \right) + 6 \ce{Cl^-} \left( aq \right) + \ce{Cu_3(PO_4)_2} \left( s \right)\nonumber$
Notice that the balance of the equation is carried through when writing the dissociated ions. For example, there are six chloride ions on the reactant side because the coefficient of 3 is multiplied by the subscript of 2 on the copper (II) chloride formula. The spectator ions, $\ce{K^+}$ and $\ce{Cl^-}$, can be eliminated.
Net ionic equation:
$3 \ce{Cu^{2+}} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \rightarrow \ce{Cu_3(PO_4)_2} \left( s \right)\nonumber$
Step 3: Think about your result.
For a precipitation reaction, the net ionic equation always shows the two ions that come together to form the precipitate. The equation is balanced by mass and charge.
Summary
• A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction.
• The net ionic equation is the chemical equation that shows only those elements, compounds, and ions that are directly involved in the chemical reaction.
• An example of writing a net ionic equation is outlined. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.18%3A_Net_Ionic_Equations.txt |
Predicting the weather is tricky business. A thorough examination of a large amount of data is needed to make the daily forecast. Wind patterns, historical data, barometric pressure—these and many other data are fed into computers that then use a set of rules to predict what will happen based on past history.
Solubility Rules
Some combinations of aqueous reactants result in the formation of a solid precipitate as a product. However, some combinations will not produce such a product. If solutions of sodium nitrate and ammonium chloride are mixed, no reaction occurs. One could write a molecular equation showing a double-replacement reaction, but both products, sodium chloride and ammonium nitrate, are soluble and would remain in the solution as ions. Every ion is a spectator ion and there is no net ionic equation at all.
It is useful to be able to predict when a precipitate will occur in a reaction. To do so, you can use a set of guidelines called the solubility rules (shown in Table $1$).
Table $1$: Solubility Rules for Ionic Compounds in Water
Soluble Compounds containing the alkali metal ions $\left( \ce{Li^+}, \: \ce{Na^+}, \: \ce{K^+}, \: \ce{Rb^+}, \: \ce{Cs^+} \right)$ and ammonium ion $\left( \ce{NH_4^+} \right)$.
Soluble Compounds containing the nitrate ion $\left( \ce{NO_3^-} \right)$, acetate ion $\left( \ce{CH_3COO^-} \right)$, chlorate ion $\left( \ce{ClO_3^-} \right)$, and bicarbonate ion $\left( \ce{HCO_3^-} \right)$.
Mostly Soluble Compounds containing the chloride ion $\left( \ce{Cl^-} \right)$, bromide ion $\left( \ce{Br^-} \right)$, and iodide ion $\left( \ce{I^-} \right)$. Exceptions are those of silver $\left( \ce{Ag^+} \right)$, mercury (I) $\left( \ce{Hg_2^{2+}} \right)$, and lead (II) $\left( \ce{Pb^{2+}} \right)$.
Mostly Soluble Compounds containing the sulfate ion $\left( \ce{SO_4^{2-}} \right)$. Exceptions are those of silver $\left( \ce{Ag^+} \right)$, calcium $\left( \ce{Ca^{2+}} \right)$, strontium $\left( \ce{Sr^{2+}} \right)$, barium $\left( \ce{Ba^{2+}} \right)$, mercury (I) $\left( \ce{Hg_2^{2+}} \right)$, and lead (II) $\left( \ce{Pb^{2+}} \right)$.
Mostly Insoluble Compounds containing the carbonate ion $\left( \ce{CO_3^{2-}} \right)$, phosphate ion $\left( \ce{PO_4^{3-}} \right)$, chromate ion $\left( \ce{CrO_4^{2-}} \right)$, sulfide ion $\left( \ce{S^{2-}} \right)$, and silicate ion $\left( \ce{SiO_3^{2-}} \right)$. Exceptions are those of the alkali metals and ammonium.
Mostly Insoluble Compounds containing the hydroxide ion $\left( \ce{OH^-} \right)$. Exceptions are those of the alkali metals and the barium ion $\left( \ce{Ba^{2+}} \right)$.
For practice using the solubility rules, predict if a precipitate will form when solutions of cesium bromide and lead (II) nitrate are mixed.
$\ce{Cs^+} \left( aq \right) + \ce{Br^-} \left( aq \right) + \ce{Pb^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \rightarrow ?\nonumber$
The potential precipitates from a double-replacement reaction are cesium nitrate and lead (II) bromide. According to the solubility rules table, cesium nitrate is soluble because all compounds containing the nitrate ion, as well as all compounds containing the alkali metal ions, are soluble. Most compounds containing the bromide ion are soluble, but lead (II) is an exception. Therefore, the cesium and nitrate ions are spectator ions and the lead (II) bromide is a precipitate. The balanced net ionic reaction is:
$\ce{Pb^{2+}} \left( aq \right) + 2 \ce{Br^-} \left( aq \right) \rightarrow \ce{PbBr_2} \left( s \right)\nonumber$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.19%3A_Predicting_Precipitates_Using_Solubility_Rules.txt |
• 17.1: Chemical Potential Energy
• 17.2: Heat
• 17.3: Exothermic and Endothermic Processes
• 17.4: Heat Capacity and Specific Heat
• 17.5: Specific Heat Calculations
• 17.6: Enthalpy
The factors influencing a reaction are complicated and varied. Since a catalyst affects activation energy, we might assume it would have some sort of impact on the amount of heat that is absorbed or released by the reaction—but it does not. The change in heat content of a reaction depends solely on the chemical compositions of the reactants and products, not on the path taken to get from one to the other.
• 17.7: Calorimetry
At one time, calories in foods were measured with a bomb calorimeter. A weighed amount of the food would be placed in the calorimeter and the system was then sealed and filled with oxygen. An electric spark ignited the food-oxygen mixture. The amount of heat released when the food burned gave an idea of the calories present within the food. Today, calories are calculated from the protein, carbohydrate, and fat content of food (all determined by chemical analysis).
• 17.8: Thermochemical Equations
Heating a home is becoming more and more expensive. The decision to use gas, oil, electricity, or wood can be multi-faceted. Part of the decision is based on which fuel will provide the highest amount of energy release when burned. Studies of thermochemistry can be very useful in getting reliable information for making these important choices.
• 17.9: Stoichiometric Calculations and Enthalpy Changes
There is a growing concern about damage to the environment done by emissions from manufacturing plants. Many companies are taking steps to reduce these harmful emissions by adding equipment that will trap the pollutants. In order to know what equipment (and the quantity) to order, studies are done to measure the amount of product currently produced. Since pollution is often both particulate and thermal, energy changes need to be determined in addition to the amounts of products released.
• 17.10: Heats of Fusion and Solidification
• 17.11: Heats of Vaporization and Condensation
Natural resources for electric power generation have traditionally been waterfalls, oil, coal, or nuclear power. Research is being carried out to look for other renewable sources to run the generators. Geothermal sites (such as geysers) are being considered because of the steam they produce. Capabilities can be estimated by knowing how much steam is released in a given time at a particular site.
• 17.12: Multi-Step Problems with Changes of State
If you have a cube of ice, which process will take more energy—the melting of that ice cube or the conversion of the water to steam? The short answer is that more energy is needed to convert the water to steam. The long answer is really a series of questions: How do you get from one point to the other? What is the temperature of the ice? What is the mass of that ice cube? A long process is involved to take the material from the starting point to the end point.
• 17.13: Heat of Solution
• 17.14: Heat of Combustion
• 17.15: Hess's Law of Heat Summation
• 17.16: Standard Heat of Formation
• 17.17: Calculating Heat of Reaction from Heat of Formation
17: Thermochemistry
Gunpowder was originally developed by the Chinese in the ninth century AD, primarily for rockets. This material is composed of charcoal, sulfur, and saltpeter (potassium nitrate). The explosive reaction that occurs involves the conversion of the charcoal to carbon dioxide, with the potassium nitrate providing the extra oxygen needed for a rapid reaction. Although sulfur was included to stabilize the product, gunpowder is still highly explosive.
Types of Energy
Two basic types of energy exist: potential energy and kinetic energy. Potential energy is stored energy. It has not yet been released, but is ready to go. Kinetic energy is the energy of motion. It causes work to be done through movement.
Chemical Potential Energy
Energy is the capacity for doing work or supplying heat. When you fill your car with gasoline, you are providing it with potential energy. Chemical potential energy is the energy stored in the chemical bonds of a substance. The various chemicals that make up gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car. The release of that energy does two things: some of the potential energy is transformed into work, which is used to move the car; at the same time, some of the potential energy is converted to heat and makes the car's engine very hot. The energy changes of a system occur as either heat or work, or some combination of both.
Dynamite is another example of chemical potential energy. The major component of dynamite is nitroglycerin, a very unstable material. By mixing it with diatomaceous earth, the stability is increased and it is less likely to explode if it receives a physical shock. When ignited, the nitroglycerin explodes rapidly, releasing large amounts of nitrogen and other gases along with a massive amount of heat.
Summary
• Potential energy is stored energy.
• Kinetic energy is the energy of motion.
• Chemical potential energy is energy available in the chemical bonds of a compound. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.01%3A_Chemical_Potential_Energy.txt |
Blacksmiths heat solid iron in order to shape it into a variety of different objects. Iron is a rigid, solid metal. At room temperature, it is extremely difficult to bend iron. However, when heated to a high enough temperature, iron can be easily worked. The heat energy in the forge is transferred to the metal, making the iron atoms vibrate more and move around more readily.
Heat is energy that is transferred from one object or substance to another because of a difference in temperature between the two. Heat always flows from an object at a higher temperature to an object at a lower temperature (see figure below). The flow of heat will continue until the two objects are at the same temperature.
Figure \(1\): Object A starts with a higher temperature than object B. No heat flows when the objects are isolated from each other. When brought into contact, heat flows from A to B until the temperatures of the two objects are the same.
Thermochemistry is the study of energy changes that occur during chemical reactions and during changes of state. When chemical reactions occur, some chemical bonds are broken, while new chemical bonds form. As a result of the rearrangement of atoms, the total chemical potential energy of the system either increases or decreases.
Summary
• Heat is transferred energy from a site of higher energy to a site of lower energy.
• Thermochemistry is the study of energy changes that occur during chemical reactions and during changes of state.
17.03: Exothermic and Endothermic Processes
A campfire is an example of basic thermochemistry. The reaction is initiated by the application of heat from a match. The reaction converting wood to carbon dioxide and water (among other things) continues, releasing heat energy in the process. This heat energy can then be used to cook food, roast marshmallows, or simply to keep warm when it's cold outside.
Exothermic and Endothermic Processes
When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe: the system and the surroundings. The system is the specific portion of matter in a given space that is being studied during an experiment or an observation. The surroundings is everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is the particular chemicals being reacted, while the surroundings is the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings.
A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings, and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter $q$. The sign of $q$ for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of $q$ for an exothermic process is negative because the system is losing heat.
Units of Heat
Heat flow is measured in one of two common units: the calorie and the joule. The joule $\left( \text{J} \right)$ is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A calorie $\left( \text{cal} \right)$ is the quantity of heat required to raise the temperature of 1 gram of water by $1^\text{o} \text{C}$. For example, raising the temperature of $100 \: \text{g}$ of water from $20^\text{o} \text{C}$ to $22^\text{o} \text{C}$ would require $100 \times 2 = 200 \: \text{cal}$.
Calories contained within food are actually kilocalories $\left( \text{kcal} \right)$. In other words, if a certain snack contains 85 food calories, it actually contains $85 \: \text{kcal}$ or $85,000 \: \text{cal}$. In order to make the distinction, the dietary calorie is written with a capital C.
$1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories}\nonumber$
To say that the snack "contains" 85 Calories means that $85 \: \text{kcal}$ of energy are released when that snack is processed by the human body.
Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below.
$1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J}\nonumber$
We can calculate the amount of heat released in kilojoules when a 400 Calorie hamburger is digested:
$400. \: \text{Cal} = 400. \: \text{kcal} \times \frac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ}\nonumber$
Summary
• The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed.
• A specific portion of matter in a given space that is being studied during an experiment or an observation is the system.
• The surroundings is everything in the universe that is not part of the system.
• A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings.
• A reaction or change is exothermic if heat is released by the system into the surroundings. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.02%3A_Heat.txt |
If a swimming pool and wading pool, both full of water at the same temperature, were subjected to the same input of heat energy, the wading pool would certainly rise in temperature more quickly than the swimming pool. The heat capacity of an object depends both on its mass and its chemical composition. Because of its much larger mass, the swimming pool of water has a larger heat capacity than the wading pool.
Heat Capacity and Specific Heat
Different substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water under the same sun exposure will not become nearly as hot. This means that water has a high heat capacity (the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$). Water is very resistant to changes in temperature, while metals generally are not. The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. The table below lists the specific heats of some common substances. The symbol for specific heat is $c_p$, with the $p$ subscript referring to the fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram per degree $\left( \text{J/g}^\text{o} \text{C} \right)$ or calories per gram per degree $\left( \text{cal/g}^\text{o} \text{C} \right)$. This text will use $\text{J/g}^\text{o} \text{C}$ for specific heat.
Substance Specific Heat $\left( \text{J/g}^\text{o} \text{C} \right)$
Table $1$: Specific Heats of Some Common Substances
Water (l) 4.18
Water (s) 2.06
Water (g) 1.87
Ammonia (g) 2.09
Ethanol (l) 2.44
Aluminum (s) 0.897
Carbon, graphite (s) 0.709
Copper (s) 0.385
Gold (s) 0.129
Iron (s) 0.449
Lead (s) 0.129
Mercury (l) 0.140
Silver (s) 0.233
Notice that water has a very high specific heat compared to most other substances. Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days.
Summary
• Heat capacity is the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$.
• The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$.
17.05: Specific Heat Calculations
Water has a high capacity for absorbing heat. In a car radiator, it serves to keep the engine cooler than it would otherwise run. As the water circulates through the engine, it absorbs heat from the engine block. When it passes through the radiator, the cooling fan and the exposure to the outside environment allow the water to cool somewhat before it makes another passage through the engine.
Specific Heat Calculations
The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat $\left( q \right)$ to specific heat $\left( c_p \right)$, mass $\left( m \right)$, and temperature change $\left( \Delta T \right)$ is shown below.
$q = c_p \times m \times \Delta T\nonumber$
The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by $\Delta T = T_f - T_i$, where $T_f$ is the final temperature and $T_i$ is the initial temperature.
Example $1$
A $15.0 \: \text{g}$ piece of cadmium metal absorbs $134 \: \text{J}$ of heat while rising from $24.0^\text{o} \text{C}$ to $62.7^\text{o} \text{C}$. Calculate the specific heat of cadmium.
Known
• Heat $= q = 134 \: \text{J}$
• Mass $= m = 15.0 \: \text{g}$
• $\Delta T = 62.7^\text{o} \text{C} - 24.0^\text{o} \text{C} = 38.7^\text{o} \text{C}$
Unknown
The specific heat equation can be rearranged to solve for the specific heat.
Step 2: Solve.
$c_p = \dfrac{q}{m \times \Delta T} = \dfrac{134 \: \text{J}}{15.0 \: \text{g} \times 38.7^\text{o} \text{C}} = 0.231 \: \text{J/g}^\text{o} \text{C}\nonumber$
Step 3: Think about your result.
The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.
Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that $60.0 \: \text{g}$ of water at $23.52^\text{o} \text{C}$ was cooled by the removal of $813 \: \text{J}$ of heat. The change in temperature can be calculated using the specific heat equation:
$\Delta T = \dfrac{q}{c_p \times m} = \dfrac{813 \: \text{J}}{4.18 \: \text{J/g}^\text{o} \text{C} \times 60.0 \: \text{g}} = 3.24^\text{o} \text{C}\nonumber$
Since the water was being cooled, the temperature decreases. The final temperature is:
$T_f = 23.52^\text{o} \text{C} - 3.24^\text{o} \text{C} = 20.28^\text{o} \text{C}\nonumber$
Summary
• The specific heat of a substance can be used to calculate the temperature change of the substance when it is heated or cooled.
• Specific heat calculations are illustrated.
17.06: Enthalpy
The factors influencing a reaction are complicated and varied. Since a catalyst affects activation energy, we might assume it would have some sort of impact on the amount of heat that is absorbed or released by the reaction—but it does not. The change in heat content of a reaction depends solely on the chemical compositions of the reactants and products, not on the path taken to get from one to the other.
Enthalpy
Heat changes in chemical reactions are often measured in the laboratory under conditions in which the reacting system is open to the atmosphere. In these cases, the system is at a constant pressure. Enthalpy $\left( H \right)$ is the heat content of a system at constant pressure. Chemists routinely measure changes in enthalpy of chemical systems as reactants are converted into products. The heat that is absorbed or released by a reaction at constant pressure is the same as the enthalpy change, and is given the symbol $\Delta H$. Unless otherwise specified, all reactions in this material are assumed to take place at constant pressure.
The change in enthalpy of a reaction is a measure of the differences in enthalpy of the reactants and products. The enthalpy of a system is determined by the energies needed to break chemical bonds and the energies needed to form chemical bonds. Energy needs to be put into the system in order to break chemical bonds—they do not come apart spontaneously in most cases. Bond formation to produce products will involve release of energy. The change in enthalpy shows the trade-offs made in these two processes. Does it take more energy to break bonds than that needed to form bonds? If so, the reaction is endothermic and the enthalpy change is positive. If more energy is produced in bond formation than that needed for bond breaking, the reaction is exothermic and the enthalpy is negative.
Several factors influence the enthalpy of a system. Enthalpy is an extensive property, determined in part by the amount of material being dealt with. The state of reactants and products (solid, liquid, or gas) influences the enthalpy value for a system. The direction of the reaction affects the enthalpy value. A reaction that takes place in the opposite direction has the same numerical enthalpy value, but the opposite sign.
Summary
• Enthalpy, the heat content of a system at constant pressure, is related to the heat of reaction.
• The enthalpy of a system is determined by the energies needed to break chemical bonds and the energies needed to form chemical bonds.
• Factors that influencing enthalpy include: amount of materials, the state of reactants and products, and direction of the reaction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.04%3A_Heat_Capacity_and_Specific_Heat.txt |
At one time, calories in foods were measured with a bomb calorimeter. A weighed amount of the food would be placed in the calorimeter and the system was then sealed and filled with oxygen. An electric spark ignited the food-oxygen mixture. The amount of heat released when the food burned gave an idea of the calories present within the food. Today, calories are calculated from the protein, carbohydrate, and fat content of food (all determined by chemical analysis).
