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Learning Objectives • Recognize bond characteristics of covalent compounds: bond length and bond polarity. • Use electronegativity values to predict bond polarity. If there is only one bond in the molecule, the bond polarity determines the molecular polarity. Any diatomic molecule in which the two atoms are the same element must be a nonpolar molecule. A diatomic molecule that consists of a polar covalent bond, such as HF, is a polar molecule where one end of the molecule is slightly positive, while the other end is slightly negative. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. Hence, a molecule with two poles has a dipole moment. For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. The figure below shows a comparison between carbon dioxide and water. Carbon dioxide $\left( \ce{CO_2} \right)$ is a linear molecule with carbon in the center and two oxygens at the terminal ends. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the $\ce{C}$ atom to each $\ce{O}$ atom. However, since the dipoles are of equal strength and pointing in opposite directions, they cancel out and the overall molecular polarity of $\ce{CO_2}$ is zero (no net dipole), therefore $\ce{CO_2}$ is a nonpolar molecule. Water has a bent molecular structure because it has four electron groups, two bonded groups and two lone electron groups on the central oxygen atom. The individual O–H bond dipoles point from the slightly positive $\ce{H}$ atoms toward the more electronegative $\ce{O}$ atom. Because of the bent shape, the dipoles, which are equal in strength, both point towards the oxygen atom and will not cancel each other out, therefore, the water molecule is polar. In the figure below, you can see that the oxygen end of the molecule is slightly negative and the hydrogen side is slightly positive, there is a separation of charge throughout the whole molecule, a net dipole (shown in blue) that points upward. Some other molecules are shown in the figure below. Notice that a tetrahedral molecule such as $\ce{CH_4}$ is nonpolar. However, if one of the peripheral $\ce{H}$ atoms is replaced with another atom that has a different electronegativity, the molecule becomes polar. A trigonal planar molecule $\left( \ce{BF_3} \right)$ may be nonpolar if all three peripheral atoms are the same, but a trigonal pyramidal molecule $\left( \ce{NH_3} \right)$ is polar. To summarize, to be polar, a molecule must: 1. Contain at least one polar covalent bond. 2. Have a molecular structure such that the sum of the vectors of each bond dipole moment do not cancel. Steps to Identify Polar Molecules 1. Draw the Lewis structure. 2. Figure out the geometry (using VSEPR theory). 3. Visualize or draw the geometry. 4. Find the net dipole moment (you don't have to actually do calculations if you can visualize it). 5. If the net dipole moment is zero, it is non-polar. Otherwise, it is polar. Properties of Polar Molecules Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure $4$). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances. While molecules can be described as "polar covalent" or "ionic", it must be noted that this is often a relative term, with one molecule simply being more polar or less polar than another. However, the following properties are typical of such molecules. Polar molecules tend to: • have higher melting points than nonpolar molecules • have higher boiling points than nonpolar molecules • be more soluble in water (dissolve better) than nonpolar molecules • have lower vapor pressures than nonpolar molecules Interactive Element: Molecular Polarity The following simulation is very useful in exploring more about bond and molecular polarity. Explore on your own or use the exercise and examples below. Example $1$: Polarity Simulations Open the molecule polarity simulation (above) and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field. Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if: 1. A and C are very electronegative and B is in the middle of the range. 2. A is very electronegative, and B and C are not. Solution 1. Molecular dipole moment points immediately between A and C. 2. Molecular dipole moment points along the A–B bond, toward A. Exercise $1$ Determine the partial charges that will give the largest possible bond dipoles. Answer The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will. 4.11: Naming Binary Molecular Compounds Learning Objectives • Name binary molecular compounds. Naming binary (two-element) covalent compounds is very similar to naming simple ionic compounds. The first element in the formula is simply listed using the name of the element. The second element is named by taking the stem of the element name and adding the suffix -ide. Unlike for ionic compounds, molecular compounds can be formed using the same elements in different ratios. Therefore, it is important to indicate the number of each type of atom, using a system of numerical prefixes, listed in Table \(1\). Normally, no prefix is added to the first element’s name if there is only one atom of the first element in a molecule. If the second element is oxygen, the trailing vowel is usually omitted from the end of a polysyllabic prefix but not a monosyllabic one (that is, we would say “monoxide” rather than “monooxide” and “trioxide” rather than “troxide”). Table \(1\): Numerical Prefixes for Naming Binary Covalent Compounds Number of Atoms in Compound Prefix on the Name of the Element 1 mono-* 2 di- 3 tri- 4 tetra- 5 penta- 6 hexa- 7 hepta- 8 octa- 9 nona- 10 deca- *This prefix is not used for the first element’s name. Let us practice by naming the compound whose molecular formula is CCl4. The name begins with the name of the first element carbon. The second element, chlorine, becomes chloride, and we attach the correct numerical prefix (“tetra-”) to indicate that the molecule contains four chlorine atoms. Putting these pieces together gives the name carbon tetrachloride for this compound. Example \(1\) Write the molecular formula for each compound. 1. chlorine trifluoride 2. phosphorus pentachloride 3. sulfur dioxide 4. dinitrogen pentoxide Solution If there is no numerical prefix on the first element’s name, we can assume that there is only one atom of that element in a molecule. 1. ClF3 2. PCl5 3. SO2 4. N2O5 (The di- prefix on nitrogen indicates that two nitrogen atoms are present.) Exercise \(1\) Write the molecular formula for each compound. 1. nitrogen dioxide 2. dioxygen difluoride 3. sulfur hexafluoride 4. selenium monoxide Answer a: a. NO2 Answer b: O2F2 Answer c: SF6 Answer d: SeO Because it is so unreactive, sulfur hexafluoride is used as a spark suppressant in electrical devices such as transformers. Example \(2\) Write the name for each compound. 1. BrF5 2. S2F2 3. CO Solution 1. bromine pentafluoride 2. disulfur difluoride 3. carbon monoxide Exercise \(2\) Write the name for each compound. 1. CF4 2. SeCl2 3. SO3 Answer a: carbon tetrafluoride Answer b: selenium dichloride Answer c: sulfur trioxide For some simple covalent compounds, we use common names rather than systematic names. We have already encountered these compounds, but we list them here explicitly: • H2O: water • NH3: ammonia • CH4: methane Methane is the simplest organic compound. Organic compounds are compounds with carbon atoms and are named by a separate nomenclature system that we will introduce later.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/04%3A_Molecular_Compounds/4.10%3A_Polar_Molecules.txt
• 5.1: Chemical Equations Chemical reactions occur when one or more chemicals combine to form one or more new chemicals. The law of conservation of matter is obeyed when writing chemical equations to describe chemical reactions. • 5.2: Balancing Chemical Equations Chemical reactions are represented by chemical equations that list reactants and products. Proper chemical equations are balanced; the same number of each element’s atoms appears on each side of the equation. • 5.3: Precipitation Reactions and Solubility Guidelines A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate. Solubility rules are used to predict whether a precipitate will form or not. • 5.4: Acids, Bases, and Neutralization Reactions The Arrhenius definition of an acid is a substance that increases the amount of H+ in an aqueous solution. The Arrhenius definition of a base is a substance that increases the amount of OH- in an aqueous solution. Neutralization is the reaction of an acid and a base, which forms water and a salt. Net ionic equations for neutralization reactions may include solid acids, solid bases, solid salts, and water. • 5.5: Redox Reactions Chemical reactions in which electrons are transferred are called oxidation-reduction, or redox, reactions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidation and reduction always occur together, even though they can be written as separate chemical equations. • 5.6: Recognizing Redox Reactions Oxidation numbers, indicating if an atom is neural, electron-rich, or electron-poor, are assigned to atoms in a redox equation. Keeping track of oxidation numbers on the reactant and product sides of a chemical equation provide information to determine what species is oxidized and what species is reduced. • 5.7: Net Ionic Equations 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. 05: Classification and Balancing of Chemical Reactions Learning Objectives • Define chemical reaction. • Understand the Law of Conservation of Matter Water (H2O) is composed of hydrogen and oxygen. Suppose we imagine a process in which we take some elemental hydrogen (H2) and elemental oxygen (O2) and let them react to make water. The statement "hydrogen and oxygen react to make water" is one way to represent that process, which is called a chemical reaction. Figure $1$ shows a rather dramatic example of this very reaction. To simplify the writing of reactions, we use formulas instead of names when we describe a reaction. We can also use symbols to represent other words in the reaction. A plus sign connects the initial substances (and final substances, if there is more than one), and an arrow (→) represents the chemical change: $\ce{H_2 + O_2 \rightarrow H_2O} \label{Eq1}$ This statement is one example of a chemical equation, an abbreviated way of using symbols to represent a chemical change. The substances on the left side of the arrow are called reactants, and the substances on the right side of the arrow are called products. It is not uncommon to include a phase label with each formula—(s) for solid, (ℓ) for liquid, (g) for gas, and (aq) for a substance dissolved in water, also known as an aqueous solution. If we included phase labels for the reactants and products, under normal environmental conditions, the reaction would be as follows: $\ce{H2(g) + O2(g) \rightarrow H2O (ℓ)} \label{Eq2}$ This equation is still not complete because it does not satisfy the law of conservation of matter. Count the number of atoms of each element on each side of the arrow. On the reactant side, there are two H atoms and two O atoms; on the product side, there are two H atoms and only one oxygen atom. The equation is not balanced because the number of oxygen atoms on each side is not the same (Figure $2$). To make this chemical equation conform to the law of conservation of matter, we must revise the amounts of the reactants and the products as necessary to get the same number of atoms of a given element on each side. Because every substance has a characteristic chemical formula, we cannot change the chemical formulas of the individual substances. For example, we cannot change the formula for elemental oxygen to O. However, we can assume that different numbers of reactant molecules or product molecules may be involved. For instance, perhaps two water molecules are produced, not just one: $\ce{H2(g) + O2 (g) \rightarrow 2H2O (ℓ)} \label{Eq3}$ The 2 preceding the formula for water is called a coefficient. It implies that two water molecules are formed. There are now two oxygen atoms on each side of the equation. This point is so important that we should repeat it. You cannot change the formula of a chemical substance to balance a chemical reaction! You must use the proper chemical formula of the substance. Unfortunately, by inserting the coefficient 2 in front of the formula for water, we have also changed the number of hydrogen atoms on the product side as well. As a result, we no longer have the same number of hydrogen atoms on each side. This can be easily fixed, however, by putting a coefficient of 2 in front of the diatomic hydrogen reactant: $\ce{2H2(g) + O2(g) \rightarrow 2H2O (ℓ)} \label{Eq4}$ Now we have four hydrogen atoms and two oxygen atoms on each side of the equation. The law of conservation of matter is satisfied because we now have the same number of atoms of each element in the reactants and in the products. We say that the reaction is now balanced (Figure $3$). Note: The diatomic oxygen has a coefficient of 1, which typically is not written but assumed in balanced chemical equations. Proper chemical equations should be balanced. Writing balanced reactions is a chemist’s way of acknowledging the law of conservation of matter. Example $1$ Is each chemical equation balanced? 1. 2Na(s) + O2(g) → 2Na2O(s) 2. CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ) 3. AgNO3(aq) + 2KCl(aq) → AgCl(s) + KNO3(aq) Solution 1. By counting, we find two sodium atoms and two oxygen atoms in the reactants and four sodium atoms and two oxygen atoms in the products. This equation is not balanced. 2. The reactants have one carbon atom, four hydrogen atoms, and four oxygen atoms. The products have one carbon atom, four hydrogen atoms, and four oxygen atoms. This equation is balanced. 3. The reactants have one silver atom, one nitrogen atom, three oxygen atoms, two potassium atoms, and two chlorine atoms. The products have one silver atom, one chlorine atom, one potassium atom, one nitrogen atom, and three oxygen atoms. Because there are different numbers of chlorine and potassium atoms, this equation is not balanced. Exercise $1$ Is each chemical equation balanced? 1. $2Hg_{(ℓ)} + O_{2(g)} \rightarrow Hg_2O_{2(s)}$ 2. $C_2H_{4(g)} + 2O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(ℓ)}$ 3. $Mg(NO_3)_{2(s)} + 2Li_{(s)} \rightarrow Mg_{(s)} + 2LiNO_{3(s)}$. Answer a: balanced Answer b: O is not balanced; the 4 atoms of oxygen on the left does not balance with the 6 oxygen atoms on the right Answer c: balanced
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.01%3A_Chemical_Equations.txt
Learning Objectives • Balance chemical equations. How does one balance a chemical equation, starting with the correct formulas of the reactants and products? Basically, a back-and-forth (or trial-and-error) approach is adopted, counting the number of atoms of one element on one side, checking the number of atoms of that element on the other side, and changing a coefficient if necessary. Then check another element, going back and forth from one side of the equation to another, until each element has the same number of atoms on both sides of the arrow. In many cases, it does not matter which element is balanced first and which is balanced last, as long as all elements have the same number of atoms on each side of the equation. Below are guidelines for writing and balancing chemical equations. 1. Determine the correct chemical formulas for each reactant and product. Write the skeleton equation. 2. 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. 3. Balance each element one at a time by placing coefficients in front of the formulas. No coefficient is written for a 1. It is best to begin by balancing elements that only appear in one chemical formula on each side of the equation. NEVER change the subscripts in a chemical formula - you can only balance equations by using coefficients. 4. Check each atom or polyatomic ion to be sure that they are equal on both sides of the equation. 5. Make sure that all coefficients are in the lowest possible ratio. If necessary, reduce to the lowest ratio. For example, to balance the equation Step 1: Write the skeleton equation with the correct formulas. $\ce{CH4 + Cl2 \rightarrow CCl4 + HCl} \label{Eq5}$ Step 2: Count the number of each atom or polyatomic ion on both sides of the equation. $\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \ 4 \: \ce{H} \: \text{ions} & 1 \: \ce{H} \: \text{ions} \ 2 \: \ce{Cl} \: \text{atom} & 5 \: \ce{Cl} \: \text{atoms} \end{array}$ Step 3: We find that both sides are already balanced with one carbon atom. So we proceed to balance the hydrogen atoms. We find that the reactant side has four hydrogen atoms, so the product side must also have four hydrogen atoms. This is balanced by putting a 4 in front of the HCl: $\ce{CH4 + Cl2 \rightarrow CCl4 + 4HCl } \label{Eq6}$ $\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \ 4 \: \ce{H} \: \text{ions} & 4 \: \ce{H} \: \text{ions} \ 2 \: \ce{Cl} \: \text{atom} & 8 \: \ce{Cl} \: \text{atoms} \end{array}$ Now each side has four hydrogen atoms. The product side has a total of eight chlorine atoms (four from the $CCl_4$ and four from the four molecules of HCl), so we need eight chlorine atoms as reactants. Because elemental chlorine is a diatomic molecule, we need four chlorine molecules to get a total of eight chlorine atoms. We add another 4 in front of the Cl2 reactant: $\ce{CH4 + 4Cl2 \rightarrow CCl4 + 4HCl } \label{Eq7}$ $\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{C} \: \text{atom} & 1 \: \ce{C} \: \text{atom} \ 4 \: \ce{H} \: \text{ions} & 4 \: \ce{H} \: \text{ions} \ 8 \: \ce{Cl} \: \text{atom} & 8 \: \ce{Cl} \: \text{atoms} \end{array}$ Step 3: Now we check: each side has one carbon atom, four hydrogen atoms, and eight chlorine atoms. The chemical equation is balanced. And, the coefficients are in the lowest possible ratio. Example $2$ 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. Solution Step 1: Write the skeleton equation with the correct formulas. $\ce{Pb(NO_3)_2} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)$ Step 2: Count the number of each atom or polyatomic ion on both sides of the equation. $\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \ 2 \: \ce{NO_3^-} \: \text{ions} & 1 \: \ce{NO_3^-} \: \text{ions} \ 1 \: \ce{Na} \: \text{atom} & 1 \: \ce{Na} \: \text{atom} \ 1 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}$ Step 3: Solve. 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: $\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{NaCl} \left( aq \right) \rightarrow 2 \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)$ Step 4: The new count for each atom and polyatomic ion becomes: $\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \ 2 \: \ce{NO_3^-} \: \text{ions} & 2 \: \ce{NO_3^-} \: \text{ions} \ 2 \: \ce{Na} \: \text{atom} & 2 \: \ce{Na} \: \text{atom} \ 2 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}$ Step 5: Think about the result. The equation is now balanced since there are equal numbers of atoms of each element on both sides of the equation. And, the coefficients are in the lowest possible ratio. Exercise $2$ Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose (C6H12O6) are converted to ethanol (C2H5OH) and carbon dioxide CO2. Write a balanced chemical reaction for the fermentation of glucose. Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible. Answer C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.02%3A_Balancing_Chemical_Equations.txt
Learning Objectives • Use solubility rules to predict whether a precipitate will form. • Write and balance chemical equations for precipitation reactions. There are many types of chemical reactions that you will learn about in chemistry. In this chapter, we will focus on some reactions of ionic compounds: precipitation reactions (a type of double-replacement reaction), acid-base neutralization reactions, and oxidation-reduction reactions. The practice of barter (trading one thing for another) has been in existence since the beginning of time. In the past, for example, items like chickens were bartered for newspapers. Person A had something that person B wanted, and vice versa. So, when person A and person B traded items, they each had something new. Some chemical reactions are like that—compounds swap parts, and the products are new materials. 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: $\ce{AB} + \ce{CD} \rightarrow \ce{AD} + \ce{CB}$ 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: $2 \ce{KI} \left( aq \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{KNO_3} \left( aq \right) + \ce{PbI_2} \left( s \right)$ 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 $1$ below). The other product of the reaction, potassium nitrate, remains soluble. To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use solubility rules, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble, or insoluble). Table $1$ lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules. If a compound is soluble, we use the (aq) label with it, indicating that it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected. Table $1$: Solubility Rules: Soluble Compounds and Their Exceptions These compounds generally dissolve in water (are soluble): Exceptions: All compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ None All compounds of NO3 and C2H3O2 None Compounds of Cl, Br, I Ag+, Hg22+, Pb2+ Compounds of SO42 Hg22+, Pb2+, Sr2+, Ba2+ Table $2$: Solubility Rules: Insoluble Compounds and Their Exceptions These compounds generally do not dissolve in water (are insoluble): Exceptions: Compounds of CO32 and PO43 Compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ Compounds of OH Compounds of Li+, Na+, K+, Rb+, Cs+, NH4+, Sr2+, and Ba2+ For example, consider the possible double-replacement reaction between Na2SO4 and SrCl2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble, except for Ag+, Hg22+, and Pb2+, which are not being considered here. Therefore, Na2SO4 and SrCl2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO4? Compounds of the sulfate ion are generally soluble, but Sr2+ is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be: $\ce{Na2SO4(aq) + SrCl2(aq) → 2NaCl(aq) + SrSO4(s)}\nonumber$ You would expect to see a visual change corresponding to SrSO4 precipitating out of solution (Figure $2$). Example $4$: Will a precipitation reaction occur? If so, identify the products. 1. Ca(NO3)2 + KBr → ? 2. NaOH + FeCl2 → ? Solution 1. According to the solubility rules, both Ca(NO3)2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr2 and KNO3. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case. 2. According to the solubility rules, both NaOH and FeCl2 are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)2. NaCl is soluble, but, according to the solubility rules, Fe(OH)2 is not. Therefore, a reaction would occur, and Fe(OH)2(s) would precipitate out of solution. The balanced chemical equation is $\ce{2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)}\nonumber$ Exercise $4$ $\ce{Sr(NO3)2 + KCl → }\nonumber$ Answer No reaction; all possible products are soluble. Key Takeaways • A single-replacement reaction replaces one element for another in a compound. • The periodic table or an activity series can help predict whether single-replacement reactions occur. • A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds. • A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate. • Solubility rules are used to predict whether some double-replacement reactions will occur.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.03%3A_Precipitation_Reactions_and_Solubility_Guidelines.txt
Learning Objectives • Identify an acid-base neutralization reaction and predict its products. Previously, you learned that an acid is any compound that produces hydrogen ions (H+) in an aqueous solution, and the chemical opposite of an acid is a base, which is a compound that produces hydroxide ions (OH) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the Arrhenius definition of an acid and a base, respectively. You may recognize that, based on the description of a hydrogen atom, an H+ ion is a hydrogen atom that has lost its lone electron; that is, H+ is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H+ ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the hydronium ion as H3O+, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way that a hydrogen ion appears in an aqueous solution, although in many chemical reactions H+ and H3O+ are treated equivalently. The reaction of an acid and a base is called a neutralization reaction. Although acids and bases have their own unique chemistries, the acid and base "cancel" each other's chemistry to produce a rather innocuous substance, water. In fact, the general reaction between an acid and a base is: $\ce{acid + base → water + salt}\nonumber$ where the term salt is generally used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. (In chemistry, the word salt refers to more than just table salt.) For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is $\ce{HCl(aq) + KOH(aq) → H2O(ℓ) + KCl(aq)}\nonumber$ where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)2(aq), additional molecules of HCl and H2O are required to balance the chemical equation: $\ce{2HCl(aq) + Mg(OH)2(aq) → 2H2O(ℓ) + MgCl2(aq)}\nonumber$ Here, the salt is MgCl2. (This is one of several reactions that take place when a type of antacid, a base, is used to treat stomach acid.) Example $1$: Write the neutralization reactions between each acid and base. 1. HNO3(aq) and Ba(OH)2(aq) 2. H3PO4(aq) and Ca(OH)2(aq) Solution First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation. 1. The expected products are water and barium nitrate, so the initial chemical reaction is $\ce{HNO3(aq) + Ba(OH)2(aq) → H2O(ℓ) + Ba(NO3)2(aq)}\nonumber$ To balance the equation, we need to realize that there will be two H2O molecules, so two HNO3 molecules are required: $\ce{2HNO3(aq) + Ba(OH)2(aq) → 2H2O(ℓ) + Ba(NO3)2(aq)}\nonumber$ This chemical equation is now balanced. 2. The expected products are water and calcium phosphate, so the initial chemical equation is $\ce{H3PO4(aq) + Ca(OH)2(aq) → H2O(ℓ) + Ca3(PO4)2(s)}\nonumber$ According to the solubility rules, Ca3(PO4)2 is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions. We end up with six water molecules to balance the equation: $\ce{2H3PO4(aq) + 3Ca(OH)2(aq) → 6H2O(ℓ) + Ca3(PO4)2(s)}\nonumber$ This chemical equation is now balanced. Exercise $1$ Write the neutralization reaction between H2SO4(aq) and Sr(OH)2(aq). Answer $\ce{H2SO4(aq) + Sr(OH)2(aq) → 2H2O(ℓ) + SrSO4(aq)}\nonumber$ Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)3(s) still proceeds according to the equation: $\ce{3HCl(aq) + Fe(OH)3(s) → 3H2O(ℓ) + FeCl3(aq)}\nonumber$ even though Fe(OH)3 is not soluble. When one realizes that Fe(OH)3(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids; the neutralization reaction produces products that are soluble and wash away. (Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!) Key Takeaways • The Arrhenius definition of an acid is a substance that increases the amount of H+ in an aqueous solution. • The Arrhenius definition of a base is a substance that increases the amount of OH in an aqueous solution. • Neutralization is the reaction of an acid and a base, which forms water and a salt.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.04%3A_Acids_Bases_and_Neutralization_Reactions.txt
Learning Objectives • To identify a chemical reaction as an oxidation-reduction reaction. When zinc metal is submerged into a quantity of aqueous $\ce{HCl}$, the following reaction occurs (Figure $1$): $\ce{Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2(aq)} \label{eq:1}$ This is one example of what is sometimes called a single replacement reaction because $\ce{Zn}$ replaces $\ce{H}$ in combination with $\ce{Cl}$. Because some of the substances in this reaction are aqueous, we can separate them into ions: $\ce{Zn (s) + 2H^{+} (aq) + 2Cl^{−} (aq) → H2(g) + Zn^{2+} (aq) + 2Cl^{−} (aq)} \nonumber$ Viewed this way, the net reaction seems to be a charge transfer between zinc and hydrogen atoms. (There is no net change experienced by the chloride ion.) In fact, electrons are being transferred from the zinc atoms to the hydrogen atoms (which ultimately make a molecule of diatomic hydrogen), changing the charges on both elements. To understand electron-transfer reactions like the one between zinc metal and hydrogen ions, chemists separate them into two parts: one part focuses on the loss of electrons, and one part focuses on the gain of electrons. The loss of electrons is called oxidation. The gain of electrons is called reduction. Because any loss of electrons by one substance must be accompanied by a gain in electrons by something else, oxidation and reduction always occur together. As such, electron-transfer reactions are also called oxidation-reduction reactions, or simply redox reactions. The atom that loses electrons is oxidized, and the atom that gains electrons is reduced. Also, because we can think of the species being oxidized as causing the reduction, the species being oxidized is called the reducing agent, and the species being reduced is called the oxidizing agent. Because batteries are used as sources of electricity (that is, of electrons), all batteries are based on redox reactions. Although the two reactions occur together, it can be helpful to write the oxidation and reduction reactions separately as half reactions. In half reactions, we include only the reactant being oxidized or reduced, the corresponding product species, any other species needed to balance the half reaction, and the electrons being transferred. Electrons that are lost are written as products; electrons that are gained are written as reactants. For example, in our earlier equation, now written without the chloride ions, $\ce{Zn (s) + 2H^{+} (aq) → Zn^{2+}(aq) + H2(g)} \nonumber$ zinc atoms are oxidized to Zn2+. The half reaction for the oxidation reaction, omitting phase labels, is as follows: $\ce{Zn → Zn^{2+} + 2e^{−}} \nonumber$ This half reaction is balanced in terms of the number of zinc atoms, and it also shows the two electrons that are needed as products to account for the zinc atom losing two negative charges to become a 2+ ion. With half reactions, there is one more item to balance: the overall charge on each side of the reaction. If you check each side of this reaction, you will note that both sides have a zero net charge. Hydrogen is reduced in the reaction. The balanced reduction half reaction is as follows: $\ce{2H^{+} + 2e^{−} → H2} \nonumber$ There are two hydrogen atoms on each side, and the two electrons written as reactants serve to neutralize the 2+ charge on the reactant hydrogen ions. Again, the overall charge on both sides is zero. The overall reaction is simply the combination of the two half reactions and is shown by adding them together. Because we have two electrons on each side of the equation, they can be canceled. This is the key criterion for a balanced redox reaction: the electrons have to cancel exactly. If we check the charge on both sides of the equation, we see they are the same—2+. (In reality, this positive charge is balanced by the negative charges of the chloride ions, which are not included in this reaction because chlorine does not participate in the charge transfer.) Redox reactions are often balanced by balancing each individual half reaction and then combining the two balanced half reactions. Sometimes a half reaction must have all of its coefficients multiplied by some integer for all the electrons to cancel. The following example demonstrates this process. Example $1$: Reducing Silver Ions Write and balance the redox reaction that has silver ions and aluminum metal as reactants and silver metal and aluminum ions as products. Identify the substance oxidized, substance reduced, reducing agent and reducing agent. Solution We start by using symbols of the elements and ions to represent the reaction: $\ce{Ag^{+} + Al → Ag + Al^{3+}} \nonumber$ The equation looks balanced as it is written. However, when we compare the overall charges on each side of the equation, we find a charge of +1 on the left but a charge of +3 on the right. This equation is not properly balanced. To balance it, let us write the two half reactions. Silver ions are reduced, and it takes one electron to change Ag+ to Ag: Reduction half-reaction: $\ce{Ag^{+} + e^{−} → Ag} \nonumber$ Aluminum is oxidized, losing three electrons to change from Al to Al3+: Oxidation half-reaction: $\ce{Al → Al^{3+} + 3e^{−}} \nonumber$ To combine these two half reactions and cancel out all the electrons, we need to multiply the silver reduction reaction by 3: Now the equation is balanced, not only in terms of elements but also in terms of charge. • The substance oxidized is the reactant that had undergone oxidation: Al • The substance reduced is the reactant that had undergone reduction: Ag+ • The reducing agent is the same as the substance oxidized: Al • The oxidizing agent is the same as the substance reduced: Ag+ Exercise $1$ Write and balance the redox reaction that has calcium ions and potassium metal as reactants and calcium metal and potassium ions as products. Identify the substance oxidized, substance reduced, reducing agent and reducing agent. Answer Reduction: Ca2+ + 2e→ Ca Oxidation: 2 (K → K+ + e) Combined: Ca2+ + 2K → Ca + 2K+ • The substance oxidized is the reactant that had undergone oxidation: K • The substance reduced is the reactant that had undergone reduction: Ca2+ • The reducing agent is the same as the substance oxidized: K • The oxidizing agent is the same as the substance reduced: Ca2+ Potassium has been used as a reducing agent to obtain various metals in their elemental form. To Your Health: Redox Reactions and Pacemaker Batteries All batteries use redox reactions to supply electricity because electricity is basically a stream of electrons being transferred from one substance to another. Pacemakers—surgically implanted devices for regulating a person’s heartbeat—are powered by tiny batteries, so the proper operation of a pacemaker depends on a redox reaction. Pacemakers used to be powered by NiCad batteries, in which nickel and cadmium (hence the name of the battery) react with water according to this redox reaction: $\ce{Cd(s) + 2NiOOH(s) + 2H2O(ℓ) → Cd(OH)2(s) + 2Ni(OH) 2(s)} \nonumber$ The cadmium is oxidized, while the nickel atoms in NiOOH are reduced. Except for the water, all the substances in this reaction are solids, allowing NiCad batteries to be recharged hundreds of times before they stop operating. Unfortunately, NiCad batteries are fairly heavy batteries to be carrying around in a pacemaker. Today, the lighter lithium/iodine battery is used instead. The iodine is dissolved in a solid polymer support, and the overall redox reaction is as follows: $\ce{2Li(s) + I2(s) → 2LiI (s)} \nonumber$ Lithium is oxidized, and iodine is reduced. Although the lithium/iodine battery cannot be recharged, one of its advantages is that it lasts up to 10 years. Thus, a person with a pacemaker does not have to worry about periodic recharging; about once per decade a person requires minor surgery to replace the pacemaker/battery unit. Lithium/iodine batteries are also used to power calculators and watches. Oxidation and reduction can also be defined in terms of changes in composition. The original meaning of oxidation was “adding oxygen,” so when oxygen is added to a molecule, the molecule is being oxidized. The reverse is true for reduction: if a molecule loses oxygen atoms, the molecule is being reduced. For example, the acetaldehyde ($\ce{CH3CHO}$) molecule takes on an oxygen atom to become acetic acid ($\ce{CH3COOH}$). $\ce{2CH3CHO + O2 → 2CH_3COOH} \nonumber$ Thus, acetaldehyde is being oxidized. Similarly, reduction and oxidation can be defined in terms of the gain or loss of hydrogen atoms. If a molecule adds hydrogen atoms, it is being reduced. If a molecule loses hydrogen atoms, the molecule is being oxidized. For example, in the conversion of acetaldehyde into ethanol ($\ce{CH3CH2OH}$), hydrogen atoms are added to acetaldehyde, so the acetaldehyde is being reduced: $\ce{CH3CHO + H2 → CH3CH2OH} \nonumber$ Table $1$: Oxidation-Reduction Reactions and the Changes in Oxygen and Hydrogen. Process Change in oxygen (some reactions) Change in hydrogen (some reactions) Oxidation gain lose Reduction lose gain Example $2$ In each conversion, indicate whether oxidation or reduction is occurring. 1. N2 → NH3 2. CH3CH2OHCH3 → CH3COCH3 3. HCHO → HCOOH Solution 1. Hydrogen is being added to the original reactant molecule, so reduction is occurring. 2. Hydrogen is being removed from the original reactant molecule, so oxidation is occurring. 3. Oxygen is being added to the original reactant molecule, so oxidation is occurring. Exercise $2$ In each conversion, indicate whether oxidation or reduction is occurring. 1. CH4 → CO2 + H2O 2. NO2 → N2 3. CH2=CH2 → CH3CH3 Answer a: Oxygen is being added. Oxidation is occurring. Answer b: Oxygen is being removed. Reduction is occurring. Answer a: Hydrogen is being added. Reduction is occurring. Key Takeaway Chemical reactions in which electrons are transferred are called oxidation-reduction, or redox, reactions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidation and reduction always occur together, even though they can be written as separate chemical equations.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.05%3A_Redox_Reactions.txt
Learning Objectives • Assign oxidation numbers to atoms in simple compounds. • Recognize a reaction as an oxidation-reduction reaction. Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate. Assigning Oxidation Numbers The rules for assigning oxidation numbers to atoms are as follows: 1. Atoms in their elemental state are assigned an oxidation number of 0. 2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges. 3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion [H], in which case rule 2 prevails). 4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral). Here are some examples for practice. In H2, both H atoms have an oxidation number of 0 by rule 1. In MgCl2, magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1 by rule 2. In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H2O2) has an oxidation number of +1, while each O atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO2. Each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound. Example $1$ Assign oxidation numbers to the atoms in each substance. 1. Cl2 2. GeO2 3. Ca(NO3)2 Solution 1. Cl2 is the elemental form of chlorine. Rule 1 states each atom has an oxidation number of 0. 2. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4. 3. Ca(NO3)2 can be separated into two parts: the Ca2+ ion and the NO3 ion. Considering these separately, the Ca2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3 ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation x + 3(−2) = −1 where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for x, x + (−6) = −1x = +5 Thus the oxidation number on the N atom in the NO3 ion is +5. Exercise $1$: Phosphoric Acid Assign oxidation numbers to the atoms in H3PO4. Answer H: +1; O: −2; P: +5 All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively. Example $2$: Identify what is being oxidized and reduced in this redox reaction. $\ce{2Na + Br2 → 2NaBr} \nonumber\nonumber$ Solution Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br ions have an oxidation number of −1. $2\underset{0}{Na}+\underset{0}{Br_{2}}\rightarrow 2\underset{+1 -1}{NaBr} \nonumber\nonumber$ Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced: Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom). Exercise $2$ Identify what is being oxidized and reduced in this redox reaction. $\ce{C + O2 → CO2}\nonumber\nonumber$ Answer C is being oxidized from 0 to +4; O is being reduced from 0 to −2. Oxidation reactions can become quite complex, as attested by the following redox reaction: $6H^{+}(aq)+2\underset{+7}{MnO_{4}^{-}}(aq)+5\underset{-1}{H_{2}O_{2}}(l)\rightarrow 2\underset{+2}{Mn^{2+}}(aq)+5\underset{0}{O_{2}}(g)+8H_{2}O(l)\nonumber$ To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? Food and Drink Application: Fortifying Food with Iron Iron is an essential mineral in our diet; iron-containing compounds like the heme protein in hemoglobin could not function without it. Most biological iron has the form of the Fe2+ ion; iron with other oxidation numbers is almost inconsequential in human biology (although the body does contain an enzyme to reduce Fe3+ to Fe2+, so Fe3+ must have some biological significance, albeit minor). To ensure that we ingest enough iron, many foods are enriched with iron. Although Fe2+ compounds are the most logical substances to use, some foods use "reduced iron" as an ingredient (bread and breakfast cereals are the most well-known examples). Reduced iron is simply iron metal; iron is added as a fine metallic powder. The metallic iron is oxidized to Fe2+ in the digestive system and then absorbed by the body, but the question remains: Why are we ingesting metallic iron? Why not just use Fe2+ salts as an additive? Although it is difficult to establish conclusive reasons, a search of scientific and medical literature suggests a few reasons. One reason is that fine iron filings do not affect the taste of the product. The size of the iron powder (several dozen micrometers) is not noticeable when chewing iron-supplemented foods, and the tongue does not detect any changes in flavor that can be detected when using Fe2+ salts. Fe2+ compounds can affect other properties of foodstuffs during preparation and cooking, like dough pliability, yeast growth, and color. Finally, of the common iron substances that might be used, metallic iron is the least expensive. These factors appear to be among the reasons why metallic iron is the supplement of choice in some foods. Key Takeaways • Oxidation numbers are used to keep track of electrons in atoms. • There are rules for assigning oxidation numbers to atoms. • Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number (gain of electrons).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.06%3A_Recognizing_Redox_Reactions.txt