Calorimetry
Calorimetry is the measurement of the transfer of heat into or out of a system during a chemical reaction or physical process. A calorimeter is an insulated container that is used to measure heat changes. The majority of reactions that can be analyzed in a calorimetry experiment are either liquids or aqueous solutions. A frequently used and inexpensive calorimeter is a set of nested foam cups fitted with a lid to limit the heat exchange between the liquid in the cup and the air in the surroundings (see figure below). In a typical calorimetry experiment, specific volumes of the reactants are dispensed into separate containers and the temperature of each is measured. They are then mixed into the calorimeter, which starts the reaction. The reactant mixture is stirred until the reaction is complete, while the temperature of the reaction is continuously monitored.
The key to all calorimetry experiments is the assumption that there is no heat exchange between the insulated calorimeter and the room. Consider the case of a reaction taking place between aqueous reactants: the water in which the solids have been dissolved is the surroundings, while the dissolved substances are the system. The temperature change that is measured is the temperature change that is occurring in the surroundings. If the temperature of the water increases as the reaction occurs, the reaction is exothermic. Heat was released by the system into the surrounding water. An endothermic reaction absorbs heat from the surroundings, so the temperature of the water decreases as heat leaves the surroundings to enter the system.
The temperature change of the water is measured in the experiment and the specific heat of water can be used to calculate the heat absorbed by the surroundings $\left( q_\text{surr} \right)$.
$q_\text{surr} = m \times c_p \times \Delta T\nonumber$
In the equation, $m$ is the mass of the water, $c_p$ is the specific heat of the water, and $\Delta T$ is $T_f - T_i$. The heat absorbed by the surroundings is equal, but opposite in sign, to the heat released by the system. Because the heat change is determined at constant pressure, the heat released by the system $\left( q_\text{sys} \right)$ is equal to the enthalpy change $\left( \Delta H \right)$.
$q_\text{sys} = \Delta H = -q_\text{surr} = - \left( m \times c_p \times \Delta T \right)\nonumber$
The sign of $\Delta H$ is positive for an endothermic reaction and negative for an exothermic reaction.
Example $1$
In an experiment, $25.0 \: \text{mL}$ of $1.00 \: \text{M} \: \ce{HCl}$ at $25.0^\text{o} \text{C}$ is added to $25.0 \: \text{mL}$ of $1.00 \: \text{M} \: \ce{NaOH}$ at $25.0^\text{o} \text{C}$ in a foam cup calorimeter. A reaction occurs and the temperature rises to $32.0^\text{o} \text{C}$. Calculate the enthalpy change $\left( \Delta H \right)$ in $\text{kJ}$ for this reaction. Assume the densities of the solutions are $1.00 \: \text{g/mL}$ and that their specific heat is the same as that of water.
Known
• $c_p = 4.18 \: \text{J/g}^\text{o} \text{C}$
• $V_\text{final} = 25.0 \: \text{mL} + 25.0 \: \text{mL} = 50.0 \: \text{mL}$
• $\Delta T = 32.0^\text{o} \text{C} - 25.0^\text{o} \text{C} = 7.0^\text{o} \text{C}$
• Density $= 1.00 \: \text{g/mL}$
Unknown
The volume and density can be used to find the mass of the solution after mixing. Then calculate the change in enthalpy by using $\Delta H = q_\text{sys} = -q_\text{surr} = - \left( m \times c_p \times \Delta T \right)$.
Step 2: Solve.
\begin{align*} m &= 50.0 \: \text{mL} \times \frac{1.00 \: \text{g}}{\text{mL}} = 50.0 \: \text{g} \ \Delta H &= - \left( m \times c_p \times \Delta T \right) = - \left( 50.0 \: \text{g} \times 4.18 \: \text{J/g}^\text{o} \text{C} \times 7.0^\text{o} \text{C} \right) = -1463 \: \text{J} = -1.5 \: \text{kJ} \end{align*}\nonumber
Step 3: Think about the result.
The enthalpy change is negative because the reaction releases heat to the surroundings, resulting in an increase in the temperature of the water.
Summary
• Calorimetry is the measurement of the transfer of heat into or out of a system during a chemical reaction or physical process.
• A calorimeter is an insulated container that is used to measure heat changes.
• Calculations involving enthalpy changes are illustrated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.07%3A_Calorimetry.txt |
Heating a home is becoming more and more expensive. The decision to use gas, oil, electricity, or wood can be multi-faceted. Part of the decision is based on which fuel will provide the highest amount of energy release when burned. Studies of thermochemistry can be very useful in getting reliable information for making these important choices.
Thermochemical Equation
When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of $1 \: \text{mol}$ of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of a balanced equation:
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) + 890.4 \: \text{kJ}\nonumber \nonumber$
The equation tells us that $1 \: \text{mol}$ of methane combines with $2 \: \text{mol}$ of oxygen to produce $1 \: \text{mol}$ of carbon dioxide and $2 \: \text{mol}$ of water. In the process, $890.4 \: \text{kJ}$ is released and so it is written as a product of the reaction. A thermochemical equation is a chemical equation that includes the enthalpy change of the reaction. The process in the above thermochemical equation can be shown visually in the figure below.
In the combustion of methane example, the enthalpy change is negative because heat is being released by the system. Therefore, the overall enthalpy of the system decreases. The heat of reaction is the enthalpy change for a chemical reaction. In the case above, the heat of reaction is $-890.4 \: \text{kJ}$. The thermochemical reaction can also be written in this way:
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ}\nonumber \nonumber$
Heats of reaction are typically measured in kilojoules. It is important to include the physical states of the reactants and products in a thermochemical equation, as the value of the $\Delta H$ depends on those states.
Endothermic reactions absorb energy from the surroundings as the reaction occurs. When $1 \: \text{mol}$ of calcium carbonate decomposes into $1 \: \text{mol}$ of calcium oxide and $1 \: \text{mol}$ of carbon dioxide, $177.8 \: \text{kJ}$ of heat is absorbed. The process is shown visually in the figure above (B). The thermochemical reaction is shown below.
$\ce{CaCO_3} \left( s \right) + 177.8 \: \text{kJ} \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber \nonumber$
Because the heat is absorbed by the system, the $177.8 \: \text{kJ}$ is written as a reactant. The heat of reaction is positive for an endothermic reaction.
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = 177.8 \: \text{kJ}\nonumber \nonumber$
The way in which a reaction is written influences the value of the enthalpy change for the reaction. Many reactions are reversible, meaning that the product(s) of the reaction are capable of combining and reforming the reactant(s). If a reaction is written in the reverse direction, the sign of the $\Delta H$ changes. For example, we can write an equation for the reaction of calcium oxide with carbon dioxide to form calcium carbonate:
$\ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \rightarrow \ce{CaCO_3} \left( s \right) + 177.8 \: \text{kJ}\nonumber \nonumber$
The reaction is exothermic and thus the sign of the enthalpy change is negative.
$\ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \rightarrow \ce{CaCO_3} \left( s \right) \: \: \: \: \: \Delta H = -177.8 \: \text{kJ}\nonumber \nonumber$
Summary
• A thermochemical equation is a chemical equation that includes the enthalpy change of the reaction.
• The heat of reaction is the enthalpy change for a chemical reaction.
17.09: Stoichiometric Calculations and Enthalpy Changes
There is a growing concern about damage to the environment done by emissions from manufacturing plants. Many companies are taking steps to reduce these harmful emissions by adding equipment that will trap the pollutants. In order to know what equipment (and the quantity) to order, studies are done to measure the amount of product currently produced. Since pollution is often both particulate and thermal, energy changes need to be determined in addition to the amounts of products released.
Stoichiometric Calculations and Enthalpy Changes
Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems. Refer again to the combustion reaction of methane. Since the reaction of $1 \: \text{mol}$ of methane released $890.4 \: \text{kJ}$, the reaction of $2 \: \text{mol}$ of methane would release $2 \times 890.4 \: \text{kJ} = 1781 \: \text{kJ}$. The reaction of $0.5 \: \text{mol}$ of methane would release $\frac{890,4 \: \text{kJ}}{2} = 445.2 \: \text{kJ}$. As with other stoichiometry problems, the moles of a reactant or product can be linked to mass or volume.
Example $1$
Sulfur dioxide gas reacts with oxygen to form sulfur trioxide in an exothermic reaction, according to the following thermochemical equation.
$2 \ce{SO_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right) + 198 \: \text{kJ}\nonumber$
Calculate the enthalpy change that occurs when $58.0 \: \text{g}$ of sulfur dioxide is reacted with excess oxygen.
Step 1: List the known quantities and plan the problem.
• Mass $\ce{SO_2} = 58.0 \: \text{g}$
• Molar mass $\ce{SO_2} = 64.07 \: \text{g/mol}$
• $\Delta H = -198 \: \text{kJ}$ for the reaction of $2 \: \text{mol} \: \ce{SO_2}$
Unknown
The calculation requires two steps. The mass of $\ce{SO_2}$ is converted to moles. Then the moles of $\ce{SO_2}$ is multiplied by the conversion factor of $\left( \frac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} \right)$.
Step 2: Solve.
$\Delta H = 58.0 \: \text{g} \: \ce{SO_2} \times \frac{1 \: \text{mol} \: \ce{SO_2}}{64.07 \: \text{g} \: \ce{SO_2}} \times \frac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} = 89.6 \: \text{kJ}\nonumber$
Step 3: Think about your result.
The mass of sulfur dioxide is slightly less than $1 \: \text{mol}$. Since $198 \: \text{kJ}$ is released for every $2 \: \text{mol}$ of $\ce{SO_2}$ that reacts, the heat released when about $1 \: \text{mol}$ reacts is one half of 198. The $89.6 \: \text{kJ}$ is slightly less than half of 198. The sign of $\Delta H$ is negative because the reaction is exothermic.
Summary
• Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems.
• Calculations of energy changes in enthalpy equations are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.08%3A_Thermochemical_Equations.txt |
Suppose that you are holding an ice cube in your hand. It feels cold because heat energy leaves your hand and enters the ice cube. What happens to the ice cube? It melts. However, the temperature during a phase change remains constant. So the heat that is being lost by your hand does not raise the temperature of the ice above its melting temperature of $0^\text{o} \text{C}$. Rather, all the heat goes into the change of state. Energy is absorbed during the process of changing ice into water. The water that is produced also remains at $0^\text{o} \text{C}$ until all of the ice is melted.
Heats of Fusion and Solidification
All solids absorb heat as they melt to become liquids. The gain of heat in this endothermic process goes into changing the state, rather than changing the temperature. The molar heat of fusion $\left( \Delta H_\text{fus} \right)$ of a substance is the heat absorbed by one mole of that substance as it is converted from a solid to a liquid. Since the melting of any substance absorbs heat, it follows that the freezing of any substance releases heat. The molar heat of solidification $\left( \Delta H_\text{solid} \right)$ of a substance is the heat released by one mole of that substance as it is converted from a liquid to a solid. Since fusion and solidification of a given substance are the exact opposite processes, the numerical value of the molar heat of fusion is the same as the numerical value of the molar heat of solidification, but opposite in sign. In other words, $\Delta H_\text{fus} = - \Delta H_\text{solid}$. The figure below shows all of the possible changes of state along with the direction of heat flow during each process.
Every substance has a unique value for its molar heat of fusion, depending on the amount of energy required to disrupt the intermolecular forces present in the solid. When $1 \: \text{mol}$ of ice at $0^\text{o} \text{C}$ is converted to $1 \: \text{mol}$ of liquid water at $0^\text{o} \text{C}$, $6.01 \: \text{kJ}$ of heat are absorbed from the surroundings. When $1 \: \text{mol}$ of water at $0^\text{o} \text{C}$ freezes to ice at $0^\text{o} \text{C}$, $6.01 \: \text{kJ}$ of heat is released into the surroundings.
$\begin{array}{ll} \ce{H_2O} \left( s \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H_\text{fus} = 6.01 \: \text{kJ/mol} \ \ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( s \right) & \Delta H_\text{solid} = -6.01 \: \text{kJ/mol} \end{array}\nonumber$
The molar heats of fusion and solidification of a given substance can be used to calculate the heat absorbed or released when various amounts are melted or frozen.
Example $1$
Calculate the heat absorbed when $31.6 \: \text{g}$ of ice at $0^\text{o} \text{C}$ is completely melted.
Known
• Mass $= 31.6 \: \text{g}$ ice
• Molar mass $\ce{H_2O} = 18.02 \: \text{g/mol}$
• Molar heat of fusion $= 6.01 \: \text{kJ/mol}$
Unknown
The mass of ice is first converted to moles. This is then multiplied by the conversion factor of $\left( \frac{6.01 \: \text{kJ}}{1 \: \text{mol}} \right)$ in order to find the $\text{kJ}$ of heat absorbed.
Step 2: Solve.
$31.6 \: \text{g ice} \times \frac{1 \: \text{mol ice}}{18.02 \: \text{g ice}} \times \frac{6.01 \: \text{kJ}}{1 \: \text{mol ice}} = 10.5 \: \text{kJ}\nonumber$
Step 3: Think about your result.
The given quantity is a bit less than 2 moles of ice, and so just less than $12 \: \text{kJ}$ of heat is absorbed by the melting process.
Summary
• The molar heat of fusion $\left( \Delta H_\text{fus} \right)$ of a substance is the heat absorbed by one mole of that substance as it is converted from a solid to a liquid.
• The molar heat of solidification $\left( \Delta H_\text{solid} \right)$ of a substance is the heat released by one mole of that substance as it is converted from a liquid to a solid.
• Calculations of heat changes during fusion and solidification are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.10%3A_Heats_of_Fusion_and_Solidification.txt |
Natural resources for electric power generation have traditionally been waterfalls, oil, coal, or nuclear power. Research is being carried out to look for other renewable sources to run the generators. Geothermal sites (such as geysers) are being considered because of the steam they produce. Capabilities can be estimated by knowing how much steam is released in a given time at a particular site.
Heat of Vaporization and Condensation
Energy is absorbed in the process of converting a liquid at its boiling point into a gas. As with the melting point of a solid, the temperature of a boiling liquid remains constant and the input of energy goes into changing the state. The molar heat of vaporization $\left( \Delta H_\text{vap} \right)$ of a substance is the heat absorbed by one mole of that substance as it is converted from a liquid to a gas. As a gas condenses to a liquid, heat is released. The molar heat of condensation $\left( \Delta H_\text{cond} \right)$ of a substance is the heat released by one mole of that substance as it is converted from a gas to a liquid. Since vaporization and condensation of a given substance are the exact opposite processes, the numerical value of the molar heat of vaporization is the same as the numerical value of the molar heat of condensation, but opposite in sign. In other words, $\Delta H_\text{vap} = -\Delta H_\text{cond}$.
When $1 \: \text{mol}$ of water at $100^\text{o} \text{C}$ and $1 \: \text{atm}$ pressure is converted to $1 \: \text{mol}$ of water vapor at $100^\text{o} \text{C}$, $40.7 \: \text{kJ}$ of heat is absorbed from the surroundings. When $1 \: \text{mol}$ of water vapor at $100^\text{o} \text{C}$ condenses to liquid water at $100^\text{o} \text{C}$, $40.7 \: \text{kJ}$ of heat is released into the surroundings.
$\begin{array}{ll} \ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right) & \Delta H_\text{vap} = 40.7 \: \text{kJ/mol} \ \ce{H_2O} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H_\text{cond} =-40.7 \: \text{kJ/mol} \end{array}\nonumber$
Other substances have different values for their molar heats of fusion and vaporization; these substances are summarized in the table below.
Table $1$: Molar Heats of Fusion and Vaporization
Substance $\Delta H_\text{fus}$ $\left( \text{kJ/mol} \right)$ $\Delta H_\text{vap}$ $\left( \text{kJ/mol} \right)$
Ammonia $\left( \ce{NH_3} \right)$ 5.65 23.4
Ethanol $\left( \ce{C_2H_5OH} \right)$ 4.60 43.5
Methanol $\left( \ce{CH_3OH} \right)$ 3.16 35.3
Oxygen $\left( \ce{O_2} \right)$ 0.44 6.82
Water $\left( \ce{H_2O} \right)$ 6.01 40.7
Notice that for all substances, the heat of vaporization is substantially higher than the heat of fusion. Much more energy is required to change the state from a liquid to a gas than from a solid to a liquid. This is because of the large separation of the particles in the gas state. The values of the heats of fusion and vaporization are related to the strength of the intermolecular forces. All of the substances in the table above, with the exception of oxygen, are capable of hydrogen bonding. Consequently, the heats of fusion and vaporization of oxygen are far lower than the others.
Example $1$
What mass of methanol vapor condenses to a liquid as $20.0 \: \text{kJ}$ of heat is released?
Known
• $\Delta H = 20.0 \: \text{kJ}$
• $\Delta H_\text{cond} = -35.3 \: \text{kJ/mol}$
• Molar mass $\ce{CH_3OH} = 32.05 \: \text{g/mol}$
Unknown
First the $\text{kJ}$ of heat released in the condensation is multiplied by the conversion factor $\left( \frac{1 \: \text{mol}}{-35.3 \: \text{kJ}} \right)$ to find the moles of methanol that condensed. Then, moles are converted to grams.
Step 2: Solve.
$-20.0 \: \text{kJ} \times \frac{1 \: \text{mol} \: \ce{CH_3OH}}{-35.3 \: \text{kJ}} \times \frac{32.05 \: \text{g} \: \ce{CH_3OH}}{1 \: \text{mol} \: \ce{CH_3OH}} = 18.2 \: \text{g} \: \ce{CH_3OH}\nonumber$
Step 3: Think about your result.
Condensation is an exothermic process, so the enthalpy change is negative. Slightly more than one-half mole of methanol is condensed.
Summary
• The molar heat of vaporization $\left( \Delta H_\text{vap} \right)$ is the heat absorbed by one mole of a substance as it is converted from a liquid to a gas.
• The molar heat of condensation $\left( \Delta H_\text{cond} \right)$ is the heat released by one mole of a substance as it is converted from a gas to a liquid.
• Examples of calculations involving the molar heat of vaporization and condensation are illustrated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.11%3A_Heats_of_Vaporization_and_Condensation.txt |
If you have a cube of ice, which process will take more energy—the melting of that ice cube or the conversion of the water to steam? The short answer is that more energy is needed to convert the water to steam. The long answer is really a series of questions: How do you get from one point to the other? What is the temperature of the ice? What is the mass of that ice cube? A long process is involved to take the material from the starting point to the end point.
Multi-Step Problems with Changes of State
Heating curves show the phase changes that a substance undergoes as heat is continuously absorbed.
The specific heat of a substance allows us to calculate the heat absorbed or released as the temperature of the substance changes. It is possible to combine that type of problem with a change of state to solve a problem involving multiple steps. The figure above shows ice at $-30^\text{o} \text{C}$ being converted in a five-step process to gaseous water (steam) at $140^\text{o} \text{C}$. It is now possible to calculate the heat absorbed during that entire process. The process and the required calculations are summarized below.
1. Ice is heated from $-30^\text{o} \text{C}$ to $0^\text{o} \text{C}$. The heat absorbed is calculated by using the specific heat of ice and the equation $\Delta H = c_p \times m \times \Delta T$.
2. Ice is melted at $0^\text{o} \text{C}$. The heat absorbed is calculated by multiplying the moles of ice by the molar heat of fusion.
3. Water at $0^\text{o} \text{C}$ is heated to $100^\text{o} \text{C}$. The heat absorbed is calculated by using the specific heat of water and the equation $\Delta H = c_p \times m \times \Delta T$.
4. Water is vaporized to steam at $100^\text{o} \text{C}$. The heat absorbed is calculated by multiplying the moles of water by the molar heat of vaporization.
5. Steam is heated from $100^\text{o} \text{C}$ to $140^\text{o} \text{C}$. The heat absorbed is calculated by using the specific heat of steam and the equation $\Delta H = c_p \times m \times \Delta T$.
Example $1$
Calculate the total amount of heat absorbed (in $\text{kJ}$) when $2.00 \: \text{mol}$ of ice at $-30^\text{o} \text{C}$ is converted to steam at $140.0^\text{o} \text{C}$. The required specific heats can be found in the table in "Heat Capacity and Specific Heat".
Known
• $2.00 \: \text{mol}$ ice $= 36.04 \: \text{g}$ ice
• $c_p$ (ice) $= 2.06 \: \text{J/g}^\text{o} \text{C}$
• $c_p$ (water) $= 4.18 \: \text{J/g}^\text{o} \text{C}$
• $c_p$ (steam) $= 1.87 \: \text{J/g}^\text{o} \text{C}$
• $\Delta H_\text{fus} = 6.01 \: \text{kJ/mol}$
• $\Delta H_\text{vap} = 40.7 \: \text{kJ/mol}$
Unknown
Follow the steps previously described. Note that the mass of the water is needed for the calculations that involve the specific heat, while the moles of water is needed for the calculations that involve changes of state. All heat quantities must be in kilojoules so that they can be added together to get a total for the five-step process.
Step 2: Solve.
1. $\Delta H_1 = 2.06 \: \text{J/g}^\text{o} \text{C} \times 36.04 \: \text{g} \times 30^\text{o} \text{C} \times \frac{1 \: \text{kJ}}{1000 \: \text{J}} = 2.23 \: \text{kJ}\nonumber$
2. $\Delta H_2 = 2.00 \: \text{mol} \times \frac{6.01 \: \text{kJ}}{1 \: \text{mol}} = 12.0 \: \text{kJ}\nonumber$
3. $\Delta H_3 = 4.18 \: \text{J/g}^\text{o} \text{C} \times 36.04 \: \text{g} \times 100^\text{o} \text{C} \times \frac{1 \: \text{kJ}}{1000 \: \text{kJ}} = 15.1 \: \text{kJ}\nonumber$
4. $\Delta H_4 = 2.00 \: \text{mol} \times \frac{40.7 \: \text{kJ}}{1 \: \text{mol}} = 81.4 \: \text{kJ}\nonumber$
5. $\Delta H_5 = 1.87 \: \text{J/g}^\text{o} \text{C} \times 36.04 \: \text{g} \times 40^\text{o} \text{C} \times \frac{1 \: \text{kJ}}{1000 \: \text{J}} = 2.70 \: \text{kJ}\nonumber$
$\Delta H_\text{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5 = 113.4 \: \text{kJ}\nonumber$
Step 3: Think about your result.
The total heat absorbed as the ice at $-30^\text{o} \text{C}$ is heated to steam at $140^\text{o} \text{C}$ is $133.4 \: \text{kJ}$. The largest absorption of heat comes during the vaporization of the liquid water. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.12%3A_Multi-Step_Problems_with_Changes_of_State.txt |
When preparing dilutions of concentrated sulfuric acid, the directions usually call for adding the acid slowly to water with frequent stirring. When this acid is mixed with water, a great deal of heat is released in the dissolution process. If water were added to acid, the water would quickly heat and splatter, causing harm to the person making the solution.
Heat of Solution
Enthalpy changes also occur when a solute undergoes the physical process of dissolving into a solvent. Hot packs and cold packs (see figure below) use this property. Many hot packs use calcium chloride, which releases heat when it dissolves, according to the equation below.
$\ce{CaCl_2} \left (s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right) + 82.8 \: \text{kJ}\nonumber$
The molar heat of solution $\left( \Delta H_\text{soln} \right)$ of a substance is the heat absorbed or released when one mole of the substance is dissolved in water. For calcium chloride, $\Delta H_\text{soln} = -82.8 \: \text{kJ/mol}$.
Many cold packs use ammonium nitrate, which absorbs heat from the surroundings when it dissolves.
$\ce{NH_4NO_3} \left( s \right) + 25.7 \: \text{kJ} \rightarrow \ce{NH_4^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right)\nonumber$
Cold packs are typically used to treat muscle strains and sore joints. The cold pack is activated and applied to the affected area. As the ammonium nitrate dissolves, it absorbs heat from the body and helps to limit swelling. For ammonium nitrate, $\Delta H_\text{soln} = 25.7 \: \text{kJ/mol}$.
Example $1$
The molar heat of solution, $\Delta H_\text{soln}$, of $\ce{NaOH}$ is $-44.51 \: \text{kJ/mol}$. In a certain experiment, $50.0 \: \text{g}$ of $\ce{NaOH}$ is completely dissolved in $1.000 \: \text{L}$ of $20.0^\text{o} \text{C}$ water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.
Known
• Mass $\ce{NaOH} = 50.0 \: \text{g}$
• Molar mass $\ce{NaOH} = 40.00 \: \text{g/mol}$
• $\Delta H_\text{soln} \: \left( \ce{NaOH} \right) = -44.51 \: \text{kJ/mol}$
• Mass $\ce{H_2O} = 1.000 \: \text{kg} = 1000. \: \text{g}$ (assumes density $= 1.00 \: \text{g/mL}$)
• $T_\text{initial} \: \left( \ce{H_2O} \right) = 20.0^\text{o} \text{C}$
• $c_p \: \left( \ce{H_2O} \right) = 4.18 \: \text{J/g}^\text{o} \text{C}$
Unknown
This is a multiple-step problem:
1) Grams $\ce{NaOH}$ is converted to moles.
2) Moles is multiplied by the molar heat of solution.
3) The joules of heat released in the dissolution process is used with the specific heat equation and the total mass of the solution to calculate the $\Delta T$.
4) The $T_\text{final}$ is determined from $\Delta T$.
Step 2: Solve.
$50.0 \: \text{g} \: \ce{NaOH} \times \frac{1 \: \text{mol} \: \ce{NaOH}}{40.00 \: \text{g} \: \ce{NaOH}} \times \frac{-44.51 \: \text{kJ}}{1 \: \text{mol} \: \ce{NaOH}} \times \frac{1000 \: \text{J}}{1 \: \text{kJ}} = -5.56 \times 10^4 \: \text{J}\nonumber$
$\Delta T = \frac{\Delta H}{c_p \times m} = \frac{-5.56 \times 10^4 \: \text{J}}{4.18 \: \text{J/g}^\text{o} \text{C} \times 1050 \: \text{g}} = 13.2^\text{o} \text{C}\nonumber$
$T_\text{final} = 20.0^\text{o} \text{C} + 13.2^\text{o} \text{C} = 33.2^\text{o} \text{C}\nonumber$
Step 3: Think about your result.
The dissolution process releases a large amount of heat, which causes the temperature of the solution to rise. Care must be taken when preparing concentrated solutions of sodium hydroxide, because of the large amounts of heat released.
Summary
• The molar heat of solution $\left( \Delta H_\text{soln} \right)$ of a substance is the heat absorbed or released when one mole of the substance is dissolved in water.
• Sample calculations using molar heat of solution are given.
17.14: Heat of Combustion
In efforts to reduce gas consumption from oil, ethanol is often added to regular gasoline. It has a high octane rating and burns more slowly than regular gas. This "gasohol" is widely used in many countries. It produces somewhat lower carbon monoxide and carbon dioxide emissions, but does increase air pollution from other materials.
Molar Heat of Combustion
Many chemical reactions are combustion reactions. It is often important to know the energy produced in such a reaction so that we can determine which fuel might be the most efficient for a given purpose. The molar heat of combustion $\left( He \right)$ is the heat released when one mole of a substance is completely burned.
Typical combustion reactions involve the reaction of a carbon-containing material with oxygen to form carbon dioxide and water as products. If methanol is burned in air, we have:
$\ce{CH_3OH} + \ce{O_2} \rightarrow \ce{CO_2} + 2 \ce{H_2O} \: \: \: \: \: He = 890 \: \text{kJ/mol}\nonumber$
In this case, one mole of oxygen reacts with one mole of methanol to form one mole of carbon dioxide and two moles of water.
It should be noted that inorganic substances can also undergo a form of combustion reaction:
$2 \ce{Mg} + \ce{O_2} \rightarrow 2 \ce{MgO}\nonumber$
In this case, there is no water and no carbon dioxide formed. For the purposes of this chapter, these reactions are generally not considered in the discussion of combustion reactions.
Example $1$
Heats of combustion are usually determined by burning a known amount of the material in a bomb calorimeter with an excess of oxygen. By measuring the temperature change, the heat of combustion can be determined.
A 1.55 gram sample of ethanol is burned and produced a temperature increase of $55^\text{o} \text{C}$ in 200 grams of water. Calculate the molar heat of combustion.
Known
• Mass of ethanol $= 1.55 \: \text{g}$
• Molar mass of ethanol $= 46.1 \: \text{g/mol}$
• Mass of water $= 200 \: \text{g}$
• $c_p$ water $= 4.18 \: \text{J/g}^\text{o} \text{C}$
• Temperature increase $= 55^\text{o} \text{C}$
Step 2: Solve.
Amount of ethanol used: $\frac{1.55 \: \text{g}}{46.1 \: \text{g/mol}} = 0.0336 \: \text{mol}\nonumber$
Energy generated: $4.184 \: \text{J/g}^\text{o} \text{C} \times 200 \: \text{g} \times 55^\text{o} \text{C} = 46024 \: \text{J} = 46.024 \: \text{kJ}\nonumber$
Molar heat of combustion: $\frac{46.024 \: \text{kJ}}{0.0336 \: \text{mol}} = 1370 \: \text{kJ/mol}\nonumber$
Step 3: Think about your result.
The burning of ethanol produces a significant amount of heat.
Summary
• The molar heat of combustion $\left( He \right)$ is the heat released when one mole of a substance is completely burned.
• Calculations using the molar heat of combustion are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.13%3A_Heat_of_Solution.txt |
Calculating the energy involved in the operation of an acetylene torch is no simple matter. Since there is a complex series of reactions taking place, simple methods for determining the heat of reaction will not work. We need to develop new approaches to these calculations.
Adding Heats of Reaction
It is sometimes very difficult or even impossible to measure the enthalpy change for a reaction directly in the laboratory. Some reactions take place extremely slowly, making a direct measurement unfeasible. In other cases, a given reaction may be an intermediate step in a series of reactions. Some reactions may be difficult to isolate because multiple side reactions may occur at the same time. Fortunately, it is possible to measure the enthalpy change for a reaction with an indirect method. Hess's law of heat summation states that if two or more thermochemical equations can be added together to give a final equation, then the heats of reaction can also be added to give a heat of reaction for the final equation.
An example will illustrate how Hess's law can be used. Acetylene $\left( \ce{C_2H_2} \right)$ is a gas that burns at an extremely high temperature $\left( 3300^\text{o} \text{C} \right)$ and is used in welding. On paper, acetylene gas can be produced by the reaction of solid carbon (graphite) with hydrogen gas.
$2 \ce{C} \left( s, graphite \right) + \ce{H_2} \left( g \right) \rightarrow \ce{C_2H_2} \left( g \right) \: \: \: \: \: \Delta H = ?\nonumber$
Unfortunately, this reaction would be virtually impossible to perform in the laboratory because carbon would react with hydrogen to form many different hydrocarbon products simultaneously. There is no way to create conditions under which only acetylene would be produced.
However, enthalpy changes for combustion reactions are relatively easy to measure. The heats of combustion for carbon, hydrogen, and acetylene are shown below, along with each balanced equation.
$\begin{array}{ll} \ce{C} \left( s, graphite \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) & \Delta H = -393.5 \: \text{kJ} \ \ce{H_2} \left( g \right) + \frac{1}{2} \ce{O_2} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H = -285.8 \: \text{kJ} \ \ce{C_2H_2} \left( g \right) + \frac{5}{2} \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) & \Delta H = -1301.1 \: \text{kJ} \end{array}\nonumber$
To use Hess's law, we need to determine how the three equations above can be manipulated so that they can be added together to result in the desired equation (the formation of acetylene from carbon and hydrogen).
In order to do this, we will go through the desired equation, one substance at a time—choosing the combustion reaction from the equations above that contains that substance. It may be necessary to either reverse a combustion reaction, or multiply it by some factor in order to make it "fit" to the desired equation. The first reactant is carbon and in the equation for the desired reaction, the coefficient of the carbon is a 2. So, we will write the first combustion reaction, doubling all of the coefficients and the $\Delta H$:
$2 \ce{C} \left( s, graphite \right) + 2 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = 2 \left( -393.5 \right) = -787.0 \: \text{kJ}\nonumber$
The second reactant is hydrogen and its coefficient is a 1, as it is in the second combustion reaction. Therefore, that reaction will be used as written.
$\ce{H_2} \left( g \right) + \frac{1}{2} \ce{O_2} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -285.8 \: \text{kJ}\nonumber$
The product of the reaction is $\ce{C_2H_2}$ and its coefficient is also a 1. In combustion reaction #3, the acetylene is a reactant. Therefore, we will reverse reaction 3, changing the sign of the $\Delta H$:
$2 \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{C_2H_2} \left( g \right) + \frac{5}{2} \ce{O_2} \left( g \right) \: \: \: \: \: \Delta H = 1301.1 \: \text{kJ}\nonumber$
Now, these three reactions can be summed together. Any substance that appears in equal quantities as a reactant in one equation and a product in another equation cancels out algebraically. The values for the enthalpy changes are likewise added.
$\begin{array}{rccclclrcl} 2 \ce{C} \left( s, graphite \right) & + & \cancel{2 \ce{O_2} \left( g \right)} & \rightarrow & \cancel{2 \ce{CO_2} \left( g \right)} & & & \Delta H & = & -787.0 \: \text{kJ} \ \ce{H_2} \left( g \right) & + & \cancel{\frac{1}{2} \ce{O_2} \left( g \right)} & \rightarrow & \cancel{\ce{H_2O} \left( l \right)} & & & \Delta H & = & -285.8 \: \text{kJ} \ \cancel{2 \ce{CO_2} \left( g \right)} & + & \cancel{\ce{H_2O} \left( l \right)} & \rightarrow & \ce{C_2H_2} \left( g \right) & + & \cancel{\frac{5}{2} \ce{O_2} \left( g \right)} & \Delta H & = & 1301.1 \: \text{kJ} \ \hline 2 \ce{C} \left( s, graphite \right) & + & \ce{H_2} \left( g \right) & \rightarrow & \ce{C_2H_2} \left( g \right) & & & \Delta H & = & 228.3 \: \text{kJ} \end{array}\nonumber$
So, the heat of reaction for the combination of carbon with hydrogen to produce acetylene is $228.3 \: \text{kJ}$. When one mole of acetylene is produced, $228.3 \: \text{kJ}$ of heat is absorbed, making the reaction endothermic.
Summary
• Hess's law of heat summation states that if two or more thermochemical equations can be added together to give a final equation, then the heats of reaction can also be added to give a heat of reaction for the final equation.
• Hess's law is used to calculate the heat of reaction for processes that cannot be measured directly. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.15%3A_Hess%27s_Law_of_Heat_Summation.txt |
The Hope diamond, a form of carbon, is a very expensive piece of jewelry—currently worth about \$350,000,000. A graphite pencil, comprised partly of carbon, can be purchased for less than a dollar. Both items contain carbon, but there is a big difference in how that carbon is organized. The diamond was formed under very different reaction conditions than the graphite, so it has a different heat of formation.
Standard Heat of Formation
A relatively straightforward chemical reaction is one in which elements are combined to form a compound. Sodium and chlorine react to form sodium chloride. Hydrogen and oxygen combine to form water. Like other reactions, these are accompanied by either the absorption or release of heat. The standard heat of formation $\left( \Delta H^\text{o}_\text{f} \right)$ is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states. The standard conditions for thermochemistry are $25^\text{o} \text{C}$ and $101.3 \: \text{kPa}$. Therefore, the standard state of an element is its state at $25^\text{o} \text{C}$ and $101.3 \: \text{kPa}$. For example, iron is a solid, bromine is a liquid, and oxygen is a gas under those conditions. The standard heat of formation of an element in its standard state by definition is equal to zero. The $\Delta H^\text{o}_\text{f} = 0$ for the diatomic elements, $\ce{H_2} \left( g \right)$, $\ce{N_2} \left( g \right)$, $\ce{O_2} \left( g \right)$, $\ce{F_2} \left( g \right)$, $\ce{Cl_2} \left( g \right)$, $\ce{Br_2} \left( g \right)$, and $\ce{I_2} \left( g \right)$. The graphite form of solid carbon is in its standard state with $\Delta H^\text{o}_\text{f} = 0$, while diamond is not its standard state. Some standard heats of formation are listed in the table below.