Learning Objectives • Write ionic and net ionic equations for chemical reactions involving ions. 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 written to show the ions that are dissolved in solution: $\ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber$ 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/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.07%3A_Net_Ionic_Equations.txt
Learning Objectives • Calculate formula masses for covalent and ionic compounds. • Define the amount unit mole and the related quantity Avogadro’s number. • Calculate molar mass of a compound from the molecular formula. We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances. Formula Mass In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance, which is equal to the sum of the atomic masses for all the atoms represented in the substance’s formula. Formula Mass for Covalent Substances For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl3), a covalent compound once used as a surgical anesthetic and now primarily used in the production of tetrafluoroethylene, the building block for the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure $1$ outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu. Likewise, the molecular mass of an aspirin molecule, C9H8O4, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure $2$). Example $1$: Computing Molecular Mass for a Covalent Compound Ibuprofen, C13H18O2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound? Solution Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore: Exercise $1$ Acetaminophen, C8H9NO2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound? Answer 151.16 amu Formula Mass for Ionic Compounds Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.” As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na+, and chloride anions, Cl, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (Figure $3$). Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses. Example $2$: Computing Formula Mass for an Ionic Compound Aluminum sulfate, Al2(SO4)3, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound? Solution The formula for this compound indicates it contains Al3+ and SO42 ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al2S3O12. Following the approach outlined above, the formula mass for this compound is calculated as follows: Exercise $2$ Calcium phosphate, $\ce{Ca3(PO4)2}$, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate? Answer 310.18 amu The Mole So far, we have been talking about chemical substances in terms of individual atoms and molecules. Yet we do not typically deal with substances one atom or molecule at a time; we work with millions, billions, and trillions of atoms and molecules at a time. We need a way to deal with macroscopic, rather than microscopic, amounts of matter. We need a unit of amount that relates quantities of substances on a scale that we can interact with. Chemistry uses a unit called mole. The mole (mol) is an counting term similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth. The number of entities composing a mole has been experimentally determined to be $6.02214179 \times 10^{23}$, a fundamental constant named Avogadro’s number ($N_A$) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being $6.022 \times 10^{23}/\ce{mol}$. How big is a mole? It is very large. Suppose you had a mole of dollar bills that need to be counted. If everyone on earth (about 6 billion people) counted one bill per second, it would take about 3.2 million years to count all the bills. A mole of sand would fill a cube about 32 km on a side. A mole of pennies stacked on top of each other would have about the same diameter as our galaxy, the Milky Way. Atoms and molecules are very tiny, so one mole of carbon atoms would make a cube that is 1.74 cm on a side, small enough to carry in your pocket. One mole of water molecules is approximately 18 mL or just under 4 teaspoons of water. Example $3$ How many molecules are present in 2.76 mol of H2O? How many atoms is this? Solution The definition of a mole is an equality that can be used to construct a conversion factor. Also, because we know that there are three atoms in each molecule of H2O, we can also determine the number of atoms in the sample. $2.76\, \cancel{mol\, H_{2}O}\times \frac{6.022\times 10^{23}molecules\, H_{2}O}{\cancel{mol\, H_{2}O}}=1.66\times 10^{24}molecules\, H_{2}O \nonumber\nonumber$ To determine the total number of atoms, we have $1.66\times 10^{24}\cancel{molecules\, H_{2}O}\times \frac{3\, atoms}{1\, molecule}=4.99\times 10^{24}\, atoms \nonumber\nonumber$ Exercise $3$ How many molecules are present in 4.61 × 10−2 mol of $\ce{O2}$? Answer 2.78 × 1022 molecules Molar Mass Why is the mole unit so important? It represents the link between the microscopic and the macroscopic, especially in terms of mass. A mole of a substance has the same mass in grams as one unit (atom or molecules) has in atomic mass units. The mole unit allows us to express amounts of atoms and molecules in visible amounts that we can understand. For example, we already know that, by definition, a mole of carbon has a mass of exactly 12 g. This means that exactly 12 g of C has 6.022 × 1023 atoms: 12 g C = 6.022 × 1023 atoms C We can use this equality as a conversion factor between the number of atoms of carbon and the number of grams of carbon. How many grams are there, say, in 1.50 × 1025 atoms of carbon? This is a one-step conversion: $1.50\times 10^{25}\cancel{atoms\, C}\times \frac{12.0000\, g\, C}{6.022\times 10^{23}\cancel{atoms\, C}}=299\, g\, C\nonumber$ But it also goes beyond carbon. Previously we defined atomic and molecular masses as the number of atomic mass units per atom or molecule. Now we can do so in terms of grams. The atomic mass of an element is the number of grams in 1 mol of atoms of that element, while the molecular mass of a compound is the number of grams in 1 mol of molecules of that compound. Sometimes these masses are called molar masses to emphasize the fact that they are the mass for 1 mol of things. (The term molar is the adjective form of mole and has nothing to do with teeth.) Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure $1$). Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12C contains 1 mole of 12C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12C. Table $1$: Mass of one mole of elements Element Average Atomic Mass (amu) Molar Mass (g/mol) Atoms/Mole C 12.01 12.01 $6.022 \times 10^{23}$ H 1.008 1.008 $6.022 \times 10^{23}$ O 16.00 16.00 $6.022 \times 10^{23}$ Na 22.99 22.99 $6.022 \times 10^{23}$ Cl 33.45 35.45 $6.022 \times 10^{23}$ While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water after a rainfall. Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules. The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass. Here are some examples. The mass of 1 hydrogen atom is 1.0079 u; the mass of 1 mol of hydrogen atoms is 1.0079 g. Elemental hydrogen exists as a diatomic molecule, H2. One molecule has a mass of 1.0079 u + 1.0079 u = 2.0158 u, while 1 mol of H2 has a mass of 1.0079 g + 1.0079 g = 2.0158 g. One molecule of H2O has a mass of about 18.01 u; 1 mol H2O has a mass of 18.01 g. A single unit of NaCl has a mass of 58.45 u; NaCl has a molar mass of 58.45 g. In each of these moles of substances, there are 6.022 × 1023 units: 6.022 × 1023 atoms of H, 6.022 × 1023 molecules of H2 and H2O, 6.022 × 1023 units of NaCl ions. These relationships give us plenty of opportunities to construct conversion factors for simple calculations. Example $4$: Sugar What is the molar mass of sugar ($\ce{C6H12O6}$)? Solution To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula; but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table. 6 C = 6 × 12.011 = 72.066 12 H = 12 × 1.0079 = 12.0948 6 O = 6 × 15.999 = 95.994 TOTAL = 180.155 g/mol Per convention, the unit grams per mole is written as a fraction. Exercise $4$ What is the molar mass of $\ce{AgNO3}$? Answer 169.87 g/mol Summary The mole is a key unit in chemistry. The molar mass of a substance, in grams, is numerically equal to one atom's or molecule's mass in atomic mass units.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/06%3A_Chemical_Reactions_-_Mole_and_Mass_Relationships/6.01%3A_The_Mole_and_Avogadros_Number.txt
Learning Objectives • To convert between mass units and mole units. As we just discussed, molar mass is defined as the mass (in grams) of 1 mole of substance (or Avogadro's number of molecules or formula units). The simplest type of manipulation using molar mass as a conversion factor is a mole-gram conversion (or its reverse, a gram-mole conversion). We also established that 1 mol of Al has a mass of 26.98 g (Example $1$). Stated mathematically, 1 mol Al = 26.98 g Al We can divide both sides of this expression by either side to get one of two possible conversion factors: $\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\quad and \quad \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \nonumber$ The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.” Example $1$ What is the mass of 3.987 mol of Al? Solution The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the $\mathrm{\dfrac{26.98\: g\: Al}{1\: mol\: Al}}$ conversion factor. We start with the given quantity and multiply by the conversion factor: $\mathrm{3.987\: mol\: Al\times\dfrac{26.98\: g\: Al}{1\: mol\: Al}}$ Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives $\mathrm{3.987\: mol\: Al\times \dfrac{26.98\: g\: Al}{1\: mol\: Al}=107.6\: g\: Al}$ Our final answer is expressed to four significant figures. Exercise $1$ How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.) Answer $\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}$ Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure $1$ is a chart for determining what conversion factor is needed, and Figure $2$ is a flow diagram for the steps needed to perform a conversion. Example $2$ A biochemist needs 0.00655 mol of bilirubin (C33H36N4O6) for an experiment. How many grams of bilirubin will that be? Solution To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula: 33 C molar mass: 33 × 12.01 g = 396.33 g 36 H molar mass: 36 × 1.01 g = 36.36 g 4 N molar mass: 4 × 14.01 g = 56.04 g 6 O molar mass: 6 × 16.00 g = 96.00 g Total:   584.73 g The molar mass of bilirubin is 584.73 g. Using the relationship 1 mol bilirubin = 584.73 g bilirubin we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure $2$: $\mathrm{0.00655\: mol\: bilirubin \times \dfrac{584.73\: g\: bilirubin}{mol\: bilirubin}=3.83\: g\: bilirubin}$ The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin. Exercise $2$ A chemist needs 457.8 g of KMnO4 to make a solution. How many moles of KMnO4 is that? Answer $\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}$ To Your Health: Minerals For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission. The US Department of Agriculture has established some recommendations for the RDIs of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet. Table $1$: Essential Minerals and their Composition in Humans Mineral Male (age 19–30 y) Female (age 19–30 y) Ca 1,000 mg 0.025 mol 1,000 mg 0.025 mol Cr 35 µg 6.7 × 10−7 mol 25 µg 4.8 × 10−7 mol Cu 900 µg 1.4 × 10−5 mol 900 µg 1.4 × 10−5 mol F 4 mg 2.1 × 10−4 mol 3 mg 1.5 × 10−4 mol I 150 µg 1.2 × 10−6 mol 150 µg 1.2 × 10−6 mol Fe 8 mg 1.4 × 10−4 mol 18 mg 3.2 × 10−4 mol K 3,500 mg 9.0 × 10−2 mol 3,500 mg 9.0 × 10−2 mol Mg 400 mg 1.6 × 10−2 mol 310 mg 1.3 × 10−2 mol Mn 2.3 mg 4.2 × 10−5 mol 1.8 mg 3.3 × 10−5 mol Mo 45 mg 4.7 × 10−7 mol 45 mg 4.7 × 10−7 mol Na 2,400 mg 1.0 × 10−1 mol 2,400 mg 1.0 × 10−1 mol P 700 mg 2.3 × 10−2 mol 700 mg 2.3 × 10−2 mol Se 55 µg 7.0 × 10−7 mol 55 µg 7.0 × 10−7 mol Zn 11 mg 1.7 × 10−4 mol 8 mg 1.2 × 10−4 mol Table $1$ illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women. Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly. Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/06%3A_Chemical_Reactions_-_Mole_and_Mass_Relationships/6.02%3A_Gram-Mole_Conversions.txt
Learning Objectives • To use a balanced chemical reaction to determine molar relationships between the substances. Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). Here, we will extend the meaning of the coefficients in a chemical equation. Consider the simple chemical equation $\ce{2H_2 + O_2 → 2H_2O}\nonumber$ The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as $\ce{4H_2 + 2O_2 → 4H_2O}\nonumber$ The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as $\ce{22H_2 + 11O_2 → 22H_2O}\nonumber$ because 22:11:22 also reduces to 2:1:2. Suppose we want to use larger numbers. Consider the following coefficients: $12.044 \times 10^{23}\; H_2 + 6.022 \times 10^{23}\; O_2 → 12.044 \times 10^{23}\; H_2O\nonumber$ These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 1023 is 1 mol, while 12.044 × 1023 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as $\ce{2 \;mol\; H_2 + 1\; mol\; O_2 → 2 \;mol\; H_2O}\nonumber$ We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is $\ce{2H_2 + O_2 → 2H_2O}\nonumber$ Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.” 2 molecules H2 1 molecule O2 2 molecules H2O 2 moles H2 1 mole O2 2 moles H2O 2 x 2.02 g=4.04 g H2 32.0 g O2 2 x 18.02 g=36.04 g H2O Figure $1$: This representation of the production of water from oxygen and hydrogen show several ways to interpret the quantitative information of a chemical reaction. By the same token, the ratios we constructed to describe molecules reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios: $\mathrm{\dfrac{2\: mol\: H_2}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2}}\nonumber$ $\mathrm{\dfrac{2\: mol\: H_2O}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2O}}\nonumber$ $\mathrm{\dfrac{2\: mol\: H_2}{2\: mol\: H_2O}\: or\: \dfrac{2\: mol\: H_2O}{2\: mol\: H_2}}\nonumber$ We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry. The ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products is called the stoichiometric factor. Example $1$ How many moles of oxygen react with hydrogen to produce 27.6 mol of H2O? The balanced equation is as follows: $\ce{2H2 + O2 -> 2H2O} \nonumber \nonumber$ Solution Because we are dealing with quantities of H2O and O2, we will use the stoichiometric ratio that relates those two substances. Because we are given an amount of H2O and want to determine an amount of O2, we will use the ratio that has H2O in the denominator (so it cancels) and O2 in the numerator (so it is introduced in the answer). Thus, $\mathrm{27.6\: mol\: H_2O\times\dfrac{1\: mol\: O_2}{2\: mol\: H_2O}=13.8\: mol\: O_2}$ To produce 27.6 mol of H2O, 13.8 mol of O2 react. Exercise $1$ Using 2H2 + O2 → 2H2O, how many moles of hydrogen react with 3.07 mol of oxygen to produce H2O? Answer $\mathrm{3.07\: mol\: O_2\times\dfrac{2\: mol\: H_2}{1\: mol\: O_2}=6.14\: mol\: H_2}$ Key Takeaway • The balanced chemical reaction can be used to determine molar relationships between substances.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/06%3A_Chemical_Reactions_-_Mole_and_Mass_Relationships/6.03%3A_Mole_Relationships_and_Chemical_Equations.txt
Learning Objectives • To convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction. We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence: Collectively, these conversions are called mole-mass calculations. As an example, consider the balanced chemical equation $\ce{Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3} \nonumber \nonumber$ If we have 3.59 mol of Fe2O3, how many grams of SO3 can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3 in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3 needed. Then using the molar mass of SO3 as a conversion factor, we determine the mass that this number of moles of SO3 has. The first step resembles the exercises we did in Section 6.4. As usual, we start with the quantity we were given: $\mathrm{3.59\: mol\: Fe_2O_3\times\dfrac{3\: mol\: SO_3}{1\: mol\: Fe_2O_3}=10.77\: mol\: SO_3} \nonumber \nonumber$ The mol Fe2O3 units cancel, leaving mol SO3 unit. Now, we take this answer and convert it to grams of SO3, using the molar mass of SO3 as the conversion factor: $\mathrm{10.77\: mol\: SO_3\times\dfrac{80.07\: g\: SO_3}{1\: mol\: SO_3}=862.4\: g\: SO_3} \nonumber \nonumber$ Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3 will react with 3.59 mol of Fe2O3. Many problems of this type can be answered in this manner. The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows: We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time. Example $1$ How many grams of CO2 are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation? $\ce{CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O} \nonumber$ Solution Our strategy will be to convert from moles of HCl to moles of CO2 and then from moles of CO2 to grams of CO2. We will need the molar mass of CO2, which is 44.01 g/mol. Performing these two conversions in a single-line gives 46.0 g of CO2: The molar ratio between CO2 and HCl comes from the balanced chemical equation. Exercise How many grams of glucose (C6H12O6) are produced if 17.3 mol of H2O are reacted according to this balanced chemical equation? $\ce{6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2} \nonumber$ Answer $\mathrm{17.3\: mol\: H_2O\times\dfrac{1\: mol\: C_6H_{12}O_6}{6\: mol\: H_2O}\times\dfrac{180.18\: g\: C_6H_{12}O_6}{1\: mol\: C_6H_{12}O_6}=520\: g\: C_6H_{12}O_6}$ It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows: This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques. Example $2$: Chlorination of Carbon Methane can react with elemental chlorine to make carbon tetrachloride ($\ce{CCl4}$). The balanced chemical equation is as follows: $\ce{CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl} \nonumber$ How many grams of HCl are produced by the reaction of 100.0 g of $\ce{CCl4}$? Solution First, let us work the problem in stepwise fashion. We begin by converting the mass of $\ce{CCl4}$ to moles of $\ce{CCl4}$, using the molar mass of $\ce{CCl4}$ (16.05 g/mol) as the conversion factor: $\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}=6.231\: mol\: CH_4}$ Note that we inverted the molar mass so that the gram units cancel, giving us an answer in moles. Next, we use the balanced chemical equation to determine the ratio of moles $\ce{CCl4}$ and moles HCl and convert our first result into moles of HCl: $\mathrm{6.231\: mol\: CH_4\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}=24.92\: mol\: HCl}$ Finally, we use the molar mass of HCl (36.46 g/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl: $\mathrm{24.92\: mol\: HCl\times\dfrac{36.46\: g\: HCl}{1\: mol\: HCl}=908.5\: g\: HCl}$ In each step, we have limited the answer to the proper number of significant figures. If desired, we can do all three conversions on a single line: $\mathrm{100.0\: g\: CH_4\times\dfrac{1\: mol\: CH_4}{16.05\: g\: CH_4}\times\dfrac{4\: mol\: HCl}{1\: mol\: CH_4}\times\dfrac{36.46\: g\: HCl}{1\: mol\: HCl}=908.7\: g\: HCl}$ This final answer is slightly different from our first answer because only the final answer is restricted to the proper number of significant figures. In the first answer, we limited each intermediate quantity to the proper number of significant figures. As you can see, both answers are essentially the same. Exercise $2$: Oxidation of Propanal The oxidation of propanal (CH3CH2CHO) to propionic acid (CH3CH2COOH) has the following chemical equation: CH3CH2CHO + 2K2Cr2O7 → CH3CH2COOH + other products How many grams of propionic acid are produced by the reaction of 135.8 g of K2Cr2O7? Answer $\mathrm{135.8\: g\: K_2Cr_2O_7\times\dfrac{1\: mol\: K_2Cr_2O_7}{294.20\: g\: K_2Cr_2O_7}\times\dfrac{1\: mol\: CH_3CH_2COOH}{2\: mol\: K_2Cr_2O_7}\times\dfrac{74.09\: g\: CH_3CH_2COOH}{1\: mol\: CH_3CH_2COOH}=17.10\: g\: CH_3CH_2COOH}$ To Your Health: The Synthesis of Taxol Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree (Taxus brevifolia). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of $\ce{C47H51NO14}$. Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory. After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed. Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug. Key Takeaway • A balanced chemical equation can be used to relate masses or moles of different substances in a reaction.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/06%3A_Chemical_Reactions_-_Mole_and_Mass_Relationships/6.04%3A_Mass_Relationships_and_Chemical_Equations.txt
Learning Objectives • Define and determine theoretical yields, actual yields, and percent yields. • Identify a limiting reagent from a set of reactants. • Calculate how much product will be produced from the limiting reagent. • Calculate how much reactant(s) remains when the reaction is complete. Yield In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions: 1. The reaction goes exactly as written. 2. The reaction proceeds completely. In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO2. However, we will continue to ignore side reactions, unless otherwise noted. The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons. When we calculate an amount of product assuming that all the reactant reacts, we calculate the theoretical yield, an amount that is theoretically produced as calculated using the balanced chemical reaction. In many cases, however, this is not what really happens. In many cases, less—sometimes, much less—of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the actual yield. By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made. Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The percent yield is a comparison between the actual yield and the theoretical yield and is defined as $\text{percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \label{yield}$ It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments. A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible. Example $1$: A worker reacts 30.5 g of Zn with nitric acid and evaporates the remaining water to obtain 65.2 g of Zn(NO3)2. What are the theoretical yield, the actual yield, and the percent yield? $\ce{Zn(s) + 2HNO_3(aq) → Zn(NO_3)_2(aq) + H_2(g)} \nonumber$ Solution A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g/mol) and Zn(NO3)2 (189.41 g/mol). In three steps, the mass-mass calculation is: $30.5\cancel{g\, Zn}\times \frac{1\, \cancel{mol\, Zn}}{65.39\cancel{g\, Zn}}\times \frac{1\, \cancel{mol\, Zn(NO_{3})_{2}}}{1\cancel{mol\, Zn}}\times \frac{189.41\, g\, Zn(NO_{3})_{2}}{1\cancel{mol\,Zn(NO_{3})_{2}}}=88.3\, g\, Zn(NO_{3})_{2}\nonumber$ Thus, the theoretical yield is 88.3 g of Zn(NO3)2. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO3)2. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100 (Equation \ref{yield}): $\frac{65.2\, g\, Zn(NO_{3})_{2}}{88.3\, g\,Zn(NO_{3})_{2}}\times 100\%=73.8\%\nonumber$ The worker achieved almost three-fourths of the possible yield. Exercise $1$ A synthesis produced 2.05 g of NH3 from 16.5 g of N2. What is the theoretical yield and the percent yield? $N_2(g) + 3H_2(g) → 2NH_3(g)\nonumber$ *Technically, this is a reversible reaction (with double arrows), but for this exercise consider it irreversible (single arrow). Answer theoretical yield = 20.1 g; percent yield = 10.2% Chemistry is Everywhere: Actual Yields in Drug Synthesis and Purification Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste. Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C13H21NO2; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do: impure albuterol → intermediate A percent yield = 70% intermediate A → intermediate B percent yield = 100% intermediate B → intermediate C percent yield = 40% intermediate C → intermediate D percent yield = 72% intermediate D → purified albuterol percent yield = 35% overall percent yield = 70% × 100% × 40% × 72% × 35% = 7.5% That is, only about one-fourteenth of the original material was turned into the purified drug. This demonstrates one reason why some drugs are so expensive—a lot of material is lost in making a high-purity pharmaceutical. Limiting Reagent In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess. Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus $1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{3.7.1}$ If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies. Figure $1$: The Concept of a Limiting Reactant in the Preparation of Brownies A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reagent; the other reactant or reactants are considered to be in excess. A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reagent and which is in excess. The key to recognizing which reactant is the limiting reagent is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reagent. What we need to do is determine an amount of one product (either moles or mass), assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reagent. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful. For example, consider this reaction: $4As(s) + 3O_2(g) → 2As_2O_3(s)\nonumber$ Suppose we start a reaction with 50.0 g of As and 50.0 g of O2. Which one is the limiting reagent? We need to perform two mole-mass calculations, each assuming that each reactant reacts completely. Then we compare the amount of the product produced by each and determine which is less. The calculations are as follows: $50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{2\, mol\, As_{2}O_{3}}{4\cancel{mol\, As}}=0.334\, mol\, As_{2}O_{3}\nonumber$ $50.0\cancel{g\, O_{2}}\times \frac{1\cancel{mol\, O_{2}}}{32.00\cancel{g\, O_{2}}}\times \frac{2\, mol\, As_{2}O_{3}}{3\cancel{mol\, O_{2}}}=1.04\, mol\, As_{2}O_{3}\nonumber$ Comparing these two answers, it is clear that 0.334 mol of As2O3 is less than 1.04 mol of As2O3, so arsenic is the limiting reagent. If this reaction is performed under these initial conditions, the arsenic will run out before the oxygen runs out. We say that the oxygen is "in excess." Identifying the limiting reagent, then, is straightforward. However, there are usually two associated questions: (1) what mass of product (or products) is then actually formed? and (2) what mass of what reactant is left over? The first question is straightforward to answer: simply perform a conversion from the number of moles of product formed to its mass, using its molar mass. For As2O3, the molar mass is 197.84 g/mol; knowing that we will form 0.334 mol of As2O3 under the given conditions, we will get $0.334\cancel{mol\, As_{2}O_{3}}\times \frac{197.84\, g\, As_{2}}{\cancel{1\, mol\, As_{2}O_{3}}}=66.1\, g\, As_{2}O_{3}\nonumber$ The second question is somewhat more convoluted to answer. First, we must do a mass-mass calculation relating the limiting reagent (here, As) to the other reagent (O2). Once we determine the mass of O2 that reacted, we subtract that from the original amount to determine the amount left over. According to the mass-mass calculation, $50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{3\cancel{mol\, O_{2}}}{4\cancel{mol\, As}}\times \frac{32.00\, g\, O_{2}}{\cancel{1\, mol\, O_{2}}}=16.0\, g\, O_{2}\; reacted\nonumber$ Because we reacted 16.0 g of our original O2, we subtract that from the original amount, 50.0 g, to get the mass of O2 remaining: 50.0 g O2 − 16.0 g O2 reacted = 34.0 g O2 left over You must remember to perform this final subtraction to determine the amount remaining; a common error is to report the 16.0 g as the amount remaining. Example $1$: A 5.00 g quantity of Rb is combined with 3.44 g of MgCl2 according to this chemical reaction: $2R b(s) + MgCl_2(s) → Mg(s) + 2RbCl(s) \nonumber\nonumber$ What mass of Mg is formed, and what mass of what reactant is left over? Solution Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less. $5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, Mg}}{2\cancel{mol\, Rb}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.711\, g\, Mg \nonumber$ $3.44\cancel{g\, MgCl_{2}}\times \frac{1\cancel{mol\, MgCl_{2}}}{95.21\cancel{g\, MgCl_{2}}}\times \frac{1\cancel{mol\, Mg}}{1\cancel{mol\, MgCl_{2}}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.878\, g\, Mg \nonumber$ The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb, and then subtract the amount reacted from the original amount. $5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, MgCl_{2}}}{2\cancel{mol\, Rb}}\times \frac{95.21\, g\, Mg}{\cancel{1\, mol\, MgCl_{2}}}=2.78\, g\, MgCl_{2}\: \: reacted \nonumber$ Because we started with 3.44 g of MgCl2, we have 3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left Exercise $1$ Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent? $\underbrace{22.7\, g}_{MgO(s)}+\underbrace{17.9\, g}_{H_2S}\rightarrow MgS(s)+H_{2}O(l) \nonumber$ Answer H2S is the limiting reagent; 1.5 g of MgO are left over. Summary Theoretical yield is the calculated yield using the balanced chemical reaction. Actual yield is what is actually obtained in a chemical reaction. Percent yield is a comparison of the actual yield with the theoretical yield. The limiting reagent is the reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced and how much of the other reactants remain.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/06%3A_Chemical_Reactions_-_Mole_and_Mass_Relationships/6.05%3A_Limiting_Reagent_and_Percent_Yield.txt
• 7.1: Energy and Chemical Bonds Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules. • 7.2: Heat Changes during Chemical Reactions Heat transfer is related to temperature change. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat. • 7.3: Exothermic and Endothermic Reactions Atoms are held together by a certain amount of energy called bond energy. Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively. • 7.4: Why Do Chemical Reactions Occur? Free Energy One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe. This is not particularly useful and a criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy. • 7.5: How Do Chemical Reactions Occur? Reaction Rates • 7.6: Effects of Temperature, Concentration, and Catalysts on Reaction Rates By their nature, some reactions occur very quickly, while others are very slow. However, certain changes in the reaction conditions can have an effect on the rate of a given chemical reaction. Collision theory can be utilized to explain these rate effects. • 7.7: Reversible Reactions and Chemical Equilibrium • 7.8: Equilibrium Equations and Equilibrium Constants Every chemical equilibrium can be characterized by an equilibrium constant, known as Keq. The Keq and KP expressions are formulated as amounts of products divided by amounts of reactants; each amount (either a concentration or a pressure) is raised to the power of its coefficient in the balanced chemical equation. Solids and liquids do not appear in the expression for the equilibrium constant. • 7.9: Le Chatelier’s Principle- The Effect of Changing Conditions on Equilibria Le Chatelier's principle addresses how an equilibrium shifts when the conditions of an equilibrium are changed. The direction of shift can be predicted for changes in concentrations, temperature, or pressure. Catalysts do not affect the position of an equilibrium; they help reactions achieve equilibrium faster. 07: Chemical Reactions - Energy Rates and Equilibrium Learning Objectives • Define energy and distinguish the different types of energy, potential and kinetic. Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure \(1\)). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges. Over 90% of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world’s energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter. Energy Energy can be defined as the capacity to supply heat or do work. One type of work (w) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire—we move matter (the air in the pump) against the opposing force of the air surrounding the tire. Like matter, energy comes in different types. One scheme classifies energy into two types: potential energy, the energy an object has because of its relative position, composition, or condition, often referred to as stored energy, and kinetic energy, the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure \(2\)). A battery has potential energy because the chemicals within it can produce electricity that can do work. Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.) When one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car’s engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic energy) when they move the cylinders’ pistons. According to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry. To encompass both chemical and nuclear changes, we combine these laws into one statement: The total quantity of matter and energy in the universe is fixed.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.01%3A_Energy_and_Chemical_Bonds.txt
Learning Objectives • Define bond dissociation energy. • Determine if a chemical process is exothermic or endothermic. A general statement, based on countless observations over centuries of study, is that all objects tend to move spontaneously to a position of minimum energy unless acted on by some other force or object. Bond Dissociation Energy Atoms bond together to form compounds because in doing so they attain lower energies than they possess as individual atoms. A quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. That is, the bonded atoms have a lower energy than the individual atoms do. When atoms combine to make a compound, energy is always given off, and the compound has a lower overall energy. In making compounds, atoms act like a basketball on a playground slide; they move in the direction of decreasing energy. We can reverse the process by putting energy into a molecule, which causes its bonds to break, separating the molecule into individual atoms. Bonds between certain specific elements usually have a characteristic energy, called the bond dissociation energy, that is needed to break the bond. The same amount of energy was liberated when the atoms made the chemical bond in the first place. The term bond dissociation energy is usually used to describe the strength of interactions between atoms that make covalent bonds. A C–C bond has an approximate bond energy of 80 kcal/mol, while a C=C has a bond energy of about 145 kcal/mol. The C=C bond is stronger than C-C (as discussed in relation to bond length in Section 4.4). For atoms in ionic compounds attracted by opposite charges, the term lattice energy is used. For now, we will deal with covalent bonds in molecules. Although each molecule has its own characteristic bond dissociation energy, some generalizations are possible. For example, although the exact value of a C–H bond energy depends on the particular molecule, all C–H bonds have a bond energy of roughly the same value because they are all C–H bonds. It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond dissociation energy of a C–H bond as being about 100 kcal/mol. Table $1$ lists the approximate bond dissociation energies of various covalent bonds. Table $1$: Approximate Bond Dissociation Energies Bond Bond Dissociation Energy (kcal/mol) C–H 100 C–O 86 C=O 190 C–N 70 C–C 85 C=C 145 C≡C 200 N–H 93 H–H 105 Br-Br 46 Cl–Cl 58 O–H 110 O=O 119 H–Br 87 H–Cl 103 When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change. The energy change, for a given reaction can be calculated using the bond dissociation energy values from Table $1$. Enthalpy Change or Heat of Reaction, ΔH During a chemical reaction, bonds are broken and new bonds are formed. Breaking chemical bonds is endothermic, a process that requires an input of energy or absorption of heat. The reverse process of bond breaking is bond formation, which is exothermic, meaning it releases energy or gives off heat. The bond dissociation energy values listed in the above table give the amount of energy required to break a specific bond. When that same bond is reformed, an identical amount of energy is released. The numerical value of energy is the same for breaking and forming a bond, but the sign, or direction of the process is different. The overall energy change of a specific bond breaking and reforming would be zero, in other words energy is neither created or destroyed, following the law of conservation of energy. In a chemical reaction, the bonds breaking are often different than the bonds reforming, sometimes there is more heat absorbed (more bonds are broken) and sometimes more heat is released (more bonds are formed). The measured difference between the total heat absorbed and the total heat released during a chemical reaction (performed at constant pressure) is called the heat of reaction or enthalpy change, and is represented by the symbol ΔH (where the Δ stands for change and the H represents enthalpy). $\text{enthalpy change} ≈ \Sigma\ (\text{bond dissociation energies}_{reactants}) - \Sigma\ (\text{bond dissociation energies}_{products})$ The ≈ sign is used because we are adding together average bond dissociation energies; hence this approach does not give exact values for the enthalpy change, ΔH. Let’s consider the reaction of 2 mols of hydrogen gas (H2) with 1 mol of oxygen gas (O2) to give 2 mol water: $2H_2(g)+O_2(g) \rightarrow 2H_2O(g)$ H–H = 105 kcal/mol O=O = 119 kcal/mol O–H = 110 kcal/mol In this reaction, 2 H–H bonds and 1 O=O bonds from the reactant side are broken, while 4 O–H bonds (two for each H2O) are formed on the product side. The energy changes can be tabulated and calculated as follows: Reactant Bond Dissociation Energy (kcal/mol) Product Bond Dissociation Energy (kcal/mol) 2 H–H 2 mol x 105 kcal/mol = 210 kcal 4 O–H 4 mols x 110 kcal/mol = 440 kcal 1 O=O 1 mol x 119 kcal/mol = 119 kcal Total = 329 kcal   Total = 440 kcal $\: \: \: \: \: \Delta H ≈ \Sigma\ (\text{bond dissociation energies}_{reactants}) - \Sigma\ (\text{bond dissociation energies}_{products})$ $\: \: \: \: \: \Delta H ≈ 329 \: \text{kcal} - 440 \: \text{kcal}$ $\: \: \: \: \: \Delta H ≈ −111 \: \text{kcal}$ The enthalpy change (ΔH) of the reaction is approximately −111 kcal/mol. This means that bonds in the products (440 kcal) are stronger than the bonds in the reactants (329 kcal) by about 111 kcal/mol. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it absorbs. This excess energy is released as heat, so the reaction is exothermic. Hence, we can re-write the reaction with the heat released (111 kcal) on the product side of the equation, as follows: We can also re-write the reaction equation with the ΔH information (see below). Note that an exothermic reaction has a negative ΔH value. $2H_2(g)+O_2(g) \rightarrow 2H_2O(g)\ \: \: \: \: \: \Delta H = -111 \: \text{kcal}$ Example $1$ What is the enthalpy change for this reaction? Is the reaction exothermic or endothermic? $H_2(g)+Br_2(g) \rightarrow 2HBr(g)$ Solution Step 1- First look at the equation and identify which bonds exist on in the reactants (bonds broken). • one H-H bond and • one Br-Br bond Step 2- Do the same for the products (bonds formed) • two H-Br bonds Step 3- Identify the bond dissociation energies of these bonds from Table $1$: • H-H bonds: 105 kcal/mol • Br-Br bonds: 46 kcal/mol Step 4- Set up the table (see below) and apply the formula for enthalpy change. Reactant Bond Dissociation Energy (kcal/mol) Product Bond Dissociation Energy (kcal/mol) 1 H–H 1 mol x 105 kcal/mol = 105 kcal 2 H–Br 2 mols x 87 kcal/mol = 174 kcal 1 Br–Br 1 mol x 46 kcal/mol = 46 kcal Total = 151 kcal   Total = 174 kcal $\: \: \: \: \: \Delta H ≈ 151 \: \text{kcal} - 174 \: \text{kcal}$ $\: \: \: \: \: \Delta H ≈ −23 \: \text{kcal}$ Step 5- Since ΔH is negative (−23 kcal), the reaction is exothermic. Exercise $1$ Using the bond dissociation energies given in the chart above, find the enthalpy change for the thermal decomposition of water: $H_2(g)+Cl_2(g) \rightarrow 2HCl(g)$ Is the reaction written above exothermic or endothermic? Explain. Answer ΔH = −43 kcal Since ΔH is negative (−43 kcal), the reaction is exothermic.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.02%3A_Heat_Changes_during_Chemical_Reactions.txt