Standard Heats of Formation of Selected Substances
Table $1$: Standard Heats of Formation of Selected Substances
Substance $\Delta H^\text{o}_\text{f}$ $\left( \text{kJ/mol} \right)$ Substance $\Delta H^\text{o}_\text{f}$ $\left( \text{kJ/mol} \right)$
$\ce{Al_2O_3} \left( s \right)$ -1669.8 $\ce{H_2O_2} \left( l \right)$ -187.6
$\ce{BaCl_2} \left( s \right)$ -860.1 $\ce{KCl} \left( s \right)$ -435.87
$\ce{Br_2} \left( g \right)$ 30.91 $\ce{NH_3} \left( g \right)$ -46.3
$\ce{C} \left( s, graphite \right)$ 0 $\ce{NO} \left( g \right)$ 90.4
$\ce{C} \left( s, diamond \right)$ 1.90 $\ce{NO_2} \left( g \right)$ 33.85
$\ce{CH_4} \left( g \right)$ -74.85 $\ce{NaCl} \left( s \right)$ -411.0
$\ce{C_2H_5OH} \left( l \right)$ -276.98 $\ce{O_3} \left( g \right)$ 142.2
$\ce{CO} \left( g \right)$ -110.5 $\ce{P} \left( s, white \right)$ 0
$\ce{CO_2} \left( g \right)$ -393.5 $\ce{P} \left( s, red \right)$ -18.4
$\ce{CaO} \left( s \right)$ -635.6 $\ce{PbO} \left( s \right)$ -217.86
$\ce{CaCO_3} \left( s \right)$ -1206.9 $\ce{S} \left( rhombic \right)$ 0
$\ce{HCl} \left( g \right)$ -92.3 $\ce{S} \left( monoclinic \right)$ 0.30
$\ce{CuO} \left( s \right)$ -155.2 $\ce{SO_2} \left( g \right)$ -296.1
$\ce{CuSO_4} \left( s \right)$ -769.86 $\ce{SO_3} \left( g \right)$ -395.2
$\ce{Fe_2O_3} \left( s \right)$ -822.2 $\ce{H_2S} \left( s \right)$ -20.15
$\ce{H_2O} \left( g \right)$ -241.8 $\ce{SiO_2} \left( s \right)$ -859.3
$\ce{H_2O} \left( l \right)$ -285.8 $\ce{ZnCl_2} \left( s \right)$ -415.89
Summary
• The standard heat of formation $\left( \Delta H^\text{o}_\text{f} \right)$ is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states.
• The standard conditions for thermochemistry are $25^\text{o} \text{C}$ and $101.3 \: \text{kPa}$. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.16%3A_Standard_Heat_of_Formation.txt |
Natural diamonds are mined from sites around the world. However, the price of natural diamonds is carefully controlled, so other sources of diamonds are being explored. Several different methods for producing synthetic diamonds are available, usually involving treating carbon at very high temperatures and pressures. The diamonds produced are now of high quality, but are primarily used in industrial applications. Diamonds are one of the hardest materials available and are widely used for cutting and grinding tools.
Calculating Heat of Reaction from Heat of Formation
An application of Hess's law allows us to use standard heats of formation to indirectly calculate the heat of reaction for any reaction that occurs at standard conditions. An enthalpy change that occurs specifically under standard conditions is called the standard enthalpy (or heat) of reaction and is given the symbol $\Delta H^\text{o}$. The standard heat of reaction can be calculated by using the following equation.
$\Delta H^\text{o} = \sum n \Delta H^\text{o}_\text{f} \: \text{(products)} - \sum n \Delta H^\text{o}_\text{f} \: \text{(reactants)}\nonumber$
The symbol $\Sigma$ is the Greek letter sigma and means "the sum of". The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants. The symbol "$n$" signifies that each heat of formation must first be multiplied by its coefficient in the balanced equation.
Standard Heats of Formation of Selected Substances
Table $1$: Standard Heats of Formation of Selected Substances
Substance $\Delta H^\text{o}_\text{f}$ $\left( \text{kJ/mol} \right)$ Substance $\Delta H^\text{o}_\text{f}$ $\left( \text{kJ/mol} \right)$
$\ce{Al_2O_3} \left( s \right)$ -1669.8 $\ce{H_2O_2} \left( l \right)$ -187.6
$\ce{BaCl_2} \left( s \right)$ -860.1 $\ce{KCl} \left( s \right)$ -435.87
$\ce{Br_2} \left( g \right)$ 30.91 $\ce{NH_3} \left( g \right)$ -46.3
$\ce{C} \left( s, graphite \right)$ 0 $\ce{NO} \left( g \right)$ 90.4
$\ce{C} \left( s, diamond \right)$ 1.90 $\ce{NO_2} \left( g \right)$ 33.85
$\ce{CH_4} \left( g \right)$ -74.85 $\ce{NaCl} \left( s \right)$ -411.0
$\ce{C_2H_5OH} \left( l \right)$ -276.98 $\ce{O_3} \left( g \right)$ 142.2
$\ce{CO} \left( g \right)$ -110.5 $\ce{P} \left( s, white \right)$ 0
$\ce{CO_2} \left( g \right)$ -393.5 $\ce{P} \left( s, red \right)$ -18.4
$\ce{CaO} \left( s \right)$ -635.6 $\ce{PbO} \left( s \right)$ -217.86
$\ce{CaCO_3} \left( s \right)$ -1206.9 $\ce{S} \left( rhombic \right)$ 0
$\ce{HCl} \left( g \right)$ -92.3 $\ce{S} \left( monoclinic \right)$ 0.30
$\ce{CuO} \left( s \right)$ -155.2 $\ce{SO_2} \left( g \right)$ -296.1
$\ce{CuSO_4} \left( s \right)$ -769.86 $\ce{SO_3} \left( g \right)$ -395.2
$\ce{Fe_2O_3} \left( s \right)$ -822.2 $\ce{H_2S} \left( s \right)$ -20.15
$\ce{H_2O} \left( g \right)$ -241.8 $\ce{SiO_2} \left( s \right)$ -859.3
$\ce{H_2O} \left( l \right)$ -285.8 $\ce{ZnCl_2} \left( s \right)$ -415.89
Example $1$
Calculate the standard heat of reaction $\left( \Delta H^\text{o} \right)$ for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas.
Known
• $\Delta H^\text{o}_\text{f}$ for $\ce{NO} \left( g \right) = 90.4 \: \text{kJ/mol}$
• $\Delta H^\text{o}_\text{f}$ for $\ce{O_2} \left( g \right) = 0$ (element)
• $\Delta H^\text{o}_\text{f}$ for $\ce{NO_2} \left( g \right) = 33.85 \: \text{kJ/mol}$
Unknown
First write the balanced equation for the reaction. Then apply the equation to calculate the standard heat of reaction from the standard heats of formation.
Step 2: Solve.
The balanced equation is: $2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)$
Applying the equation from the text:
\begin{align*} \Delta H^\text{o} &= \left[ 2 \: \text{mol} \: \ce{NO_2} \left( 33.85 \: \text{kJ/mol} \right) \right] - \left[ 2 \: \text{mol} \: \ce{NO} \left( 90.4 \: \text{kJ/mol} \right) + 1 \: \text{mol} \: \ce{O_2} \left( 0 \: \text{kJ/mol} \right) \right] \ &= -113 \: \text{kJ} \end{align*}\nonumber
The standard heat of reaction is $-113 \: \text{kJ}\nonumber \] Step 3: Think about your result. The reaction is exothermic, which makes sense because it is a combustion reaction and combustion reactions always release heat. Summary • An enthalpy change that occurs specifically under standard conditions is called the standard enthalpy (or heat) of reaction and is given the symbol \(\Delta H^\text{o}$.
• Standard heats of reaction can be calculated from standard heats of formation. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.17%3A_Calculating_Heat_of_Reaction_from_Heat_of_Formation.txt |
Drag racing is a sport that involves two cars starting from a dead stop, and driving as fast as they can down a quarter-mile strip. At the end of the strip are timers that determine both elapsed time (how long it took for the cars to cover the quarter mile) and top speed (how fast they were going as they went through the timer chute). Both pieces of data are important. One car may accelerate faster and get ahead that way, while the other car may be slower off the line, but can get up to a higher top speed at the end of the run.
Chemical Reaction Rate
Chemical reactions vary widely in terms of the speed with which they occur. Some reactions occur very quickly. If a lit match is brought in contact with lighter fluid or another flammable liquid, it erupts into flames instantly and burns fast. Other reactions occur very slowly. A container of milk in the refrigerator will be good to drink for weeks before it begins to turn sour. Millions of years were required for dead plants under Earth's surface to accumulate and eventually turn into fossil fuels such as coal and oil.
Chemists need to be concerned with the rates at which chemical reactions occur. Rate is another word for speed. If a sprinter takes $11.0 \: \text{s}$ to run a $100 \: \text{m}$ dash, his rate is given by the distance traveled divided by the time.
$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{100 \; \text{m}}{11.0 \: \text{s}} = 9.09 \: \text{m/s}\nonumber$
The sprinter's average running rate for the race is $9.09 \: \text{m/s}$. We say that it is his average rate because he did not run at that speed for the entire race. At the very beginning of the race, while coming from a standstill, his rate must be slower until he is able to get up to his top speed. His top speed must then be greater than $9.09 \: \text{m/s}$ so that, taken over the entire race, the average ends up at $9.09 \: \text{m/s}$.
Chemical reactions can't be measured in units of meters per second, as that would not make any sense. A reaction rate is the change in concentration of a reactant or product with time. Suppose that a simple reaction were to take place in which a $1.00 \: \text{M}$ aqueous solution of substance $\ce{A}$ was converted to substance $\ce{B}$.
$\ce{A} \left( aq \right) \rightarrow \ce{B} \left( aq \right)\nonumber$
Suppose that after 20.0 seconds, the concentration of $\ce{A}$ had dropped from $1.00 \: \text{M}$ to $0.72 \: \text{M}$ as $\ce{A}$ was slowly being converted to $\ce{B}$. We can express the rate of this reaction as the change in concentration of $\ce{A}$ divided by time.
$\text{rate} = -\frac{\Delta \left[ \ce{A} \right]}{\Delta t} = -\frac{\left[ \ce{A} \right]_\text{final} - \left[ \ce{A} \right]_\text{initial}}{\Delta t}\nonumber$
A bracket around a symbol or formula means the concentration in molarity of that substance. The change in concentration of $\ce{A}$ is its final concentration minus its initial concentration. Because the concentration of $\ce{A}$ is decreasing over time, the negative sign is used. Thus, the rate for the reaction is positive and the units are molarity per second or $\text{M/s}$.
$\text{rate} = -\frac{0.72 \: \text{M} - 1.00 \: \text{M}}{20.0 \: \text{s}} = -\frac{-0.28 \: \text{M}}{20.0 \: \text{s}} = 0.014 \: \text{M/s}\nonumber$
The molarity of $\ce{A}$ decreases by an average rate of $0.014 \: \text{M}$ every second. In summary, the rate of a chemical reaction is measured by the change in concentration over time for a reactant or product. The unit of measurement for a reaction rate is molarity per second $\left( \text{M/s} \right)$.
Summary
• Chemists need to be concerned with the rates at which chemical reactions occur.
• The reaction rate indicates how fast the reaction proceeds. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.01%3A_Chemical_Reaction_Rate.txt |
Car damage can be very expensive, especially if the driver who did the damage does not have car insurance. Many people have had the experience of backing up while parallel parking and hearing that "bump". Fortunately, cars are often not going fast enough to do any damage. Every once in a while, though, there is a rearrangement of the body parts of a car when it is hit with sufficient speed; then things need to be fixed.
Collision Theory
The behavior of the atoms, molecules, or ions that comprise the reactants is responsible for the rates of a given chemical reaction. Collision theory is a set of principles that states that the reacting particles can form products when they collide with one another, provided those collisions have enough kinetic energy and the correct orientation. Particles that lack the necessary kinetic energy may collide, but the particles will simply bounce off one another unchanged. The figure below illustrates the difference. In the first collision, the particles bounce off one another, and no rearrangement of atoms has occurred. The second collision occurs with greater kinetic energy, and so the bond between the two red atoms breaks. One red atom bonds with the other molecule as one product, while the single red atom is the other product. The first collision is called an ineffective collision, while the second collision is called an effective collision.
Summary
• Collision theory explains how materials can collide and become new materials.
• Effective collisions result in product formation.
• Ineffective collisions do not result in product formation.
18.03: Activation Energy
The sight of fireworks cascading across the night sky is a hallmark of special occasions. These materials, invented hundreds of years ago, can be dangerous if not handled properly. The chemicals do not react until the fuse burns down and heat is applied to the system. Then, the rocket is launched and explodes high in the sky.
Activation Energy
Why do some chemical reactions occur readily while others require input of heat in order to take place? If we mix metallic sodium with water, a reaction occurs immediately, releasing a great deal of heat (enough to ignite the hydrogen gas that is formed). Group II metals, such as calcium, react at a much slower rate. Unlike the extremely vigorous reaction with sodium, the reaction with calcium is slow enough that we can trap the hydrogen gas released.
Supplying reactant particles with energy causes the bonds between the atoms to vibrate with a greater frequency. This increase in vibrational energy makes a chemical bond more likely to break, and a chemical reaction more likely to occur, when those particles collide with other particles. The activation energy for a reaction is the minimum energy that colliding particles must have in order to undergo a reaction. Some reactions occur readily at room temperature because the reacting particles already have the requisite activation energy at that temperature. Other reactions only occur when heated because the particles do not have enough energy unless an external source of heat provides the particles with more kinetic energy. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.02%3A_Collision_Theory.txt |
Sisyphus was a mythological being who was a very evil king. As a punishment for his misdoings, he was supposed to roll a large stone up to the top of a long hill. A spell had been placed on the stone so that it would roll back down before reaching the top, never to complete the task. Sisyphus was condemned to an eternity of trying to get to the top of the hill, but never succeeding.
Potential Energy Diagrams
The energy changes that occur during a chemical reaction can be shown in a diagram called a potential energy diagram, or sometimes called a reaction progress curve. A potential energy diagram shows the change in potential energy of a system as reactants are converted into products. The figure below shows basic potential energy diagrams for an endothermic (A) and an exothermic (B) reaction. Recall that the enthalpy change $\left( \Delta H \right)$ is positive for an endothermic reaction and negative for an exothermic reaction. This can be seen in the potential energy diagrams. The total potential energy of the system increases for the endothermic reaction as the system absorbs energy from the surroundings. The total potential energy of the system decreases for the exothermic reaction as the system releases energy to the surroundings.
The activation energy for a reaction is illustrated in the potential energy diagram by the height of the hill between the reactants and the products. For this reason, the activation energy of a reaction is sometimes referred to as the activation energy barrier. Reacting particles must have enough energy so that when they collide, they can overcome that barrier (see figure below).
Summary
• A potential energy diagram shows the change in potential energy of a system as reactants are converted into products.
• Potential energy diagrams for endothermic and exothermic reactions are described.
• Diagrams of activation energy and reaction progress are given.
18.05: Activated Complex
Velcro is a synthetic material that allows fabric (among other things) to stick together. Another, more unusual, use for Velcro is the sport of "Velcro-jumping". The participant wears clothing made of Velcro and jumps at a Velcro-covered wall. Sometimes the collision with the wall will result in the person sticking to the wall. Other times, the person simply bounces off the wall and does not connect.
Activated Complex
Reactant particles sometimes collide with one another and remain unchanged by the collision. Other times, the collision leads to the formation of products. The state of the particles that is in between the reactants and products is called the activated complex. An activated complex is an unstable arrangement of atoms that exists momentarily at the peak of the activation energy barrier. Because of its high energy, the activated complex exists for an extremely short period of time (about $10^{-13} \: \text{s}$). There is equal likelihood that the activated complex either reforms the original reactants or goes on to form products. The figure below shows the formation of a possible activated complex between colliding hydrogen and oxygen molecules. Because of their unstable nature and brief existence, very little is known about the exact structures of many activated complexes.
Summary
• An activated complex is an unstable arrangement of atoms that exists momentarily at the peak of the activation energy barrier.
• The role of the activated complex in reactions is described.
18.06: Factors Affecting Reaction Rate
Driving on a crowded freeway can be stressful. Lots of cars, drivers who aren't paying attention, people who speed or who drive too slow—the chances of a collision are rather high. A lot of cars in a particular amount of space means a high car concentration and many opportunities for unwanted connections with other cars.
Factors Affecting Reaction Rates
According to their nature, some reactions occur very quickly, while others are very slow. However, certain changes in the reacting conditions can have an effect on the rate of a given chemical reaction. Collision theory can be utilized to explain these rate effects.
Concentration
An increase in the concentration of one or more reacting substances results in an increase in the rate of reaction. When more particles are present in a given amount of space, a greater number of collisions will naturally occur between those particles. Since the rate of a reaction is dependent on the number of collisions occurring between reactants, the rate increases as the concentration increases.
Pressure
When the pressure of a gas is increased, its particles are forced closer together, decreasing the amount of empty space between the particles. Therefore, an increase in the pressure of a gas is also an increase in the concentration of the gas. For gaseous reactions, an increase in pressure increases the rate of reaction due to a greater number of collisions between reacting particles.
Surface Area
A large log placed in a fire will burn relatively slowly. If the same mass of wood were added to the fire in the form of small twigs, the twigs would burn much more quickly. This is because the twigs provide a greater surface area than the log does. An increase in the surface area of a reactant increases the rate of a reaction. Surface area is larger when a given amount of a solid is present as smaller particles. A powdered reactant has a greater surface area than the same reactant as a solid chunk. In order to increase the surface area of a substance, it may be ground into smaller particles or dissolved into a liquid. In solution, the dissolved particles are separated from each other and will react more quickly with other reactants.
Temperature
Raising the temperature of a chemical reaction usually results in a higher rate of reaction. When the reactant particles are heated, they move faster and faster. This results in a greater frequency of collisions. A more important effect of the temperature increase is that the collisions occur with a greater force, and are thus more likely to surmount the activation energy barrier and go on to form products. Increasing the temperature of a reaction increases the number of effective collisions between reacting particles, so the reaction rate increases.
Summary
• Factors that affect (and generally increase) reaction rate are:
• Concentration of reactants
• Pressure (in the case of a gas)
• Surface area
• Temperature | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams.txt |
Gasoline-powered vehicles emit various harmful materials. Nitrogen oxides are formed when atmospheric nitrogen reacts with oxygen at the high temperatures found in a car engine. Carbon monoxide is a by-product of the incomplete combustion of hydrocarbons. Evaporated and unused fuel releases volatile hydrocarbons into the atmosphere to help form smog. The presence of a catalytic converter in the exhaust system of the car causes these materials to react and be changed into less harmful products.
Catalysts
Sometimes a particular substance added to a chemical reaction will cause that reaction to undergo a dramatic increase in rate. Hydrogen peroxide is used as a disinfectant for scrapes and cuts, and is found in many medicine cabinets as a $3\%$ aqueous solution. Hydrogen peroxide naturally decomposes to produce water and oxygen gas, but the reaction is very slow. A bottle of hydrogen peroxide will last for several years before it needs to be replaced. However, the addition of just a small amount of manganese (IV) oxide to hydrogen peroxide will cause it to decompose completely in just a matter of minutes. A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without being used up in the reaction. After the reaction occurs, a catalyst returns to its original state; so catalysts can be used over and over again. Because it is neither a reactant nor a product, a catalyst is shown in a chemical equation by being written above the yield arrow.