Learning Objectives • Use bond dissociation energies to calculate enthalpy change or heat of reaction. • Determine if a chemical process is exothermic or endothermic. Endothermic and Exothermic Reactions Endothermic and exothermic reactions can be thought of as having energy as either a reactant of the reaction or a product. Endothermic reactions require energy, so energy is a reactant. Heat flows from the surroundings to the system (reaction mixture) and the enthalpy of the system increases ($\Delta H$ is positive). As discussed in the previous section, heat is released (considered a product) in an exothermic reaction, and the enthalpy of the system decreases ($\Delta H$ is negative). In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases (gets cold). A chemical reaction is exothermic if heat is released by the system into the surroundings. Because the surroundings is gaining heat from the system, the temperature of the surroundings increases. See Figure $1$. Endothermic Reaction: 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. Because the heat is absorbed by the system, the $177.8 \: \text{kJ}$ is written as a reactant. The $\Delta H$ is positive for an endothermic reaction. $\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = +177.8 \: \text{kJ}$ Exothermic Reaction: 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 the balanced equation in two ways. First, the amount of heat released can be written in the product side of the reaction. Another way is to write the $\Delta H$ information with a negative sign, $-890.4 \: \text{kJ}$. $\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ}$ Example $1$ Is each chemical reaction exothermic or endothermic? 1. CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ) + 213 kcal 2. N2(g) + O2(g) + 45 kcal → 2NO(g) Solution 1. Because energy (213 kcal) is a product, energy is given off by the reaction. Therefore, this reaction is exothermic. 2. Because energy (45 kcal) is a reactant, energy is absorbed by the reaction. Therefore, this reaction is endothermic. Exercise $1$ Is each chemical reaction exothermic or endothermic? 1. H2(g) + F2(g) → 2HF (g) + 130 kcal 2. 2C(s) + H2(g) + 5.3 kcal → C2H2(g) Answer a. The energy (130 kcal) is produced, hence the reaction is exothermic b. The energy (5.3 kcal) is supplied or absorbed to react, hence, the reaction is endothermic Energy Diagrams Endothermic and exothermic reactions can be visually represented by energy-level diagrams like the ones in Figure $2$. In endothermic reactions, the reactants have higher bond energy (stronger bonds) than the products. Strong bonds have lower potential energy than weak bonds. Hence, the energy of the reactants is lower than that of the products. This type of reaction is represented by an "uphill" energy-level diagram shown in Figure $\PageIndex{2A}$. For an endothermic chemical reaction to proceed, the reactants must absorb energy from their environment to be converted to products. In an exothermic reaction, the bonds in the product have higher bond energy (stronger bonds) than the reactants. In other words, the energy of the products is lower than the energy of the reactants, hence is energetically downhill, shown in Figure $\PageIndex{2B}$. Energy is given off as reactants are converted to products. The energy given off is usually in the form of heat (although a few reactions give off energy as light). In the course of an exothermic reaction, heat flows from the system to its surroundings, and thus, gets warm. Table $1$: Endothermic and Exothermic Reactions Endothermic Reactions Exothermic Reactions Heat is absorbed by reactants to form products. Heat is released. Heat is absorbed from the surroundings; as a result, the surroundings get cold. Heat is released by the reaction to surroundings; surroundings feel hot. $\Delta H_{rxn}$ is positive $\Delta H_{rxn}$ is negative The bonds broken in the reactants are stronger than the bonds formed in the products. The bonds formed in the products are stronger than the bonds broken in the reactants. The reactants are lower in energy than the products. The products are lower in energy than the reactants. Represented by an "uphill" energy diagram. Represented by an "downhill" energy diagram Concept Review Exercises 1. What is the connection between energy and chemical bonds? 2. Why does energy change during the course of a chemical reaction? 3. Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and which is exothermic? Explain. 4. Classify "burning paper" as endothermic or exothermic processes. Answers 1. Chemical bonds have a certain energy that is dependent on the elements in the bond and the number of bonds between the atoms. 2. Energy changes because bonds rearrange to make new bonds with different energies. 3. Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold. 4. "Burning paper" is exothermic because burning (also known as combustion) releases heat Key Takeaways • Atoms are held together by a certain amount of energy called bond energy. • Energy is required to break bonds. Energy is released when chemical bonds are formed because atoms become more stable. • Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.03%3A_Exothermic_and_Endothermic_Reactions.txt
Learning Outcomes • Describe the meaning of a spontaneous reaction in terms of enthalpy and entropy changes. • Define free energy. • Determine the spontaneity of a reaction based on the value of its change in free energy at high and low temperatures. Spontaneous Reactions A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring. A roaring bonfire (see Figure $1$ below) is an example of a spontaneous reaction. A fire is exothermic, which means a decrease in the energy of the system as energy is released to the surroundings as heat. The products of a fire are composed mostly of gases such as carbon dioxide and water vapor, so the entropy of the system increases during most combustion reactions. This combination of a decrease in energy and an increase in entropy means that combustion reactions occur spontaneously. A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, one or both of the driving forces must favor the reactants over the products. In other words, the reaction is endothermic, is accompanied by a decrease in entropy, or both. Out atmosphere is composed primarily of a mixture of nitrogen and oxygen gases. One could write an equation showing these gases undergoing a chemical reaction to form nitrogen monoxide. $\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO} \left( g \right)$ Fortunately, this reaction is nonspontaneous at normal temperatures and pressures, it is a highly endothermic reaction. However, nitrogen monoxide is capable of being produced at very high temperatures, and this reaction has been observed to occur as a result of lightning strikes. One must be careful not to confuse the term spontaneous with the notion that a reaction occurs rapidly. A spontaneous reaction is one in which product formation is favored, even if the reaction is extremely slow. You do not have to worry about a piece of paper on your desk suddenly bursting into flames, although its combustion is a spontaneous reaction. What is missing is the required activation energy to get the reaction started. If the paper were to be heated to a high enough temperature, it would begin to burn, at which point the reaction would proceed spontaneously until completion. Entropy as a Driving Force An example of a very simple spontaneous process is that of a melting ice cube. Energy is transferred from the room to the ice cube, causing it to change from the solid to the liquid state. $\ce{H_2O} \left( s \right) + 6.01 \: \text{kJ} \rightarrow \ce{H_2O} \left( l \right)$ The solid state of water, ice, is highly ordered because its molecules are fixed in place. The melting process frees the water molecules from their hydrogen-bonded network and allows them a greater degree of movement. Water is more disordered than ice. The change from the solid to the liquid state of any substance corresponds to an increase in the disorder of the system. The tendency in nature for systems to proceed toward a state of greater disorder or randomness is called entropy, which is symbolized by S, and expressed in units of Joules per mole-kelvin, $\mathrm{J/(mol\cdot K)}$. Larger values of S indicate that the particles in a substance have more disorder or randomness. In the example above, the particles in the ice cube (solid water) have lower freedom of motion, they are less random. As the ice melts to liquid water, the particles become more disordered and entropy increases. If the liquid water was heated further, the particles would even gain more freedom of motion, become more disordered, and eventually change to a gas, which has even higher entropy. Chemical reactions also tend to proceed in such a way as to increase the total entropy of the system, measured by the entropy change ($\Delta S$) between reactants and products. How can you tell if a certain reaction shows an increase or a decrease in entropy? The states of the reactants and produces 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 dissolving 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 more 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 nongaseous products are formed from gaseous reactants. The examples below will 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. Gibbs Free Energy Many chemical reactions and physical processes release energy that can be used to do other things. When the fuel in a car is burned, some of the released energy is used to power the vehicle. Free energy is energy that is available to do work. Spontaneous reactions release free energy as they proceed. The determining factors for spontaneity of a reaction depend on both the enthalpy and entropy changes that occur for the system. The free energy change ($\Delta G$) of a reaction is a mathematical combination of the enthalpy change and the entropy change. $\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o}$ The symbol for free energy is $G$, in honor of American scientist Josiah Gibbs (1839 - 1903), who made many contributions to thermodynamics. The change in Gibbs free energy is equal to the change in enthalpy minus the mathematical product of the change in entropy multiplied by the Kelvin temperature. Each thermodynamic quantity in the equation is for substances in their standard states, as indicated by the $^\text{o}$ superscripts. A spontaneous reaction is one that releases free energy, and so the sign of $\Delta G$ must be negative. Since both $\Delta H$ and $\Delta S$ can be either positive or negative, depending on the characteristics of the particular reaction, there are four different possible combinations. The outcomes for $\Delta G$ based on the signs of $\Delta H$ and $\Delta S$ are outlined in the table below. Recall that $- \Delta \text{H}$ indicates that the reaction is exothermic and a $+ \Delta \text{H}$ means the reaction is endothermic. For entropy, $+ \Delta \text{S}$ means the entropy is increasing and the system is becoming more disordered. A $- \Delta \text{S}$ means that entropy is decreasing and the system is becoming less disordered (more ordered). A process that releases free energy, ($-\Delta G$), is said to be exergonic. Processes that require free energy ($+\Delta G$), are endergonic. These terms are used when considering chemical reactions that occur in living systems. Table $1$: Enthalpy, Entropy, and Free Energy Changes. $\Delta H$ $\Delta S$ $\Delta G$ negative positive always negative positive positive negative at higher temperatures, positive at lower temperatures negative negative negative at lower temperatures, positive at higher temperatures positive negative always positive Keep in mind that the temperature in the Gibbs free energy equation is the Kelvin temperature, so it can only have a positive value. When $\Delta H$ is negative and $\Delta S$ is positive, the sign of $\Delta G$ will always be negative, and the reaction will be spontaneous at all temperatures. This corresponds to both driving forces being in favor of product formation. When $\Delta H$ is positive and $\Delta S$ is negative, the sign of $\Delta G$ will always be positive, and the reaction can never be spontaneous. This corresponds to both driving forces working against product formation. When one driving force favors the reaction, but the other does not, it is the temperature that determines the sign of $\Delta G$. Consider first an endothermic reaction (positive $\Delta H$) that also displays an increase in entropy (positive $\Delta S$). It is the entropy term that favors the reaction. Therefore, as the temperature increases, the $T \Delta S$ term in the Gibbs free energy equation will begin to predominate and $\Delta G$ will become negative. A common example of a process which falls into this category is the melting of ice (see figure below). At a relatively low temperature (below $273 \: \text{K}$), the melting is not spontaneous because the positive $\Delta H$ term "outweighs" the $T \Delta S$ term. When the temperature rises above $273 \: \text{K}$, the process becomes spontaneous because the larger $T$ value has tipped the sign of $\Delta G$ over to being negative. When the reaction is exothermic (negative $\Delta H$) but undergoes a decrease in entropy (negative $\Delta S$), it is the enthalpy term which favors the reaction. In this case, a spontaneous reaction is dependent upon the $T \Delta S$ term being small relative to the $\Delta H$ term, so that $\Delta G$ is negative. The freezing of water is an example of this type of process. It is spontaneous only at a relatively low temperature. Above $273. \: \text{K}$, the larger $T \Delta S$ value causes the sign of $\Delta G$ to be positive, and freezing does not occur. Contributors and Attributions • Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.04%3A_Why_Do_Chemical_Reactions_Occur_Free_Energy.txt
Learning Outcomes • Define reaction rate. • Explain the concept of activation energy. • Label a diagram with reactants, products, enthalpy of forward and reverse reactions, activation energy of forward and reverse reactions, and activated complex. Chemical kinetics is the study of the rates of chemical reactions. In this lesson, you will learn how to express the rate of a chemical reaction and about various factors that influence reaction rates. Expressing Reaction Rate Chemical reactions vary widely in the speeds with which they occur. Some reactions occur very quickly. It a lighted match is brought into contact with lighter fluid or another flammable liquid, it erupts into flame 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 seconds $\left( s \right)$ to run a 100 meter (\left( m \right)\) dash, his rate or speed is given by the distance traveled divided by the time (see figure below). $\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{100 \: \text{m}}{11.0 \: \text{s}} = 9.09 \: \text{m/s}$ 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 the 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 molar $\left( \text{M} \right)$ aqueous solution of substance $\ce{A}$ was converted to substance $\ce{B}$. $\ce{A} \left( aq \right) \rightarrow \ce{B} \left( aq \right)$ Suppose that after 20.0 seconds, the concentration of $\ce{A}$ had dropped from $1.00 \: \text{M}$ to $0.72 \: \text{M}$ as it was being converted to substance $\ce{B}$. We can express the rate of this reaction as the change in concentration of $\ce{A}$ divided by the time. $\text{rate} = -\frac{\Delta \left[ \ce{A} \right]}{\Delta t} = -\frac{\left[ \ce{A} \right]_\text{final} - \left[ \ce{A} \right]_\text{initial}}{\Delta t}$ 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 $\text{M/s}$. $\text{rate} = -\frac{0.72 \: \text{M} - 1.00 \: \text{M}}{20.0 \: \text{s}} = -\frac{-0.28 \: \text{M}}{20.0 \: \text{s}} = 0.041 \: \text{M/s}$ Over the first 20.0 seconds of this reaction, the molarity of $\ce{A}$ decreases by an average rate of $0.041 \: \text{M}$ 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 $\left( \text{M/s} \right)$. Collision Theory The behavior of the reactant atoms, molecules, or ions is responsible for the rates of a given chemical reaction. Collision theory is a set of principles based around the idea that reactant particles form products when they collide with one another, but only when those collisions have enough kinetic energy and the correct orientation to cause a reaction. 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 produce. The first collision is called an ineffective collision, while the second collision is called an effective collision. 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. Additionally, more energetic particles have more forceful collisions, which also increases the likelihood that a rearrangement of atoms will take place. 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 most of 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 to react unless more is provided by an external source of heat. Potential Energy Diagrams Then energy changes that occur during a chemical reaction can be shown in a diagram called a potential energy diagram, sometimes called a reaction progress curve. A potential energy diagram shows the change in the potential energy of a system as reactants are converted into products. The figure below shows basic potential energy diagrams for an endothermic (left) and an exothermic (right) 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 this barrier (see figure below). As discussed earlier, reactant particles sometimes collide with one another and yet 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 only for an extremely short period of time (about $10^{-13} \: \text{s}$). The activated complex is equally likely to either reform the original reactants or go on to form the 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 most activated complexes.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.05%3A_How_Do_Chemical_Reactions_Occur_Reaction_Rates.txt
Learning Outcomes • Describe how temperatures, concentration of reactant, and a catalyst affect the reaction rate. By their nature, some reactions occur very quickly, while others are very slow. However, certain changes in the reaction conditions can have an effect on the rate of a given chemical reaction. Collision theory can be utilized to explain these rate effects. Concentration Increasing the concentration of one or more of the reacting substances generally increases the reaction rate. 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 frequency of collisions between the reactants, the rate increases as the concentration increases. Temperature Raising the temperature of a chemical reaction results in a higher reaction rate. When the reactant particles are heated, they move faster and faster, resulting in a greater frequency of collisions. An even more important effect of the temperature increase is that the collisions occur with a greater force, which means the reactants are more likely to surmount the activation energy barrier and go on to form products. Increasing the temperature of a reaction increases not only the frequency of collisions, but also the percentage of those collisions that are effective, resulting in an increased reaction rate. Paper is certainly a highly combustible material, but paper does not burn at room temperature because the activation energy for the reaction is too high. The vast majority of collisions between oxygen molecules and the paper are ineffective. However, when the paper is heated by the flame from a match, it reaches a point where the molecules now have enough energy to react. The reaction is very exothermic, so the heat released by the initial reaction will provide enough energy to allow the reaction to continue, even if the match is removed. The paper continues to burn rapidly until it is gone. Catalysts The rates of some chemical reactions can be increased dramatically by introducing certain other substances into the reaction mixture. Hydrogen peroxide is used as a disinfectant for scrapes and cuts, and it can be 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 without being used up in the reaction. It accomplishes this task by providing an alternate reaction pathway that has a lower activation energy barrier. 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. $2 \ce{H_2O_2} \left( aq \right) \overset{\ce{MnO_2}}{\rightarrow} 2 \ce{H_2O} \left( l \right) + \ce{O_2} \left( g \right)$ A catalyst works by changing the mechanism of the reaction, which can be though of as the specific set of smaller steps by which the reactants become products. The important point is that the use of a catalyst lowers the overall activation energy of the reaction (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 extremely important parts of many chemical reactions. Enzymes in your body act 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. 7.07: Reversible Reactions and Chemical Equilibrium Learning Objectives • Define chemical equilibrium. • Recognize chemical equilibrium as a dynamic process. Consider the following reaction occurring in a closed container (so that no material can go in or out): H2 + I2 → 2HI This is simply the reaction between elemental hydrogen and elemental iodine to make hydrogen iodide. The way the equation is written, we are led to believe that the reaction goes to completion, that all the H2 and the I2 react to make HI. However, this is not the case because it is a reversible reaction, meaning it can go in either direction. As soon as there is enough product formed, the HI can react and the reverse chemical reaction occurs essentially "undoing" the first reaction: 2HI → H2 + I2 Eventually, the reverse reaction proceeds so quickly that it matches the speed of the forward reaction. When that happens, the concentration of the reactants and products remains constant, there is no further change; the reaction has reached chemical equilibrium (sometimes just spoken as equilibrium; plural equilibria), the point at which the forward and reverse processes balance each other's progress. Because two opposing processes are occurring at once, it is conventional to represent an equilibrium using a double arrow, like this: $H_{2}+I_{2}\rightleftharpoons 2HI\nonumber$ The double arrow implies that the reaction is going in both directions. Note that the reaction must still be balanced. Example $1$ Write the equilibrium equation that exists between calcium carbonate as a reactant and calcium oxide and carbon dioxide as products. Solution As this is an equilibrium situation, a double arrow is used. The equilibrium equation is written as follows: $CaCO_{3}+\rightleftharpoons CaO+CO_{2}\nonumber$ Exercise $1$ Write the equilibrium equation between elemental hydrogen and elemental oxygen as reactants and water as the product. Answer $2H_{2}+O_{2}+\rightleftharpoons 2H_{2}O\nonumber$ One thing to note about equilibrium is that the reactions do not stop; both the forward reaction and the reverse reaction continue to occur. They both occur at the same rate, so any overall change by one reaction is canceled by the reverse reaction. We say that chemical equilibrium is dynamic, rather than static. Also, because both reactions are occurring simultaneously, the equilibrium can be written backward. For example, representing an equilibrium as $H_{2}+I_{2}\rightleftharpoons 2HI\nonumber$ is the same thing as representing the same equilibrium as $2HI\rightleftharpoons H_{2}+I_{2}\nonumber$ The reaction must be at equilibrium for this to be the case, however. Key Takeaways • Chemical reactions eventually reach equilibrium, a point at which forward and reverse reactions balance each other's progress. • Chemical equilibria are dynamic: the chemical reactions are always occurring; they just cancel each other's progress.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.06%3A_Effects_of_Temperature_Concentration_and_Catalysts_on_.txt
Learning Objectives • Define the equilibrium constant. • Construct an equilibrium constant expression for a chemical reaction. In the mid 1860s, Norwegian scientists C. M. Guldberg and P. Waage noted a peculiar relationship between the amounts of reactants and products in an equilibrium. No matter how many reactants they started with, a certain ratio of reactants and products was achieved at equilibrium. Today, we call this observation the law of mass action. It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution, $aA+bB\rightleftharpoons cC+dD\nonumber$ the equilibrium constant, also known as Keq, is defined by the following expression: $K_{eq}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\nonumber$ where [A] is the molar concentration of species A at equilibrium, and so forth. The coefficients a, b, c, and d in the chemical equation become exponents in the expression for Keq. The Keq is a characteristic numerical value for a given reaction at a given temperature; that is, each chemical reaction has its own characteristic Keq. The concentration of each reactant and product in a chemical reaction at equilibrium is related; the concentrations cannot be random values, but they depend on each other. The numerator of the expression for Keq has the concentrations of every product (however many products there are), while the denominator of the expression for Keq has the concentrations of every reactant, leading to the common products over reactants definition for the Keq. Let us consider a simple example. Suppose we have this equilibrium: $A\rightleftharpoons B\nonumber$ There is one reactant, one product, and the coefficients on each are just 1 (assumed, not written). The Keq expression for this equilibrium is $K_{eq}=\frac{[B]}{[A]}\nonumber$ (Exponents of 1 on each concentration are understood.) Suppose the numerical value of Keq for this chemical reaction is 2.0. If [B] = 4.0 M, then [A] must equal 2.0 M so that the value of the fraction equals 2.0: $K_{eq}=\frac{[B]}{[A]}=\frac{4.0}{2.0}=2.0\nonumber$ By convention, the units are understood to be M and are omitted from the Keq expression. Suppose [B] were 6.0 M. For the Keq value to remain constant (it is, after all, called the equilibrium constant), then [A] would have to be 3.0 M at equilibrium: $K_{eq}=\frac{[B]}{[A]}=\frac{6.0}{3.0}=2.0\nonumber$ If [A] were not equal to 3.0 M, the reaction would not be at equilibrium, and a net reaction would occur until that ratio was indeed 2.0. At that point, the reaction is at equilibrium, and any net change would cease. (Recall, however, that the forward and reverse reactions do not stop because chemical equilibrium is dynamic.) The issue is the same with more complex expressions for the Keq; only the mathematics become more complex. Generally speaking, given a value for the Keq and all but one concentration at equilibrium, the missing concentration can be calculated. Example $1$ Given the following reaction: $\ce{H2 + I2 <=> 2HI}$ If the equilibrium [HI] is 0.75 M and the equilibrium [H2] is 0.20 M, what is the equilibrium [I2] if the Keq is 0.40? Solution We start by writing the Keq expression. Using the products over reactants approach, the Keq expression is as follows: $K_{eq}=\frac{[HI]^{2}}{[H_{2}][I_{2}]}\nonumber$ Note that [HI] is squared because of the coefficient 2 in the balanced chemical equation. Substituting for the equilibrium [H2] and [HI] and for the given value of Keq: $0.40=\frac{(0.75)^{2}}{(0.20)[I_{2}]}\nonumber$ To solve for [I2], we have to do some algebraic rearrangement: divide the 0.40 into both sides of the equation and multiply both sides of the equation by [I2]. This brings [I2] into the numerator of the left side and the 0.40 into the denominator of the right side: $[I_{2}]=\frac{(0.75)^{2}}{(0.20)(0.40)}\nonumber$ Solving, [I2] = 7.0 M The concentration unit is assumed to be molarity. This value for [I2] can be easily verified by substituting 0.75, 0.20, and 7.0 into the expression for Keq and evaluating: you should get 0.40, the numerical value of Keq (and you do). Exercise $1$ Given the following reaction: $\ce{H2 + I2 <=> 2HI}$ If the equilibrium [HI] is 0.060 M and the equilibrium [I2] is 0.90 M, what is the equilibrium [H2] if the Keq is 0.40? Answer 0.010 M In some types of equilibrium problems, square roots, cube roots, or even higher roots need to be analyzed to determine a final answer. Make sure you know how to perform such operations on your calculator; if you do not know, ask your instructor for assistance. Example $2$ The following reaction is at equilibrium: $\ce{N2 + 3H2 <=> 2NH3}$ The Keq at a particular temperature is 13.7. If the equilibrium [N2] is 1.88 M and the equilibrium [NH3] is 6.62 M, what is the equilibrium [H2]? Solution We start by writing the Keq expression from the balanced chemical equation: $K_{eq}=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\nonumber$ Substituting for the known equilibrium concentrations and the Keq, this becomes $13.7=\frac{(6.62)^{2}}{(1.88)[H_{2}]^{3}}\nonumber$ Rearranging algebraically and then evaluating the numerical expression, we get $[H_{2}]^{3}=\frac{(6.62)^{2}}{(1.88)(13.7)}=1.7015219754\nonumber$ To solve for [H2], we need to take the cube root of the equation. Performing this operation, we get [H2] = 1.19 M You should verify that this is correct using your own calculator to confirm that you know how to do a cube root correctly. Exercise $2$ The following reaction is at equilibrium: $\ce{N2 + 3H2 <=> 2NH3}$ The Keq at a particular temperature is 13.7. If the equilibrium [N2] is 0.055 M and the equilibrium [H2] is 1.62 M, what is the equilibrium [NH3]? Answer 1.79 M The Keq was defined earlier in terms of concentrations. For gas-phase reactions, the Keq can also be defined in terms of the partial pressures of the reactants and products, Pi. For the gas-phase reaction $aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)\nonumber$ the pressure-based equilibrium constant, KP, is defined as follows: $K_{P}=\frac{P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}}\nonumber$ where PA is the partial pressure of substance A at equilibrium in atmospheres, and so forth. As with the concentration-based equilibrium constant, the units are omitted when substituting into the expression for KP. Example $3$ What is the KP for this reaction, given the equilibrium partial pressures of 0.664 atm for NO2 and 1.09 for N2O4? $\ce{2NO2(g) <=> N2O4(g)}$ Solution Write the KP expression for this reaction: $K_{P}=\frac{P_{N_{2}O_{4}}}{P_{NO_{2}}^{2}}\nonumber$ Then substitute the equilibrium partial pressures into the expression and evaluate: $K_{P}=\frac{(1.09)}{(0.664)^{2}}=2.47\nonumber$ Exercise $3$ What is the KP for this reaction, given the equilibrium partial pressures of 0.44 atm for H2, 0.22 atm for Cl2, and 2.98 atm for HCl? $\ce{H2 + Cl2 <=> 2HCl}$ Answer 91.7 There is a simple relationship between Keq (based on concentration units) and KP (based on pressure units): $K_{P}=K_{eq}\cdot \left ( RT\right )^{\Delta n}\nonumber$ where R is the ideal gas law constant (in units of L·atm/mol·K), T is the absolute temperature, and Δn is the change in the number of moles of gas in the balanced chemical equation, defined as ngas,prodsngas,rcts. Note that this equation implies that if the number of moles of gas are the same in reactants and products, Keq = KP. Example $4$ What is the KP at 25°C for this reaction if the Keq is 4.2 × 10−2? $\ce{N2 + 3H2 <=> 2NH3}$ Solution Before we use the relevant equation, we need to do two things: convert the temperature to kelvins and determine Δn. Converting the temperature is easy: T = 25 + 273 = 298 K To determine the change in the number of moles of gas, take the number of moles of gaseous products and subtract the number of moles of gaseous reactants. There are 2 mol of gas as product and 4 mol of gas of reactant: Δn = 2 − 4 = −2 mol Note that Δn is negative. Now we can substitute into our equation, using R = 0.08205 L·atm/mol·K. The units are omitted for clarity: KP = (4.2 × 10−2)(0.08205)(298)−2 Solving, KP = 7.0 × 10−5 .Exercise $4$ What is the KP at 25°C for this reaction if the Keq is 98.3?,- $\ce{I2(g) <=> 2I(g)}$ Answer 2.40 × 103 Finally, we recognize that many chemical reactions involve substances in the solid or liquid phases. For example, a particular chemical reaction is represented as follows: $\ce{2NaHCO3(s) <=> Na2CO3(s) + CO2(g) + H2O(l)}$ This chemical equation includes all three phases of matter. This kind of equilibrium is called a heterogeneous equilibrium because there is more than one phase present. The rule for heterogeneous equilibria is as follows: Do not include the concentrations of pure solids and pure liquids in Keq expressions. Only partial pressures for gas-phase substances or concentrations in solutions are included in the expressions of equilibrium constants. As such, the equilibrium constant expression for this reaction would simply be $K_{P}=P_{CO_{2}}\nonumber$ because the two solids and one liquid would not appear in the expression. Key Takeaways • Every chemical equilibrium can be characterized by an equilibrium constant, known as Keq. • The Keq and KP expressions are formulated as amounts of products divided by amounts of reactants; each amount (either a concentration or a pressure) is raised to the power of its coefficient in the balanced chemical equation. • Solids and liquids do not appear in the expression for the equilibrium constant.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.08%3A_Equilibrium_Equations_and_Equilibrium_Constants.txt
Learning Objectives • Define Le Chatelier's principle. • Predict the direction of shift for an equilibrium under stress. Once equilibrium is established, the reaction is over, right? Not exactly. An experimenter has some ability to affect the equilibrium. Chemical equilibria can be shifted by changing the conditions that the system experiences. We say that we "stress" the equilibrium. When we stress the equilibrium, the chemical reaction is no longer at equilibrium, and the reaction starts to move back toward equilibrium in such a way as to decrease the stress. The formal statement is called Le Chatelier's principle: If an equilibrium is stressed, then the reaction shifts to reduce the stress. Effect of Changes in Concentration There are several ways to stress an equilibrium. One way is to add or remove a product or a reactant in a chemical reaction at equilibrium. When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss. Example $1$ Given this reaction at equilibrium: $N_{2}+3H_{2}\rightleftharpoons 2NH_{3}\nonumber$ In which direction—toward reactants or toward products-—does the reaction shift if the equilibrium is stressed by each change? 1. H2 is added. 2. NH3 is added. 3. NH3 is removed. Solution 1. If H2 is added, there is now more reactant, so the reaction will shift toward products to reduce the added H2. 2. If NH3 is added, there is now more product, so the reaction will shift toward reactants to reduce the added NH3. 3. If NH3 is removed, there is now less product, so the reaction will shift toward products to replace the product removed. Exercise $1$ Given this reaction at equilibrium: $CO(g)+Br_{2}(g)\rightleftharpoons COBr_{2}(g)\nonumber$ In which direction—toward reactants or toward products—does the reaction shift if the equilibrium is stressed by each change? 1. Br2 is removed. 2. COBr2 is added. Answers 1. toward reactants 2. toward reactants It is worth noting that when reactants or products are added or removed, the value of the Keq does not change. The chemical reaction simply shifts, in a predictable fashion, to reestablish concentrations so that the Keq expression reverts to the correct value. Effect of Changes in Pressure and Temperature How does an equilibrium react to a change in pressure? Pressure changes do not markedly affect the solid or liquid phases. However, pressure strongly impacts the gas phase. Le Chatelier's principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas. If the number of moles of gas is the same on both sides of the reaction, pressure has no effect. Example $2$ What is the effect on this equilibrium if pressure is increased? $N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)\nonumber$ Solution According to Le Chatelier's principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts toward the products side. Exercise $2$ What is the effect on this equilibrium if pressure is decreased? $3O_{2}(g)\rightleftharpoons 2O_{3}(g)\nonumber$ Answer Reaction shifts toward reactants. What is the effect of temperature changes on an equilibrium? It depends on whether the reaction is endothermic or exothermic. Recall that endothermic means that energy is absorbed by a chemical reaction, while exothermic means that energy is given off by the reaction. As such, energy can be thought of as a reactant or a product, respectively, of a reaction: endothermic reaction: energy + reactants → products exothermic reaction: reactants → products + energy Because temperature is a measure of the energy of the system, increasing temperature can be thought of as adding energy. The reaction will react as if a reactant or a product is being added and will act accordingly by shifting to the other side. For example, if the temperature is increased for an endothermic reaction, essentially a reactant is being added, so the equilibrium shifts toward products. Decreasing the temperature is equivalent to decreasing a reactant (for endothermic reactions) or a product (for exothermic reactions), and the equilibrium shifts accordingly. Example $3$ Predict the effect of increasing the temperature on this equilibrium. $PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}+60kJ\nonumber$ Solution Because energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts back toward reactants. Exercise $3$ Predict the effect of decreasing the temperature on this equilibrium. $N_{2}O_{4}+57kJ\rightleftharpoons 2NO_{2}\nonumber$ Answer Equilibrium shifts toward reactants. In the case of temperature, the value of the equilibrium has changed because the Keq is dependent on temperature. That is why equilibria shift with changes in temperature. A catalyst is a substance that increases the speed of a reaction. Overall, a catalyst is not a reactant and is not used up, but it still affects how fast a reaction proceeds. However, a catalyst does not affect the extent or position of a reaction at equilibrium. It helps a reaction achieve equilibrium faster. Chemistry is Everywhere: Equilibria in the Garden Hydrangeas are common flowering plants around the world. Although many hydrangeas are white, there is one common species (Hydrangea macrophylla) whose flowers can be either red or blue, as shown in the accompanying figure. How is it that a plant can have different colored flowers like this? Interestingly, the color of the flowers is due to the acidity of the soil that the hydrangea is planted in. An astute gardener can adjust the pH of the soil and actually change the color of the flowers. However, it is not the H+ or OH ions that affect the color of the flowers. Rather, it is the presence of aluminum that causes the color change. The solubility of aluminum in soil, and the ability of plants to absorb it, is dependent upon the acidity of the soil. If the soil is relatively acidic, the aluminum is more soluble, and plants can absorb it more easily. Under these conditions, hydrangea flowers are blue, as Al ions interact with anthocyanin pigments in the plant. In more basic soils, aluminum is less soluble, and under these conditions the hydrangea flowers are red. Gardeners who change the pH of their soils to change the color of their hydrangea flowers are therefore employing Le Chatelier's principle: the amount of acid in the soil changes the equilibrium of aluminum solubility, which in turn affects the color of the flowers. Key Takeaways • Le Chatelier's principle addresses how an equilibrium shifts when the conditions of an equilibrium are changed. • The direction of shift can be predicted for changes in concentrations, temperature, or pressure. • Catalysts do not affect the position of an equilibrium; they help reactions achieve equilibrium faster. Exercise $1$ 1. Define Le Chatelier's principle. 2. What is meant by a stress? What are some of the ways an equilibrium can be stressed? 3. Given this equilibrium, predict the direction of shift for each stress. $H_{2}(g)+I_{2}(s)+53kJ\rightleftharpoons 2HI(g)\nonumber$ 1. decreased temperature 2. increased pressure 3. removal of HI 4. Given this equilibrium, predict the direction of shift for each stress. $H_{2}(g)+F_{2}(g)\rightleftharpoons 2HF(g)+546kJ\nonumber$ 1. increased temperature 2. addition of H2 3. decreased pressure 5. Given this equilibrium, predict the direction of shift for each stress. $2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)+196kJ\nonumber$ 1. removal of SO3 2. addition of O2 3. decreased temperature 6. Given this equilibrium, predict the direction of shift for each stress. $CO_{2}(g)+C(s)+171kJ\rightleftharpoons 2CO(g)\nonumber$ 1. addition of CO 2. increased pressure 3. addition of a catalyst 7. The synthesis of NH3 uses this chemical reaction. $N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)+92kJ\nonumber$ Identify three stresses that can be imposed on the equilibrium to maximize the amount of NH3. 8. The synthesis of CaCO3 uses this chemical reaction. $CaO(s)+CO_{2}(g)\rightleftharpoons CaCO_{3}(s)+180kJ\nonumber$ Identify three stresses that can be imposed on the equilibrium to maximize the amount of CaCO3. Answers 1. When an equilibrium is stressed, the equilibrium shifts to minimize that stress. 1. toward reactants • toward reactants • toward products 1. toward products 2. toward products 3. toward products • increased pressure, decreased temperature, removal of NH3