$\ce{H_2O_2} \left( aq \right) \overset{\ce{MnO_2}}{\rightarrow} 2 \ce{H_2O} \left( l \right) + \ce{O_2} \left( g \right)\nonumber$
A catalyst works by changing the specific way in which the reaction occurs, called its mechanism. The important outcome from the use of a catalyst is that the overall activation energy of the reaction is lowered (see figure below). With a lower activation energy barrier, a greater percentage of reactant molecules are able to have effective collisions, and the reaction rate increases.
Catalysts are an extremely important part of many chemical reactions. Enzymes in your body acts as nature's catalysts, allowing important biochemical reactions to occur at reasonable rates. Chemical companies constantly search for new and better catalysts to make reactions go faster and thus make the company more profitable.
Summary
• A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without being used up in the reaction.
• Catalysts are an important part of many chemical reactions.
18.08: Rate Law and Specific Rate Constant
Where are people moving from and where are they moving to? How fast is the population changing in different areas? These are important considerations for those individuals or companies who decide where to build schools or hospitals, or where to open new businesses. If an area is growing rapidly, action needs to be taken quickly to accommodate the growth. Rates of change affect many decisions.
Rate Law and Specific Rate Constant
Consider a simple chemical reaction in which reactant $\ce{A}$ is converted into product $\ce{B}$, according to the equation below.
$\ce{A} \rightarrow \ce{B}\nonumber$
The rate of reaction is given by the change in concentration of $\ce{A}$ as a function of time. The rate of disappearance of $\ce{A}$ is also proportional to the concentration of $\ce{A}$.
$-\frac{\Delta \left[ \ce{A} \right]}{\Delta t} \propto \left[ \ce{A} \right]\nonumber$
Since the rate of a reaction generally depends upon collision frequency, it stands to reason that as the concentration of $\ce{A}$ increases, the rate of reaction increases. Likewise, as the concentration of $\ce{A}$ decreases, the rate of reaction decreases. The expression for the rate of the reaction can be shown as follows:
$\text{rate} = -\frac{\Delta \left[ \ce{A} \right]}{\Delta t} \: \: \: \text{or} \: \: \: \text{rate} = k \left[ \ce{A} \right]\nonumber$
The proportionality between the rate and $\left[ \ce{A} \right]$ becomes an equal sign by the insertion of a constant $\left( k \right)$. A rate law is an expression showing the relationship of the reaction rate to the concentrations of each reactant. The specific rate constant $\left( k \right)$ is the proportionality constant relating the rate of the reaction to the concentrations of reactants. The rate law and the specific rate constant for any chemical reaction must be determined experimentally. The value of the rate constant is temperature dependent. A large value of the rate constant means that the reaction is relatively fast, while a small value of the rate constant means that the reaction is relatively slow.
Summary
• A rate law is an expression showing the relationship of the reaction rate to the concentrations of each reactant.
• The specific rate constant $\left( k \right)$ is the proportionality constant relating the rate of the reaction to the concentrations of reactants.
• The rate law and the specific rate constant for any chemical reaction must be determined experimentally. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.07%3A_Catalysts.txt |
Forest fires cause extensive damage; both plant and animal life are harmed. The severity of a forest fire depends on how much plant life is available to burn—the more available dry plant material, the more serious the fire, and the more rapidly it will spread.
Order of Reaction
In the reaction $\ce{A} \rightarrow \ce{B}$, the rate of the reaction is directly proportional to the concentration of $\ce{A}$ raised to the first power. That is to say, $\left[ \ce{A} \right] = \left[ \ce{A} \right]^1$. A first-order reaction is a reaction in which the rate is proportional to the concentration of only one reactant. As a first-order reaction proceeds, the rate of reaction decreases because the concentration of the reactant decreases (see figure below). The graph of concentration versus time is curved. The reaction rate $\left( \frac{\Delta \left[ \ce{A} \right]}{\Delta t} \right)$ can be determined graphically by the slope of a tangent to the curve at any point. The rate of the reaction at the time shown with the red triangle is given by:
$\text{rate} = -\frac{\left[ \ce{A} \right]_\text{final} - \left[ \ce{A} \right]_\text{initial}}{\Delta t} = -\frac{0.35 \: \text{M} - 0.63 \: \text{M}}{3.0 \: \text{s} - 1.0 \: \text{s}} = 0.14 \: \text{M/s}\nonumber$
The rates of some reactions depend on the concentrations of more than one reactant. Consider a reaction in which a molecule of $\ce{A}$ collides with a molecule of $\ce{B}$ to form product $\ce{C}$.
$\ce{A} + \ce{B} \rightarrow \ce{C}\nonumber$
Doubling the concentration of $\ce{A}$ alone would double the reaction rate. Likewise, doubling the concentration of $\ce{B}$ alone would also double the rate. The rate law must reflect the rate dependence on both reactants.
$\text{rate} = k \left[ \ce{A} \right] \left[ \ce{B} \right]\nonumber$
This reaction is said to be first order with respect to $\ce{A}$ and first order with respect to $\ce{B}$. Overall, it is a second-order reaction. The rate law and the order of a reaction must be determined experimentally.
Summary
• A first-order reaction is a reaction in which the rate is proportional to the concentration of only one reactant.
• As a first-order reaction proceeds, the rate of reaction decreases because the concentration of the reactant decreases (see figure below).
18.10: Determining the Rate Law from Experimental Data
Determining the amount of time a process requires calls for a timer. These devices can be simple kitchen timers (not very precise) or complex systems that can measure to a fraction of a second. Accurate time measurement is essential in kinetics studies for assessing rates of chemical reactions.
Determining the Rate Law from Experimental Data
In order to experimentally determine a rate law, a series of experiments must be performed with various starting concentrations of reactants. The initial rate law is then measured for each of the reactions. Consider the reaction between nitrogen monoxide gas and hydrogen gas to form nitrogen gas and water vapor:
$2 \ce{NO} \left( g \right) + 2 \ce{H_2} \left( g \right) \rightarrow \ce{N_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\nonumber$
The following data were collected for this reaction at $1280^\text{o} \text{C}$ (see table below).
Reaction between nitrogen monoxide gas and hydrogen gas to form nitrogen gas and water vapor
Table $1$
Experiment $\left[ \ce{NO} \right]$ $\left[ \ce{H_2} \right]$ Initial Rate $\left( \text{M/s} \right)$
1 0.0050 0.0020 $1.25 \times 10^{-5}$
2 0.010 0.0020 $5.00 \times 10^{-5}$
3 0.010 0.0040 $1.00 \times 10^{-4}$
Notice that the starting concentrations of $\ce{NO}$ and $\ce{H_2}$ were varied in a specific way. In order to compare the rates of reaction and determine the order with respect to each reactant, the initial concentration of each reactant must be changed while the other is held constant.
Comparing experiments 1 and 2: The concentration of $\ce{NO}$ was doubled, while the concentration of $\ce{H_2}$ was held constant. The initial rate of the reaction quadrupled, since $\frac{5.00 \times 10^{-5}}{1.25 \times 10^{-5}} = 4$. Therefore, the order of the reaction with respect to $\ce{NO}$ is 2. In other words, $\text{rate} \propto \left[ \ce{NO} \right]^2$. Because $2^2 = 4$, the doubling of $\left[ \ce{NO} \right]$ results in a rate that is four times greater.
Comparing experiments 2 and 3: The concentration of $\ce{H_2}$ was doubled while the concentration of $\ce{NO}$ was held constant. The initial rate of the reaction doubled, since $\frac{1.00 \times 10^{-4}}{5.00 \times 10^{-5}} = 2$. Therefore, the order of the reaction with respect to $\ce{H_2}$ is 1, or $\text{rate} \propto \left[ \ce{H_2} \right]^1$. Because $2^1 = 2$, the doubling of $\ce{H_2}$ results in a rate that is twice as great. The overall rate law then includes both of these results.
$\text{rate} = k \left[ \ce{NO} \right]^2 \left[ \ce{H_2} \right]\nonumber$
The sum of the exponents is $2 + 1 = 3$, making the reaction third-order overall. Once the rate law for a reaction is determined, the specific rate constant can be found by substituting the data for any of the experiments into the rate law and solving for $k$.
$k = \frac{\text{rate}}{\left[ \ce{NO} \right]^2 \left[ \ce{H_2} \right]} = \frac{1.25 \times 10^{-5} \: \text{M/s}}{\left( 0.0050 \: \text{M} \right)^2 \left( 0.0020 \: \text{M} \right)} = 250 \: \text{M}^{-2} \text{s}^{-1}\nonumber$
Notice that the rate law for the reaction does not relate to the balanced equation for the overall reaction. The coefficients of $\ce{NO}$ and $\ce{H_2}$ are both 2, while the order of the reaction with respect to the $\ce{H_2}$ is only one. The units for the specific rate constant vary with the order of the reaction. So far, we have seen reactions that are first or second order with respect to a given reactant. Occasionally, the rate of a reaction may not depend on the concentration of one of the reactants at all. In this case, the reaction is said to be zero-order with respect to that reactant.
Summary
• The process of using experimental data to determine a rate law is described.
• Once the rate law for a reaction is determined, the specific rate constant can be found by substituting the data for any of the experiments.
• Occasionally, the rate of a reaction may not depend on the concentration of one of the reactants at all; the reaction is said to be zero-order with respect to that reactant. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.09%3A_Order_of_Reaction.txt |
What most people conceptualize as an airplane is the final product. They do not see the long, complex assembly line that begins with a few parts and gradually becomes the finished product. Assembly lines are intricate organizations that go through a series of complicated steps to bring hundreds (maybe thousands) of parts together into the completed aircraft.
Reaction Mechanisms and the Elementary Step
Chemical reactions rarely occur in one simple step. The overall balanced equation for a chemical reaction does not always tell us how a reaction actually proceeds. In many cases, the overall reaction takes place in a series of small steps. An elementary step (or elementary reaction) is one step in a series of simple reactions that show the progress of a reaction at the molecular level. A reaction mechanism is the sequence of elementary steps that together comprise an entire chemical reaction. As an analogy, consider the route that you might take while driving to the grocery store. That route may consist of several turns, similar to the elementary steps. The overall reaction specifies only the beginning point (your house) and the final destination (the store), with no information about the specifics in between.
Summary
• An elementary step (or elementary reaction) is one step in a series of simple reactions that show the progress of a reaction at the molecular level.
• A reaction mechanism is the sequence of elementary steps that together comprise an entire chemical reaction.
18.12: Reaction Intermediate
Ozone $\left( \ce{O_3} \right)$ depletion in the atmosphere is of significant concern. This gas serves as a protection against the ultraviolet rays of the sun. Ozone is naturally depleted in addition to the depletion caused by human-made chemicals. The depletion reaction is a two-step process:
$\ce{O_3} + \text{ultraviolet light} \rightarrow \ce{O_2} + \ce{O} \cdot \: \text{(free radical) slow reaction}\nonumber$
$\ce{O} \cdot + \ce{O_3} \rightarrow 2 \ce{O_2} \: \text{fast reaction}\nonumber$
The free radical is not a part of the overall equation, but can be detected in the lab.
Intermediate
Reaction mechanisms describe how the material in a chemical reaction gets from the initial reactants to the final products. One reaction that illustrates a reaction mechanism is the reaction between nitrogen monoxide and oxygen to form nitrogen dioxide:
$2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\nonumber$
It may seem as though this reaction would occur as the result of a collision between two $\ce{NO}$ molecules with one $\ce{O_2}$ molecule. However, careful analysis of the reaction has detected the presence of $\ce{N_2O_2}$ during the reaction. A proposed mechanism for the reaction consists of two elementary steps:
Step 1: $2 \ce{NO} \left( g \right) \rightarrow \ce{N_2O_2} \left( g \right)$
Step 2: $\ce{N_2O_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)$
In the first step, two molecules of $\ce{NO}$ collide to form a molecule of $\ce{N_2O_2}$. In the second step, that molecule of $\ce{N_2O_2}$ collides with a molecule of $\ce{O_2}$ to produce two molecules of $\ce{NO_2}$. The overall chemical reaction is the sum of the two elementary steps:
\begin{align*} 2 \ce{NO} \left( g \right) &\rightarrow \cancel{\ce{N_2O_2} \left( g \right)} \ \cancel{\ce{N_2O_2} \left( g \right)} + \ce{O_2} \left( g \right) &\rightarrow 2 \ce{NO_2} \left( g \right) \ \hline 2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) &\rightarrow 2 \ce{NO_2} \left( g \right) \end{align*}\nonumber
The $\ce{N_2O_2}$ molecule is not part of the overall reaction. It was produced in the first elementary step, then reacts in the second elementary step. An intermediate is a species which appears in the mechanism of a reaction, but not in the overall balanced equation. An intermediate is always formed in an early step in the mechanism and consumed in a later step.
Summary
• Reaction mechanisms describe how the material in a chemical reaction gets from the initial reactants to the final products.
• An intermediate is a species which appears in the mechanism of a reaction, but not in the overall balanced equation. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.11%3A_Reaction_Mechanisms_and_the_Elementary_Step.txt |
Putting jigsaw puzzles together is an enjoyable hobby for some. Often these puzzles come in a box, so all the pieces must be spread out before starting the puzzle. Various internet sites sell jigsaw puzzles. Buyers can choose the level of difficulty, the shape of the pieces—and can time themselves to see how well they do compared to others that tried the puzzle. The puzzle often looks complicated, with the final product generally comprised of hundreds of pieces. However, in its assembly, there is a series of elementary steps, and the puzzle goes together one piece at a time.
Molecularity of a Reaction
The molecularity of a reaction is the number of molecules reacting in an elementary step. A unimolecular reaction is one in which only one reacting molecule participates in the reaction. Two reactant molecules collide with one another in a bimolecular reaction. A termolecular reaction involves three reacting molecules in one elementary step. Termolecular reactions are relatively rare because they involve the simultaneous collision of three molecules in the correct orientation, a rare event. When termolecular reactions do occur, they tend to be very slow.
Given the reaction:
$2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\nonumber$
We might guess that the reaction was termolecular since it appears that three molecules of reactants are involved. However, our definition of molecularity states that we need to look at an elementary step, and not the overall reaction. Data on the reaction mechanism demonstrates that the reaction occurs in two steps:
Step 1: $2 \ce{NO} \left( g \right) \rightarrow \ce{N_2O_2} \left( g \right)$
Step 2: $\ce{N_2O_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)$
So we see that each elementary step is bimolecular and not termolecular. Notice that the colliding molecules may be the same (as in step 1 above) or different (as in step 2 above).
Another reaction involves the conversion of ozone $\left( \ce{O_3} \right)$ to oxygen $\left( \ce{O_2} \right)$ with ultraviolet light. The two elementary steps are as follows:
$\ce{O_3} + \text{ultraviolet light} \rightarrow \ce{O_2} + \ce{O} \cdot \: \text{(free radical)}\nonumber$
$\ce{O} \cdot + \ce{O_3} \rightarrow 2 \ce{O_2}\nonumber$
The first step is unimolecular (one molecule of ozone reacts) and the second step is bimolecular (one ozone free radical and one ozone molecule react together).
Summary
• The molecularity of a reaction is the number of molecules reacting in an elementary step.
• Reactions can be unimolecular (one reacting molecule), bimolecular (two reacting molecules) or termolecular (three reacting molecules).
18.14: Rate-Determining Step
Airline travel can be very frustrating. Travelers usually have to get to the airport two hours before their flight leaves. They stand in line to check their baggage and get a boarding pass. Then they stand in line for security screening. Finally, travelers wait in line to board the plane. Since there are only so many ticket agents, not everybody can be waited on immediately. The same with the security screen—only so many body scanners are available. And getting on the plane involves going one-by-one down a very narrow aisle to get to a designated seat. All these limits slow down the airline travel process.
Rate-Determining Step
The determination of a reaction mechanism can only be made in the laboratory. When a reaction occurs in a sequence of elementary steps, the overall reaction rate is governed by whichever one of those steps is the slowest. The rate-determining step is the slowest step in the sequence of steps in a reaction mechanism. Imagine driving on a one-lane road where it is not possible to pass another vehicle. The rate of flow of traffic on such a road would be dictated by whichever car is traveling at the lowest speed. The rate-determining step is a similar concept to this slow car determining the traffic flow rate—the overall reaction rate is determined by the slowest part of the process.
Decomposition of Hydrogen Peroxide
Recall that a catalyst is a substance that increases the rate of a chemical reaction without being consumed. Catalysts lower the overall activation energy for a reaction by providing an alternative mechanism for the reaction to follow. One such catalyst for the decomposition of hydrogen peroxide is the iodide ion $\left( \ce{I^-} \right)$.
$2 \ce{H_2O_2} \left( aq \right) \overset{\ce{I^-}}{\rightarrow} 2 \ce{H_2O} \left( l \right) + \ce{O_2} \left( g \right)\nonumber$
By experiment, the rate of reaction is found to be first-order with respect to both $\ce{H_2O_2}$ and $ce{I^-}$, and second order overall.
$\text{rate} = k \left[ \ce{H_2O_2} \right] \left[ \ce{I^-} \right]\nonumber$
The reaction cannot occur in one step corresponding to the overall balanced equation. If it did, the reaction would be second-order with respect to $\ce{H_2O_2}$, since the coefficient of the $\ce{H_2O_2}$ in the balanced equation is a 2. A reaction mechanism that accounts for the rate law, and for the detection of the $\ce{IO^-}$ ion as an intermediate, can be constructed. It consists of two bimolecular elementary steps:
Step 1: $\ce{H_2O_2} \left( aq \right) + \ce{I^-} \left( aq \right) \rightarrow \ce{H_2O} \left( l \right) + \ce{IO^-} \left( aq \right)$
Step 2: $\ce{H_2O_2} \left( aq \right) + \ce{IO^-} \left( aq \right) \rightarrow \ce{H_2O_2} \left( l \right) + \ce{O_2} \left( g \right) + \ce{I^-} \left( aq \right)$
If step 2 is the rate-determining step, then the rate law for that step will be the rate law for the overall reaction.
$\text{rate} = k \left[ \ce{H_2O_2} \right] \left[ \ce{I^-} \right]\nonumber$
The rate law for the slow step of the proposed mechanism agrees with the overall experimentally determined rate law. The $\ce{IO^-}$ is present as an intermediate in the reaction. The iodide catalyst also appears in the mechanism. It is consumed in the first elementary step and then is regenerated in the second step. This is the requirement for a catalyst—to not be used up in the reaction.
Summary
• The rate-determining step in a reaction is defined.
• The process for determining the rate-determining step is described.
18.15: Mechanisms and Potential Energy Diagrams
For some, a roller coaster is the epitome of excitement, while for others it is a torture device to be avoided at all costs (and some folks are in-between). For a roller coaster, the rate-limiting step is the climb up to the top at the beginning of the ride, where the cars are hauled slowly to the peak of the ride. This peak must be the highest because the potential energy at this point must be enough to cause the roller coaster to move through the smaller peaks without stopping.