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/07%3A_Chemical_Reactions_-_Energy_Rates_and_Equilibrium/7.09%3A_Le_Chateliers_Principle-_The_Effect_of_Changing_Condit.txt
• 8.1: States of Matter and Their Changes Another way that we can describe the properties of matter is the state (also called phase). The amount of energy in molecules of matter determines the state of matter. Matter can exist in one of several different states, including a gas, liquid, or solid state. • 8.2: Intermolecular Forces A phase is a form of matter that has the same physical properties throughout. Molecules interact with each other through various forces: ionic and covalent bonds, dipole-dipole interactions, hydrogen bonding, and dispersion forces. • 8.3: Gases and the Kinetic-Molecular Theory By integrating the knowledge of gaseous behavior from the gas laws and kinetic theory, we gain deeper insights into gases behavior. • 8.4: Pressure Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. • 8.5: Boyle’s Law - The Relation between Volume and Pressure The behavior of gases can be modeled with gas laws. Boyle's law relates a gas's pressure and volume at constant temperature and amount. • 8.6: Charles’s Law- The Relation between Volume and Temperature What happens to the volume of a gas as its heated? From experience, you probably know the answer. This article will explore the connection between volume and temperature. • 8.7: Gay-Lussac's Law- The Relationship Between Pressure and Temperature According to Gay-Lussac’s law, for a given amount of gas held at constant volume, the pressure is proportional to the absolute temperature. • 8.8: The Combined Gas Law The combined gas law relates pressure, volume, and temperature of a fixed amount of gas. • 8.9: Avogadro’s Law - The Relation between Volume and Molar Amount Avogadro showed that the volume of a gas is directly proportional to the number of moles of gas (Avogadro’s law). • 8.10: The Ideal Gas Law The ideal gas law relates the four independent physical properties of a gas at any time. The ideal gas law can be used in stoichiometry problems with chemical reactions that involve gases. • 8.11: Partial Pressure and Dalton's Law The pressure of a gas in a gas mixture is termed the partial pressure. Dalton's law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures. • 8.12: Liquids The intermolecular interactions between molecules in a liquid can be used to describe properties such as boiling point, vapor pressure, and surface tension. • 8.13: Solids Solids can be divided into amorphous solids and crystalline solids. Crystalline solids can be ionic, molecular, covalent network, or metallic. • 8.14: Changes of State Calculations There is an energy change associated with any phase change. There is an energy change associated with any phase change. 08: Gases Liquids and Solids Learning Objectives • Review the states of matter and their properties • Describe how change in temperature will affect the state of matter. Previously, you were introduced to the three states, also called phases, of matter; solid, liquid, and gas. A phase is a certain form of matter that includes a specific set of physical properties. That is, the atoms, the molecules, or the ions that make up the phase do so in a consistent manner throughout the phase. Science recognizes three stable phases: the solid phase, in which individual particles can be thought of as in contact and held in place; the liquid phase, in which individual particles are in contact but moving with respect to each other; and the gas phase, in which individual particles are separated from each other by relatively large distances (see Figure \(1\)). The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the attractive forces between molecules, called intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces draw the particles together. A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces. We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO2, as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure \(2\). Energy Changes That Accompany Phase Changes Phase changes are always accompanied by a change in the enthalpy, \(\Delta H\), of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; the \(\Delta H\) is positive (endothermic). Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; the \(\Delta H\) is negative (exothermic). The energy change associated with each common phase change is shown in Figure \(2\). \(\Delta H\) is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state. Previously, we defined the enthalpy changes associated with various chemical and physical processes. The molar enthalpy of fusion (\(ΔH_{fus}\)), is the energy required to convert a solid to a liquid, a process known as fusion (or melting). As noted above, the process of melting requires energy and therefore, the (\(ΔH_{fus}\)) is positive. The reverse process of freezing would release energy making the (\(ΔH_{fus}\)) negative. The molar enthalpy of vaporization (\(ΔH_{vap}\)), is the energy required to convert a liquid to a gas, known as vaporization. Melting points, enthalpies of fusion, boiling points, and enthalpies of vaporization for selected compounds are listed in Table \(1\). Table \(1\): Melting and Boiling Points and Enthalpies of Fusion and Vaporization for Selected Substances. Values given under 1 atm. of external pressure. Substance Melting Point (°C) ΔHfus (kJ/mol) Boiling Point (°C) ΔHvap (kJ/mol) N2 −210.0 0.71 −195.8 5.6 HCl −114.2 2.00 −85.1 16.2 Br2 −7.2 10.6 58.8 30.0 CCl4 −22.6 2.56 76.8 29.8 CH3CH2OH (ethanol) −114.1 4.93 78.3 38.6 CH3(CH2)4CH3 (n-hexane) −95.4 13.1 68.7 28.9 H2O 0 6.01 100 40.7 Na 97.8 2.6 883 97.4 NaF 996 33.4 1704 176.1 The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.01%3A_States_of_Matter_and_Their_Changes.txt
Learning Objective • Identify the different types of intermolecular forces. • Relate the physical properties of a substance to the strength of attractive forces. Why does a substance exist as a solid, liquid, or a gas at specific temperatures? Why do some substances evaporate quickly or melt more easily? These questions can be answered by considering the balance between the energy of the particles and intermolecular forces (or intermolecular interactions) between the particles. If the forces between particles are strong enough, the substance is a liquid or, if stronger, a solid. If the forces between particles are weak and sufficient energy is present, the particles separate from each other, so the gas phase is the preferred phase. The energy of the particles is mostly determined by temperature, so temperature is the main variable that determines what phase is stable at any given point. There are three types of intermolecular forces: London Dispersion, dipole-dipole, and hydrogen bonding, collectively termed van der Waals forces, that will be introduced below. London Dispersion Forces There are forces between all molecules that are caused by electrons being in different places in a molecule at any one time, which sets up a temporary separation of charge (a temporary dipole moment) that disappears almost as soon as it appears. These very weak intermolecular interactions are called dispersion forces (or London dispersion forces, named for the physicist Fritz London, who first described this force in the early 1900s). Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close. Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table $1$. Table $1$: Melting and Boiling Points of the Halogens Halogen Molar Mass Atomic Radius Melting Point Boiling Point fluorine, F2 38 g/mol 72 pm 53 K 85 K chlorine, Cl2 71 g/mol 99 pm 172 K 238 K bromine, Br2 160 g/mol 114 pm 266 K 332 K iodine, I2 254 g/mol 133 pm 387 K 457 K astatine, At2 420 g/mol 150 pm 575 K 610 K The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces. Example $1$ Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning. Solution Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/mol, and 123 g/mol, respectively. Therefore, CH4 is expected to have the lowest boiling point and SnH4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be CH4 < SiH4 < GeH4 < SnH4 A graph of the actual boiling points of these compounds versus the period of the group 14 elements shows this prediction to be correct: Exercise $1$ Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10. Answer All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore C2H6 < C3H8 < C4H10. Dipole-Dipole Intermolecular Forces Recall that a polar molecule will have a net unequal distribution of electrons in its covalent bonds resulting in a partial positive charge on one side of the molecule and a partial negative charge on the other side of the molecule—a separation of charge called a dipole. The electrostatic attraction between oppositely charged ends of polar molecules are called dipole-dipole interactions, (as illustrated in Figure $1$). Dipole-dipole attractions are between permanent dipoles and are therefore generally stronger than dispersion forces, which are between temporary dipoles. Thus, a polar molecule such as CH2Cl2 has a significantly higher boiling point (313 K, or 40°C) than a nonpolar molecule like CF4 (145 K, or −128°C), even though it has a lower molar mass (85 g/mol vs. 88 g/mol). Example $2$ Predict which will have the higher boiling point: N2 or CO. Explain your reasoning. Solution CO and N2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N2 molecules, so CO is expected to have the higher boiling point. Exercise $2$ Predict which will have the higher boiling point: $\ce{ICl}$ or $\ce{Br2}$. Explain your reasoning. Answer ICl. ICl and Br2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point. Hydrogen Bonds An unusually strong form of dipole-dipole interaction is called hydrogen bonding. Hydrogen bonding is found in molecules with an H atom bonded to an N atom, an O atom, or an F atom. Such covalent bonds are very polar, and the dipole-dipole interaction between these bonds in two or more molecules is strong enough to create a new category of intermolecular force. Hydrogen bonding is the reason water has unusual properties. For such a small molecule (its molar mass is only 18 g/mol), H2O has relatively high melting and boiling points. Its boiling point is 373 K (100°C), while the boiling point of a similar molecule, H2S, is 233 K (−60°C). This is because H2O molecules experience hydrogen bonding, while H2S molecules do not. This strong attraction between H2O molecules requires additional energy to separate the molecules in the condensed phase, so its boiling point is higher than would be expected. Hydrogen bonding is also responsible for water's ability as a solvent, its high heat capacity, and its ability to expand when freezing; the molecules line up in such a way that there is extra space between the molecules, increasing its volume in the solid state (Figure $2$). Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, but are generally much stronger than other dipole-dipole attractions and dispersion forces. Hydrogen Bonding and DNA Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure $3$. Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure $4$ The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication. Applications: Geckos and Intermolecular Forces Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way. Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure 8.1.12, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight. In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications. Boiling Points and Bonding Types In order for a substance to enter the gas phase, its particles must completely overcome the intermolecular forces holding them together. Therefore, a comparison of boiling points is essentially equivalent to comparing the strengths of the attractive intermolecular forces exhibited by the individual molecules. For small molecular compounds, London dispersion forces are the weakest intermolecular forces. Dipole-dipole forces are somewhat stronger, and hydrogen bonding is a particularly strong form of dipole-dipole interaction. However, when the mass of a nonpolar molecule is sufficiently large, its dispersion forces can be stronger than the dipole-dipole forces in a lighter polar molecule. Thus, nonpolar Cl2 has a higher boiling point than polar HCl. Table $3$: Intermolecular Forces and Boiling Points Substance Strongest Intermolecular Force Boiling Point $\left( ^\text{o} \text{C} \right)$ $\ce{H_2}$ dispersion -253 $\ce{Ne}$ dispersion -246 $\ce{O_2}$ dispersion -183 $\ce{Cl_2}$ dispersion -34 $\ce{HCl}$ dipole-dipole -85 $\ce{HBr}$ dipole-dipole -66 $\ce{H_2S}$ dipole-dipole -61 $\ce{NH_3}$ hydrogen bonding -33 $\ce{HF}$ hydrogen bonding 20 $\ce{H_2O}$ hydrogen bonding 100 Example $3$ Consider the compounds dimethylether (CH3OCH3), ethanol (CH3CH2OH), and propane (CH3CH2CH3). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning. Solution The shapes of CH3OCH3, CH3CH2OH, and CH3CH2CH3 are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH3CH2CH3 is nonpolar, it may exhibit only dispersion forces. Because CH3OCH3 is polar, it will also experience dipole-dipole attractions. Finally, CH3CH2OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH3CH2CH3 < CH3OCH3 < CH3CH2OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C. Exercise $3$ Ethane (CH3CH3) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH3NH2). Explain your reasoning. Answer The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH3CH3 and CH3NH2 are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of −6 °C. Example $4$: Intermolecular Forces What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? 1. potassium chloride (KCl) 2. ethanol (C2H5OH) 3. bromine (Br2) Solution 1. Potassium chloride is composed of ions, so the intermolecular interaction in potassium chloride is ionic forces. Because ionic interactions are strong, it might be expected that potassium chloride is a solid at room temperature. 2. Ethanol has a hydrogen atom attached to an oxygen atom, so it would experience hydrogen bonding. If the hydrogen bonding is strong enough, ethanol might be a solid at room temperature, but it is difficult to know for certain. (Ethanol is actually a liquid at room temperature.) 3. Elemental bromine has two bromine atoms covalently bonded to each other. Because the atoms on either side of the covalent bond are the same, the electrons in the covalent bond are shared equally, and the bond is a nonpolar covalent bond. Thus, diatomic bromine does not have any intermolecular forces other than dispersion forces. It is unlikely to be a solid at room temperature unless the dispersion forces are strong enough. Bromine is a liquid at room temperature. Exercise $4$ What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? 1. methylamine (CH3NH2) 2. calcium sulfate (CaSO4) 3. carbon monoxide (CO) Answer a. dipole-dipole, hydrogen bonding b. ionic forces (solid at room temperature) c. dipole-dipole Concept Review Exercise 1. What types of intermolecular interactions can exist in compounds? 2. What is the difference between covalent network and covalent molecular compounds? Answer 1. ionic bonding, network covalent, dispersion forces, dipole-dipole interactions, and hydrogen bonding. 2. Covalent network compounds contain atoms that are covalently bonded to other individual atoms in a giant 3-dimensional network. Covalent molecular compounds contain individual molecules that are attracted to one another through dispersion, dipole-dipole or hydrogen bonding. Key Takeaways • A phase is a form of matter that has the same physical properties throughout. • Molecules interact with each other through various forces: dipole-dipole interactions, hydrogen bonding, and dispersion forces. • Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one dipolar molecule for the partial positive end of another. • Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N. • The temporary dipole that results from the motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Exercises 1. List the three common phases in the order you are likely to find them—from lowest temperature to highest temperature. 2. List the three common phases in the order they exist from lowest energy to highest energy. 3. List these intermolecular interactions from weakest to strongest: London forces, hydrogen bonding, and ionic interactions. 4. List these intermolecular interactions from weakest to strongest: covalent network bonding, dipole-dipole interactions, and dispersion forces. 5. What type of intermolecular interaction is predominate in each substance? 1. water (H2O) 2. sodium sulfate (Na2SO4) 3. decane (C10H22) 6. What type of intermolecular interaction is predominate in each substance? 1. diamond (C, crystal) 2. helium (He) 3. ammonia (NH3) 7. Explain how a molecule like carbon dioxide (CO2) can have polar covalent bonds but be nonpolar overall. 8. Sulfur dioxide (SO2) has a formula similar to that of carbon dioxide (see Exercise 7) but is a polar molecule overall. What can you conclude about the shape of the SO2 molecule? 9. What are some of the physical properties of substances that experience covalent network bonding? 10. What are some of the physical properties of substances that experience only dispersion forces? Answers 1. solid, liquid, and gas 2. solid, liquid, and gas 1. London forces, hydrogen bonding, and ionic interactions 4. dispersion, dipole-dipole, network covalent 1. hydrogen bonding 2. ionic interactions 3. dispersion forces 6. a. network covalent b. dispersion c. hydrogen bonding 1. The two covalent bonds are oriented in such a way that their dipoles cancel out. 8. SO2 is not a linear molecule. It has a bent or V-shape. 9. very hard, high melting point 10. very soft, very low melting point
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.02%3A_Intermolecular_Forces.txt
Learning Objectives • State the major concepts behind the kinetic theory of gases. • Relate the general properties of gases to the kinetic theory. Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all-encompassing theory. Today, that theory is the kinetic theory of gases. The kinetic theory of gases is based on the following statements: 1. Gases consist of tiny particles of matter that are in constant motion. 2. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic—that is, there is no net loss of energy from the collisions. 3. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them. 4. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas. 5. The average speed of gas particles is dependent on the temperature of the gas. Figure \(1\) shows a representation of how we mentally picture the gas phase. This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other. An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements, and the kinetic theory of gases is well accepted by the scientific community. • The physical behavior of gases is explained by the kinetic theory of gases. • An ideal gas adheres exactly to the kinetic theory of gases. 8.04: Pressure Learning Objectives • To describe and measure the pressure of a gas. At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: temperature (expressed in kelvins), volume (expressed in liters), amount (expressed in moles), and pressure (in atmospheres). As we explain in this section, these variables are not independent. If we know the values of any three of these quantities, we can calculate the fourth and thereby obtain a full physical description of the gas. Temperature, volume, and amount have been discussed in previous chapters. We now discuss pressure and its units of measurement. Units of Pressure Any object, whether it is your computer, a person, or a sample of gas, exerts a force on any surface with which it comes in contact. The air in a balloon, for example, exerts a force against the interior surface of the balloon, and a liquid injected into a mold exerts a force against the interior surface of the mold, just as a chair exerts a force against the floor because of its mass and the effects of gravity. If the air in a balloon is heated, the increased kinetic energy of the gas eventually causes the balloon to burst because of the increased pressure (an increase in the collisions between gas molecules) inside the balloon. Pressure (P) is defined as the amount of force (F) per unit area (A): $P=\dfrac{F}{A}$ Pressure is dependent on both the force exerted and the size of the area to which the force is applied. We know from experience that applying the same force to a smaller area produces a higher pressure. When we use a hose to wash a car, for example, we can increase the pressure of the water by reducing the size of the opening of the hose with a thumb. The units of pressure are derived from the units used to measure force and area. The SI unit for pressure, derived from the SI units for force (newtons) and area (square meters), is the newton per square meter ($N/m^2$), which is called the Pascal (Pa), after the French mathematician Blaise Pascal (1623–1662): $\rm 1 \;Pa=1\;N/m^2$ Example $1$ Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is 1. lying flat? 2. standing on edge in a bookcase? Given: mass and dimensions of object Asked for: pressure Strategy: 1. Calculate the force exerted by the book and then compute the area that is in contact with a surface. 2. Substitute these two values into Equation $\ref{10.2.1}$ to find the pressure exerted on the surface in each orientation. Solution: The force exerted by the book does not depend on its orientation. Recall that the force exerted by an object is F = ma, where m is its mass and a is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s2 at Earth’s surface). In SI units, the force exerted by the book is therefore $F = ma = 2.00 \;\rm kg\times 9.8067 \dfrac{\rm m}{\rm s^2} = 19.6 \dfrac{\rm kg·m}{\rm s^2} = 19.6\;\rm N \nonumber$ A We calculated the force as 19.6 N. When the book is lying flat, the area is $A=\rm0.270 \;m\times0.210 \;m= 0.0567 \;m^2. \nonumber$ B The pressure exerted by the text lying flat is thus $P=\dfrac{F}{A}=\dfrac{19.6\;\rm N}{0.0567\;\rm m^2}=3.46\times10^2 \rm Pa \nonumber$ A If the book is standing on its end, the force remains the same, but the area decreases: $\rm A=\rm21.0 \;cm\times4.5 \;cm = 0.210 \;m\times0.045 \;m = 9.5 \times 10^{−3} \;\rm m^2 \nonumber$ B The pressure exerted by the text lying flat is thus $P=\dfrac{19.6\;\rm N}{9.5\times10^{-3}\;\rm m^2}=2.06\times10^3 \;\rm Pa \nonumber$ Exercise $1$ What pressure does a 60.0 kg student exert on the floor 1. when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm2)? 2. as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm2)? Answer a 3.27 × 104 Pa Answer b 5.9 × 106 Pa Atmospheric Pressure Just as we exert pressure on a surface because of gravity, so does our atmosphere. We live at the bottom of an ocean of gases that becomes progressively less dense with increasing altitude. Approximately 99% of the mass of the atmosphere lies within 30 km of Earth’s surface, and half of it is within the first 5.5 km (Figure $1$). Every point on Earth’s surface experiences a net pressure called atmospheric pressure. The pressure exerted by the atmosphere is considerable: a 1.0 m2 column, measured from sea level to the top of the atmosphere, has a mass of about 10,000 kg, which gives a pressure of about 100 kPa. Atmospheric pressure can be measured using a barometer, a device invented in 1643 by one of Galileo’s students, Evangelista Torricelli (1608–1647). A barometer may be constructed from a long glass tube that is closed at one end. It is filled with mercury and placed upside down in a dish of mercury without allowing any air to enter the tube. Some of the mercury will run out of the tube, but a relatively tall column remains inside (Figure $2$). Why doesn’t all the mercury run out? Gravity is certainly exerting a downward force on the mercury in the tube, but it is opposed by the pressure of the atmosphere pushing down on the surface of the mercury in the dish, which has the net effect of pushing the mercury up into the tube. Because there is no air above the mercury inside the tube in a properly filled barometer (it contains a vacuum), there is no pressure pushing down on the column. Thus the mercury runs out of the tube until the pressure exerted by the mercury column itself exactly balances the pressure of the atmosphere. Under normal weather conditions at sea level, the two forces are balanced when the top of the mercury column is approximately 760 mm above the level of the mercury in the dish, as shown in Figure $2$. This value varies with meteorological conditions and altitude. In Denver, Colorado, for example, at an elevation of about 1 mile, or 1609 m (5280 ft), the height of the mercury column is 630 mm rather than 760 mm. Mercury barometers have been used to measure barometric pressure for so long that they have their own unit for pressure: the millimeter of mercury (mmHg), often called the torr, after Torricelli. Standard barometric pressure is the barometric pressure required to support a column of mercury exactly 760 mm tall; this pressure is also referred to as 1 atmosphere (atm). These units are also related to the pascal: \begin{align} \rm 1\; atm &= 760 \; mmHg \[4pt] &= 760 \; torr \[4pt] &= 1.01325 \times 10^5 \; Pa \[4pt] &= 101.325 \; kPa\label{10.2.3} \end{align} Thus a pressure of 1 atm equals 760 mmHg exactly. We are so accustomed to living under this pressure that we never notice it. Instead, what we notice are changes in the pressure, such as when our ears pop in fast elevators in skyscrapers or in airplanes during rapid changes in altitude. We make use of barometric pressure in many ways. We can use a drinking straw because sucking on it removes air and thereby reduces the pressure inside the straw. The barometric pressure pushing down on the liquid in the glass then forces the liquid up the straw. Example $2$: Barometric Pressure One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the barometric pressure is only 454 mmHg. Convert this pressure to 1. atmospheres (atm). 2. bar. Given: pressure in millimeters of mercury Asked for: pressure in atmospheres and bar Strategy: Use the conversion factors in Equation $\ref{10.2.3}$ to convert from millimeters of mercury to atmospheres and kilopascals. Solution: From Equation $\ref{10.2.3}$, we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus \begin{align} P &=\rm 454 \;mmHg\times\dfrac{1\;atm}{760\;mmHg} \[4pt] &= 0.597\;atm \nonumber \end{align} \nonumber The pressure in bar is given by \begin{align} P&=\rm 0.597\;atm\times\dfrac{1.01325\;bar}{1\;atm}\[4pt] &= 0.605\;bar \nonumber \end{align} \nonumber Exercise $2$: Barometric Pressure Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal barometric pressure at this altitude is about 0.308 atm. Convert this pressure to 1. millimeters of mercury. 2. bar. Answer a 234 mmHg; Answer b 0.312 bar Barometers measure atmospheric pressure, but manometers measure the pressures of samples of gases contained in an apparatus. The key feature of a manometer is a U-shaped tube containing mercury (or occasionally another nonvolatile liquid). A closed-end manometer is shown schematically in part (a) in Figure $3$. When the bulb contains no gas (i.e., when its interior is a near vacuum), the heights of the two columns of mercury are the same because the space above the mercury on the left is a near vacuum (it contains only traces of mercury vapor). If a gas is released into the bulb on the right, it will exert a pressure on the mercury in the right column, and the two columns of mercury will no longer be the same height. The difference between the heights of the two columns is equal to the pressure of the gas. If the tube is open to the atmosphere instead of closed, as in the open-end manometer shown in part (b) in Figure $3$, then the two columns of mercury have the same height only if the gas in the bulb has a pressure equal to the barometric pressure. If the gas in the bulb has a higher pressure, the mercury in the open tube will be forced up by the gas pushing down on the mercury in the other arm of the U-shaped tube. The pressure of the gas in the bulb is therefore the sum of the barometric pressure (measured with a barometer) and the difference in the heights of the two columns. If the gas in the bulb has a pressure less than that of the atmosphere, then the height of the mercury will be greater in the arm attached to the bulb. In this case, the pressure of the gas in the bulb is the barometric pressure minus the difference in the heights of the two columns. Summary Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m2). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area on which the force is exerted. The pressure exerted by Earth’s atmosphere, called barometric pressure, is about 101 kPa or 14.7 lb/in.2 at sea level. barometric pressure can be measured with a barometer, a closed, inverted tube filled with mercury. The height of the mercury column is proportional to barometric pressure, which is often reported in units of millimeters of mercury (mmHg), also called torr. Standard barometric pressure, the pressure required to support a column of mercury 760 mm tall, is yet another unit of pressure: 1 atmosphere (atm). A manometer is an apparatus used to measure the pressure of a sample of a gas.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.03%3A_Gases_and_the_Kinetic-Molecular_Theory.txt
Learning Objectives • Define the relationship between gas volume and pressure, Boyle's Law. • Use Boyle's Law to calculate changes in pressure or volume of a gas. When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas was kept constant, pressure and volume were related: As one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related. There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant (k) for a given amount of gas at a constant temperature: $P \times V=k\; (at\; constant\; n\; and\; T)\nonumber$ If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled P1 and V1 and the new conditions are labeled P2 and V2, we have $P_{1}V_{1}= k = P_{2}V_{2}\; \nonumber$ where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply $P_{1}V_{1}= P_{2}V_{2}\; (at\; constant\; n\; and\; T)\nonumber$ This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's law, after the English scientist Robert Boyle, who first announced it in 1662. Figure $1$ shows two representations of how Boyle's law works. Boyle's law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle's law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won't matter what the unit is, but the unit must be the same on both sides of the equation. Example $1$ A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant? Solution First, determine what quantities we are given. We are given an initial pressure and an initial volume, so let these values be P1 and V1: P1 = 2.44 atm and V1 = 4.01 L We are given another quantity, final pressure of 1.93 atm, but not a final volume. This final volume is the variable we will solve for. P2 = 1.93 atm and V2 = ? L Substituting these values into Boyle's law, we get (2.44 atm)(4.01 L) = (1.93 atm)V2 To solve for the unknown variable, we isolate it by dividing both sides of the equation by 1.93 atm—both the number and the unit: $\frac{(2.44\, atm)(4.01\, L)}{1.93\, atm}=\frac{(1.93\, atm)\, V_{2}}{1.93\, atm}\nonumber$ Note that, on the left side of the equation, the unit atm is in the numerator and the denominator of the fraction. They cancel algebraically, just as a number would. On the right side, the unit atm and the number 1.93 are in the numerator and the denominator, so the entire quantity cancels: $\frac{(2.44\, \cancel{atm})(4.01\, L)}{1.93\, \cancel{atm}}=\frac{(1.93\, \cancel{atm})\, V_{2}}{1.93\, \cancel{atm}}\nonumber$ What we have left is $\frac{(2.44)(4.01\, L)}{1.93}=V_{2}\nonumber$ Now we simply multiply and divide the numbers together and combine the answer with the $L$ unit, which is a unit of volume. Doing so, we get $V_2 = 5.07\, L$ Does this answer make sense? We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle's law. Exercise $1$ If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2? Answer 124 mL As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units. Example $2$ A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure? Solution We can still use Boyle's law to answer this, but now the two volume quantities have different units. It does not matter which unit we change, as long as we perform the conversion correctly. Let us change the 0.663 L to milliliters: $0.663\, L\times \frac{1000\, ml}{1\, L}=663\, ml\nonumber$ Now that both volume quantities have the same units, we can substitute into Boyle's law: $(722\, torr)(88.8\, ml)=P_{2}(663\, ml)\nonumber$ $\frac{(722\, torr)(88.8)\, ml}{(663\, ml)}=P_{2}\nonumber$ The mL units cancel, and we multiply and divide the numbers to get P2 = 96.7 torr The volume is increasing, and the pressure is decreasing, which is as expected for Boyle's law. Exercise $2$ If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2? Answer 119 mL To Your Health: Breathing and Boyle’s Law What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure $2$). Summary • The behavior of gases can be modeled with gas laws. • Boyle's law relates a gas's pressure and volume at constant temperature and amount.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.05%3A_Boyles_Law_-_The_Relation_between_Volume_and_Pressure.txt
Learning Objectives • Define the relationship between gas temperature and volume, Charles's Law. • Use Charles's Law to calculate changes in temperature or volume of a gas. In addition to pressure and volume, another measurable characteristics of a gas is temperature (T). Perhaps one can vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume (V). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related. Kelvin Temperature Scale A mathematical relationship between V and T should be possible except for one thought: what temperature scale should we use? We know from previous chapters that scientists uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin, not its temperature in degrees Celsius. If the temperature of a gas is expressed in Kelvins, then experiments show that the ratio of volume to temperature is a constant (k): $\frac{V}{T}=k\nonumber$ Charles's Law We can modify this equation as we modified Boyle's law: the initial conditions V1 and T1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V2 and T2, as long as pressure and the amount of the gas remain constant. Thus, we have another gas law: $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\; (at\; constant\; P\; and\; n)\nonumber$ This gas law is commonly referred to as Charles's law, after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle's law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale: K = °C + 273 where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius. Example $1$ A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas. Solution First, we assign the given values to their variables. The initial volume is V1, so V1 = 34.8 mL, and the initial temperature is T1, so T1 = 315 K. The temperature is increased to 559 K, so the final temperature T2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles's law yields $\frac{34.8\, ml}{315\, K}=\frac{V_{2}}{559\, K}\nonumber$ We solve for V2 by algebraically isolating the V2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side: $\frac{(559\cancel{K})(34.8\, ml)}{315\, \cancel{K}}=\frac{V_{2}(\cancel{559\, K})}{\cancel{559\, K}}\nonumber$ The expression simplifies to $\frac{(559)(34.8\, ml)}{315}=V_{2}\nonumber$ By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is V2 = 61.8 mL Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does. Exercise $1$ If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K? Answer 1.82 L It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles's law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles's law. That is, rather than write it as $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\nonumber$ write the equation as $\frac{T_{1}}{V_{1}}=\frac{T_{2}}{V_{2}}\nonumber$ It is still an equality and a correct form of Charles's law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler. Example $2$ A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must the temperature of the gas be for its volume to be 25.0 L? Solution Here, we are looking for a final temperature, so we will use the reciprocal form of Charles's law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins: −67°C + 273 = 206 K In using the gas law, we must use T1 = 206 K as the temperature. Substituting into the reciprocal form of Charles's law, we get $\frac{206\, K}{34.8\, L}=\frac{T_{2}}{25.0\, L}\nonumber$ Bringing the 25.0 L quantity over to the other side of the equation, we get $\frac{(25.0\cancel{L})(206\, K)}{34.8\cancel{L}}=T_{2}\nonumber$ The L units cancel, so our final answer is T2 = 148 K This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example. Exercise $2$ If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2? Answer 235 K, or −38°C Summary • Charles's law relates a gas's volume and temperature at constant pressure and amount. • In gas laws, temperatures must always be expressed in kelvins.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.06%3A_Charless_Law-_The_Relation_between_Volume_and_Temperature.txt
Learning Objectives • Define the relationship between gas pressure and temperature, Gay-Lussac's Law. A third gas law may be derived as a corollary to Boyle's and Charles's laws. Suppose we double the thermodynamic temperature of a sample of gas. According to Charles’s law, the volume should double. Now, how much pressure would be required at the higher temperature to return the gas to its original volume? According to Boyle’s law, we would have to double the pressure to halve the volume. Thus, if the volume of gas is to remain the same, doubling the temperature will require doubling the pressure. This law was first stated by the Frenchman Joseph Gay-Lussac (1778 to 1850). According to Gay-Lussac’s law, for a given amount of gas held at constant volume, the pressure is proportional to the absolute temperature. Mathematically, $P\propto T\ce{\:or\:} P=k×T\ce{\:or\:} \dfrac{P}{T}=k \nonumber$ where $\propto$ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas. In terms of two sets of data: $\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$. This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.) Gay-Lussac’s law tells us that it may be dangerous to heat a gas in a closed container. The increased pressure might cause the container to explode, as you can see in the video below. The video shows very, very cold nitrogen gas in a bottle being warmed by the air. Since the bottle's volume is relatively constant, as the temperature of the nitrogen gas (formed when the liquid nitrogen boils) increases, so does the pressure inside the bottle until, finally, BOOM! Example $1$: Temperature A container is designed to hold a pressure of 2.5 atm. The volume of the container is 20.0 cm3, and it is filled with air at room temperature (20°C) and normal atmospheric pressure. Would it be safe to throw the container into a fire where temperatures of 600°C would be reached? Solution Using the common-sense method, we realize that the pressure will increase at the higher temperature, and so: $P_{\text{2}}=\text{1}\text{.0 atm }\times \frac{\text{(273}\text{.15 + 600) K}}{\text{(273}\text{.15 + 20) K}}=\text{3}\text{.0 atm} \nonumber$ This would exceed the safe strength of the container. Note that the volume of the container was not needed to solve the problem. This concept works in reverse, as well. For instance, if we subject a gas to lower temperatures than their initial state, the external atmosphere can actually force the container to shrink. The following video demonstrates how a sample of hot gas, when cooled will collapse a container. A syringe barrel is filled with hot steam (vaporized water) and a plunger placed to cap off the end. The syringe is then placed in a beaker of ice water to cool the internal gas. When the temperature of the water vapor decreases, the pressure exerted by the vapor decreases as well. This leads to a difference in pressure between the vapor inside the barrel and the atmosphere. Atmospheric pressure then pushes the plunger into the barrel. Summary • Gay-Lussac's law relates a gas's temperature and pressure at constant volume and amount. • In gas laws, temperatures must always be expressed in kelvins. 8.08: The Combined Gas Law Learning Objectives • Use the combined gas law to determine the relationships between pressure, volume, and temperature of a gas. One thing we notice about all gas laws, collectively, is that volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the combined gas law, and its mathematical form is: $\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\; at\; constant\; n\nonumber$ This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in kelvins. Notice that each of the previous gas laws introduced, can be derived from the combined gas law: $\text{At constant T,}\; \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\; gives\; \text{Boyle's Law:}\; P_{1}V_{1}=P_{2}V_{2}\;$ $\text{At constant P,}\; \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\; gives\; \text{Charles's Law:}\; \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\;$ $\text{At constant V,}\; \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\; gives\; \text{Gay-Lussac's Law:}\; \frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\;$ In other words, if you know the equation for the combined gas law, you can calculate relationships between pressure, volume, or temperature of a fixed amount of gas. Example $2$ A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas? Solution We can use the combined gas law directly; all the units are consistent with each other, and the temperatures are given in Kelvin. Substituting, $\frac{(1.82\, atm)(8.33\, L)}{286\, K}=\frac{P_{2}(5.72\, L)}{355\, K}\nonumber$ We rearrange this to isolate the P2 variable all by itself. When we do so, certain units cancel: $\frac{(1.82\, atm)(8.33\, \cancel{L})(355\, \cancel{K})}{(286\, \cancel{K})(5.72\, \cancel{L})}=P_{2}\nonumber$ Multiplying and dividing all the numbers, we get $P_2 = 3.29\, atm \nonumber$ Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing. Exercise $2$ If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2? Answer 105 mL Summary • There are gas laws that relate any two physical properties of a gas. • The combined gas law relates pressure, volume, and temperature of a gas.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.07%3A_Gay-Lussac%27s_Law-_The_Relationship_Between_Pressure_and_Temperature.txt
Learning Objectives • Describe the relationship between the amount and volume of a gas, Avogadro's Law. • Define the conditions of standard temperature and pressure. The Relationship between Amount and Volume: Avogadro's Law We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $1$). This is the historic “Avogadro’s hypothesis.” Because the number of particles is related to the number of moles (1 mol = 6.022 × 1023 particles), Avogadro's law essentially states that equal volumes of different gases, at the same temperature and pressure, contain the same amount (moles, particles) of gas. At constant temperature and pressure, the volume (V) of a sample of gas is directly proportional to the number of moles (n) of gas in the sample. Stated mathematically, $V\propto n\ce{\:or\:} V=k×n\ce{\:or\:} \dfrac{V}{n}=k \nonumber$ where $\propto$ means “is proportional to,” and k is a proportionality constant that is the same for all gases. In terms of two sets of data: $\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}$. This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures. Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n versus T. Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws). Standard Temperature and Pressure It should be obvious by now that some physical properties of gases depend strongly on the conditions. What we need is a set of standard conditions so that properties of gases can be properly compared to each other. Standard Temperature and Pressure (STP) is defined as exactly 100 kPa of pressure (0.986 atm) and 273 K (0°C). For simplicity, we will use 1 atm as standard pressure. Defining STP allows us to more directly compare the properties of gases that differ from one another. One property shared among gases is a molar volume. The molar volume is the volume of 1 mol of a gas. At STP, the molar volume of a gas can be easily determined by using the ideal gas law: $(1\, atm)V=(1\, mol) \left(0.08205\dfrac{L.atm}{mol.K}\right)(273\, K)\nonumber$ All the units cancel except for L, the unit of volume. So V = 22.4 L Note that we have not specified the identity of the gas; we have specified only that the pressure is 1 atm and the temperature is 273 K. This makes for a very useful approximation: any gas at STP has a volume of 22.4 L per mole of gas; that is, the molar volume at STP is 22.4 L/mol (Figure $1$). This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at STP. If the conditions are not at STP, a molar volume of 22.4 L/mol is not applicable. However, if the conditions are at STP, the combined gas law can be used to calculate what the volume of the gas would be if at STP; then the 22.4 L/mol molar volume can be used. Example $4$ How many moles of $\ce{Ar}$ are present in 38.7 L at STP? Solution We can use the molar volume, 22.4 L/mol, as a conversion factor, but we need to reverse the fraction so that the L units cancel and mol units are introduced. It is a one-step conversion: $38.7\, \cancel{L}\times \frac{1\, mol}{22.4\cancel{L}}=1.73\, mol\nonumber$ Exercise $4$ What volume does 4.87 mol of $\ce{Kr}$ have at STP? Answer 109 L Example $5$ What volume of $\ce{H2}$ is produced at STP when 55.8 g of $\ce{Zn}$ metal react with excess $\ce{HCl}$? $\ce{Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)} \nonumber$ Solution This is a stoichiometry problem with a twist: we need to use the molar volume of a gas at STP to determine the final answer. The first part of the calculation is the same as in a previous example: $55.8\cancel{g\, Zn}\times \frac{1\cancel{mol\, Zn}}{65.41\cancel{g\, Zn}}\times \frac{1\, mol\, H_{2}}{1\cancel{mol\, Zn}}=0.853\, H_{2}\nonumber$ Now we can use the molar volume, 22.4 L/mol, because the gas is at STP: $0.853\cancel{mol\, H_{2}}\times \frac{22.4\, L}{1\cancel{mol\, H_{2}}}=19.1\, L\, H_{2}\nonumber$ Alternatively, we could have applied the molar volume as a third conversion factor in the original stoichiometry calculation. Exercise $5$ What volume of $\ce{HCl}$ is generated if 3.44 g of $\ce{Cl2}$ are reacted at STP? $\ce{H2(g) + Cl2(g) → 2HCl(g)} \nonumber$ Answer 2.17 L Summary • Avogadro's law states that the volume of gas is directly proportional to the number of moles of gas. • Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases. • At STP, gases have a volume of 22.4 L per mole.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.09%3A_Avogadros_Law_-_The_Relation_between_Volume_and_Molar_Amount.txt
Learning Objectives • Describe the ideal gas law. • Use the ideal gas law to calculate pressure, volume, temperature, or moles of an ideal gas. So far, the gas laws we have considered have all required that the gas change its conditions; then we can predict a resulting change in one of its properties. Are there any gas laws that relate the physical properties of a gas at any given time? Consider a further extension of the combined gas law to include n. By analogy to Avogadro's law, n is positioned in the denominator of the fraction, opposite the volume. So, $\frac{PV}{nT}=constant\nonumber$ Because pressure, volume, temperature, and amount are the only four independent physical properties of a gas, the constant in the above equation is truly a constant. Indeed, because we do not need to specify the identity of a gas to apply the gas laws, this constant is the same for all gases. We define this constant with the symbol R, so the previous equation is written as $\dfrac{PV}{nT}=R\nonumber$ which is usually rearranged as $PV = nRT\nonumber$ This equation is called the ideal gas law. It relates the four independent properties of a gas at any time. The constant $R$ is called the ideal gas law constant. Its value depends on the units used to express pressure and volume. Table $1$: Values of the Ideal Gas Law Constant lists the numerical values of $R$. Numerical Value Units 0.08205 $\dfrac{L \cdot atm}{mol \cdot K}$ 62.36 $\dfrac{L \cdot torr}{mol \cdot K}=\dfrac{L \cdot mmHg}{mol \cdot K}$ 8.314 $\dfrac{J}{mol \cdot K}$ The ideal gas law is used like any other gas law, with attention paid to the unit and expression of the temperature in kelvin. However, the ideal gas law does not require a change in the conditions of a gas sample. The ideal gas law implies that if you know any three of the physical properties of a gas, you can calculate the fourth property. Example $1$ A 4.22 mol sample of $\ce{Ar}$ has a pressure of 1.21 atm and a temperature of 34°C. What is its volume? Solution The first step is to convert temperature to kelvins: $34 + 273 = 307\, \ce{K} \nonumber$ Now we can substitute the conditions into the ideal gas law: $(1.21atm)(V)=(4.22\, mol)\left(0.08205\dfrac{L.atm}{mol.K}\right)(307\, K)\nonumber$ The atm unit is in the numerator of both sides, so it cancels. On the right side of the equation, the mol and K units appear in the numerator and the denominator, so they cancel as well. The only unit remaining is L, which is the unit of volume that we are looking for. We isolate the volume variable by dividing both sides of the equation by 1.21: $V=\dfrac{(4.22)(0.08205)(307)}{1.21}L\nonumber$ Then solving for volume, we get V = 87.9 L Exercise $1$ A 0.0997 mol sample of $\ce{O2}$ has a pressure of 0.692 atm and a temperature of 333 K. What is its volume? Answer 3.94 L Example $2$ At a given temperature, 0.00332 g of Hg in the gas phase has a pressure of 0.00120 mmHg and a volume of 435 L. What is its temperature? Solution We are not given the number of moles of Hg directly, but we are given a mass. We can use the molar mass of Hg to convert to the number of moles. $0.00332\cancel{g\, Hg}\times \frac{1\, mol\, Hg}{200.59\cancel{g\, \, Hg}}=0.0000165\, mol=1.65\times 10^{-5}mol\nonumber$ Pressure is given in units of millimeters of mercury. We can either convert this to atmospheres or use the value of the ideal gas constant that includes the mmHg unit. We will take the second option. Substituting into the ideal gas law, $(0.00332\, mm\, Hg)(435\, L)=(1.65\times 10^{-5}mol)(62.36\frac{L.mmHg}{mol.K})T\nonumber$ The mmHg, L, and mol units cancel, leaving the K unit, the unit of temperature. Isolating T on one side, we get $T=\frac{(0.00332)(435)}{(1.65\times 10^{-5})(62.36)}K\nonumber$ Then solving for K, we get T = 1,404 K. Exercise $2$ For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume? Answer 8.21 L The ideal gas law can also be used in stoichiometry problems. Example $3$ What volume of $\ce{H2}$ is produced at 299 K and 1.07 atm when 55.8 g of $\ce{Zn}$ metal react with excess $\ce{HCl}$? $\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}\nonumber$ Solution Here we have a stoichiometry problem where we need to find the number of moles of H2 produced. Then we can use the ideal gas law, with the given temperature and pressure, to determine the volume of gas produced. First, the number of moles of H2 is calculated: $55.8\cancel{g\, Zn}\times \frac{1\cancel{mol\, Zn}}{65.41\cancel{g\, Zn}}\times \dfrac{1\, mol\, H_{2}}{1\cancel{mol\, Zn}}=0.853\, H_{2}\nonumber$ Now that we know the number of moles of gas, we can use the ideal gas law to determine the volume, given the other conditions: $(1.07atm)V=(0.853\, mol)\left(0.08205\dfrac{L.atm}{mol.K}\right)(299\, K)\nonumber$ All the units cancel except for L, for volume, which means V = 19.6 L Exercise $3$ What pressure of $\ce{HCl}$ is generated if 3.44 g of $\ce{Cl2}$ are reacted in 4.55 L at 455 K? $\ce{H2(g) + Cl2(g) → 2HCl(g)}\nonumber$ Answer 0.796 atm Summary • The ideal gas law relates the four independent physical properties of a gas at any time. • The ideal gas law can be used in stoichiometry problems with chemical reactions that involve gases.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.10%3A_The_Ideal_Gas_Law.txt
Learning Objective • Define Dalton's law of partial pressures. • Use Dalton's law to calculate partial pressure of a gas in a mixture. One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix. In gas mixtures, each component in the gas phase can be treated separately because, according to the kinetic theory of gases, there are little to no attractive or repulsive forces between particles. Each component of a gas mixture shares the same temperature and volume, (remember that gases expand to fill the volume of their container; gases in a mixture continue to do that as well). However, each individual gas will have its own pressure within a mixture. This is called the partial pressure of a gas, $P_i$. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture. Dalton's law of partial pressures states that the total pressure of a gas mixture, $P_{tot}$, is equal to the sum of the partial pressures of the components, $P_i$: which is expressed algebraically as $P_{total}=P_1+P_2+P_3 ... = \sum_i P_i \nonumber$ or, equivalently $P_{total} = \dfrac{RT}{V} \sum_i n_i \nonumber$ where $i$ counts over all gases in mixture. Although this law may seem trivial, it reinforces the idea that gases behave independently of each other. Example $1$ A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container? Solution Dalton's law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together: Ptot = 2.33 atm + 0.77 atm = 3.10 atm Exercise $1$ N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2? Answer 152 torr Example $2$ A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers? Solution Because gases act independently of each other, we can determine the resulting final pressures using Boyle's law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle's law to determine the final pressure of H2: (2.50 atm)(2.00 L) = P2(7.00 L) Solving for P2, we get P2 = 0.714 atm = partial pressure of H2. Now we do that same thing for the O2: (1.90 atm)(5.00 L) = P2(7.00 L)P2 = 1.36 atm = partial pressure of O2 The total pressure is the sum of the two resulting partial pressures: Ptot = 0.714 atm + 1.36 atm = 2.07 atm Exercise $2$ If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the containers are opened, what is the resulting total pressure? Answer 0.51 atm Dalton’s law states that in a gas mixture ($P_{total}$) each gas will exert a pressure independent of the other gases ($P_n$) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure ($P_{total}$) by the gas percentage (%). $P_{Total} = P_1 + P_2 + P_3 + P_4 + ... + P_n \nonumber$ or $P_n = \dfrac{\text{% of individual gas}_n}{P_{Total}} \nonumber$ Table $1$: Partial Pressures for the gases in air on a typical day Gas Partial Pressure (mm Hg) Percentage (%) Nitrogen, (N_2\) $P_{N_2}$ = 594 78 Oxygen, $O_2$ $P_{O_2}$= 160 21 Carbon Dioxide, $CO_2$ $P_{CO_2}$ = 0.25 0.033 Water Vapor, $H_2O$ $P_{H_2O}$ = 5.7 0.75 Other trace gases $P_{Other}$ = 0.05 0.22 Total air $P_{Total}$ = 760 1 Application of Dalton's Law: Collecting Gases over Water A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part. The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H2O. The partial pressure of H2O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas. Example $3$ Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of $O_2$ produced. Solution From Dalton's law, $P_{O_2} = P_{total} – P_{H_2O} = 754 – 19.8 = 734 \; torr = 0.966\; atm \nonumber$ Now use the Ideal Gas Law to convert to moles $n =\dfrac{PV}{RT} = \dfrac{(0.966\; atm)(0.155\;L)}{(0.082\; L atm mol^{-1} K^{-1})(295\; K)}= 0.00619 \; mol \nonumber$ Exercise $3$ $\ce{CO2}$, generated by the decomposition of $\ce{CaCO3}$, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of $\ce{CO2}$ were generated? Answer 0.129 mol Food and Drink Application: Carbonated Beverages Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have $\ce{CO2}$ gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher. All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of $\ce{CO2}$ gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly-sealed package (usually a bottle or a can) and sold. When the container is opened, the $\ce{CO2}$ pressure is released, resulting in the well-known hiss of an opening container, and $\ce{CO2}$ bubbles come out of solution. This must be done with care: if the $\ce{CO2}$ comes out too violently, a mess can occur! The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates $\ce{CO2}$ as a digestion product. This process is called fermentation. The overall reaction is $\ce{C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(aq)}\nonumber$ When this process occurs in a closed container, the $\ce{CO2}$ produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of $\ce{CO2}$ gas. Summary • The pressure of a gas in a gas mixture is termed the partial pressure. • Dalton's law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.11%3A_Partial_Pressure_and_Dalton%27s_Law.txt
Learning Objective • Define the vapor pressure of liquids. • Explain the origin of both surface tension and capillary action. There are some properties that all liquids, including water, have. All liquids have a certain portion of particles with enough energy to enter the gas phase, and if these particles are at the surface of the liquid, they do so (Figure \(1\)). The formation of a gas from a liquid at temperatures below the boiling point is called evaporation. At these temperatures, the material in the gas phase is called vapor, rather than gas; the term gas is reserved for when the gas phase is the stable phase. If the available volume is large enough, eventually all the liquid will become vapor. But if the available volume is not enough, eventually some of the vapor particles will reenter the liquid phase (Figure \(2\) Equilibrium). At some point, the number of particles entering the vapor phase will equal the number of particles leaving the vapor phase, so there is no net change in the amount of vapor in the system. We say that the system is at equilibrium. The partial pressure of the vapor at equilibrium is called the vapor pressure of the liquid. Understand that the liquid has not stopped evaporating. The reverse process—condensation—is occurring as fast as evaporation, so there is no net change in the amount of vapor in the system. The term dynamic equilibrium represents a situation in which a process still occurs, but the opposite process also occurs at the same rate, so that there is no net change in the system. The vapor pressure for a substance is dependent on the temperature of the substance; as the temperature increases, so does the vapor pressure. Figure \(3\) - Plots of Vapor Pressure versus Temperature for Several Liquids, is a plot of vapor pressure versus temperature for several liquids. Having defined vapor pressure, we can also redefine the boiling point of a liquid: the temperature at which the vapor pressure of a liquid equals the surrounding environmental pressure. The normal boiling point, then, is the temperature at which the vapor pressure is 760 torr, or exactly 1 atm. Thus boiling points vary with surrounding pressure, a fact that can have large implications on cooking foods at lower- or higher-than-normal elevations. Atmospheric pressure varies significantly with altitude. Example \(1\) Use Figure \(3\) to estimate the boiling point of water at 500 torr, which is the approximate atmospheric pressure at the top of Mount Everest. Solution See the accompanying figure. Five hundred torr is between 400 and 600, so we extend a line from that point on the y-axis across to the curve for water and then drop it down to the x-axis to read the associated temperature. It looks like the point on the water vapor pressure curve corresponds to a temperature of about 90°C, so we conclude that the boiling point of water at 500 torr is 90°C. By reading the graph properly, you can estimate the boiling point of a liquid at different temperatures. Exercise \(1\) Use Figure \(3\) to estimate the boiling point of ethanol at 400 torr. Answer about 65°C The vapor pressure curve for water is not exactly zero at the melting point 0°C. Even ice has a vapor pressure—that is why it sublimes over time. However, the vapor pressures of solids are typically much lower than that of liquids. At −1°C, the vapor pressure of ice is 4.2 torr. At a freezer temperature of 0°F (−17°C), the vapor pressure of ice is only 1.0 torr; so-called deep freezers can get down to −23°C, where the vapor pressure of ice is only 0.6 torr. All liquids share some other properties as well. Surface tension is an effect caused by an imbalance of forces on the atoms at the surface of a liquid, as shown in Figure \(4\). The blue particle in the bulk of the liquid experiences intermolecular forces from all around, as illustrated by the arrows. However, the yellow particle on the surface does not experience any forces above it because there are no particles above it. This leads to an imbalance of forces, called surface tension. Surface tension is responsible for several well-known behaviors of liquids, including water. Liquids with high surface tension tend to bead up when present in small amounts (Figure \(5\)). Surface tension causes liquids to form spheres in free fall or zero gravity. Surface tension is also responsible for the fact that small insects can "walk" on water. Because of surface tension, it takes energy to break the surface of a liquid, and if an object (such as an insect) is light enough, there is not enough force due to gravity for the object to break through the surface, so the object stays on top of the water (Figure \(6\)). Carefully done, this phenomenon can also be illustrated with a thin razor blade or a paper clip. The fact that small droplets of water bead up on surfaces does not mean that water—or any other liquid—does not interact with other substances. Sometimes the attraction can be very strong. Adhesion is the tendency of a substance to interact with other substances because of intermolecular forces, while cohesion is the tendency of a substance to interact with itself. If cohesive forces within a liquid are stronger than adhesive forces between a liquid and another substance, then the liquid tends to keep to itself; it will bead up. However, if adhesive forces between a liquid and another substance are stronger than cohesive forces, then the liquid will spread out over the other substance, trying to maximize the interface between the other substance and the liquid. It is said that the liquid wets the other substance. Adhesion and cohesion are important for other phenomena as well. In particular, if adhesive forces are strong, then when a liquid is introduced to a small-diameter tube of another substance, the liquid will move up or down in the tube, as if ignoring gravity. Because tiny tubes are called capillaries, this phenomenon is called capillary action. For example, one type of capillary action, capillary rise, is seen when water or water-based liquids rise up in thin glass tubes (like the capillaries sometimes used in blood tests), forming an upwardly curved surface called a meniscus. Capillary action is also responsible for the "wicking" effect that towels and sponges use to dry wet objects; the matting of fibers forms tiny capillaries that have good adhesion with water. Cotton is a good material for this; polyester and other synthetic fabrics do not display similar capillary action, which is why you seldom find rayon bath towels. A similar effect is observed with liquid fuels or melted wax and their wicks. Capillary action is thought to be at least partially responsible for transporting water from the roots to the tops of trees, even tall ones. On the other hand, some liquids have stronger cohesive forces than adhesive forces. In this case, in the presence of a capillary, the liquid is forced down from its surface; this is an example of a type of capillary action called capillary depression. In this case, the meniscus curves downward. Mercury has very strong cohesive forces; when a capillary is placed in a pool of mercury, the surface of the mercury liquid is depressed (Figure \(7\)). Chemistry is Everywhere: Waxing a Car Responsible car owners are encouraged to wax their cars regularly. In addition to making the car look nicer, it also helps protect the surface, especially if the surface is metal. Why? The answer has to do with cohesion and adhesion (and, to a lesser extent, rust). Water is an important factor in the rusting of iron, sometimes used extensively in outer car bodies. Keeping water away from the metal is one way to minimize rusting. A coat of paint helps with this. However, dirty or scratched paint can attract water, and adhesive forces will allow the water to wet the surface, maximizing its contact with the metal and promoting rust. Wax is composed of long hydrocarbon molecules that do not interact well with water. (Hydrocarbons are compounds with C and H atoms; for more information on hydrocarbons, see Chapter 16). That is, a thin layer of wax will not be wetted by water. A freshly waxed car has low adhesive forces with water, so water beads up on the surface, as a consequence of its cohesion and surface tension. This minimizes the contact between water and metal, thus minimizing rust. Summary • All liquids evaporate. • If volume is limited, evaporation eventually reaches a dynamic equilibrium, and a constant vapor pressure is maintained. • All liquids experience surface tension, an imbalance of forces at the surface of the liquid. • All liquids experience capillary action, demonstrating either capillary rise or capillary depression in the presence of other substances.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.12%3A_Liquids.txt
Learning Objective • Describe the general properties of a solid. • Describe the six different types of solids. A solid is like a liquid in that particles are in contact with one another. Solids are unlike liquids in that the intermolecular forces are strong enough to hold the particles in place. At low enough temperatures, all substances are solids (helium is the lone exception), but the temperature at which the solid state becomes the stable phase varies widely among substances, from 20 K (−253°C) for hydrogen to over 3,900 K (3,600°C) for carbon. The solid phase has several characteristics. First, solids maintain their shape. They do not fill their entire containers like gases do, and they do not adopt the shape of their containers like liquids do. They cannot be easily compressed like gases can, and they have relatively high densities. Solids may also demonstrate a variety of properties. For example, many metals can be beaten into thin sheets or drawn into wires, while compounds such as NaCl will shatter if they are struck. Some metals, such as sodium and potassium, are rather soft, while others, such as diamond, are very hard and can easily scratch other substances. Appearances differ as well: most metals are shiny and silvery, but sulfur (a nonmetal) is yellow, and ionic compounds can take on a rainbow of colors. Solid metals conduct electricity and heat, while ionic solids do not. Many solids are opaque, but some are transparent. Some dissolve in water, but some do not. Figure \(1\), shows two solids that exemplify the similar and dissimilar properties of solids. Solids can have a wide variety of physical properties. We will review the different types of solids and the bonding that gives them their properties. First, we must distinguish between two general types of solids. An amorphous solid is a solid with no long-term structure or repetition. Examples include glass and many plastics, both of which are composed of long chains of molecules with no order from one molecule to the next. A crystalline solid is a solid that has a regular, repeating three-dimensional structure. A crystal of NaCl (Figure \(1\)) is one example: at the atomic level, NaCl is composed of a regular three-dimensional array of Na+ ions and Cl ions. There is only one type of amorphous solid. However, there are several different types of crystalline solids, depending on the identity of the units that compose the crystal. An ionic solid is a crystalline solid composed of ions (even if the ions are polyatomic). NaCl is an example of an ionic solid (Figure \(2\) - An Ionic Solid). The Na+ ions and Cl ions alternate in three dimensions, repeating a pattern that goes on throughout the sample. The ions are held together by the attraction of opposite charges—a very strong force. Hence most ionic solids have relatively high melting points; for example, the melting point of NaCl is 801°C. Ionic solids are typically very brittle. To break them, the very strong ionic attractions need to be broken; a displacement of only about 1 × 10−10 m will move ions next to ions of the same charge, which results in repulsion. Ionic solids do not conduct electricity in their solid state; however, in the liquid state and when dissolved in some solvent, they do conduct electricity. This fact originally promoted the idea that some substances exist as ionic particles. A molecular solid is a crystalline solid whose components are covalently bonded molecules. Many molecular substances, especially when carefully solidified from the liquid state, form solids where the molecules line up with a regular fashion similar to an ionic crystal, but they are composed of molecules instead of ions. Because the intermolecular forces between molecules are typically less strong than in ionic solids, molecular solids typically melt at lower temperatures and are softer than ionic solids. Ice is an example of a molecular solid. In the solid state, the molecules line up in a regular pattern (Figure \(3\)). Some very large molecules, such as biological molecules, will form crystals only if they are very carefully solidified from the liquid state or, more often, from a dissolved state; otherwise, they will form amorphous solids. Some solids are composed of atoms of one or more elements that are covalently bonded together in a seemingly never-ending fashion. Such solids are called covalent network solids. Each piece of the substance is essentially one huge molecule, as the covalent bonding in the crystal extends throughout the entire crystal. The two most commonly known covalent network solids are carbon in its diamond form and silicon dioxide (SiO2). Figure \(4\) - Covalent Network Solids, shows the bonding in a covalent network solid. Generally, covalent network solids are poor conductors of electricity, although their ability to conduct heat is variable: diamond is one of the most thermally conductive substances known, while SiO2 is about 100 times less thermally conductive. Most covalent network solids are very hard, as exemplified by diamond, which is the hardest known substance. Covalent network solids have high melting points by virtue of their network of covalent bonds, all of which would have to be broken for them to transform into a liquid. Indeed, covalent network solids are among the highest-melting substances known: the melting point of diamond is over 3,500°C, while the melting point of SiO2 is around 1,650°C. These characteristics are explained by the network of covalent bonds throughout the sample. A metallic solid is a solid with the characteristic properties of a metal: shiny and silvery in color and a good conductor of heat and electricity. A metallic solid can also be hammered into sheets and pulled into wires. A metallic solid exhibits metallic bonding, a type of intermolecular interaction caused by the sharing of the s valence electrons by all atoms in the sample. It is the sharing of these valence electrons that explains the ability of metals to conduct electricity and heat well. It is also relatively easy for metals to lose these valence electrons, which explains why metallic elements usually form cations when they make compounds. Example \(1\) Predict the type of crystal exhibited by each solid. 1. MgO 2. Ag 3. CO2 Solution 1. A combination of a metal and a nonmetal makes an ionic compound, so MgO would exist as ionic crystals in the solid state. 2. Silver is a metal, so it would exist as a metallic solid in the solid state. 3. CO2 is a covalently bonded molecular compound. In the solid state, it would form molecular crystals. (You can actually see the crystals in dry ice with the naked eye.) Exercise \(1\) Predict the type of crystal exhibited by each solid. 1. I2 2. Ca(NO3)2 Answers 1. molecular crystals 2. ionic crystals Food and Drink Application: The Rocks We Eat The foods and beverages we eat and drink all have different phases: solid, liquid, and gas. (How do we ingest gases? Carbonated beverages have gas, which sometimes cause a person to belch.) However, among the solids we eat, three in particular are, or are produced from, rocks. Yes, rocks! The first one is NaCl, or common salt. Salt is the only solid that we ingest that is actually mined as a rock (hence the term rock salt; it really is a rock). Salt provides both Na+ ions and Cl ions, both of which are necessary for good health. Salt preserves food, a function that was much more important before the days of modern food preparation and storage. The fact that saltiness is one of the major tastes the tongue can detect suggests a strong evolutionary link between ingesting salt and survival. There is some concern today that there is too much salt in the diet; it is estimated that the average person consumes at least three times as much salt daily as is necessary for proper bodily function. The other two rocks we eat are related: sodium bicarbonate (NaHCO3) and sodium carbonate (Na2CO3). However, we do not mine these substances directly from the ground; we mine trona, whose chemical formula is Na3H(CO3)2. This substance is dissolved in water and treated with CO2 gas to make either Na2CO3 or NaHCO3. Another process, called the Solvay process, is also used to make Na2CO3. In the Solvay process, NH3 and CO2 are added to solutions of NaCl to make NaHCO3 and NH4Cl; the NaHCO3 precipitates and is heated to produce Na2CO3. Either way, we get these two products from the ground (i.e., rocks). NaHCO3 is also known as baking soda, which is used in many baked goods. Na2CO3 is used in foods to regulate the acid balance. It is also used in laundry (where it is called washing soda) to interact with other ions in water that tend to reduce detergent efficiency. Summary • Solids can be divided into amorphous solids and crystalline solids. • Crystalline solids can be ionic, molecular, covalent network, or metallic.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.13%3A_Solids.txt