Potential Energy Diagrams
The potential energy diagram can illustrate the mechanism for a reaction by showing each elementary step of the reaction with distinct activation energy (see figure below).
The reaction whose potential energy diagram is shown in the figure is a two-step reaction. The activation energy for each step is labeled $E_{a1}$ and $E_{a2}$. Each elementary step has its own activated complex, labeled $AC_1$ and $AC_2$. Note that the overall enthalpy change of the reaction $\left( \Delta H \right)$ is unaffected by the individual steps, since it depends only on the initial and final states.
In this example, the rate-limiting step in the reaction is the first step, leading to the formation of the activated complex $AC_1$. The activation energy is higher for this step than for step two, which has a considerably lower activation energy. If the rate-limiting step were the second step, the peak labeled $AC_2$ would be the higher peak than $AC_1$ and $E_{a2}$ would be greater than $E_{a1}$. The same approach can be taken for a potential energy diagram with more than two peaks. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.13%3A_Molecularity.txt |
A solution of cobalt chloride in water is pink due to the presence of the solvated $\ce{Co^{2+}}$ ion. If sufficient $\ce{HCl}$ is added, the solution turns blue as the $\ce{CoCl_4^{2-}}$ ion forms. The reaction can be shifted back to the pink form if more water is added to the solution.
Chemical Reversibility
Up until this point, we have written the equations for chemical reactions in a way that would seem to indicate that all reactions proceed completely until all the reactants have been converted into products. In reality, a great many chemical reactions do not proceed entirely to completion. A reversible reaction is a reaction in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously. One example of a reversible reaction is the reaction of hydrogen gas and iodine vapor to form hydrogen iodide. The forward and reverse reactions can be written as follows.
$\text{Forward reaction:} \: \: \: \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\nonumber$
$\text{Reverse reaction:} \: \: \: 2 \ce{HI} \left( g \right) \rightarrow \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right)\nonumber$
In the forward reaction, hydrogen and iodine combine to form hydrogen iodide. In the reverse reaction, hydrogen iodide decomposes back into hydrogen and iodine. The two reactions can be combined into one equation by the use of a double arrow:
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightleftharpoons 2 \ce{HI} \left( g \right)\nonumber$
The double arrow is the indication that the reaction is reversible.
When hydrogen and iodine gases are mixed in a sealed container, they begin to react and form hydrogen iodide. At first, only the forward reaction occurs because no $\ce{HI}$ is present. As the forward reaction proceeds, it begins to slow down as the concentrations of the $\ce{H_2}$ and the $\ce{I_2}$ decrease. As soon as some $\ce{HI}$ has formed, it begins to decompose back into $\ce{H_2}$ and $\ce{I_2}$. The rate of the reverse reaction starts out slow because the concentration of $\ce{HI}$ is low. Gradually, the rate of the forward reaction decreases while the rate of the reverse reaction increases. Eventually the rate of combination of $\ce{H_2}$ and $\ce{I_2}$ to produce $\ce{HI}$ becomes equal to the rate of decomposition of $\ce{HI}$ into $\ce{H_2}$ and $\ce{I_2}$. When the rates of the forward and reverse reactions have become equal to one another, the reaction has achieved a state of balance.
19.02: Chemical Equilibrium
A tug of war involves two teams at the ends of a rope. The goal is to pull the other team over a line in the middle. At first, there is a great deal of tension on the rope, but no apparent movement. A bystander might think that there is nothing happening. In reality, there is a great deal of tension on the rope as the two teams pull in opposite directions at the same time.
Chemical Equilibrium
Hydrogen and iodine gases react to form hydrogen iodide, according to the following reaction:
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightleftharpoons 2 \ce{HI} \left( g \right)\nonumber$
$\text{Forward reaction:} \: \: \: \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\nonumber$
$\text{Reverse reaction:} \: \: \: 2 \ce{HI} \left( g \right) \rightarrow \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right)\nonumber$
Initially, only the forward reaction occurs because no $\ce{HI}$ is present. As soon as some $\ce{HI}$ has formed, it begins to decompose back into $\ce{H_2}$ and $\ce{I_2}$. Gradually, the rate of the forward reaction decreases while the rate of the reverse reaction increases. Eventually, the rate of combination of $\ce{H_2}$ and $\ce{I_2}$ to produce $\ce{HI}$ becomes equal to the rate of decomposition of $\ce{HI}$ into $\ce{H_2}$ and $\ce{I_2}$. When the rates of the forward and reverse reactions have become equal to one another, the reaction has achieved a state of balance. Chemical equilibrium is the state of a system in which the rate of the forward reaction is equal to the rate of the reverse reaction.
Chemical equilibrium can be attained whether the reaction begins with all reactants and no products, all products and no reactants, or some of both. The figure below shows changes in concentration of $\ce{H_2}$, $\ce{I_2}$, and $\ce{HI}$ for two different reactions. In the reaction depicted by the graph on the left (A), the reaction begins with only $\ce{H_2}$ and $\ce{I_2}$ present. There is no $\ce{HI}$ initially. As the reaction proceeds toward equilibrium, the concentrations of the $\ce{H_2}$ and $\ce{I_2}$ gradually decrease, while the concentration of the $\ce{HI}$ gradually increases. When the curve levels out and the concentrations all become constant, equilibrium has been reached. At equilibrium, concentrations of all substances are constant. In reaction B, the process begins with only $\ce{HI}$ and no $\ce{H_2}$ or $\ce{I_2}$. In this case, the concentration of $\ce{HI}$ gradually decreases while the concentrations of $\ce{H_2}$ and $\ce{I_2}$ gradually increase until equilibrium is again reached. Notice that in both cases, the relative position of equilibrium is the same, as shown by the relative concentrations of reactants and products. The concentration of $\ce{HI}$ at equilibrium is significantly higher than the concentrations of $\ce{H_2}$ and $\ce{I_2}$. This is true whether the reaction began with all reactants or all products. The position of equilibrium is a property of the particular reversible reaction and does not depend upon how equilibrium was achieved.
Conditions for Equilibrium
It may be tempting to think that once equilibrium has been reached, the reaction stops. Chemical equilibrium is a dynamic process. The forward and reverse reactions continue to occur even after equilibrium has been reached. However, because the rates of the reactions are the same, there is no change in the relative concentrations of reactants and products for a reaction that is at equilibrium. The conditions and properties of a system at equilibrium are summarized below.
1. The system must be closed, meaning no substances can enter or leave the system.
2. Equilibrium is a dynamic process. Even though we don't necessarily see the reactions, both the forward and reverse reactions are taking place.
3. The rates of the forward and reverse reactions must be equal.
4. The amount of reactants and products do not have to be equal. However, after equilibrium is attained, the amounts of reactants and products will be constant.
The description of equilibrium in this concept refers primarily to equilibrium between reactants and products in a chemical reaction. Other types of equilibrium include phase equilibrium and solution equilibrium. A phase equilibrium occurs when a substance is in equilibrium between two states. For example, a stoppered flask of water attains equilibrium when the rate of evaporation is equal to the rate of condensation. A solution equilibrium occurs when a solid substance is in a saturated solution. At this point, the rate of dissolution is equal to the rate of recrystallization. Although these are all different types of transformations, most of the rules regarding equilibrium apply to any situation in which a process occurs reversibly.
Summary
• Chemical equilibrium is the state of a system in which the rate of the forward reaction is equal to the rate of the reverse reaction.
• Conditions for chemical equilibrium are listed.
• A phase equilibrium occurs when a substance is in equilibrium between two states.
• A solution equilibrium occurs when a solid substance is in a saturated solution.
19.03: Equilibrium Constant
Red blood cells transport oxygen to tissues so that they can function. In the absence of oxygen, cells cannot carry out their biochemical responsibilities. Oxygen moves to the cells attached to hemoglobin, a protein found in red blood cells.
The protein inside (a) red blood cells that carries oxygen to cells and carbon dioxide to the lungs is (b) hemoglobin. Hemoglobin is made up of four symmetrical subunits and four heme groups. Iron associated with the heme binds oxygen. It is the iron in hemoglobin that gives blood its red color. (CC_BY Openstax)
In cases of carbon monoxide poisoning, $\ce{CO}$ binds much more strongly to the hemoglobin, blocking oxygen attachment and lowering the amount of oxygen reaching the cells. Treatment involves the patient breathing pure oxygen to displace the carbon monoxide. The equilibrium reaction shown below illustrates the shift toward the right when excess oxygen is added to the system:
$\ce{Hb(CO)_4} \left( aq \right) + 4 \ce{O_2} \left( g \right) \rightleftharpoons \ce{Hb(O_2)_4} \left( aq \right) + 4 \ce{CO} \left( g \right)\nonumber$
Equilibrium Constant
Consider the hypothetical reversible reaction in which reactants $\ce{A}$ and $\ce{B}$ react to form products $\ce{C}$ and $\ce{D}$. This equilibrium can be shown as demonstrated below, where the lower case letters represent the coefficients of each substance.
$a \ce{A} + b \ce{B} \rightleftharpoons c \ce{C} + d \ce{D}\nonumber$
As we have established, the rates of the forward and reverse reactions are the same at equilibrium, and so the concentrations of all of the substances are constant. Since that is the case, it stands to reason that a ratio of the concentrations for any given reaction at equilibrium maintains a constant value. The equilibrium constant $\left( K_\text{eq} \right)$ is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For the general reaction above, the equilibrium constant expression is written as follows:
$K_\text{eq} = \frac{\left[ \ce{C} \right]^c \left[ \ce{D} \right]^d}{\left[ \ce{A} \right]^a \left[ \ce{B} \right]^b}\nonumber$
The concentrations of each substance, indicated by the square brackets around the formula, are measured in molarity units $\left( \text{mol/L} \right)$.
The value of the equilibrium constant for any reaction is only determined by experiment. As detailed in the previous section, the position of equilibrium for a given reaction does not depend on the starting concentrations, and so the value of the equilibrium constant is truly constant. It does, however, depend on the temperature of the reaction. This is because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant. For any reaction in which a $K_\text{eq}$ is given, the temperature should be specified. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.01%3A_Reversible_Reaction.txt |
Iron is an important component of red blood cells. Patients who have low iron will usually be anemic and have a lower than normal number of red blood cells. One way to assess serum iron concentration is with the use of Ferrozine, a complex organic molecule. Ferrozine forms a product with $\ce{Fe^{3+}}$, producing a pink color. In order to determine factors affecting the reaction, we need to measure the equilibrium constant. If the equilibrium does not lie far in the direction of products, precautions need to be taken when using this material to measure iron in serum.
Calculations with Equilibrium Constants
The general value of the equilibrium constant gives us information about whether the reactants or the products are favored at equilibrium. Since the product concentrations are in the numerator of the equilibrium expression, a $K_\text{eq} > 1$ means that the products are favored over the reactants. A $K_\text{eq} < 1$ means that the reactants are favored over the products.
Though it would often seem that the $K_\text{eq}$ value would have various units depending on the values of the exponents in the expression, the general rule is that any units are dropped. All $K_\text{eq}$ values will be reported as having no units.
Example $1$
Equilibrium occurs when nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide gas:
$2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightleftharpoons 2 \ce{NO_2} \left( g \right)\nonumber$
At equilibrium at $230^\text{o} \text{C}$, the concentrations are measured to be $\left[ \ce{NO} \right] = 0.0542 \: \text{M}$, $\left[ \ce{O_2} \right] = 0.127 \: \text{M}$, and $\left[ \ce{NO_2} \right] = 15.5 \: \text{M}$. Calculate the equilibrium constant at this temperature.
Known
• $\left[ \ce{NO} \right] = 0.0542 \: \text{M}$
• $\left[ \ce{O_2} \right] = 0.127 \: \text{M}$
• $\left[ \ce{NO_2} \right] = 15.5 \: \text{M}$
Unknown
The equilibrium expression is first written according to the general form in the text. The equilibrium values are substituted into the expression and the value calculated.
Step 2: Solve.
$K_\text{eq} = \frac{\left[ \ce{NO_2} \right]^2}{\left[ \ce{NO} \right]^2 \left[ \ce{O_2} \right]}\nonumber$
Substituting in the concentrations at equilibrium:
$K_\text{eq} = \frac{\left( 15.5 \right)^2}{\left( 0.0542 \right)^2 \left( 0.127 \right)} = 6.44 \times 10^5\nonumber$
Step 3: Think about your result.
The equilibrium concentration of the product $\ce{NO_2}$ is significantly higher than the concentrations of the reactants $\ce{NO}$ and $\ce{O_2}$. As a result, the $K_\text{eq}$ value is much larger than 1, an indication that the product is favored at equilibrium.
The equilibrium expression only shows those substances whose concentrations are variable during the reaction. A pure solid or a pure liquid does not have a concentration that will vary during a reaction. Therefore, an equilibrium expression omits pure solids and liquids, and only shows the concentrations of gases and aqueous solutions. The decomposition of mercury (II) oxide can be shown by the following equation, followed by its equilibrium expression.
$2 \ce{HgO} \left( s \right) \rightleftharpoons 2 \ce{Hg} \left( l \right) + \ce{O_2} \left( g \right) \: \: \: \: \: K_\text{eq} = \left[ \ce{O_2} \right]\nonumber$
The stoichiometry of an equation can also be used in a calculation of an equilibrium constant. At $40^\text{o} \text{C}$, solid ammonium carbamate decomposes to ammonia and carbon dioxide gases.
$\ce{NH_4CO_2NH_2} \left( s \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) + \ce{CO_2} \left( g \right)\nonumber$
At equilibrium, the $\left[ \ce{CO_2} \right]$ is found to be $4.71 \times 10^{-3} \: \text{M}$. Can the $K_\text{eq}$ value be calculated from that information alone? Because the ammonium carbamate is a solid, it is not present in the equilibrium expression.
$K_\text{eq} = \left[ \ce{NH_3} \right]^2 \left[ \ce{CO_2} \right]\nonumber$
The stoichiometry of the chemical equation indicates that as the ammonium carbamate decomposes, $2 \: \text{mol}$ of ammonia gas is produced for every $1 \: \text{mol}$ of carbon dioxide. Therefore, at equilibrium, the concentration of the ammonia will be twice the concentration of carbon dioxide. So,
$\left[ \ce{NH_3} \right] = 2 \times \left( 4.71 \times 10^{-3} \right) = 9.42 \times 10^{-3} \: \text{M}$
Substituting these values into the $K_\text{eq}$ expression:
$K_\text{eq} = \left( 9.42 \times 10^{-3} \right)^2 \left( 4.71 \times 10^{-3} \right) = 4.18 \times 10^{-7}\nonumber$
Using Equilibrium Constants
The equilibrium constants are known for a great many reactions. Hydrogen and bromine gases combine to form hydrogen bromide gas. At $730^\text{o} \text{C}$, the equation and $K_\text{eq}$ are given below.
$\ce{H_2} \left( g \right) + \ce{Br_2} \left( g \right) \rightleftharpoons 2 \ce{HBr} \left( g \right) \: \: \: \: \: K_\text{eq} = 2.18 \times 10^6\nonumber$
A certain reaction is begun with only $\ce{HBr}$. When the reaction mixture reaches equilibrium at $730^\text{o} \text{C}$, the concentration of bromine gas is measured to be $0.00243 \: \text{M}$. What is the concentration of the $\ce{H_2}$ and the $\ce{HBr}$ at equilibrium?
Since the reaction begins with only $\ce{HBr}$ and the mole ratio of $\ce{H_2}$ to $\ce{Br_2}$ is 1:1, the concentration of $\ce{H_2}$ at equilibrium is also $0.00243 \: \text{M}$. The equilibrium expression can be rearranged to solve for the concentration of $\ce{HBr}$ at equilibrium:
\begin{align*} K_\text{eq} &= \frac{\left[ \ce{HBr} \right]^2}{\left[ \ce{H_2} \right] \left[ \ce{Br_2} \right]} \ \left[ \ce{HBr} \right] &= \sqrt{K_\text{eq} \left[ \ce{H_2} \right] \left[ \ce{Br_2} \right]} \ &= \sqrt{2.18 \times 10^6 \left( 0.00243 \right) \left( 0.00243 \right)} = 3.59 \: \text{M} \end{align*}\nonumber
Since the value of the equilibrium constant is very high, the concentration of $\ce{HBr}$ is much greater than that of $\ce{H_2}$ and $\ce{Br_2}$ at equilibrium.
19.05: Le Chatelier's Principle
There are some who enjoy going up in an airplane, strapping on a parachute, and diving out the door to free-fall before opening the chute, and dropping to the ground. This stressful activity (so they say) relieves the stress of everyday life. The release of adrenaline caused by this stressful behavior is said to promote a mood enhancement that helps you deal better with other stresses in your daily life.
Le Chatelier's Principle
Chemical equilibrium was studied by French chemist Henri Le Chatelier (1850-1936) and his description of how a system responds to a stress on equilibrium has become known as Le Chatelier's principle: When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentration of reactants or products, changes in the temperature of the system, or changes in the pressure of the system. We will discuss each of these stresses separately. The change to the equilibrium position in every case is either a favoring of the forward reaction or a favoring of the reverse reaction. When the forward reaction is favored, the concentrations of products increase, while the concentrations of reactants decrease. When the reverse reaction is favored, the concentrations of products decrease, while the concentrations of reactants increase.
Le Chatelier's Principle
Table $1$
Original Equilibrium Favored Reaction Result
$\ce{A} \rightleftharpoons \ce{B}$ Forward: $\ce{A} \rightarrow \ce{B}$ $\left[ \ce{A} \right]$ decreases; $\left[ \ce{B} \right]$ increases
$\ce{A} \rightleftharpoons \ce{B}$ Reverse: $\ce{A} \leftarrow \ce{B}$ $\left[ \ce{A} \right]$ increases; $\left[ \ce{B} \right]$ decreases | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.04%3A_Calculations_with_Equilibrium_Constants.txt |
Phenolphthalein is a chemical that has one structure in a high acid environment, and another structure in a low acid environment. If the hydrogen ion concentration is high, the compound is colorless, but turns red if the hydrogen ion concentration is low. By adding hydrogen ions to the solution, or removing them through a chemical reaction, we can vary the color of the dye.
Effect of Concentration
A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases:
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right)\nonumber$
If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more $\ce{N_2}$ is added, the forward reaction will be favored because the forward reaction uses up $\ce{N_2}$ and converts it to $\ce{NH_3}$. The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more $\ce{NH_3}$ is produced. The concentration of $\ce{NH_3}$ increases, while the concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substances. As can be seen in the figure below, if more $\ce{N_2}$ is added, a new equilibrium is achieved by the system. The new concentration of $\ce{NH_3}$ is higher because of the favoring of the forward reaction. The new concentration of the $\ce{H_2}$ is lower. The concentration of $\ce{N_2}$ is higher than in the original equilibrium, but went down slightly following the addition of the $\ce{N_2}$ that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, $K_\text{eq}$, does not change as a result of the stress to the system.
If more $\ce{NH_3}$ were added, the reverse reaction would be favored. This "favoring" of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of $\ce{NH_3}$ would result in increased formation of the reactants, $\ce{N_2}$ and $\ce{H_2}$.
An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, $\ce{NH_3}$ is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more $\ce{NH_3}$ will be produced. The concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease. Continued removal of $\ce{NH_3}$ will eventually force the reaction to go to completion until all of the reactants are used up. If either $\ce{N_2}$ or $\ce{H_2}$ were removed from the equilibrium system, the reverse reaction would be favored, and the concentration of $\ce{NH_3}$ would decrease.
The effects of changes in concentration on an equilibrium system, according to Le Chatelier's principle, are summarized in the table below.
Effects of changes in concentration on an equilibrium system, according to Le Chatelier's principle
Table $1$
Stress Response
Addition of reactant Forward reaction favored.
Addition of product Reverse reaction favored.
Removal of reactant Reverse reaction favored.
Removal of product Forward reaction favored.
19.07: Effect of Temperature
Carbon monoxide is often thought of as nothing more than a hazardous gas produced from incomplete combustion of carbon products. However, there is a large market for industrially-manufactured carbon monoxide that is used to synthesize most of the acetic acid produced in the world. One reaction that leads to $\ce{CO}$ formation involves its formation by passing air over excess carbon at high temperatures. The initial product (carbon dioxide) equilibrates with the remaining hot carbon, forming carbon monoxide. At lower temperatures, $\ce{CO_2}$ formation is favored, while $\ce{CO}$ is the predominant product above $800^\text{o} \text{C}$.
Effect of Temperature
Increasing or decreasing the temperature of a system at equilibrium is also a stress to the system. The equation for the Haber-Bosch process is written again below, as a thermochemical equation.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) + 91 \: \text{kJ}\nonumber$
The forward reaction is the exothermic direction: the formation of $\ce{NH_3}$ releases heat. The reverse reaction is the endothermic direction: as $\ce{NH_3}$ decomposes to $\ce{N_2}$ and $\ce{H_2}$, heat is absorbed. An increase in the temperature of a system favors the direction of the reaction that absorbs heat, the endothermic direction. Absorption of heat in this case is a relief of the stress provided by the temperature increase. For the Haber-Bosch process, an increase in temperature favors the reverse reaction. The concentration of $\ce{NH_3}$ in the system decreases, while the concentrations of $\ce{N_2}$ and $\ce{H_2}$ increase.
A decrease in the temperature of a system favors the direction of the reaction that releases heat, the exothermic direction. For the Haber-Bosch process, a decrease in temperature favors the forward reaction. The concentration of $\ce{NH_3}$ in the system increases, while the concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease.
For changes in concentration, the system responds in such a way that the value of the equilibrium constant, $K_\text{eq}$, is unchanged. However, a change in temperature shifts the equilibrium, and the $K_\text{eq}$ value either increases or decreases. As discussed in the previous section, values of $K_\text{eq}$ are dependent on the temperature. When the temperature of the system for the Haber-Bosch process is increased, the resultant shift in equilibrium toward the reactants means that the $K_\text{eq}$ value decreases. When the temperature is decreased, the shift in equilibrium toward the products means that the $K_\text{eq}$ value increases.
Le Chatelier's principle, as related to temperature changes, can be illustrated by the reaction in which dinitrogen tetroxide is in equilibrium with nitrogen dioxide:
$\ce{N_2O_4} \left( g \right) + \text{heat} \rightleftharpoons 2 \ce{NO_2} \left( g \right)\nonumber$
Dinitrogen tetroxide $\left( \ce{N_2O_4} \right)$ is colorless, while nitrogen dioxide $\left( \ce{NO_2} \right)$ is dark brown in color. When $\ce{N_2O_4}$ breaks down into $\ce{NO_2}$, heat is absorbed according to the forward reaction above. Therefore, an increase in temperature of the system will favor the forward reaction. Conversely, a decrease in temperature will favor the reverse reaction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.06%3A_Effect_of_Concentration.txt |
The ammonia storage tank in the picture below does two things. First, it stores ammonia at high pressure to minimize the reverse reaction that would lead to less ammonia, and more nitrogen and hydrogen. Secondly, it sends an important message. Ammonia is used to make methamphetamine, a dangerous drug of abuse. Locks and other safety mechanisms built into the tanks help stop the theft of ammonia to be used for this illicit activity.
Effect of Pressure
Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again to the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston, as shown in the figure below.
On the far left, the reaction system contains primarily $\ce{N_2}$ and $\ce{H_2}$, with only one molecule of $\ce{NH_3}$ present. As the piston is pushed inwards, the pressure of the system increases according to Boyle's Law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to Le Chatelier's principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored, in which one molecule of $\ce{N_2}$ combines with three molecules of $\ce{H_2}$ to form two molecules of $\ce{NH_3}$. The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which produces fewer total moles of gas. In this case, it is the forward reaction that is favored.
A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which $\ce{NH_3}$ decomposes to $\ce{N_2}$ and $\ce{H_2}$. This is because the overall number of gas molecules would increase, and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction that produces more total moles of gas. This is summarized in the table below.
Table $1$
Stress Response
Pressure increase Reaction produces fewer gas molecules.
Pressure decrease Reaction produces more gas molecules.
Like changes in concentration, the $K_\text{eq}$ value for a given reaction is unchanged by a change in pressure.
It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. Calcium carbonate decomposes according to the equilibrium reaction:
$\ce{CaCO_3} \left( s \right) \rightleftharpoons \ce{CaO} \left( s \right) + \ce{O_2} \left( g \right)\nonumber$
Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of $\ce{CaCO_3}$ because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of $\ce{HCl}$ from $\ce{H_2}$ and $\ce{Cl_2}$.
$\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightleftharpoons 2 \ce{HCl} \left( g \right)\nonumber$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.08%3A_Effect_of_Pressure.txt |
Fires are a part of life. Some fires clear the land and allow new growth. Other fires provide warmth on a cold night. Unfortunately, many fires are destructive, leaving damage in their wake. All fires leave the environment changed, never to revert back to its original state. The carbon dioxide and water generated by a fire go off into the atmosphere and do not return. The change is permanent and irreversible.
Going to Completion
When one of the products of a reaction is removed from the chemical equilibrium system as soon as it is produced, the reverse reaction cannot establish itself and equilibrium is never reached. Reactions such as these are said to go to completion. These processes are often referred to as non-reversible reactions. Reactions which go to completion tend to produce one of three types of products: (1) an insoluble precipitate, (2) a gas, (3) a molecular compound such as water. Examples of these reactions are shown below.
1. Formation of a precipitate:
$\ce{AgNO_3} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber$
2. Formation of a gas:
$\ce{Mg} \left( s \right) + 2 \ce{HCl} \left( aq \right) \rightarrow \ce{MgCl_2} \left( aq \right) + \ce{H_2} \left( g \right)\nonumber$
3. Formation of water:
$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)\nonumber$
If we look at these reactions in more detail, we can see some things that are not apparent the way the equations are written. Looking at the first equation, we do not see a double arrow between reactants and products, because the reaction is considered essentially irreversible. However, if we consider the net ionic equation
$\ce{Ag^+} + \ce{Cl^-} \rightarrow \ce{AgCl}\nonumber$
then the reverse reaction would be:
$\ce{AgCl} \rightarrow \ce{Ag^+} + \ce{Cl^-}\nonumber$
The $K_\text{eq}$ for the reverse reaction is $1.8 \times 10^{-10}$. For all practical purposes, the reaction goes to completion.
Formation of a gas in an open system is essentially irreversible since the gas escapes into the atmosphere. Looking at the activity series, we see that $\ce{Mg}$ is much higher in the series than hydrogen. So the reaction would be expected to go strongly in the indicated direction.
The third reaction gets a little more complicated. In solution, the reactants $\ce{HCl}$ and $\ce{NaOH}$ will be ionized completely, as does the $\ce{NaCl}$ product. Water exists in an equilibrium with $\ce{H^+}$ and $\ce{OH^-}$, with the dissociation constant for water being $1 \times 10^{-14}$. So, in the solution resulting from the reaction given here, the $\left[ \ce{H^+} \right]$ is $1 \times 10^{-7} \: \text{M}$, a very insignificant amount. For all practical purposes, this reaction can be said to go to completion.
19.10: Le Chatelier's Principle and the Equilibrium Constant
With online banking, management of personal finances can become less complicated in some ways. You can automatically deposit paychecks, pay bills, and designate how much goes into savings or other special accounts each month. If you want to maintain $10\%$ of your bank account in savings, you can set up a program that moves money in and out of the account when you get a paycheck or pay bills. The amount of money in savings will change as the money comes in and out of the bank, but the ratio of savings to checking will always be constant.
Le Chatelier's Principle and the Equilibrium Constant
Occasionally, when students apply Le Chatelier's principle to an equilibrium problem involving a change in concentration, they assume that $K_\text{eq}$ must change. This seems logical since we talk about "shifting" the equilibrium in one direction or the other. However, $K_\text{eq}$ is a constant, for a given equilibrium at a given temperature, so it must not change. Here is an example of how this works. Consider the simplified equilibrium below:
$\ce{A} \rightleftharpoons \ce{B}\nonumber$
Let's say we have a 1.0 liter container. At equilibrium, the following amounts are measured:
\begin{align*} \ce{A} &= 0.50 \: \text{mol} \ \ce{B} &= 1.0 \: \text{mol} \end{align*}\nonumber
The value of $K_\text{eq}$ is given by:
$K_\text{eq} = \frac{\left[ \ce{B} \right]}{\left[ \ce{A} \right]} = \frac{1.0 \: \text{M}}{0.50 \: \text{M}} = 2.0\nonumber$
Now we will disturb the equilibrium by adding 0.50 moles of $\ce{A}$ to the mixture. The equilibrium will shift towards the right, forming more $\ce{B}$. Immediately after the addition of $\ce{A}$ and before any response, we how have $1.0 \: \text{mol}$ of $\ce{A}$ and $1.0 \: \text{mol}$ of $\ce{B}$. The equilibrium then shifts in the forward direction. We will introduce a variable $\left( x \right)$, which will represent the change in concentrations as the reaction proceeds. Since the mole ratio of $\ce{A}$:$\ce{B}$ is 1:1, as $\left[ \ce{A} \right]$ decreases by the amount $x$, the $\left[ \ce{B} \right]$ increases by the amount $x$. We set up an analysis called ICE, which stands for Initial, Change, and Equilibrium. The values in the table below represent molar concentrations.
$\begin{array}{l|ll} & \ce{A} & \ce{B} \ \hline \text{Initial} & 1.0 & 1.0 \ \text{Change} & -x & +x \ \text{Equilibrium} & 1.0 - x & 1.0 + x \end{array}\nonumber$
At the new equilibrium position, the values for $\ce{A}$ and $\ce{B}$ as a function of $x$ can be set equal to the value of the $K_\text{eq}$. Then, one can solve for $x$.
$K_\text{eq} = 2.0 = \frac{\left[ \ce{B} \right]}{\left[ \ce{A} \right]} = \frac{1.0 + x}{1.0 - x}\nonumber$
Solving for $x$:
\begin{align*} 2.0 \left( 1.0 - x \right) &= 1.0 + x \ 2.0 - 2.0x &= 1.0 + x \ 3.0x &= 1.0 \ x &= 0.33 \end{align*}\nonumber
This value for $x$ is now plugged back in to the Equilibrium line of the table, and used to find the final concentrations of $\ce{A}$ and $\ce{B}$ after the reaction.
$\left[ \ce{A} \right] = 1.0 - x = 0.67 \: \text{M}\nonumber$
$\left[ \ce{B} \right] = 1.0 + x = 1.33 \: \text{M}\nonumber$
The value of $K_\text{eq}$ has been maintained since $\frac{1.33}{0.67} = 2.0$. This shows that even though a change in concentration of one of the substances in equilibrium causes a shift in the equilibrium position, the value of the equilibrium constant does not change. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.09%3A_Nonreversible_Reactions.txt |
At one time, a major analytical technique was gravimetric analysis. Gravimetric analysis involves an ion being precipitated out of solution, purified, and weighed to determine the amount of that ion in the original material. As an example, measurement of $\ce{Ca^{2+}}$ involved dissolving the sample in water, precipitating the calcium as calcium oxalate, purifying the precipitate, drying it, and weighing the final product. Although this approach can be very accurate (atomic weights for many elements were determined this way), the process is slow, tedious, and prone to a number of errors in technique. Newer methods are now available that measure minute amounts of calcium ions in solution without the long, involved gravimetric approach.
Solubility Product Constant
Ionic compounds have widely differing solubilities. Sodium chloride has a solubility of about $360 \: \text{g}$ per liter of water at $25^\text{o} \text{C}$. Salts of alkali metals tend to be quite soluble. On the other end of the spectrum, the solubility of zinc hydroxide is only $4.2 \times 10^{-4} \text{g/L}$ of water at the same temperature. Many ionic compounds containing hydroxide are relatively insoluble.
Most ionic compounds that are considered to be insoluble will still dissolve to a small extent in water. These "mostly insoluble" compounds are considered to be strong electrolytes, because the portion of the compound that dissolves also dissociates. As an example, silver chloride dissociates to a small extent into silver ions and chloride ions upon being added to water.
$\ce{AgCl} \left( s \right) \rightleftharpoons \ce{Ag^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber$
The process is written as an equilibrium because the dissociation occurs only to a small extent. Therefore, an equilibrium expression can be written for the process. Keep in mind that the solid silver chloride does not have a variable concentration, and so is not included in the expression.
$K_\text{sp} = \left[ \ce{Ag^+} \right] \left[ \ce{Cl^-} \right]\nonumber$
This equilibrium constant is called the solubility product constant $\left( K_\text{sp} \right)$ and is equal to the mathematical product of the ions each raised to the power of the coefficient of the ion in the dissociation equation.
The stoichiometry of the formula of the ionic compound dictates the form of the $K_\text{sp}$ expression. For example, the formula of calcium phosphate is $\ce{Ca_3(PO_4)_2}$. The dissociation equation and $K_\text{sp}$ expression are shown below:
$\ce{Ca_3(PO_4)_2} \left( s \right) \rightleftharpoons 3 \ce{Ca^{2+}} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \: \: \: K_\text{sp} = \left[ \ce{Ca^{2+}} \right]^3 \left[ \ce{PO_4^{3-}} \right]^2\nonumber$
The table below lists solubility product constants for some common nearly insoluble ionic compounds.
Solubility Product Constants $\left( 25^\text{o} \text{C} \right)$
Table $1$: Solubility Product Constants $\left( 25^\text{o} \text{C} \right)$
Compound $K_\text{sp}$ Compound $K_\text{sp}$
$\ce{AgBr}$ $5.0 \times 10^{-13}$ $\ce{CuS}$ $8.0 \times 10^{-37}$
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Fe(OH)_2}$ $7.9 \times 10^{-16}$
$\ce{Al(OH)_3}$ $3.0 \times 10^{-34}$ $\ce{Mg(OH)_2}$ $7.1 \times 10^{-12}$
$\ce{BaCO_3}$ $5.0 \times 10^{-9}$ $\ce{PbCl_2}$ $1.7 \times 10^{-5}$
$\ce{BaSO_4}$ $1.1 \times 10^{-10}$ $\ce{PbCO_3}$ $7.4 \times 10^{-14}$
$\ce{CaCO_3}$ $4.5 \times 10^{-9}$ $\ce{PbI_2}$ $7.1 \times 10^{-9}$
$\ce{Ca(OH)_2}$ $6.5 \times 10^{-6}$ $\ce{PbSO_4}$ $6.3 \times 10^{-7}$
$\ce{Ca_3(PO_4)_2}$ $1.2 \times 10^{-26}$ $\ce{Zn(OH)_2}$ $3.0 \times 10^{-16}$
$\ce{CaSO_4}$ $2.4 \times 10^{-5}$ $\ce{ZnS}$ $3.0 \times 10^{-23}$
Summary
• The solubility product constant is equal to the mathematical product of ions each raised to the power of their coefficients in a dissociation equation.
• Calculations using solubility product constants are illustrated.
19.12: Conversion of Solubility to (K textsp)
Baking soda (sodium bicarbonate) is prepared by bubbling carbon dioxide gas through a solution of ammonia and sodium chloride. Ammonium carbonate is first formed, which then reacts with the $\ce{NaCl}$ to form sodium bicarbonate and ammonium chloride. The sodium bicarbonate is less soluble than the other materials, so it will precipitate out of solution.
Conversion of Solubility to $K_\text{sp}$
Solubility is normally expressed in $\text{g/L}$ of saturated solution. However, solubility can also be expressed in moles per liter. Molar solubility is the number of moles of solute in one liter of saturated solution. In other words, the molar solubility of a given compound represents the highest molarity solution that is possible for that compound. The molar mass of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of $\ce{Zn(OH)_2}$ is $4.2 \times 10^{-4} \: \text{g/L}$, the molar solubility can be calculated as shown below:
$\frac{4.2 \times 10^{-4} \: \cancel{\text{g}}}{\text{L}} \times \frac{1 \: \text{mol}}{99.41 \: \cancel{\text{g}}} = 4.2 \times 10^{-6} \: \text{mol/L} \: \left( \text{M} \right)\nonumber$
Solubility data can be used to calculate the $K_\text{sp}$ for a given compound. The following steps need to be taken.
1. Convert from solubility to molar solubility.
2. Use the dissociation equation to determine the concentration of each of the ions in $\text{mol/L}$.
3. Apply the $K_\text{sp}$ equation.
Example $1$
The solubility of lead (II) fluoride is found experimentally to be $0.533 \: \text{g/L}$. Calculate the $K_\text{sp}$ for lead (II) fluoride.
Known
• Solubility of $\ce{PbF_2} = 0.533 \: \text{g/L}$
• Molar mass $= 245.20 \: \text{g/mol}$
Unknown
The dissociation equation for $\ce{PbF_2}$ and the corresponding $K_\text{sp}$ expression:
$\ce{PbF_2} \left( s \right) \rightleftharpoons \ce{Pb^{2+}} \left( aq \right) + 2 \ce{F^-} \left( aq \right) \: \: \: K_\text{sp} = \left[ \ce{Pb^{2+}} \right] \left[ \ce{F^-} \right]^2\nonumber$
The steps above will be followed to calculate the $K_\text{sp}$ for $\ce{PbF_2}$.