Learning Objectives • Define enthalpy of fusion and enthalpy of vaporization. • Calculate the energy changes that accompany phase changes. • Interpret heating and cooling curves. Everyday we take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO2, as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine the changes in energy that occur during these changes in state or phase changes. As introduced in an earlier section, changes of state involve changes in enthalpy $ΔH$, or changes in energy. Melting (changing a solid to liquid) and evaporating (changing a liquid to a gas) are both endothermic processes, requiring the addition of heat to break the intermolecular interactions between molecules. The reverse processes, condensation (changing a gas to a liquid) and freezing (changing a liquid to a solid), are both exothermic, meaning heat is given off or released when intermolecular interactions are reformed. The specific amount of energy absorbed or released when one gram of substance changes between a solid and a liquid (at the melting point) is called the enthalpy of fusion or heat of fusion, ($ΔH_{fus}$). At the boiling point, when one gram of substance changes between a liquid and a gas, the energy change is called the enthalpy of vaporization or heat of vaporization, ($ΔH_{vap}$). Heating Curves It is important to understand the difference between the energy associated with changes in temperature and changes in state where the temperature remains constant. This difference is most clearly depicted in a heating curve, which is a plot of the temperature versus heating time (or heat added). The heating curve in figure $1$ is the temperature change for a 75 g sample of water as heat is added over time. The sample is initially solid (ice) at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat (Cs) of ice, which is the amount of energy (in Joules or calories) required to raise the temperature of 1 g of ice by 1°C. As the temperature increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does not increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of liquid water is greater than that of solid water. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a faster rate than seen in the other phases because the heat capacity of steam is less than that of ice or water. It is worth noting again that the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils. The temperature of a sample does not change during a phase change. If heat is added at a constant rate, as in Figure $1$, then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure $1$, the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion. A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form. Cooling Curves The cooling curve, in Figure $2$ plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled (heat is released). Although we might expect the cooling curve to be the mirror image of the heating curve in Figure $1$, the cooling curve is not an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system. Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a nucleus) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during seeding (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO2 (dry ice) into the cloud from an airplane. Solid CO2 sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO2 sublimes, it absorbs heat from the cloud, often with the desired results. A Video Discussing the Thermodynamics of Phase Changes. Video Source: The Thermodynamics of Phase Changes, YouTube(opens in new window) [youtu.be] Example $1$: Cooling Tea If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol. Given: mass, volume, initial temperature, density, specific heats, and $ΔH_{fus}$ Asked for: final temperature Strategy Substitute the given values into the general equation relating heat gained (by the ice) to heat lost (by the tea) to obtain the final temperature of the mixture. Solution When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by $q=mC_sΔT \nonumber$ where $q$ is heat, $m$ is mass, $C_s$ is the specific heat, and $ΔT$ is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C. The amount of heat gained by the ice cube as it melts is determined by its enthalpy of fusion in kJ/mol: $q=nΔH_{fus} \nonumber$ For our 50.0 g ice cube: \begin{align*} q_{ice} &= 50.0 g⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \[4pt] &= 16.7\, kJ \end{align*} \nonumber Thus, when the ice cube has just melted, it has absorbed 16.7 kJ of heat from the tea. We can then substitute this value into the first equation to determine the change in temperature of the tea: $q_{tea} = - 16,700 J = 500 mL⋅\dfrac{1.00\: g}{1\: mL}⋅4.184 J/(g•°C) ΔT \nonumber$ $ΔT = - 7.98 °C = T_f - T_i \nonumber$ $T_f = 12.02 °C \nonumber$ This would be the temperature of the tea when the ice cube has just finished melting; however, this leaves the melted ice still at 0.0°C. We might more practically want to know what the final temperature of the mixture of tea will be once the melted ice has come to thermal equilibrium with the tea. To determine this, we can add one more step to the calculation by plugging in to the general equation relating heat gained and heat lost again: \begin{align*} q_{ice} &= - q_{tea} \[4pt] q_{ice} &= m_{ice}C_sΔT = 50.0g⋅4.184 J/(g•°C)⋅(T_f - 0.0°C) \[4pt] &= 209.2 J/°C⋅T_f \end{align*} \nonumber $q_{tea} = m_{tea}C_sΔT = 500g⋅4.184 J/(g•°C)⋅(T_f - 12.02°C) = 2092 J/°C⋅T_f - 25,150 J \nonumber$ $209.2 J/°C⋅T_f = - 2092 J/°C⋅T_f + 25,150 J \nonumber$ $2301.2 J/°C⋅T_f = 25,150 J \nonumber$ $T_f = 10.9 °C \nonumber$ The final temperature is in between the initial temperatures of the tea (12.02 °C) and the melted ice (0.0 °C), so this answer makes sense. In this example, the tea loses much more heat in melting the ice than in mixing with the cold water, showing the importance of accounting for the heat of phase changes! Exercise $1$: Death by Freezing Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example $1$ Answer 200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow. Sublimation There is also a phase change where a solid goes directly to a gas: $\text{solid} \rightarrow \text{gas} \label{Eq3}$ This phase change is called sublimation. Each substance has a characteristic heat of sublimation associated with this process. For example, the heat of sublimation (ΔHsub) of H2O is 620 cal/g. We encounter sublimation in several ways. You may already be familiar with dry ice, which is simply solid carbon dioxide (CO2). At −78.5°C (−109°F), solid carbon dioxide sublimes, changing directly from the solid phase to the gas phase: $\mathrm{CO_2(s) \xrightarrow{-78.5^\circ C} CO_2(g)} \label{Eq4}$ Solid carbon dioxide is called dry ice because it does not pass through the liquid phase. Instead, it does directly to the gas phase. (Carbon dioxide can exist as liquid but only under high pressure.) Dry ice has many practical uses, including the long-term preservation of medical samples. Even at temperatures below 0°C, solid H2O will slowly sublime. For example, a thin layer of snow or frost on the ground may slowly disappear as the solid H2O sublimes, even though the outside temperature may be below the freezing point of water. Similarly, ice cubes in a freezer may get smaller over time. Although frozen, the solid water slowly sublimes, redepositing on the colder cooling elements of the freezer, which necessitates periodic defrosting (frost-free freezers minimize this redeposition). Lowering the temperature in a freezer will reduce the need to defrost as often. Under similar circumstances, water will also sublime from frozen foods (e.g., meats or vegetables), giving them an unattractive, mottled appearance called freezer burn. It is not really a “burn,” and the food has not necessarily gone bad, although it looks unappetizing. Freezer burn can be minimized by lowering a freezer’s temperature and by wrapping foods tightly so water does not have any space to sublime into. Summary • Phase changes are associated with energy changes. • The enthalpy of fusion, $ΔH_{fus}$, is the energy absorbed or released when a substance changes between a solid and a liquid. • The enthalpy of vaporization, $ΔH_{vap}$, is the energy absorbed or released when a substance changes between a liquid and a gas. • Heating and cooling curves are useful tools to show how temperature, energy, and phases are related.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/08%3A_Gases_Liquids_and_Solids/8.14%3A_Changes_of_State_Calculations.txt
• 9.1: Mixtures and Solutions Solutions form because a solute and a solvent experience similar intermolecular interactions. • 9.2: The Solution Process When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other. • 9.3: Solubility The ability of a solute to dissolve in a solvent, its solubility, is dependent on the strength of attraction between the different molecules. A saturated solution will have the maximum amount of solute dissolved. • 9.4: The Effect of Temperature on Solubility Temperature can increase or decrease the solubility of a solute. • 9.5: The Effect of Pressure on Solubility - Henry’s Law Changes in the partial pressure are directly proportional to the solubility of dissolved gas according to Henry's Law. • 9.6: Units of Concentration Various concentration units are used to express the amounts of solute in a solution. Concentration units can be used as conversion factors in stoichiometry problems. New concentrations can be easily calculated if a solution is diluted. • 9.7: Dilution A solution can be diluted by adding additional solvent to a fixed amount of solute, decreasing the concentration. • 9.8: Ions in Solution - Electrolytes Solutions containing ions can conduct electricity, therefore they are called electrolytes. • 9.9: Properties of Solutions Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. • 9.10: Osmosis and Osmotic Pressure The total concentration of solute particles in a solution determines its osmotic pressure. • 9.11: Dialysis Dialysis is similar to osmosis but can allow both solute and solvent particles to pass through a semipermeable membrane. 09: Solutions Learning Objectives • Describe the difference between homogenous and heterogenous mixtures. • Distinguish a homogenous mixture as a solution or colloid. As introduced previously, mixtures are combinations of two or more substances that each retain their individual physical properties. A mixture can be classified as either heterogenous or homogenous. In a heterogeneous mixture, the composition is not uniform throughout the sample, and sometimes the individual substances that make up the mixture can be differentiated by eye. Trail mix, salad, and blood (which is also called a suspension) are examples of heterogenous mixtures. Homogenous mixtures are uniform and have the same composition throughout. Air, simple syrup, and seawater are examples of homogenous mixtures. Homogenous mixtures can be further classified, based on the size of their particles, as solutions or colloids. Solutions are composed of particles the size of an ion or small molecule, ~0.1-2.0 nm. The examples provided above are all considered solutions; air is a solution of small gas molecules, simple syrup is a solution of sucrose in water, and seawater is a solution of ions and water. Homogenous mixtures with larger particles, ~2.0-500 nm, are classified as colloids. Milk, fog, and butter are all considered colloids. Mixtures with particles larger than 500 nm are called suspensions and are considered to be heterogenous mixtures in which the particles will settle upon standing. Many medications are classified as suspensions that need to be re-mixed before taking in order to redistribute the particles throughout the mixture. Some examples and distinguishing characteristics of solutions, colloids, and suspensions are listed in Table \(1\) below. Table \(1\): Properties of Liquid Solutions, Colloids, and Suspensions Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint The major component of a solution, called the solvent, is typically the same phase as the solution itself. Each minor component of a solution (and there may be more than one) is called the solute. In most of the solutions we will describe in this textbook, there will be no ambiguity about whether a component is the solvent or the solute. For example, in a solution of salt in water, the solute is salt, and solvent is water. Solutions come in all phases, and the solvent and the solute do not have to be in the same phase to form a solution (such as salt and water). For example, air is a gaseous solution of about 80% nitrogen and about 20% oxygen, with some other gases present in much smaller amounts. An alloy is a solid solution consisting of a metal (like iron) with some other metals or nonmetals dissolved in it. Steel, an alloy of iron and carbon and small amounts of other metals, is an example of a solid solution. Table \(2\) lists some common types of solutions, with examples of each. Table \(2\): Types of Solutions Solvent Phase Solute Phase Example gas gas air liquid gas carbonated beverages liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages) liquid solid saltwater solid gas H2 gas absorbed by Pd metal solid liquid Hg(ℓ) in dental fillings solid solid steel alloys
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.01%3A_Mixtures_and_Solutions.txt
Learning Objectives • Predict solubility based on interactions between solute and solvent. What occurs at the molecular level to cause a solute to dissolve in a solvent? The answer depends in part on the strength of attractions between solute and solvent particles. A good rule of thumb is to use is like dissolves like, which means that substances must have similar intermolecular attractions to form solutions. A substance can dissolve in a solvent, and form a solution, if the solute and solvent are attracted to each other. For example, water molecules that are held together by hydrogen bonding will dissolve solutes that can also hydrogen bond, like ethanol (CH3CH2OH). The new hydrogen bonds between the water and the ethanol molecules (solvent-solute attractions) are nearly as strong as the hydrogen bonds in water (solvent-solvent) and ethanol (solute-solute) alone, making the process of solution formation (also called dissolution or dissolving) favorable. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, are surrounded by solvent molecules. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called solvation and is illustrated in Figure \(1\). When the solvent is water, the word hydration, rather than solvation, is used. When a solute and solvent that do not have similar intermolecular interactions are mixed, a solution is not formed because the solute-solute or solvent-solvent attractions are stronger than any favorable interactions between solute and solvent. For example when water and oil are mixed, stay in separate layers, i.e., they will not mix to yield solutions and the water molecules remain hydrogen bonded to water molecules while the oil molecules stay together (Figure \(2\)). Hydrogen bonding is the dominant intermolecular attractive force present in liquid water; the nonpolar hydrocarbon molecules of cooking oils are not capable of hydrogen bonding, instead being held together by dispersion forces. Forming an oil-water solution would require overcoming the very strong hydrogen bonding in water, as well as the significantly strong dispersion forces between the relatively large oil molecules. And, since the polar water molecules and nonpolar oil molecules would not experience very strong intermolecular attraction, very little energy would be released by solvation. Ionic Compounds and Covalent Compounds as Solutes In the case of molecular solutes like glucose, the solute particles are individual molecules. However, if the solute is ionic, the individual ions separate from each other and become surrounded by solvent particles. The positively charged cations are attracted to the neg and anions of an ionic solute separate when the solute dissolves. This process is referred to as dissociation (Figure \(1\)). The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as electrolytes. Many ionic compounds dissociate completely and are therefore called strong electrolytes. Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called weak electrolytes. Acetic acid (CH3COOH), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called nonelectrolytes. Table sugar (C12H22O11) is an example of a nonelectrolyte. The term electrolyte is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na+, K+, Ca2+, Mg2+, and Cl. Example \(1\) The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte. 1. potassium chloride (KCl) 2. fructose (C6H12O6) 3. isopropyl alcohol [CH3CH(OH)CH3] 4. magnesium hydroxide [Mg(OH)2] Solution Each substance can be classified as an ionic solute or a nonionic solute. Ionic solutes are electrolytes, and nonionic solutes are nonelectrolytes. 1. Potassium chloride is an ionic compound; therefore, when it dissolves, its ions separate, making it an electrolyte. 2. Fructose is a sugar similar to glucose. (In fact, it has the same molecular formula as glucose.) Because it is a molecular compound, we expect it to be a nonelectrolyte. 3. Isopropyl alcohol is an organic molecule containing the alcohol functional group. The bonding in the compound is all covalent, so when isopropyl alcohol dissolves, it separates into individual molecules but not ions. Thus, it is a nonelectrolyte 4. Magnesium hydroxide is an ionic compound, so when it dissolves it dissociates. Thus, magnesium hydroxide is an electrolyte. Exercise \(1\) The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte. 1. acetone (CH3COCH3) 2. iron(III) nitrate [Fe(NO3)3] 3. elemental bromine (Br2) 4. sodium hydroxide (NaOH) Answer a. nonelectrolyte b. electrolyte c. nonelectrolyte d. electrolyte Electrolytes in Body Fluids Our body fluids are solutions of electrolytes and many other things. The combination of blood and the circulatory system is the river of life, because it coordinates all the life functions. When the heart stops pumping in a heart attack, the life ends quickly. Getting the heart restarted as soon as one can is crucial in order to maintain life. The primary electrolytes required in the body fluid are cations (of calcium, potassium, sodium, and magnesium) and anions (of chloride, carbonates, aminoacetates, phosphates, and iodide). These are nutritionally called macrominerals. Electrolyte balance is crucial to many body functions. Here's some extreme examples of what can happen with an imbalance of electrolytes: elevated potassium levels may result in cardiac arrhythmias; decreased extracellular potassium produces paralysis; excessive extracellular sodium causes fluid retention; and decreased plasma calcium and magnesium can produce muscle spasms of the extremities. When a patient is dehydrated, a carefully prepared (commercially available) electrolyte solution is required to maintain health and well being. In terms of child health, oral electrolyte is given when a child is dehydrated due to diarrhea. The use of oral electrolyte maintenance solutions, which is responsible for saving millions of lives worldwide over the last 25 years, is one of the most important medical advances in protecting the health of children in the century, explains Juilus G.K. Goepp, MD, assistant director of the Pediatric Emergency Department of the Children's Center at Johns Hopkins Hospital. If a parent provides an oral electrolyte maintenance solution at the very start of the illness, dehydration can be prevented. The functionality of electrolyte solutions is related to their properties, and interest in electrolyte solutions goes far beyond chemistry. Sports drinks are designed to rehydrate the body after excessive fluid depletion. Electrolytes in particular promote normal rehydration to prevent fatigue during physical exertion. Are they a good choice for achieving the recommended fluid intake? Are they performance and endurance enhancers like they claim? Who should drink them? Typically, eight ounces of a sports drink provides between fifty and eighty calories and 14 to 17 grams of carbohydrate, mostly in the form of simple sugars. Sodium and potassium are the most commonly included electrolytes in sports drinks, with the levels of these in sports drinks being highly variable. The American College of Sports Medicine says a sports drink should contain 125 milligrams of sodium per 8 ounces as it is helpful in replenishing some of the sodium lost in sweat and promotes fluid uptake in the small intestine, improving hydration. Gatorade In the summer of 1965, the assistant football coach of the University of Florida Gators requested scientists affiliated with the university study why the withering heat of Florida caused so many heat-related illnesses in football players and provide a solution to increase athletic performance and recovery post-training or game. The discovery was that inadequate replenishment of fluids, carbohydrates, and electrolytes was the reason for the “wilting” of their football players. Based on their research, the scientists concocted a drink for the football players containing water, carbohydrates, and electrolytes and called it “Gatorade.” In the next football season the Gators were nine and two and won the Orange Bowl. The Gators’ success launched the sports-drink industry, which is now a multibillion-dollar industry that is still dominated by Gatorade. University of Florida football player Chip Hinton testing Gatorade in 1965, pictured next to the leader of its team of inventors, Robert Cade. Concept Review Exercise 1. Explain how the solvation process describes the dissolution of a solute in a solvent. Answer 1. Each particle of the solute is surrounded by particles of the solvent, carrying the solute from its original phase. Key Takeaway • When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other. Exercises 1. Describe what happens when an ionic solute like Na2SO4 dissolves in a polar solvent. 2. Describe what happens when a molecular solute like sucrose (C12H22O11) dissolves in a polar solvent. 3. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent. 1. NH4NO3 2. CO2 3. NH2CONH2 4. HCl 4. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent. 1. CH3CH2CH2OH 2. Ca(CH3CO2)2 3. I2 4. KOH 5. Will solutions of each solute conduct electricity when dissolved? 1. AgNO3 2. CHCl3 3. BaCl2 4. Li2O 6. Will solutions of each solute conduct electricity when dissolved? 1. CH3COCH3 2. N(CH3)3 3. CH3CO2C2H5 4. FeCl2 Answers 1. Each ion of the ionic solute is surrounded by particles of solvent, carrying the ion from its associated crystal. 2. Each sucrose molecule is surrounded by solvent molecules (attracted to each other via intermolecular forces of attraction). 1. electrolyte 2. nonelectrolyte 3. nonelectrolyte 4. electrolyte 4. • nonelectrolyte • electrolyte • nonelectrolyte • electrolyte 5. 1. yes 2. no 3. yes 4. yes 6. a. no b. no c. no d. yes
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.02%3A_The_Solution_Process.txt
Learning Objectives • Describe solutions as saturated or unsaturated by understanding solubility. To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors. Solubility There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubility of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table $1$ lists the solubilities of various solutes in water. Solubilities vary with temperature, so Table $1$ includes the temperature at which the solubility was determined. Table $1$: Solubilities of Various Solutes in Water at 25°C (Except as Noted) Substance Solubility (g in 100 mL of H2O) AgCl(s) 0.019 C6H6(ℓ) (benzene) 0.178 CH4(g) 0.0023 CO2(g) 0.150 CaCO3(s) 0.058 CaF2(s) 0.0016 Ca(NO3)2(s) 143.9 C6H12O6 (glucose) 120.3 (at 30°C) KBr(s) 67.8 MgCO3(s) 2.20 NaCl(s) 36.0 NaHCO3(s) 8.41 C12H22O11 (sucrose) 204.0 (at 20°C) If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturated, and its concentration is known from information contained in Table $1$. If a solution contains less solute than the solubility limit, it is unsaturated. Under special circumstances, more solute can be dissolved even after the normal solubility limit is reached; such solutions are called supersaturated and are not stable. If the solute is solid, excess solute can easily recrystallize. If the solute is a gas, it can bubble out of solution uncontrollably, like what happens when you shake a soda can and then immediately open it. Most solutions we encounter are unsaturated, so knowing the solubility of the solute does not accurately express the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution that will be discussed in later sections. Solution Equilibrium When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (Figure $\PageIndex{1a}$). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. The equilibrium between dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows: $solute + solvent \underset{crystallization}{\stackrel{dissolution}{\longrightleftharpoons}} solution \nonumber$ 9.04: The Effect of Temperature on Solubility Learning Objectives • Describe how temperature affects solubility of different types of solute. 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 (Figure $1$ 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 $30° \text{C}$. According to the solubility curve in Figure $1$, approximately $48 \: \text{g}$ of $\ce{KNO_3}$ will dissolve at $30° \text{C}$. 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 $60° \text{C}$. According to the curve, the solubility of $\ce{KNO_3}$ at $60° \text{C}$ 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 $0° \text{C}$. The solubility at $0° \text{C}$ is about $14 \: \text{g}$, meaning that $80 - 14 = 66 \: \text{g}$ of the $\ce{KNO_3}$ will recrystallize. Supersaturated Solutions Some solutes, such as sodium acetate, do not recrystallize easily. Suppose an exactly saturated solution of sodium acetate is prepared at $50° \text{C}$. 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. Chemistry in Everyday Life: Handwarmers Recrystallization of excess solute from a supersaturated solution usually gives off energy as heat. Commercial heat packs, such as the one in Figure $2$, containing supersaturated sodium acetate (NaC2H3O2) take advantage of this phenomenon. You can probably find them at your local drugstore. Figure $2$: This hand warmer produces heat when the sodium acetate in a supersaturated solution precipitates. Precipitation of the solute is initiated by a mechanical shockwave generated when the flexible metal disk within the solution is “clicked.” (credit: modification of work by “Velela”/Wikimedia Commons)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.03%3A_Solubility.txt
Learning Objectives • Describe how pressure affects concentration of a solute in a solution. 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, see Figure $1$. 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: $\frac{C_1}{P_1} = \frac{C_2}{P_2} = k \nonumber$ $C_1$ and $P_1$ are the concentration and the pressure at an initial set of conditions; $C_2$ and $P_2$ are the concentration and pressure at another changed set of conditions; $k$ is a constant at a constant temperature. 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 $20^\text{o} \text{C}$. Solution Step 1: List the known quantities and plan the problem. Known • $C_1 = 0.745 \: \text{g/L}$ • $P_1 = 1.00 \: \text{atm}$ • $P_2 = 4.50 \: \text{atm}$ Unknown Substitute into Henry's law and solve for $C_2$. Step 2: Solve. $C_2 = \frac{C_1 \times P_2}{P_1} = \frac{0.745 \: \text{g/L} \times 4.50 \: \text{atm}}{1.00 \: \text{atm}} = 3.35 \: \text{g/L}\nonumber$ Step 3: Think about your result. The solubility is increased to 4.5 times its original value, according to the direct relationship. Exercise $1$ Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10−3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr). Answer 7.25 × 10−3 g in 100.0 mL or 0.0725 g/L Case Study: Decompression Sickness (“The Bends”) Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law. As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure $4$).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.05%3A_The_Effect_of_Pressure_on_Solubility_-_Henrys_Law.txt
Learning Objectives • Calculate concentration of a solution using different units. • Use concentration units to calculate the amount of solute in a solution. • Use molarity to determine quantities in chemical reactions. Percent Concentrations There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: $\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber$ If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration. Example $1$ A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution? Solution We can substitute the quantities given in the equation for mass/mass percent: $\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}$ Exercise $1$ A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution? Answer 7.90% For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a volume/volume percent (% v/v): the volume of a solute divided by the volume of a solution times 100: $\mathrm{\%\: v/v = \dfrac{volume\: of\: solute}{volume\: of\: solution}\times100\%} \nonumber$ Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass/volume percent (% m/v), is commonly used for intravenous (IV) fluids (Figure $1$). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100: $\mathrm{\%\: m/v = \dfrac{mass\: of\: solute\: (g)}{volume\: of\: solution\: (mL)}\times100\%} \nonumber$ Using Percent Concentration in Calculations The percent concentration can be used to produce a conversion factor between the amount of solute and the amount of solution. As such, concentrations can be useful in a variety of stoichiometry problems as discussed in Chapter 6. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor. As an example, if the given concentration is 5% v/v alcohol, this means that there are 5 mL of alcohol dissolved in every 100 mL solution. 5 mL alcohol = 100 mL solution The two possible conversion factors are written as follows: $\mathrm{\dfrac{5\: mL\: alcohol}{100\: mL\: solution}}$ or $\mathrm{\dfrac{100\: mL\: solution}{5\: mL\: alcohol}}$ Use the first conversion factor to convert from a given amount of solution to amount of solute. The second conversion factor is used to convert from a given amount of solute to amount of solution. Given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates. Example $2$ A sample of 45.0% v/v solution of ethanol (C2H5OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain? Solution A percentage concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% v/v implies the following: $\mathrm{45.0\%\: v/v \rightarrow \dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}}$ That is, there are 45 mL of C2H5OH for every 100 mL of solution. We can use this fraction as a conversion factor to determine the amount of C2H5OH in 115 mL of solution: $\mathrm{115\: mL\: solution\times\dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}=51.8\: mL\: C_2H_5OH}$ Exercise $2$ What volume of a 12.75% m/v solution of glucose (C6H12O6) in water is needed to obtain 50.0 g of C6H12O6? Answer $\mathrm{50.0\: g\: C_6H_12O_6\times\dfrac{100\: mL\: solution}{12.75\: g\: C_6H_12O_6}=392\: mL\: solution}$ Example $3$ A normal saline IV solution contains 9.0 g of NaCl in every liter of solution. What is the mass/volume percent of normal saline? Solution We can use the definition of mass/volume percent, but first we have to express the volume in milliliter units: 1 L = 1,000 mL Because this is an exact relationship, it does not affect the significant figures of our result. $\mathrm{\%\: m/v = \dfrac{9.0\: g\: NaCl}{1,000\: mL\: solution}\times100\%=0.90\%\: m/v}$ Exercise $3$ The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/volume percent of the bleach? Answer $\mathrm{\%\: m/v = \dfrac{27.0\: g\: NaOCl}{500.0\: mL\: solution}\times100\%=5.40\%\: m/v}$ Parts per Million (ppm) and Parts per Billion (ppb) In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units are mass based and are defined as follows: $\mathrm{ppm=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000} \nonumber$ $\mathrm{ppb=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000,000} \nonumber$ Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt). Concentrations of trace elements in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm. In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/L. Example $4$ If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg? Solution A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g of solution.” Written as a conversion factor, this concentration of Co is as follows: $\mathrm{21\: ppb\: Co \rightarrow \dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}}$ We can use this as a conversion factor, but first we must convert 70.0 kg to gram units: $\mathrm{70.0\: kg\times\dfrac{1,000\: g}{1\: kg}=7.00\times10^4\: g}$ Now we determine the amount of Co: $\mathrm{7.00\times10^4\: g\: solution\times\dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}=0.0015\: g\: Co}$ This is only 1.5 mg. Exercise $4$ An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million? Answer 0.14 ppm Mole/Volume Concentration: Molarity Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams. Molarity(M) is defined as the number of moles of a solute dissolved per liter of solution: $\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \nonumber$ Molarity is abbreviated M (often referred to as “molar”), and the units are often abbreviated as mol/L. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore $\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber$ which is read as “three point oh molar sodium chloride.” Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl(aq).” Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters. If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? Step 1: convert the mass of solute to moles using the molar mass of HCl (36.46 g/mol): $22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.46\cancel{gHCl}}=0.614\, mol\; HCl \nonumber$ Step 2: use the definition of molarity to determine the concentration: $M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber$ Example $5$ What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL? Solution Before we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol: $\mathrm{25.0\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=0.625\: mol\: NaOH}$ Next, we convert the volume units from milliliters to liters: $\mathrm{750\: mL\times\dfrac{1\: L}{1,000\: mL}=0.750\: L}$ Now that the quantities are expressed in the proper units, we can substitute them into the definition of molarity: $\mathrm{M=\dfrac{0.625\: mol\: NaOH}{0.750\: L}=0.833\: M\: NaOH}$ Exercise $5$ If a 350 mL cup of coffee contains 0.150 g of caffeine (C8H10N4O2), what is the molarity of this caffeine solution? Answer 0.00221 M Using Molarity in Calculations The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. As in the percent concentration, molarity can also be expressed as a conversion factor. Molarity is defined as moles solute per liter solution. There is an understood 1 in the denominator of the conversion factor. For example, a 3.0 M solution of sucrose means that there are three moles of sucrose dissolved in every liter of solution. Mathematically, this is stated as follows: 3.0 moles sucrose = 1 L solution Dividing both sides of this expression by either side, we generate two possible conversion factors: $\mathrm{\dfrac{3.0\: mol\: sucrose}{1\: L\: solution}}$ or $\mathrm{\dfrac{1\: L\: solution}{3.0\: mol\: sucrose}}$ The first conversion factor can be used to convert from volume (L) of solution to moles solute, and the second converts from moles of solute to volume (L) of solution. For example, suppose we are asked how many moles of sucrose are present in 0.108 L of a 3.0 M sucrose solution. The given volume (0.108 L) is multiplied by the first conversion factor to cancel the L units, and find that 0.32 moles of sucrose are present. $0.108\cancel{L\, solution}\times \dfrac{3.0\, mol\, sucrose}{\cancel{1L\, solution}}=0.32\, mol\, sucrose \nonumber$ How many liters of 3.0 M sucrose solution are needed to obtain 4.88 mol of sucrose? In such a conversion, we multiply the given (4.88 moles sucrose) with the second conversion factor. This cancels the moles units and converts it to liters of solution. $4.88\cancel{mol\, sucrose}\times \dfrac{1\, L\, solution}{\cancel{3.0\, mol\, sucrose}}=1.63\, L\, solution \nonumber$ Example $6$ 1. What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] is needed to obtain 0.450 mol of the compound? 2. Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution? Solution 1. To solve for the volume, multiply the "given" (0.450 mol of dimethylamine) with the molarity conversion factor (0.0753 M). Use the proper conversion factor to cancel the unit "mol" and get the unit volume (L) of solution: $\mathrm{0.450\: mol\: dimethylamine\times\dfrac{1\: L\: solution}{0.0753\: mol\: dimethylamine}=5.98\: L\: solution}$ 2. The strategy in solving this problem is to convert the given volume (5.00 L) using the 6.00 M (conversion factor) to solve for moles of ethylene glycol, which can then be converted to grams. Step 1: Convert the given volume (5.00 L) to moles ethylene glycol. $\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}=30.0\: mol\: C_2H_6O_2}$ Step 2: Convert 30.0 mols C2H6O2 to grams C2H6O2. Molar mass of C2H6O2= 62.08 g/mol $\mathrm{30.0\: mol\: C_2H_6O_2\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}$ The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows: $\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}$ The final answer is rounded off to 3 significant figures. Thus, there are 1,860 g of C2H6O2 in the specified amount of engine coolant. Note: Dimethylamine has a “fishy” odor. In fact, organic compounds called amines cause the odor of decaying fish. Exercise $6$ 1. What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH? 2. Acetic acid (HC2H3O2) is the acid in vinegar. How many grams of HC2H3O2 are in 0.565 L of a 0.955 M solution? Answer a. 9.84 L b. 32.4 g Solution Stoichiometry Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation: HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq) Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows: Figure $2$: Typical approach to solving Molarity problems In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving. Example $7$ How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows: HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq) Solution We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol: $\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=4.63\: mol\: NaOH}$ Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH: $\mathrm{4.63\: mol\: NaOH\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}=4.63\: mol\: HCl}$ Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed: $\mathrm{2.75\: M\: HCl=\dfrac{4.63\: mol\: HCl}{volume\: of\: HCl\: solution}}$ $\mathrm{volume\: of\: HCl=\dfrac{4.63\: mol\: HCl}{2.75\: M\: HCl}=1.68\: L\times\dfrac{1,000\: mL}{1\: L}=1,680\: mL}$ We need 1,680 mL of 2.75 M HCl to react with the NaOH. The same multi-step problem can also be worked out in a single line, rather than as separate steps, as follows: $\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}\times\dfrac{1\: L\: HCl\: solution}{2.75\: mol\: HCl}\times\dfrac{1000\: mL\: HCl\: solution}{1\: L\: HCl\: solution}=1,680\: mL\: HCl\: solution}$ Our final answer (rounded off to three significant figures) is 1,680 mL HCl solution. Exercise $7$ How many milliliters of a 1.04 M H2SO4 solution are needed to react with 98.5 g of Ca(OH)2? The balanced chemical equation for the reaction is as follows: $H_2SO_{4(aq)} + Ca(OH)_{2(s)} \rightarrow 2H_2O_{(ℓ)} + CaSO_{4(aq)} \nonumber$ Answer 1,280 mL The generic steps for performing stoichiometry problems such as this are shown in Figure $3$. You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure $3$ indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end. Solutions in Our Body Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table $1$ lists the typical concentrations of some of these solutes. Table $1$: Approximate Concentrations of Various Solutes in Some Solutions in the Body* Solution Solute Concentration (M) blood plasma Na+ 0.138 K+ 0.005 Ca2+ 0.004 Mg2+ 0.003 Cl 0.110 HCO3 0.030 stomach acid HCl 0.10 urine NaCl 0.15 PO43 0.05 NH2CONH2 (urea) 0.30 *Note: Concentrations are approximate and can vary widely. Looking Closer: The Dose Makes the Poison Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison. This means that what may be dangerous in some amounts may not be dangerous in other amounts. Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison. Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.06%3A_Units_of_Concentration.txt
Learning Objective • Calculate the concentration of solutions prepared by dilution. Often, a worker will need to change the concentration of a solution by changing the amount of solvent. Dilution is the addition of solvent, which decreases the concentration of the solute in the solution. Concentration is the removal of solvent, which increases the concentration of the solute in the solution. (Do not confuse the two uses of the word concentration here!) In both dilution and concentration, the amount of solute stays the same. This gives us a way to calculate what the new solution volume must be for the desired concentration of solute. From the definition of molarity, $\text{molarity} = \dfrac{\text{moles of solute}}{\text{liters of solution}}\nonumber$ we can solve for the number of moles of solute: moles of solute = Molarity (mol/L) x Volume of solution (L) A simpler way of writing this is to use M to represent molarity and V to represent volume. So the equation becomes moles of solute = M x V Because this quantity does not change before and after the change in concentration, the product MV must be the same before and after the concentration change. Using numbers to represent the initial and final conditions, we have $M_1V_1 = M_2V_2\nonumber$ as the dilution equation. The volumes must be expressed in the same units. Note that this equation gives only the initial and final conditions, not the amount of the change. The amount of change in volume is determined by subtraction. Example $1$ If 25.0 mL of a 2.19 M solution are diluted to 72.8 mL, what is the final concentration? Solution It does not matter which set of conditions is labeled 1 or 2, as long as the conditions are paired together properly. Using the dilution equation, we have (2.19 M)(25.0 mL) = M2(72.8 mL) Solving for the second concentration (noting that the milliliter units cancel), M2 = 0.752 M The concentration of the solution has decreased. In going from 25.0 mL to 72.8 mL, 72.8 − 25.0 = 47.8 mL of solvent must be added. Exercise $1$ A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution? Answer 135.4 mL Concentrating solutions involves removing solvent. Usually this is done by evaporating or boiling, assuming that the heat of boiling does not affect the solute. The dilution equation is used in these circumstances as well. Chemistry is Everywhere: Preparing IV Solutions In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl? Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation. If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then it is easy to calculate how much stock solution to use: (10%)V1 = (0.50%)(100 mL)V1 = 5 mL Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough even for medical purposes. Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be useless, harmful, or even fatal!
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.07%3A_Dilution.txt
Learning Objectives • Define solutes as strong electrolytes, weak electrolytes, or nonelectrolytes. • Calculate equivalents for an ionic solute. When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte. Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure $1$). Ionic Electrolytes Water and other polar molecules are attracted to ions, as shown in Figure $2$. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water. When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Let us consider what happens at the microscopic level when we add solid KCl to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules penetrate between individual K+ and Cl ions and surround them, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure shows. The reduction of the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution, resulting in an increase in the disorder of the system as the ions change from their fixed and ordered positions in the crystal to mobile and much more disordered states in solution. This increased disorder is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat. In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in disorder cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust). Covalent Electrolytes Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton to another molecule of water, yielding hydronium and hydroxide ions. $\ce{H_2O (l)+ H_2O (l) \rightleftharpoons H_3O^{+} (aq) + OH^{−} (aq)} \label{11.3.2}$ In some cases, we find that solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, when we dissolve hydrogen chloride in water, we find that the solution is a very good conductor. The water molecules play an essential part in forming ions: Solutions of hydrogen chloride in many other solvents, such as benzene, do not conduct electricity and do not contain ions. Hydrogen chloride is an acid, and so its molecules react with water, transferring H+ ions to form hydronium ions ($H_3O^+$) and chloride ions (Cl): This reaction is essentially 100% complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. Acids and bases will be explored further in a later chapter. Equivalents Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na+(aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/L solution of Ca2+(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.08%3A_Ions_in_Solution_-_Electrolytes.txt
Learning Objectives • To describe how the properties of solutions differ from those of pure solvents. Solutions are likely to have properties similar to those of their major component—usually the solvent. However, some solution properties differ significantly from those of the solvent. Here, we will focus on liquid solutions that have a solid solute, but many of the effects we will discuss in this section are applicable to all solutions. Colligative Properties Solutes affect some properties of solutions that depend only on the concentration of the dissolved particles. These properties are called colligative properties. Four important colligative properties that we will examine are vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure. Molecular compounds separate into individual molecules when they are dissolved, so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast, ionic compounds separate into their constituent ions when they dissolve, so 1 mol of an ionic compound will produce more than 1 mol of dissolved particles. For example, every mole of NaCl that dissolves yields 1 mol of Na+ ions and 1 mol of Cl ions, for a total of 2 mol of particles in solution. Thus, the effect on a solution’s properties by dissolving NaCl may be twice as large as the effect of dissolving the same amount of moles of glucose (C6H12O6). Vapor Pressure Depression All liquids evaporate. In fact, given enough volume, a liquid will turn completely into a vapor. If enough volume is not present, a liquid will evaporate only to the point where the rate of evaporation equals the rate of vapor condensing back into a liquid. The pressure of the vapor at this point is called the vapor pressure of the liquid. The presence of a dissolved solid lowers the characteristic vapor pressure of a liquid so that it evaporates more slowly. (The exceptions to this statement are if the solute itself is a liquid or a gas, in which case the solute will also contribute something to the evaporation process. We will not discuss such solutions here.) This property is called vapor pressure depression and is depicted in Figure $1$. Boiling Point Elevation A related property of solutions is that their boiling points are higher than the boiling point of the pure solvent. Because the presence of solute particles decreases the vapor pressure of the liquid solvent, a higher temperature is needed to reach the boiling point. This phenomenon is called boiling point elevation. For every mole of particles dissolved in a liter of water, the boiling point of water increases by about 0.51°C. $\Delta T_{boiling} = mol\ particles \times 0.51°C\nonumber$ The addition of one mole of sucrose (molecular compound) in one liter of water will raise the boiling point from 100°C to 100.51°C, but the addition of one mole of NaCl in one liter of water will raise the boiling point by 2 x 0.51°C = 1.02°C. Furthermore, the addition of one mole of $\ce{CaCl2}$ in one liter of water will raise the boiling point by 3 x 0.51°C = 1.53°C. Everyday Application: Adding Salt to Pasta Water When cooking dried pasta, many recipes call for salting the water before cooking the pasta. Some argue—with colligative properties on their side—that adding salt to the water raises the boiling point, thus cooking the pasta faster. Is there any truth to this? In order to increase the boiling temperature of 4 L of water by 1.0°C, with the presumption that dried pasta cooks noticeably faster at 101°C than at 100°C (although a 1° difference may make only a negligible change in cooking times), over 1 lb of salt (~1 cup) would be needed. In your experience, do you add almost a cup of salt to a pot of water to make pasta? Certainly not! A few pinches, perhaps one-fourth of a teaspoon, but not almost a cup! It is obvious that the little amount of salt that most people add to their pasta water is not going to significantly raise the boiling point of the water. So why do people add some salt to boiling water? There are several possible reasons, the most obvious of which is taste: adding salt adds a little bit of salt flavor to the pasta. It cannot be much because most of the salt remains in the water, not in the cooked pasta. However, it may be enough to detect with our taste buds. The other obvious reason is habit; recipes tell us to add salt, so we do, even if there is little scientific or culinary reason to do so. Freezing Point Depression The presence of solute particles has the opposite effect on the freezing point of a solution. When a solution freezes, only the solvent particles come together to form a solid phase, and the presence of solute particles interferes with that process. Therefore, for the liquid solvent to freeze, more energy must be removed from the solution, which lowers the temperature. Thus, solutions have lower freezing points than pure solvents do. This phenomenon is called freezing point depression. For every mole of particles in a liter of water, the freezing point decreases by about 1.86°C. $\Delta T_{freezing} = mol\ particles \times -1.86°C\nonumber$ Both boiling point elevation and freezing point depression have practical uses. For example, solutions of water and ethylene glycol (C2H6O2) are used as coolants in automobile engines because the boiling point of such a solution is greater than 100°C, the normal boiling point of water. In winter, salts like NaCl and $\ce{CaCl_2}$ are sprinkled on the ground to melt ice or keep ice from forming on roads and sidewalks (Figure $4$). This is because the solution made by dissolving sodium chloride or calcium chloride in water has a lower freezing point than pure water, so the formation of ice is inhibited. Example $1$ Which solution’s freezing point deviates more from that of pure water—a 1 M solution of NaCl or a 1 M solution of $\ce{CaCl_2}$? Solution Colligative properties depend on the number of dissolved particles, so the solution with the greater number of particles in solution will show the greatest deviation. When NaCl dissolves, it separates into two ions, Na+ and Cl. But when $\ce{CaCl2}$ dissolves, it separates into three ions—one Ca2+ ion and two Cl ions. Thus, mole for mole, $\ce{CaCl2}$ will have 50% more impact on freezing point depression than NaCl. Exercise $1$ Which solution’s boiling point deviates more from that of pure water—a 1 M solution of $\ce{CaCl2}$ or a 1 M solution of MgSO4? Answer $\ce{CaCl2}$ Example $2$ Estimate the boiling point of 0.2 M $\ce{CaCl2}$ solution. Solution The boiling point increases 0.51°C for every mole of solute per liter of water. For this estimation, let's assume that 1 liter of solution is roughly the same volume as 1 liter of water. A 0.2 M $\ce{CaCl_2}$ solution contains 0.2 moles of $\ce{CaCl_2}$ solution formula units per liter of solution. Each $\ce{CaCl_2}$ unit separates into three ions. $\mathrm{0.2\: mol\: CaCl_2\times\dfrac{3\: mol\: ions}{1\: mol\: CaCl_2}\times\dfrac{0.51° C}{1\: mol\: ion}=0.306° C} \nonumber$ The normal boiling point of water is 100°C, so the boiling point of the solution is raised to 100.31°C. Exercise $2$ Estimate the freezing point of 0.3 M $\ce{CaCl_2}$ solution. Answer –1.7°C
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.09%3A_Properties_of_Solutions.txt
Learning Objectives • Describe osmosis and how it relates to osmotic pressure. Osmotic Pressure The last colligative property of solutions we will consider is a very important one for biological systems. It involves osmosis, the process by which solvent molecules can pass through certain membranes but solute particles cannot. When two solutions of different concentration are present on either side of these membranes (called semipermeable membranes), there is a tendency for solvent molecules to move from the more dilute solution to the more concentrated solution until the concentrations of the two solutions are equal. This tendency is called osmotic pressure. External pressure can be exerted on a solution to counter the flow of solvent; the pressure required to halt the osmosis of a solvent is equal to the osmotic pressure of the solution. Consider the apparatus illustrated in Figure $1$, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis. When osmosis is carried out in an apparatus like that shown in Figure $1$, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure ($\pi$) of the solution. The osmotic pressure of a dilute solution can be determined in a similar way the pressure of an ideal gas is calculated using the ideal gas law: $\pi=\left( \frac{n}{V} \right) RT \nonumber$ where $n$ is the number of moles of particles in solution, $V$ is the volume, and $R$ is the universal gas constant. Osmolarity (osmol) is a way of reporting the total number of particles in a solution to determine osmotic pressure. It is defined as the molarity of a solute times the number of particles a formula unit of the solute makes when it dissolves (represented by $i$): $osmol = M \times i \nonumber$ If more than one solute is present in a solution, the individual osmolarities are additive to get the total osmolarity of the solution. Solutions that have the same osmolarity have the same osmotic pressure. If solutions of differing osmolarities are present on opposite sides of a semipermeable membrane, solvent will transfer from the lower-osmolarity solution to the higher-osmolarity solution. Counterpressure exerted on the high-osmolarity solution will reduce or halt the solvent transfer. An even higher pressure can be exerted to force solvent from the high-osmolarity solution to the low-osmolarity solution, a process called reverse osmosis. Reverse osmosis is used to make potable water from saltwater where sources of fresh water are scarce. Example $1$ A 0.50 M NaCl aqueous solution and a 0.30 M Ca(NO3)2 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow. Solution The solvent will flow into the solution of higher osmolarity. The NaCl solute separates into two ions—Na+ and Cl—when it dissolves, so its osmolarity is as follows: osmol (NaCl) = 0.50 M × 2 = 1.0 osmol The Ca(NO3)2 solute separates into three ions—one Ca2+ and two NO3—when it dissolves, so its osmolarity is as follows: osmol [Ca(NO3)2] = 0.30 M × 3 = 0.90 osmol The osmolarity of the Ca(NO3)2 solution is lower than that of the NaCl solution, so water will transfer through the membrane from the Ca(NO3)2 solution to the NaCl solution. Exercise $1$ A 1.5 M C6H12O6 aqueous solution and a 0.40 M Al(NO3)3 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow. Answer osmol C6H12O6 = 1.5; osmol Al(NO3)3 = 1.6 The solvent flows from C6H12O6 solution (lower osmolarity) to $\ce{Al(NO3)3}$ solution (higher osmolarity). Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture (Figure $\PageIndex{2a}$). This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation (Figure $\PageIndex{2c}$). Only if red blood cells are placed in isotonic solutions that have the same osmolarity as exists inside the cells are they unaffected by negative effects of osmotic pressure (Figure $\PageIndex{2b}$). Glucose solutions of about 0.31 M, or sodium chloride (NaCl) solutions of about 0.16 M, are isotonic with blood plasma. Note: Isotonic Solutions for Red Blood Cells The concentration of an red blood cell isotonic solution made with sodium chloride (NaCl) is half that of an isotonic solution made with glucose (0.16 M and 0.31 M respectively). This is because NaCl produces two ions when a formula unit dissolves, while molecular glucose produces only one particle when a formula unit dissolves. The osmolarities are therefore the same even though the concentrations of the two solutions are different. isotonic NaCl solution: $osmol = 0.16\ M \times 2 = 0.32\ osmol/L \nonumber$ isotonic glucose solution: $osmol = 0.31\ M \times 1 = 0.31\ osmol/L \nonumber$ Osmotic pressure explains why you should not drink seawater if you are abandoned in a life raft in the middle of the ocean. Its osmolarity is about three times higher than most bodily fluids. You would actually become thirstier as water from your cells was drawn out to dilute the salty ocean water you ingested. Our bodies do a better job coping with hypotonic solutions than with hypertonic ones. The excess water is collected by our kidneys and excreted. Osmotic pressure effects are used in the food industry to make pickles from cucumbers and other vegetables and in brining meat to make corned beef. It is also a factor in the mechanism of getting water from the roots to the tops of trees!
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.10%3A_Osmosis_and_Osmotic_Pressure.txt
Learning Objectives • Understand the difference between dialysis and osmosis. As introduced in a previous section, osmosis is the process by which solvent particles can pass through a semipermeable membrane. Dialysis is similar to osmosis with the difference being that the pores in the semipermeable membranes are larger, thus allowing both solvent and small solute molecules to pass through. Larger solute particles, such as colloids and proteins, that cannot pass through the pores in the membrane are left behind. The size of the pores in the semipermeable membranes can be varied to allow for separation (or purification) of specific solute molecules from a solution based on their size. Biochemists and molecular biologists use dialysis to purify proteins from mixtures or to change the concentration of solute particles for different types of experiments, as illustrated in Figure \(1\). Role of dialysis in the human body Kidneys filter the blood by dialysis, which occurs in tubular structures called nephrons (Figure \(2\). The nephrons filter the water, and small molecules, like glucose, amino acids, urea, and ions from the blood. Useful products and most of the water reabsorb later on, but urea and other waste products are excreted through urine. (You can read more about kidney structure and function here.) The main function of the kidneys is to filter the blood to remove wastes and extra water, which are then expelled from the body as urine. Some diseases rob the kidneys of their ability to perform this function, causing a buildup of waste materials in the bloodstream. If a kidney transplant is not available or desirable, a procedure called dialysis can be used to remove waste materials and excess water from the blood. In one form of dialysis, called hemodialysis, (see Figure \(3\)), a patient’s blood is passed though a length of tubing that travels through an artificial kidney machine (also called a dialysis machine). In the dialyser, a section of tubing composed of cellophane (a semipermeable membrane) is immersed in dialysate, a solution of sterile water, glucose, amino acids, and certain electrolytes. The osmotic pressure of the blood forces waste molecules, such as urea, and excess water through the membrane into the dialysate. Red and white blood cells are too large to pass through the membrane, so they remain in the blood. After being cleansed in this way, the blood is returned to the body. Dialysis is a continuous process, as the osmosis of waste materials and excess water takes time. Typically, 5–10 lb of waste-containing fluid is removed in each dialysis session, which can last 2–8 hours and must be performed several times a week. Although some patients have been on dialysis for 30 or more years, dialysis is always a temporary solution because waste materials are constantly building up in the bloodstream. A more permanent solution is a kidney transplant. Career Connection: Dialysis Technician Dialysis is a medical process of removing wastes and excess water from the blood by diffusion and ultrafiltration. When kidney function fails, dialysis must be done to artificially rid the body of wastes. This is a vital process to keep patients alive. In some cases, the patients undergo artificial dialysis until they are eligible for a kidney transplant. In others who are not candidates for kidney transplants, dialysis is a life-long necessity. Dialysis technicians typically work in hospitals and clinics. While some roles in this field include equipment development and maintenance, most dialysis technicians work in direct patient care. Their on-the-job duties, which typically occur under the direct supervision of a registered nurse, focus on providing dialysis treatments. This can include reviewing patient history and current condition, assessing and responding to patient needs before and during treatment, and monitoring the dialysis process. Treatment may include taking and reporting a patient’s vital signs and preparing solutions and equipment to ensure accurate and sterile procedures.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/09%3A_Solutions/9.11%3A_Dialysis.txt
Learning Objectives • Identify an Arrhenius acid and an Arrhenius base. • Identify a Brønsted-Lowry acid and a Brønsted-Lowry base. • Identify conjugate acid-base pairs in an acid-base reaction. There are three major classifications of substances known as acids or bases. The theory developed by Svante Arrhenius in 1883, the Arrhenius definition, states that an acid produces hydrogen ions, $H^{+}$, in solution and a base produces hydroxide ions, $OH^{-}$. Later, two more sophisticated and general theories were proposed. These theories are the Brønsted-Lowry and Lewis definitions of acids and bases. This section will cover the Arrhenius and Brønsted-Lowry theories; Lewis theory is discussed elsewhere. The Arrhenius Theory of Acids and Bases In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. An Arrhenius acid is a compound that increases the concentration of $H^{+}$ ions that are present when added to water. These $H^{+}$ ions form the hydronium ion ($H_3O^{+}$) when they combine with water molecules. This process is represented in a chemical equation by adding H2O to the reactants side. $\ce{HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq) }$ In this reaction, hydrochloric acid ($HCl$) dissociates completely into hydrogen ($H^{+}$) and chloride (\Cl^{-}\)) ions when dissolved in water, thereby releasing $H^{+}$ ions into solution. Formation of the hydronium ion equation: $\ce{ HCl(aq) + H_2O(ℓ) \rightarrow H_3O^{+}(aq) + Cl^{-}(aq)}$ An Arrhenius base is a compound that increases the concentration of $OH^{-}$ ions that are present when added to water. The dissociation is represented by the following equation: $\ce{ NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq) }$ In this reaction, sodium hydroxide (NaOH) disassociates into sodium ($Na^+$) and hydroxide ($OH^-$) ions when dissolved in water, thereby releasing $OH^{-}$ ions into solution. Arrhenius acids are substances which produce hydrogen ions in solution and Arrhenius bases are substances which produce hydroxide ions in solution. Limitations to the Arrhenius Theory The Arrhenius theory has many more limitations than the other two theories. The theory does not explain the weak base ammonia ($NH_3$), which in the presence of water, releases hydroxide ions into solution, but does not contain $OH^{-}$ itself. The Arrhenius definition of acid and base is also limited to aqueous (i.e., water) solutions. The Brønsted-Lowry Theory of Acids and Bases In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products: $\ce{NH3(aq) + H2O(ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}$ What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows: Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense. Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules. With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O: $\ce{HCl(g) + H_2O(ℓ) \rightarrow H_3O^{+}(aq) Cl^{−}(aq) +H_3O^{+}(aq) } \label{Eq2}$ We can depict this process using Lewis electron dot diagrams: Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; the hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense, but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance. Note: Acid and Base Definitions • A Brønsted-Lowry acid is a proton (hydrogen ion) donor. • A Brønsted-Lowry base is a proton (hydrogen ion) acceptor. • All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases. Example $1$ Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule, just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base. Solution C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows: $\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber$ Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid. Exercise $1$ Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation. $\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}}$ Answer Brønsted-Lowry acid: H2PO4-; Brønsted-Lowry base: H2O Exercise $2$ Which of the following compounds is a Bronsted-Lowry base? 1. HCl 2. HPO42- 3. H3PO4 4. NH4+ 5. CH3NH3+ Answer A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates $\ce{HCl}$, $\ce{H3PO4}$, $\ce{NH4^{+}}$ and $\ce{CH_3NH_3^{+}}$ because they are Bronsted-Lowry acids. They all give away protons. In the case of $\ce{HPO4^{2-}}$, consider the following equation: $\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber$ Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation: $\ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber$ In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric. Since HPO42- is the only compound from the options that can act as a base, the answer is (b) HPO42-. Conjugate Acid-Base Pair In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$: In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$. Example $2$ Identify the conjugate acid-base pairs in this equilibrium. $\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber$ Solution Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid-base pairs: • the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and • the parent base and its conjugate acid ($H_3O^+/H_2O$). Example $3$ Identify the conjugate acid-base pairs in this equilibrium. $(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber$ Solution One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base. The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base. Exercise $3$ Identify the conjugate acid-base pairs in this equilibrium. $\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber$ Answer H2O (acid) and OH (base); NH2 (base) and NH3 (acid) Contributions & Attributions This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: • Henry Agnew (UC Davis)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.01%3A_Acids_and_Bases_Definitions.txt
Learning Objectives • Define a strong and a weak acid and base. • Recognize an acid or a base as strong or weak. Strong and Weak Acids Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way. Consider $\ce{HCl(aq)}$. When $\ce{HCl}$ is dissolved in $\ce{H2O}$, it completely dissociates (separates) into H+(aq) and Cl(aq) ions; all the HCl molecules become ions: $\ce {HCl\ \overset{100\%}{\rightarrow}H^{+}(aq) + Cl^{-}(aq)}$ Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. HC2H3O2 is an example of a weak acid: $HC_{2}H_{3}O_{2}\ \overset{\sim 5\%}{\longrightarrow}H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq)$ Because this reaction does not go 100% to completion, it is more appropriate to write it as a reversible reaction: $HC_{2}H_{3}O_{2}\rightleftharpoons H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq)$ As it turns out, there are very few strong acids, which are given in Table $1$. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid. Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. Figure $1$: Some of the common strong acids and bases are listed here. Strong Acids Strong Bases $\ce{HClO4}$ perchloric acid $\ce{LiOH}$ lithium hydroxide $\ce{HCl}$ hydrochloric acid $\ce{NaOH}$ sodium hydroxide $\ce{HBr}$ hydrobromic acid $\ce{KOH}$ potassium hydroxide $\ce{HI}$ hydroiodic acid $\ce{Ca(OH)2}$ calcium hydroxide $\ce{HNO3}$ nitric acid $\ce{Sr(OH)2}$ strontium hydroxide $\ce{H2SO4}$ sulfuric acid $\ce{Ba(OH)2}$ barium hydroxide Strong and Weak Bases The issue is similar with bases: a strong base is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a weak base. There are very few strong bases (Table $1$); any base not listed is a weak base. All strong bases are OH compounds. So a base based on some other mechanism, such as NH3 (which does not contain OH ions as part of its formula), will be a weak base. Example $1$: Identifying Strong and Weak Acids and Bases Identify each acid or base as strong or weak. 1. HCl 2. Mg(OH)2 3. C5H5N Solution 1. Because HCl is listed in Table $1$, it is a strong acid. 2. Because Mg(OH)2 is listed in Table $1$, it is a strong base. 3. The nitrogen in C5H5N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base. Exercise $1$ Identify each acid or base as strong or weak. 1. $\ce{RbOH}$ 2. $\ce{HNO_2}$ Answer 1. strong base 2. weak acid Example $2$: Characterizing Base Ionization Write the balanced chemical equation for the dissociation of Ca(OH)2 and indicate whether it proceeds 100% to products or not. Solution This is an ionic compound of Ca2+ ions and OH ions. When an ionic compound dissolves, it separates into its constituent ions: $\ce{Ca(OH)2 → Ca^{2+}(aq) + 2OH^{−}(aq)} \nonumber$ Because Ca(OH)2 is listed in Table $1$, this reaction proceeds 100% to products. Exercise $2$ Write the balanced chemical equation for the dissociation of hydrazoic acid (HN3) and indicate whether it proceeds 100% to products or not. Answer a The reaction is as follows: $\ce{HN3 → H^{+}(aq) + N3^{−}(aq)} \nonumber$ It does not proceed 100% to products because hydrazoic acid is not a strong acid. Looking Closer: Household Acids and Bases Many household products are acids or bases. For example, the owner of a swimming pool may use muriatic acid to clean the pool. Muriatic acid is another name for hydrochloric acid [$\ce{HCl(aq)}$]. Vinegar has already been mentioned as a dilute solution of acetic acid [$\ce{HC2H3O2(aq)}$]. In a medicine chest, one may find a bottle of vitamin C tablets; the chemical name of vitamin C is ascorbic acid ($\ce{HC6H7O6}$). One of the more familiar household bases is ammonia ($\ce{NH3}$), which is found in numerous cleaning products. As we mentioned previously, ammonia is a base because it increases the hydroxide ion concentration by reacting with water: $\ce{NH3(aq) + H2O(ℓ) \rightarrow NH^{+}4(aq) + OH^{−}(aq)} \label{Eq3}$ Many soaps are also slightly basic because they contain compounds that act as Brønsted-Lowry bases, accepting protons from water and forming excess hydroxide ions. This is one reason that soap solutions are slippery. Perhaps the most dangerous household chemical is the lye-based drain cleaner. Lye is a common name for sodium hydroxide, although it is also used as a synonym for potassium hydroxide. Lye is an extremely caustic chemical that can react with grease, hair, food particles, and other substances that may build up and form a clog in a pipe. Unfortunately, lye can also attack tissues and other substances in our bodies. Thus, when we use lye-based drain cleaners, we must be very careful not to touch any of the solid drain cleaner or spill the water it was poured into. Safer, nonlye drain cleaners use peroxide compounds to react on the materials in the clog and clear the drain. Chemical Equilibrium in Weak Acids and Bases Ionization of weak acids or bases are reversible reactions, which means the forward and reverse reactions occur and eventually reach equilbrium. For example, the ionization of the weak acid $\ce{HC2H3O2(aq)}$ is as follows: $\ce{HC2H3O2(aq) + H2O(ℓ) \rightarrow H3O^{+}(aq) + C2H3O^{−}2(aq)} \label{Eq4}$ The reverse process also begins to occur: $\ce{H3O^{+}(aq) + C2H3O^{−}2(aq) \rightarrow HC2H3O2(aq) + H2O(ℓ)} \label{Eq5}$ Eventually, there is a balance between the two opposing processes, and no additional change occurs. The chemical reaction is better represented at this point with a double arrow: $\ce{HC2H3O2(aq) + H2O(ℓ) <=> H3O^{+}(aq) + C2H3O^{-}2(aq)} \label{Eq6}$ The $\rightleftharpoons$ implies that both the forward and reverse reactions are occurring, and their effects cancel each other out. A process at this point is considered to be at chemical equilibrium (or equilibrium). It is important to note that the processes do not stop. They balance out each other so that there is no further net change; that is, chemical equilibrium is a dynamic equilibrium. Example $3$: Partial Ionization Write the equilibrium chemical equation for the partial ionization of each weak acid or base. 1. HNO2(aq) 2. C5H5N(aq) Solution 1. HNO2(aq) + H2O(ℓ) ⇆ NO2(aq) + H3O+(aq) 2. C5H5N(aq) + H2O(ℓ) ⇆ C5H5NH+(aq) + OH(aq) Exercise $3$ Write the equilibrium chemical equation for the partial ionization of each weak acid or base. 1. $HF_{(aq)}$ 2. $AgOH_{(aq)}$ 3. CH3NH2(aq) Answer a. HF(aq) + H2O(ℓ) ⇆ F(aq) + H3O+(aq) b. AgOH(aq) ⇆ Ag+(aq) + OH(aq) c. CH3NH2(aq) + H2O(ℓ) ⇆ CH3NH3+(aq) + OH(aq) Strengths of Conjugate Acid and Base Pairs The extent to which an acid, $\ce{HA}$, donates protons to water molecules depends on the strength of the conjugate base, $\ce{A^{−}}$, of the acid. If $\ce{A^{−}}$ is a strong base, any protons that are donated to water molecules are recaptured by $\ce{A^{−}}$. Thus there is relatively little $\ce{A^{−}}$ and $\ce{H3O+}$ in solution, and the acid, $\ce{HA}$, is weak. If $\ce{A^{−}}$ is a weak base, water binds the protons more strongly, and the solution contains primarily $\ce{A^{−}}$ and $\ce{H3O^{+}}$—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure $2$). The first six acids in Figure $2$ are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. Those acids that lie between the hydronium ion and water in Figure $2$ form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and non-ionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure $2$ exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid. The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure $2$. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. Key Takeaways • Strong acids and bases are 100% ionized in aqueous solution. • Weak acids and bases are less than 100% ionized in aqueous solution.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.02%3A_Acid_and_Base_Strength.txt
Learning Objectives • Write the acid dissociation constant ($K_a$) expression. • Determine the relative strength of an acid using the ($K_a$) value. The ionization for a general weak acid, $\ce{HA}$, can be written as follows: $\ce{HA} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right) \nonumber$ Because the acid is weak, an equilibrium expression can be written. An acid ionization constant $\left( K_\text{a} \right)$ is the equilibrium constant for the ionization of an acid. $K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \nonumber$ The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of $K_\text{a}$ is a reflection of the strength of the acid. Weak acids with relatively higher $K_\text{a}$ values are stronger than acids with relatively lower $K_\text{a}$ values. Because strong acids are essentially $100\%$ ionized, the concentration of the acid in the denominator is nearly zero and the $K_\text{a}$ value approaches infinity. For this reason, $K_\text{a}$ values are generally reported for weak acids only. The table below is a listing of acid ionization constants for several acids. Note that polyprotic acids have a distinct ionization constant for each ionization step, with each successive ionization constant being smaller than the previous one. Name of Acid Ionization Equation $K_\text{a}$ Table $1$: Acid Ionization Constants at $25^\text{o} \text{C}$ Sulfuric acid $\ce{H_2SO_4} \rightleftharpoons \ce{H^+} + \ce{HSO_4^-}$ $\ce{HSO_4} \rightleftharpoons \ce{H^+} + \ce{SO_4^{2-}}$ very large $1.3 \times 10^{-2}$ Hydrofluoric acid $\ce{HF} \rightleftharpoons \ce{H^+} + \ce{F^-}$ $7.1 \times 10^{-4}$ Nitrous acid $\ce{HNO_2} \rightleftharpoons \ce{H^+} + \ce{NO_2^-}$ $4.5 \times 10^{-4}$ Benzoic acid $\ce{C_6H_5COOH} \rightleftharpoons \ce{H^+} + \ce{C_6H_5COO^-}$ $6.5 \times 10^{-5}$ Acetic acid $\ce{CH_3COOH} \rightleftharpoons \ce{H^+} + \ce{CH_3COO^-}$ $1.8 \times 10^{-5}$ Carbonic acid $\ce{H_2CO_3} \rightleftharpoons \ce{H^+} + \ce{HCO_3^-}$ $\ce{HCO_3^-} \rightleftharpoons \ce{H^+} + \ce{CO_3^{2-}}$ $4.2 \times 10^{-7}$ $4.8 \times 10^{-11}$ Hydrofluoric acid $HF_{(aq)}$ reacts directly with glass (very few chemicals react with glass). Hydrofluoric acid is used in glass etching. 10.04: Water as Both an Acid and a Base Learning Objectives • To write chemical equations for water acting as an acid and as a base. • Define and use the ion product constant for water, $K_w$, to calculate concentrations of $\ce{H3O+}$ and $\ce{OH-}$ in aqueous solutions. Water (H2O) is an interesting compound in many respects. Here, we will consider its ability to behave as an acid or a base. In some circumstances, a water molecule will accept a proton and thus act as a Brønsted-Lowry base. We saw an example in the dissolving of HCl in H2O: $HCl + H_2O(ℓ) \rightarrow H_3O^+(aq)+ Cl^−(aq) \label{Eq1}$ In other circumstances, a water molecule can donate a proton and thus act as a Brønsted-Lowry acid. For example, in the presence of the amide ion, a water molecule donates a proton, making ammonia as a product: $H_2O(ℓ) + NH^−_2(aq) \rightarrow OH^−(aq) + NH_3(aq) \label{Eq2}$ In this case, $NH^−_2$ is a Brønsted-Lowry base (the proton acceptor). So, depending on the circumstances, H2O can act as either a Brønsted-Lowry acid or a Brønsted-Lowry base. Water is not the only substance that can react as an acid in some cases or a base in others, but it is certainly the most common example—and the most important one. A substance that can either donate or accept a proton, depending on the circumstances, is called an amphiprotic compound. A water molecule can act as an acid or a base even in a sample of pure water. About 6 in every 100 million (6 in 108) water molecules undergo the following reaction: $H_2O(ℓ) + H_2O(ℓ) \rightarrow H_3O^+(aq) + OH^−(aq) \label{Eq3}$ This process is called the autoionization of water (Figure $1$) and occurs in every sample of water, whether it is pure or part of a solution. Autoionization occurs to some extent in any amphiprotic liquid. (For comparison, liquid ammonia undergoes autoionization as well, but only about 1 molecule in a million billion (1 in 1015) reacts with another ammonia molecule.) Example $1$ Identify water as either a Brønsted-Lowry acid or a Brønsted-Lowry base. 1. H2O(ℓ) + NO2(aq) → HNO2(aq) + OH(aq) 2. HC2H3O2(aq) + H2O(ℓ) → H3O+(aq) + C2H3O2(aq) Solution 1. In this reaction, the water molecule donates a proton to the NO2 ion, making OH(aq). As the proton donor, H2O acts as a Brønsted-Lowry acid. 2. In this reaction, the water molecule accepts a proton from HC2H3O2, becoming H3O+(aq). As the proton acceptor, H2O is a Brønsted-Lowry base. Exercise $2$ Identify water as either a Brønsted-Lowry acid or a Brønsted-Lowry base. 1. HCOOH(aq) + H2O(ℓ) → H3O+(aq) + HCOO(aq) 2. H2O(ℓ) + PO43(aq) → OH(aq) + HPO42(aq) Answer 1. H2O acts as the proton acceptor (Brønsted-Lowry base) 2. H2O acts as the proton donor (Brønsted-Lowry acid) Dissociation of Water As we have already seen, H2O can act as an acid or a base. Within any given sample of water, some $\ce{H2O}$ molecules are acting as acids, and other $\ce{H2O}$ molecules are acting as bases. The chemical equation is as follows: $\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto}$ Similar to a weak acid, the autoionization of water is an equilibrium process, and is more properly written as follows: $\ce{H2O(ℓ) + H2O(ℓ) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{Eq7}$ We often use the simplified form of the reaction: $\ce{H2O(l) <=> H+(aq) + OH−(aq)} \nonumber$ The equilibrium constant for the autoionization of water is referred to as the ion-product for water and is given the symbol $K_w$. $K_w = [\ce{H^{+}}][\ce{OH^{-}}] \nonumber$ The ion-product constant for water ($K_w$) is the mathematical product of the concentration of hydrogen ions and hydroxide ions. Note that H2O is not included in the ion-product expression because it is a pure liquid. The value of $K_w$ is very small, in accordance with a reaction that favors the reactants. At 25oC, the experimentally determined value of $K_w$ in pure water is 1.0×10−14. $K_w = [\ce{H^{+}}][\ce{OH^{−}}] = 1.0 \times 10^{−14} \nonumber$ In a sample of pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Pure water or any other aqueous solution in which this ratio holds is said to be neutral. To find the molarity of each ion, the square root of $K_w$ is taken. [H+] = [OH] = 1.0×10−7 The product of these two concentrations is 1.0×10−14 $\color{red}{\ce{[H^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \nonumber$ • For acids, the concentration of H+ or [H+] is greater than 1.0×10−7 M • For bases, the concentration of OH or [OH] is greater than 1.0×10−7 M. Aqueous HCl is an example of acidic solution. Hydrogen chloride (HCl) ionizes to produce H+ and Cl ions upon dissolving in water. This increases the concentration of H+ ions in the solution. According to Le Chatelier's principle, the equilibrium represented by $\ce{H2O(l) <=> H^{+}(aq) + OH^{−}(aq)} \nonumber$ $\ce{HCl(g) -> H^{+}(aq) + Cl^{−}(aq)} \nonumber$ is forced to the left, towards the reactant. As a result, the concentration of the hydroxide ion decreases. Now, consider KOH (aq), a basic solution. Solid potassium hydroxide (KOH) dissociates in water to yield potassium ions and hydroxide ions. KOH(s) → K+(aq) + OH(aq) The increase in concentration of the OH ions will cause a decrease in the concentration of the H+ ions. No matter whether the aqueous solution is an acid, a base, or neutral:and the ion-product of [H+][OH] remains constant. • For acidic solutions, [H+] is greater than [OH]. • For basic solutions, [OH−] is greater than [H+]. • For neutral solutions, [H3O+] = [OH−] = 1.0×10−7M This means that if you know $\ce{[H^{+}]}$ for a solution, you can calculate what $\ce{[OH^{−}]}$) has to be for the product to equal $1.0 \times 10^{−14}$, or if you know $\ce{[OH^{−}]}$), you can calculate $\ce{[H^{+}]}$. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of $K_w$. $K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10}$ Example $2$ Hydrochloric acid (HCl) is a strong acid, meaning it is 100% ionized in solution. What is the [H+] and the [OH] in a solution of 2.0×10−3 M HCl? Solution Step 1: List the known values and plan the problem. Known • [HCl] = 2.0×10−3 M • Kw = 1.0×10−14 Unknown • [H+]=?M • [OH]=?M Because HCl is 100% ionized, the concentration of H+ ions in solution will be equal to the original concentration of HCl. Each HCl molecule that was originally present ionizes into one H+ ion and one Cl− ion. The concentration of OH− can then be determined from the [H+] and Kw. Step 2: Solve. [H+]=2.0×10−3 M Kw = [H+][OH] = 1.0×10−14 [OH] = Kw/[H+] = 1.0×10−14/2.0×10−3 = 5.0×10−12 M Step 3: Think about your result. The [H+] is much higher than the [OH] because the solution is acidic. As with other equilibrium constants, the unit for Kw is customarily omitted. Exercise $2$ Sodium hydroxide (NaOH) is a strong base. What is the [H+] and the [OH] in a 0.001 M NaOH solution at 25 °C? Answer [OH] = 0.001M or 1 x 10-3M; [H+]=1×1011M. Concept Review Exercises 1. Explain how water can act as an acid. 2. Explain how water can act as a base. Answers 1. Under the right conditions, H2O can donate a proton, making it a Brønsted-Lowry acid. 2. Under the right conditions, H2O can accept a proton, making it a Brønsted-Lowry base. Key Takeaway • Water molecules can act as both an acid and a base, depending on the conditions. Exercises 1. Is H2O(ℓ) acting as an acid or a base? H2O(ℓ) + NH4+(aq) → H3O+(aq) + NH3(aq) 2. Is H2O(ℓ) acting as an acid or a base? CH3(aq) + H2O(ℓ) → CH4(aq) + OH(aq) 3. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some C2H3O2 solutions, the following reaction can occur: C2H3O2(aq) + H2O(ℓ) → HC2H3O2(aq) + OH(aq) Is H2O acting as an acid or a base in this reaction? 4. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some NH4+ solutions, the following reaction can occur: NH4+(aq) + H2O → NH3(aq) + H3O+(aq) Is H2O acting as an acid or a base in this reaction? 5. Why is pure water considered neutral? Answers 1. base 2. acid 1. acid 2. base 5. When water ionizes, equal amounts of H+ (acid) and OH(base) are formed, so the solution is neither acidic nor basic: H2O(ℓ) → H+(aq) + OH(aq) [SIDE NOTE: It is rare to truly have pure water. Water exposed to air will usually be slightly acidic because dissolved carbon dioxide gas, or carbonic acid, decreases the pH slightly below 7. Alternatively, dissolved minerals, like calcium carbonate (limestone), can make water slightly basic.]