Step 2: Solve.
$\text{molar solubility} = \frac{0.533 \: \cancel{\text{g}}}{\text{L}} \times \frac{1 \: \text{mol}}{245.20 \: \cancel{\text{g}}} = 2.17 \times 10^{-3} \: \text{M}\nonumber$
The dissociation equation shows that for every mole of $\ce{PbF_2}$ that dissociates, $1 \: \text{mol}$ of $\ce{Pb^{2+}}$ and $2 \: \text{mol}$ of $\ce{F^-}$ are produced. Therefore, at equilibrium, the concentrations of the ions are:
$\left[ \ce{Pb^{2+}} \right] = 2.17 \times 10^{-3} \: \text{M} \: \: \text{and} \: \: \left[ \ce{F^-} \right] = 2 \times 2.17 \times 10^{-3} = 4.35 \times 10^{-3} \: \text{M}\nonumber$
Substitute into the expression and solve for the $K_\text{sp}$.
$K_\text{sp} = \left( 2.17 \times 10^{-3} \right) \left( 4.35 \times 10^{-3} \right)^2 = 4.11 \times 10^{-8}\nonumber$
Step 3: Think about your result.
The solubility product constant is significantly less than 1 for a nearly insoluble compound such as $\ce{PbF_2}$.
Summary
• Molar solubility calculations are described.
• Calculations of $K_\text{sp}$ using molar solubility are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.11%3A_Solubility_Product_Constant_%28left%28_K_textsp_right%29%29.txt |
Purification of water for drinking and other uses is a complicated process. Heavy metals need to be removed, a process accomplished by the addition of carbonates and sulfates. Lead contamination can present major health problems, especially for children. Lead sulfates and carbonates are very insoluble, so will precipitate out of solution very easily.
Compound $K_\text{sp}$ Compound $K_\text{sp}$
Table $1$: Solubility Product Constants $\left( 25^\text{o} \text{C} \right)$
$\ce{AgBr}$ $5.0 \times 10^{-13}$ $\ce{CuS}$ $8.0 \times 10^{-37}$
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Fe(OH)_2}$ $7.9 \times 10^{-16}$
$\ce{Al(OH)_3}$ $3.0 \times 10^{-34}$ $\ce{Mg(OH)_2}$ $7.1 \times 10^{-12}$
$\ce{BaCO_3}$ $5.0 \times 10^{-9}$ $\ce{PbCl_2}$ $1.7 \times 10^{-5}$
$\ce{BaSO_4}$ $1.1 \times 10^{-10}$ $\ce{PbCO_3}$ $7.4 \times 10^{-14}$
$\ce{CaCO_3}$ $4.5 \times 10^{-9}$ $\ce{PbI_2}$ $7.1 \times 10^{-9}$
$\ce{Ca(OH)_2}$ $6.5 \times 10^{-6}$ $\ce{PbSO_4}$ $6.3 \times 10^{-7}$
$\ce{Ca_3(PO_4)_2}$ $1.2 \times 10^{-26}$ $\ce{Zn(OH)_2}$ $3.0 \times 10^{-16}$
$\ce{CaSO_4}$ $2.4 \times 10^{-5}$ $\ce{ZnS}$ $3.0 \times 10^{-23}$
The known $K_\text{sp}$ values from the table above can be used to calculate the solubility of a given compound by following the steps listed below.
1. Set up an ICE problem (Initial, Change, Equilibrium) in order to use the $K_\text{sp}$ value to calculate the concentration of each of the ions.
2. The concentration of the ions leads to the molar solubility of the compound.
3. Use the molar mass to convert from molar solubility to solubility.
The $K_\text{sp}$ of calcium carbonate is $4.5 \times 10^{-9}$. We begin by setting up an ICE table showing the dissociation of $\ce{CaCO_3}$ into calcium ions and carbonate ions. The variable $s$ will be used to represent the molar solubility of $\ce{CaCO_3}$. In this case, each formula unit of $\ce{CaCO_3}$ yields one $\ce{Ca^{2+}}$ ion and one $\ce{CO_3^{2-}}$ ion. Therefore, the equilibrium concentrations of each ion are equal to $s$.
$\begin{array}{r|ccccc} & \ce{CaCO_3} \left( s \right) & \rightleftharpoons & \ce{Ca^{2+}} \left( aq \right) & + & \ce{CO_3^{2-}} \left( aq \right) \ \hline \text{Initial} \: \left( \text{M} \right) & & & 0.00 & & 0.00 \ \text{Change} \: \left( \text{M} \right) & & & +s & & +s \ \text{Equilibrium} \: \left( \text{M} \right) & & & s & & s \end{array}\nonumber$
The $K_\text{sp}$ expression can be written in terms of $s$ and then used to solve for $s$.
\begin{align*} K_\text{sp} &= \left[ \ce{Ca^{2+}} \right] \left[ \ce{CO_3^{2-}} \right] = \left( s \right) \left( s \right) = s^2 \ s &= \sqrt{K_\text{sp}} = \sqrt{4.5 \times 10^{-9}} = 6.7 \times 10^{-5} \: \text{M} \end{align*}\nonumber
The concentration of each of the ions at equilibrium is $6.7 \times 10^{-5} \: \text{M}$. We can use the molar mass to convert from molar solubility to solubility.
$\frac{6.7 \times 10^{-5} \: \cancel{\text{mol}}}{\text{L}} \times \frac{100.09 \: \text{g}}{1 \: \cancel{\text{mol}}} = 6.7 \times 10^{-3} \: \text{g/L}\nonumber$
So, the maximum amount of calcium carbonate that is capable of dissolving in 1 liter of water at $25^\text{o} \text{C}$ is $6.7 \times 10^{-3}$ grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the $K_\text{sp}$ being equal to $s^2$. This is referred to as a formula of the type $\ce{AB}$, where $\ce{A}$ is the cation and $\ce{B}$ is the anion. Now let's consider a formula of the type $\ce{AB_2}$, such as $\ce{Fe(OH)_2}$. In this case, the setup of the ICE table would look like:
$\begin{array}{r|ccccc} & \ce{Fe(OH)_2} \left( s \right) & \rightleftharpoons & \ce{Fe^{2+}} \left( aq \right) & + & 2 \ce{OH^-} \left( aq \right) \ \hline \text{Initial} \: \left( \text{M} \right) & & & 0.00 & & 0.00 \ \text{Change} \: \left( \text{M} \right) & & & +s & & +2s \ \text{Equilibrium} \: \left( \text{M} \right) & & & s & & 2s \end{array}\nonumber$
When the $K_\text{sp}$ expression is written in terms of $s$, we get the following result for the molar solubility.
\begin{align*} K_\text{sp} &= \left[ \ce{Fe^{2+}} \right] \left[ \ce{OH^-} \right]^2 = \left( s \right) \left( 2s \right)^2 = 4s^3 \ s &= \sqrt[3]{\frac{K_\text{sp}}{4}} = \sqrt[3]{\frac{7.9 \times 10^{-16}}{4}} = 5.8 \times 10^{-6} \: \text{M} \end{align*}\nonumber
The table below shows the relationship between $K_\text{sp}$ and molar solubility based on the formula.
Compound Type Example $K_\text{sp}$ Expression Cation Anion $K_\text{sp}$ in Terms of $s$
Table $2$
$\ce{AB}$ $\ce{CuS}$ $\left[ \ce{Cu^{2+}} \right] \left[ \ce{S^{2-}} \right]$ $s$ $s$ $s^2$
$\ce{AB_2}$ or $\ce{A_2B}$ $\ce{Ag_2CrO_4}$ $\left[ \ce{Ag^+} \right]^2 \left[ \ce{CrO_4^{2-}} \right]$ $2s$ $s$ $4s^3$
$\ce{AB_3}$ or $\ce{A_3B}$ $\ce{Al(OH)_3}$ $\left[ \ce{Al^{3+}} \right] \left[ \ce{OH^-} \right]^3$ $s$ $3s$ $27s^4$
$\ce{A_2B_3}$ or $\ce{A_3B_2}$ $\ce{Ba_3(PO_4)_2}$ $\left[ \ce{Ba^{2+}} \right]^3 \left[ \ce{PO_4^{3-}} \right]^2$ $3s$ $2s$ $108s^5$
The $K_\text{sp}$ expressions in terms of $s$ can be used to solve problems in which the $K_\text{sp}$ is used to calculate the molar solubility, as in the examples above. Molar solubility can then be converted to solubility.
Summary
• The process of determining solubility using $K_\text{sp}$ values from the given table is:
• Set up an ICE problem and use the $K_\text{sp}$ value to calculate the concentration of each of the ions.
• The concentration of the ions leads to the molar solubility of the compound.
• Use the molar mass to convert from molar solubility to solubility. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.13%3A_Conversion_of_%28K_textsp%29_to_Solubility.txt |
The invention of the x-ray machine radically improved medical diagnosis and treatment. For the first time, it was possible to see inside a person's body to detect broken bones, tumors, obstructions, and other problems. Barium sulfate is often used to examine patients with problems of the esophagus, stomach, and intestines. This insoluble compound coats the inside of the tissues and absorbs x-rays, allowing a clear picture of the interior structure of these organs.
Predicting Precipitates
Knowledge of $K_\text{sp}$ values will allow you to be able to predict whether or not a precipitate will form when two solutions are mixed together. For example, suppose that a known solution of barium chloride is mixed with a known solution of sodium sulfate. Barium sulfate is a mostly insoluble compound, and so could potentially precipitate from the mixture. However, it is first necessary to calculate the ion product, $\left[ \ce{Ba^{2+}} \right] \left[ \ce{SO_4^{2-}} \right]$, for the solution. If the value of the ion product is less than the value of the $K_\text{sp}$, then the solution will remain unsaturated; no precipitate will form because the concentrations are not high enough to begin the precipitation process. If the value of the ion product is greater than the value of the $K_\text{sp}$, then a precipitate will form. The formation of the precipitate lowers the concentration of each of the ions until the ion product is exactly equal to the $K_\text{sp}$, at which point precipitation ceases.
Example $1$
Will a precipitate of barium sulfate form when $10.0 \: \text{mL}$ of $0.0050 \: \text{M} \: \ce{BaCl_2}$ is mixed with $20.0 \: \text{mL}$ of $0.0020 \: \text{M} \: \ce{Na_2SO_4}$?
Known
• Concentration of $\ce{BaCl_2} = 0.0050 \: \text{M}$
• Volume of $\ce{BaCl_2} = 10.0 \: \text{mL}$
• Concentration of $\ce{Na_2SO_4} = 0.0020 \: \text{M}$
• Volume of $\ce{Na_2SO_4} = 20.0 \: \text{mL}$
• $K_\text{sp}$ of $\ce{BaSO_4} = 1.1 \times 10^{-10}$
Unknown
• Ion product $= \left[ \ce{Ba^{2+}} \right] \left[ \ce{SO_4^{2-}} \right]$
• If a precipitate forms.
The concentration and volume of each solution that is mixed together must be used to calculate the $\left[ \ce{Ba^{2+}} \right]$ and the $\left[ \ce{SO_4^{2-}} \right]$. Each individual solution is diluted when they are mixed together. The ion product is calculated and compared to the $K_\text{sp}$ to determine if a precipitate forms.
Step 2: Solve.
The moles of each ion from the original solutions are calculated by multiplying the molarity by the volume in liters.
$\text{mol} \: \ce{Ba^{2+}} = 0.0050 \: \text{M} \times 0.010 \: \text{L} = 5.0 \times 10^{-5} \: \text{mol} \: \ce{Ba^{2+}}\nonumber$
$\text{mol} \: \ce{SO_4^{2-}} = 0.0020 \: \text{M} \times 0.020 \: \text{L} = 4.0 \times 10^{-5} \: \text{mol} \: \ce{SO_4^{2-}}\nonumber$
The concentration of each ion after dilution is then calculated by dividing the moles by the final solution volume of $0.030 \: \text{L}$.
$\left[ \ce{Ba^{2+}} \right] = \frac{5.0 \times 10^{-5} \: \text{mol}}{0.030 \: \text{L}} = 1.7 \times 10^{-3} \: \text{M}\nonumber$
$\left[ \ce{SO_4^{2-}} \right] = \frac{4.0 \times 10^{-5} \: \text{mol}}{0.030 \: \text{L}} = 1.3 \times 10^{-3} \: \text{M}\nonumber$
Now, the ion product is calculated.
$\left[ \ce{Ba^{2+}} \right] \left[ \ce{SO_4^{2-}} \right] = \left( 1.7 \times 10^{-3} \right) \left( 1.3 \times 10^{-3} \right) = 2.2 \times 10^{-6}\nonumber$
Since the ion product is greater than the $K_\text{sp}$, a precipitate of barium sulfate will form.
Step 3: Think about your result.
Two significant figures are appropriate for the calculated value of the ion product. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.14%3A_Predicting_Precipitates.txt |
Lithium carbonate is an essential component of lithium batteries, which tend to be longer-lasting than regular alkaline batteries. The material is obtained from lithium ores by adding $\ce{CO_2}$ under high pressure to form the more soluble $\ce{LiHCO_3}$. The mixture is then de-pressurized to remove the carbon dioxide, and the lithium carbonate precipitates out of solution.
Common Ion Effect
In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in solution:
$\ce{CaSO_4} \left( s \right) \rightleftharpoons \ce{Ca^{2+}} \left( aq \right) + \ce{SO_4^{2-}} \left( aq \right) \: \: \: K_\text{sp} = 2.4 \times 10^{-5}\nonumber$
Suppose that some calcium nitrate was added to this saturated solution. Immediately, the concentration of the calcium ion in the solution would increase. As a result, the ion product of the $\left[ \ce{Ca^{2+}} \right]$ times the $\left[ \ce{SO_4^{2-}} \right]$ would increase and now be greater than the $K_\text{sp}$. According to Le Chatelier's principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to the $K_\text{sp}$. Note that in the new equilibrium, the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.
This situation describes the common ion effect. A common ion is an ion that is in common to both salts in a solution. In the above example, the common ion is $\ce{Ca^{2+}}$. The common ion effect is a decrease in the solubility of an ionic compound as a result of the addition of a common ion. Adding calcium ion to the saturated solution of calcium sulfate causes additional $\ce{CaSO_4}$ to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.
Example $1$
What is the concentration of zinc ion in $1.00 \: \text{L}$ of a saturated solution of zinc hydroxide to which $0.040 \: \text{mol}$ of $\ce{NaOH}$ has been added?
Unknown
Express the concentrations of the two ions relative to the variable $s$. The concentration of the zinc ion will be equal to $s$, while the concentration of the hydroxide ion will be equal to $0.040 + 2s$.
Step 2: Solve.
The $K_\text{sp}$ expression can be written in terms of the variable $s$.
$K_\text{sp} = \left[ \ce{Zn^{2+}} \right] \left[ \ce{OH^-} \right]^2 = \left( s \right) \left( 0.040 + 2s \right)^2\nonumber$
Because the value of the $K_\text{sp}$ is so small, we can make the assumption that the value of $s$ will be very small compared to 0.040. This simplifies the mathematics involved in solving for $s$.
\begin{align*} K_\text{sp} &= \left( s \right) \left( 0.040 \right)^2 = 0.0016s = 3.0 \times 10^{-16} \ s &= \frac{K_\text{sp}}{\left[ \ce{OH^-} \right]^2} = \frac{3.0 \times 10^{-16}}{0.0016} = 1.9 \times 10^{-13} \: \text{M} \end{align*}\nonumber
The concentration of the zinc ion is equal to $s$ and so $\left[ \ce{Zn^{2+}} \right] = 1.9 \times 10^{-13} \: \text{M}$.
Step 3: Think about your result.
The relatively high concentration of the common ion, $\ce{OH^-}$, results in a very low concentration of zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in water.
Summary
• The common ion is an ion that is in common to both salts in a solution.
• The common ion effect is a decrease in the solubility of an ionic compound as a result of the addition of a common ion.
• Calculations involving the common ion effect are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.15%3A_Common_Ion_Effect.txt |
When the pieces of a jigsaw puzzle are dumped from the box, the pieces naturally hit the table in a very random pattern. In order to put the puzzle together, a great deal of work must be done to overcome the natural disorder of the pieces. The pieces need to be turned right-side up, then sorted by color or edge (some people like to put the border together first). Finally comes the challenge of finding the exact spot of each piece of the puzzle, in order to obtain the finished picture.
Entropy
There is a tendency in nature for systems to proceed toward a state of greater disorder or randomness. Entropy is a measure of the degree of randomness or disorder of a system. Entropy is an easy concept to understand when thinking about everyday situations. The entropy of a room that has been recently cleaned and organized is low. As time goes by, it likely will become more disordered and thus its entropy will increase (see figure below). The natural tendency of a system is for its entropy to increase.
Chemical reactions also tend to proceed in such a way as to increase the total entropy of the system. How can you tell if a certain reaction shows an increase or a decrease in entropy? The molecular state of the reactants and products provide certain clues. The general cases below illustrate entropy at the molecular level.
1. For a given substance, the entropy of the liquid state is greater than the entropy of the solid state. Likewise, the entropy of the gas is greater than the entropy of the liquid. Therefore, entropy increases in processes in which solid or liquid reactants form gaseous products. Entropy also increases when solid reactants form liquid products.
2. Entropy increases when a substance is broken up into multiple parts. The process of dissolution increases entropy because the solute particles become separated from one another when a solution is formed.
3. Entropy increases as temperature increases. An increase in temperature means that the particles of the substance have greater kinetic energy. The faster-moving particles have more disorder than particles that are moving slowly at a lower temperature.
4. Entropy generally increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules. An exception to this rule is when a gas is produced from nongaseous reactants.
These examples serve to illustrate how the entropy change in a reaction can be predicted:
$\ce{Cl_2} \left( g \right) \rightarrow \ce{Cl_2} \left( l \right)$
The entropy is decreasing because a gas is becoming a liquid.
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)$
The entropy is increasing because a gas is being produced and the number of molecules is increasing.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)$
The entropy is decreasing because four total reactant molecules are forming two total product molecules. All are gases.
$\ce{AgNO_3} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{AgCl} \left( s \right)$
The entropy is decreasing because a solid is formed from aqueous reactants.
$\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right)$
The entropy change is unknown (but likely not zero), because there are equal numbers of molecules on both sides of the equation, and all are gases.
Summary
• Entropy is defined.
• Situations involving entropy changes are described. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.01%3A_Entropy.txt |
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