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.03%3A_Acid_Dissociation_Constants.txt
Learning Objectives • Define the pH scale and use it to describe acids and bases. • Calculate the pH of a solution from $[H_3O^+]$ and $[OH^-]$. Knowing the amount of acid and base in solutions is extremely important for a wide variety of applications ranging from brewing beer or wine to studying the effects of ocean acidification to health and medicine. Scientists are good at calculating and measuring the concentration of hydronium in a solution, however, there is a more convenient way to make comparisons between solutions – the pH scale. The pH Scale One qualitative measure of the strength of an acid or a base solution is the pH scale, which is based on the concentration of the hydronium (or hydrogen) ion in aqueous solution. $pH = -\log[H^+] \nonumber$ or $pH = -\log[H_3O^+] \nonumber$ Figure $1$ illustrates this relationship, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example $1.3 \times 10^{-3}\,M$), the log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of $[H_3O^+]$, which will give a positive value for pH. A neutral (neither acidic nor basic) solution has a pH of 7. A pH below 7 means that a solution is acidic, with lower values of pH corresponding to increasingly acidic solutions. A pH greater than 7 indicates a basic solution, with higher values of pH corresponding to increasingly basic solutions. Thus, given the pH of several solutions, you can state which ones are acidic, which ones are basic, and which are more acidic or basic than others. These are summarized in Table $1. Table \(1$: Acidic, Basic and Neutral pH Values Classification Relative Ion Concentrations pH at 25 °C acidic [H+] > [OH] pH < 7 neutral [H+] = [OH] pH = 7 basic [H+] < [OH] pH > 7 Example $1$ Find the pH, given the $[H^+]$ of the following: 1. 1 ×10-3 M 2. 2.5 ×10-11 M 3. 4.7 ×10-9 M Solution pH = - log [H3O+] Substitute the known quantity into the equation and solve. Use a scientific calculator for b and c. 1. pH = - log [1 × 10−3 ] = 3.0 (1 decimal place since 1 has 1 significant figure) 2. pH = - log [2.5 ×10-11] = 10.60 (2 decimal places since 2.5 has 2 significant figures) 3. pH = - log [4.7 ×10-9] = 8.33 (2 decimal places since 4.7 has 2 significant figures) Note on significant figures: Because the number(s) before the decimal point in the pH value relate to the power on 10, the number of digits after the decimal point (underlined) is what determines the number of significant figures in the final answer. Exercise $1$ Find the pH, given [H+] of the following: 1. 5.8 ×10-4 M 2. 1.0×10-7 M Answer a. 3.24 b. 7.00 Figure $2$ lists the pH of several common solutions. The most acidic among the listed solutions is 1 M HCl with the lowest pH value (0.0): battery acid is the next most acidic solution with a pH value of 0.3. The most basic is 1M NaOH solution with the highest pH value of 14.0. Notice that some biological fluids (stomach acid and urine) are nowhere near neutral. You may also notice that many food products are slightly acidic. They are acidic because they contain solutions of weak acids. If the acid components of these foods were strong acids, the food would likely be inedible. Example $2$ Label each solution as acidic, basic, or neutral based only on the stated $pH$. 1. milk of magnesia, pH = 10.5 2. pure water, pH = 7 3. wine, pH = 3.0 Solution 1. With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)2.) 2. Pure water, with a pH of 7, is neutral. 3. With a pH of less than 7, wine is acidic. Exercise $2$ Identify each substance as acidic, basic, or neutral based only on the stated $pH$. 1. human blood with $pH$ = 7.4 2. household ammonia with $pH$ = 11.0 3. cherries with $pH$ = 3.6 Answer a. slightly basic b. basic c. acidic Measuring pH Tools have been developed that make the measurement of pH simple and convenient. For example, pH paper (Figure $3$)consists of strips of paper impregnated with one or more acid-base indicators, which are intensely colored organic molecules whose colors change dramatically depending on the pH of the solution. Placing a drop of a solution on a strip of pH paper and comparing its color with standards give the solution’s approximate pH. A more accurate tool, the pH meter, uses a glass electrode, a device whose voltage depends on the H+ ion concentration (Figure $3$). Acid Rain Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid: $\ce{H2O (l) + CO2(g) ⟶ H2CO3(aq)} \label{14}$ $\ce{H2CO3(aq) \rightleftharpoons H^+(aq) + HCO3^- (aq)} \label{15}$ Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: $\ce{H2O (l) + SO3(g) ⟶ H2SO4(aq)} \label{16}$ $\ce{H2SO4(aq) ⟶ H^+(aq) + HSO4^- (aq)} \label{17}$ Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $4$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. Figure $4$: (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.05%3A_Measuring_Acidity_in_Aqueous_Solutions-_The_pH_Scale.txt
Learning Objectives • Calculate pH from $[H_3O^+]$ and $[H_3O^+]$ from pH. Calculating pH from Hydronium Concentration The pH of solutions can be determined by using logarithms as illustrated in the next example for stomach acid. Stomach acid is a solution of $HCl$ with a hydronium ion concentration of $1.2 \times 10^{−3}\; M$, what is the $pH$ of the solution? \begin{align} \mathrm{pH} &= \mathrm{-\log [H_3O^+]} \nonumber \ &=-\log(1.2 \times 10^{−3}) \nonumber \ &=−(−2.92)=2.92 \nonumber \end{align} Logarithms To get the log value on your calculator, enter the number (in this case, the hydronium ion concentration) first, then press the LOG key. If the number is 1.0 x 10-5 (for [H3O+] = 1.0 x 10-5 M) you should get an answer of "-5". If you get a different answer, or an error, try pressing the LOG key before you enter the number. Example $2$: Converting Ph to Hydronium Concentration Find the pH, given the $[H_3O^+]$ of the following: 1. 1 ×10-3 M 2. 2.5 ×10-11 M 3. 4.7 ×10-9 M Solution Steps for Problem Solving Identify the "given" information and what the problem is asking you to "find." Given: 1. [H3O+] =1 × 10−3 M 2. [H3O+] =2.5 ×10-11 M 3. [H3O+] = 4.7 ×10-9 M Find: ? pH Plan the problem. Need to use the expression for pH (Equation \ref{pH}). pH = - log [H3O+] Calculate. Now substitute the known quantity into the equation and solve. 1. pH = - log [1 × 10−3 ] = 3.0 (1 decimal places since 1 has 1 significant figure) 2. pH = - log [2.5 ×10-11] = 10.60 (2 decimal places since 2.5 has 2 significant figures) 3. pH = - log [4.7 ×10-9] = 8.30 (2 decimal places since 4.7 has 2 significant figures) The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer: Exercise $2$ Find the pH, given [H3O+] of the following: 1. 5.8 ×10-4 M 2. 1.0×10-7 Answer a 3.22 Answer b 7.00 Calculating Hydronium Concentration from pH Sometimes you need to work "backwards"—you know the pH of a solution and need to find $[H_3O^+]$, or even the concentration of the acid solution. How do you do that? To convert pH into $[H_3O^+]$ we solve Equation \ref{pH} for $[H_3O^+]$. This involves taking the antilog (or inverse log) of the negative value of pH . $[\ce{H3O^{+}}] = \text{antilog} (-pH)$ or $[\ce{H_3O^+}] = 10^{-pH} \label{ph1}$ As mentioned above, different calculators work slightly differently—make sure you can do the following calculations using your calculator. Calculator Skills We have a solution with a pH = 8.3. What is [H3O+] ? With some calculators you will do things in the following order: 1. Enter 8.3 as a negative number (use the key with both the +/- signs, not the subtraction key). 2. Use your calculator's 2nd or Shift or INV (inverse) key to type in the symbol found above the LOG key. The shifted function should be 10x. 3. You should get the answer 5.0 × 10-9. Other calculators require you to enter keys in the order they appear in the equation. 1. Use the Shift or second function to key in the 10x function. 2. Use the +/- key to type in a negative number, then type in 8.3. 3. You should get the answer 5.0 × 10-9. If neither of these methods work, try rearranging the order in which you type in the keys. Don't give up—you must master your calculator! Example $3$: Calculating Hydronium Concentration from pH Find the hydronium ion concentration in a solution with a pH of 12.6. Is this solution an acid or a base? How do you know? Solution Steps for Problem Solving Identify the "given" information and what the problem is asking you to "find." Given: pH = 12.6 Find: [H3O+] = ? M Plan the problem. Need to use the expression for [H3O+] (Equation \ref{ph1}). [H3O+] = antilog (-pH) or [H3O+] = 10-pH Calculate. Now substitute the known quantity into the equation and solve. [H3O+] = antilog (12.60) = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places) or [H3O+] = 10-12.60 = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places) The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer: Exercise $3$ If moist soil has a pH of 7.84, what is [H3O+] of the soil solution? Answer 1.5 x 10-8 M
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.06%3A_Working_with_pH.txt
Learning Objectives • Identify a salt solution as acidic, basic, or neutral. A salt is an ionic compound that is formed when an acid and a base neutralize each other. While it may seem that salt solutions are always neutral, they can frequently be either acidic or basic. Consider the salt formed when the weak acid hydrofluoric acid is neutralized by the strong base sodium hydroxide. The molecular and net ionic equations are shown below. \begin{align*} &\ce{HF} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaF} \left( aq \right) + \ce{H_2O} \left( l \right) \ &\ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \end{align*}\nonumber Since sodium fluoride is soluble, the sodium ion is a spectator ion in the neutralization reaction. The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton. $\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$ The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic. Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution. Salts That Form Basic Solutions When solid sodium fluoride is dissolved into water, it completely dissociates into sodium ions and fluoride ions. The sodium ions do not have any capability of hydrolyzing, but the fluoride ions hydrolyze to produce a small amount of hydrofluoric acid and hydroxide ion. $\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber$ Salts that are derived from the neutralization of a weak acid $\left( \ce{HF} \right)$ by a strong base $\left( \ce{NaOH} \right)$ will always produce salt solutions that are basic. Salts That Form Acidic Solutions Ammonium chloride $\left( \ce{NH_4Cl} \right)$ is a salt that is formed when the strong acid $\ce{HCl}$ is neutralized by the weak base $\ce{NH_3}$. Ammonium chloride is soluble in water. The chloride ion produced is incapable of hydrolyzing because it is the conjugate base of the strong acid $\ce{HCl}$. In other words, the $\ce{Cl^-}$ ion cannot accept a proton from water to form $\ce{HCl}$ and $\ce{OH^-}$, as the fluoride ion did in the previous section. However, the ammonium ion is capable of reacting slightly with water, donating a proton and so acting as an acid. $\ce{NH_4^+} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{NH_3} \left( aq \right)\nonumber$ Salts That Form Neutral Solutions A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7. An example is sodium chloride, formed from the neutralization of $\ce{HCl}$ by $\ce{NaOH}$. A solution of $\ce{NaCl}$ in water has no acidic or basic properties, since neither ion is capable of hydrolyzing. Other salts that form neutral solutions include potassium nitrate $\left( \ce{KNO_3} \right)$ and lithium bromide $\left( \ce{LiBr} \right)$. The table below summarizes how to determine the acidity or basicity of a salt solution. Table $1$ Salt formed from: Salt Solution Strong acid $+$ Strong base Neutral Strong acid $+$ Weak base Acidic Weak acid $+$ Strong base Basic Salts formed from the reaction of a weak acid and a weak base are more difficult to analyze due to competing hydrolysis reactions between the cation and the anion. These salts are not considered in this chapter's concept. Summary • Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution. • Salts that are derived from the neutralization of a weak acid by a strong base will always produce salt solutions that are basic. • Salts that are derived from the neutralization of a strong acid by a weak base will always produce salt solutions that are acidic. • A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.09%3A_Acidity_and_Basicity_of_Salt_Solutions.txt
Learning Objectives • Describe the composition and function of acid–base buffers • Calculate the pH of a buffer before and after the addition of added acid or base A mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure $1$). A solution of acetic acid ($\ce{CH3COOH}$ and sodium acetate $\ce{CH3COONa}$) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia ($\ce{NH3(aq)}$) and ammonium chloride ($\ce{NH4Cl(aq)}$). How Buffers Work A mixture of acetic acid and sodium acetate is acidic because the Ka of acetic acid is greater than the Kb of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. Then more of the acetic acid reacts with water, restoring the hydronium ion concentration almost to its original value: $\ce{CH3CO2H}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$ The pH changes very little. If we add an acid such as hydrochloric acid, most of the hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid molecules: $\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)⟶\ce{CH3CO2H}(aq)+\ce{H2O}(l) \nonumber$ Thus, there is very little increase in the concentration of the hydronium ion, and the pH remains practically unchanged (Figure $2$). A mixture of ammonia and ammonium chloride is basic because the Kb for ammonia is greater than the Ka for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value: $\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l) \nonumber$ If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value: $\ce{H3O+}(aq)+\ce{NH3}(aq)⟶\ce{NH4+}(aq)+\ce{H2O}(l) \nonumber$ The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid. Example $1$: pH Changes in Buffered and Unbuffered Solutions Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds. 1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate. 2. Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL. Solution 1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate. To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples): 1. Determine the direction of change. The equilibrium in a mixture of H3O+, $\ce{CH3CO2-}$, and CH3CO2H is: $\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$ The equilibrium constant for CH3CO2H is not given, so we look it up in Table E1: Ka = 1.8 × 10−5. With [CH3CO2H] = $\ce{[CH3CO2- ]}$ = 0.10 M and [H3O+] = ~0 M, the reaction shifts to the right to form H3O+. 2. Determine x and equilibrium concentrations. A table of changes and concentrations follows: • Solve for x and the equilibrium concentrations. We find: $x=1.8×10^{−5}\:M \nonumber$ • and $\ce{[H3O+]}=0+x=1.8×10^{−5}\:M \nonumber$ Thus: $\mathrm{pH=−log[H_3O^+]=−log(1.8×10^{−5})} \nonumber$ $=4.74 \nonumber$ 4. Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = Ka. (b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL. First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium: 1. Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains: $\mathrm{0.0010\cancel{L}×\left(\dfrac{0.10\:mol\: NaOH}{1\cancel{L}}\right)=1.0×10^{−4}\:mol\: NaOH} \nonumber$ 2. Determine the moles of CH2CO2H. Before reaction, 0.100 L of the buffer solution contains: $\mathrm{0.100\cancel{L}×\left(\dfrac{0.100\:mol\:CH_3CO_2H}{1\cancel{L}}\right)=1.00×10^{−2}\:mol\:CH_3CO_2H} \nonumber$ 3. Solve for the amount of NaCH3CO2 produced. The 1.0 × 10−4 mol of NaOH neutralizes 1.0 × 10−4 mol of CH3CO2H, leaving: $\mathrm{(1.0×10^{−2})−(0.01×10^{−2})=0.99×10^{−2}\:mol\:CH_3CO_2H} \nonumber$ and producing 1.0 × 10−4 mol of NaCH3CO2. This makes a total of: [\mathrm{(1.0×10^{−2})+(0.01×10^{−2})=1.01×10^{−2}\:mol\:NaCH_3CO_2} \nonumber \] 4. Find the molarity of the products. After reaction, CH3CO2H and NaCH3CO2 are contained in 101 mL of the intermediate solution, so: $\ce{[CH3CO2H]}=\mathrm{\dfrac{9.9×10^{−3}\:mol}{0.101\:L}}=0.098\:M \nonumber$ $\ce{[NaCH3CO2]}=\mathrm{\dfrac{1.01×10^{−2}\:mol}{0.101\:L}}=0.100\:M \nonumber$ Now we calculate the pH after the intermediate solution, which is 0.098 M in CH3CO2H and 0.100 M in NaCH3CO2, comes to equilibrium. The calculation is very similar to that in part (a) of this example: This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution. (c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 × 10−5-M solution of HCl). The volume of the final solution is 101 mL. Solution This 1.8 × 10−5-M solution of HCl has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains: $\mathrm{0.100\:L×\left(\dfrac{1.8×10^{−5}\:mol\: HCl}{1\:L}\right)=1.8×10^{−6}\:mol\: HCl}$ As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 × 10−4 mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is: $(1.0×10^{−4})−(1.8×10^{−6})=9.8×10^{−5}\:M$ The concentration of NaOH is: $\dfrac{9.8×10^{−5}\:M\:\ce{NaOH}}{0.101\:\ce{L}}=9.7×10^{−4}\:M$ The pOH of this solution is: $\mathrm{pOH=−log[OH^- ]=−log(9.7×10^{−4})=3.01}$ The pH is: $\mathrm{pH=14.00−pOH=10.99}$ The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b). Exercise $1$ Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 × 10−5 M HCl solution from 4.74 to 3.00. Answer Initial pH of 1.8 × 10−5 M HCl; pH = −log[H3O+] = −log[1.8 × 10−5] = 4.74 Moles of H3O+ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L × 0.0010 L = 1.0 × 10−4 moles; final pH after addition of 1.0 mL of 0.10 M HCl: $\mathrm{pH=−log[H_3O^+]=−log\left(\dfrac{total\: moles\:H_3O^+}{total\: volume}\right)=−log\left(\dfrac{1.0×10^{−4}\:mol+1.8×10^{−6}\:mol}{101\:mL\left(\dfrac{1\:L}{1000\:mL}\right)}\right)=3.00} \nonumber$ Buffer Capacity Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure $3$). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion. The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion. Selection of Suitable Buffer Mixtures There are two useful rules of thumb for selecting buffer mixtures: 1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure $4$ shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration. 1. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7. Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H2CO3, and the bicarbonate ion, $\ce{HCO3-}$. When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction: $\ce{H3O+}(aq)+\ce{HCO3-}(aq)⟶\ce{H2CO3}(aq)+\ce{H2O}(l) \nonumber$ When an excess of the hydroxide ion is present, it is removed by the reaction: $\ce{OH-}(aq)+\ce{H2CO3}(aq)⟶\ce{HCO3-}(aq)+\ce{H2O}(l) \nonumber$ The pH of human blood thus remains very near 7.35, that is, slightly basic. Variations are usually less than 0.1 of a pH unit. A change of 0.4 of a pH unit is likely to be fatal. The Henderson-Hasselbalch Approximation The ionization-constant expression for a solution of a weak acid can be written as: $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}} \nonumber$ Rearranging to solve for [H3O+], we get: $\ce{[H3O+]}=K_\ce{a}×\ce{\dfrac{[HA]}{[A- ]}} \nonumber$ Taking the negative logarithm of both sides of this equation, we arrive at: $\mathrm{−log[H_3O^+]=−log\mathit{K}_a − log\dfrac{[HA]}{[A^- ]}} \nonumber$ which can be written as $\mathrm{pH=p\mathit{K}_a+log\dfrac{[A^- ]}{[HA]}} \nonumber$ where pKa is the negative of the common logarithm of the ionization constant of the weak acid (pKa = −log Ka). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch approximation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation. Lawrence Joseph Henderson and Karl Albert Hasselbalch Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition. In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born. Medicine: The Buffer System in Blood The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction: $\ce{CO2}(g)+\ce{2H2O}(l)⇌\ce{H2CO3}(aq)⇌\ce{HCO3-}(aq)+\ce{H3O+}(aq) \nonumber$ The concentration of carbonic acid, H2CO3 is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, $\ce{HCO3-}$, is around 0.024 M. Using the Henderson-Hasselbalch equation and the pKa of carbonic acid at body temperature, we can calculate the pH of blood: $\mathrm{pH=p\mathit{K}_a+\log\dfrac{[base]}{[acid]}=6.1+\log\dfrac{0.024}{0.0012}=7.4} \nonumber$ The fact that the H2CO3 concentration is significantly lower than that of the $\ce{HCO3-}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded. Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the $\ce{HCO3-}$ ion, producing H2CO3. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO2 from the blood through the lungs driving the equilibrium reaction such that [H3O+] is lowered. If the blood is too alkaline, a lower breath rate increases CO2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H+] and restoring an appropriate pH. Summary A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base). Key Equations • pKa = −log Ka • pKb = −log Kb • $\mathrm{pH=p\mathit{K}_a+\log\dfrac{[A^- ]}{[HA]}}$ Glossary buffer capacity amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit) buffer mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added Henderson-Hasselbalch equation equation used to calculate the pH of buffer solutions
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.10%3A_Buffer_Solutions.txt
Learning Objectives • Describe a titration experiment. • Explain what an indicator does. • Perform a titration calculation correctly. The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a titration. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions. In a titration, one reagent has a known concentration or amount, while the other reagent has an unknown concentration or amount. Typically, the known reagent (the titrant) is added to the unknown quantity and is dissolved in solution. The unknown amount of substance (the analyte) may or may not be dissolved in solution (but usually is). The titrant is added to the analyte using a precisely calibrated volumetric delivery tube called a burette (also spelled buret; see Figure $1$). The burette has markings to determine how much volume of solution has been added to the analyte. When the reaction is complete, it is said to be at the equivalence point; the number of moles of titrant can be calculated from the concentration and the volume, and the balanced chemical equation can be used to determine the number of moles (and then concentration or mass) of the unknown reactant. For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted: # mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl We also have the balanced chemical reaction between HCl and NaOH: HCl + NaOH → NaCl + H2O So we can construct a conversion factor to convert to number of moles of NaOH reacted: $0.002766\cancel{mol\, HCl}\times \frac{1\, mol\, NaOH}{1\cancel{mol\, HCl}}=0.002766\, mol\, NaOH\nonumber$ Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol): $0.002766\cancel{mol\, HCl}\times \frac{40.00\, g\, NaOH}{1\cancel{mol\, HCl}}=0.1106\, g\, NaOH\nonumber$ This type of calculation is performed as part of a titration. Example $1$ What mass of Ca(OH)2 is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO3? The balanced chemical equation is as follows: 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O Solution In liters, the volume is 0.04402 L. We calculate the number of moles of titrant: # moles HNO3 = (0.04402 L)(0.0885 M) = 0.00390 mol HNO3 Using the balanced chemical equation, we can determine the number of moles of Ca(OH)2 present in the analyte: $0.00390\cancel{mol\, HNO_{3}}\times \frac{1\, mol\, Ca(OH)_{2}}{2\cancel{mol\, HNO_{3}}}=0.00195\, mol\, Ca(OH)_{2}\nonumber$ Then we convert this to a mass using the molar mass of Ca(OH)2: $0.00195\cancel{mol\, Ca(OH)_{2}}\times \frac{74.1\, g\, Ca(OH)_{2}}{\cancel{mol\, Ca(OH)_{2}}}=0.144\, g\, Ca(OH)_{2}\nonumber$ Exercise $1$ What mass of H2C2O4 is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows: $\ce{H2C2O4 + 2NaOH → Na2C2O4 + 2H2O} \nonumber$ Answer 0.182 g How does one know if a reaction is at its equivalence point? Usually, the person performing the titration adds a small amount of an indicator, a substance that changes color depending on the acidity or basicity of the solution. Because different indicators change colors at different levels of acidity, choosing the correct one is important in performing an accurate titration.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/10%3A_Acids_and_Bases/10.11%3A_Titration.txt
Most chemists pay little attention to the nucleus of an atom except to consider the number of protons it contains because that determines an element’s identity. However, in nuclear chemistry, the composition of the nucleus and the changes that occur there are very important. Applications of nuclear chemistry may be more widespread than you realize. Many people are aware of nuclear power plants and nuclear bombs, but nuclear chemistry also has applications ranging from smoke detectors to medicine, from the sterilization of food to the analysis of ancient artifacts. In this chapter, we will examine some of the basic concepts of nuclear chemistry and some of the nuclear reactions that are important in our everyday lives. • 11.1: Nuclear Reactions Nuclear reactions are very different from chemical reactions. In chemical reactions, atoms become more stable by participating in a transfer of electrons or by sharing electrons with other atoms. In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. • 11.2: The Discovery and Nature of Radioactivity In 1896, Henri Becquerel found that a uranium compound placed near a photographic plate made an image on the plate and reasoned that the compound was emitting some kind of radiation. Further investigations showed that the radiation was a combination of particles and electromagnetic rays, with its ultimate source as the atomic nucleus. These emanations were ultimately called, collectively, radioactivity. The major types of radioactivity include alpha particles, beta particles, and gamma rays. • 11.3: Stable and Unstable Isotopes In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. A radioisotope is an isotope of an element that is unstable and undergoes radioactive decay. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. Unlike chemical reactions, nuclear reactions are not noticeably affected by changes in environmental conditions, such as temperature or pressure. • 11.4: Nuclear Decay Unstable nuclei spontaneously emit radiation in the form of particles and energy. This generally changes the number of protons and/or neutrons in the nucleus, resulting in a more stable nuclide. One type of a nuclear reaction is radioactive decay, a reaction in which a nucleus spontaneously disintegrates into a slightly lighter nucleus, accompanied by the emission of particles, energy, or both. • 11.5: Radioactive Half-Life Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively. The amount of material left over after a certain number of half-lives can be easily calculated. • 11.6: Ionizing Radiation The effects of radiation on matter are determined primarily by the energy of the radiation. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. • 11.7: Detecting and Measuring Radiation We previously used mass to indicate the amount of radioactive substance present. However, this is only one of several units used to express amounts of radiation. Some units describe the number of radioactive events occurring per unit time, while others express the amount of a person's exposure to radiation. • 11.8: Artificial Transmutation Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another artificially. The conversion of one element to another is the process of transmutation. • 11.9: Nuclear Fission and Nuclear Fusion Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy. 11: Nuclear Chemistry Learning Objectives • Describe nuclear structure in terms of protons, neutrons, and electrons • Identify a nuclear reaction • Identify the key characteristics separating nuclear and chemical reactions Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^1_1H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation: $\ce{^{A}_{Z}X} \label{Eq1}$ where • $X$ is the symbol for the element, • $A$ is the mass number, and • $Z$ is the atomic number. Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.” The element in this example, represented by the symbol C, is carbon. Its atomic number, 6, is the lower left subscript on the symbol and is the number of protons in the atom. The mass number, the superscript to the upper left of the symbol, is the sum of the number of protons and neutrons in the nucleus of this particular isotope. In this case, the mass number is 14, which means that the number of neutrons in the atom is 14 − 6 = 8 (that is, the mass number of the atom minus the number of protons in the nucleus equals the number of neutrons). Occasionally, the atomic number is omitted in this notation because the symbol of the element itself conveys its characteristic atomic number. The two isotopes of hydrogen, 2H and 3H, are given their own names: deuterium (D) and tritium (T), respectively. Another way of expressing a particular isotope is to list the mass number after the element name, like carbon-12 or hydrogen-3. Nuclear reactions are very different from chemical reactions. In chemical reactions, atoms become more stable by participating in a transfer of electrons or by sharing electrons with other atoms. In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. Unlike chemical reactions, nuclear reactions are not noticeably affected by changes in environmental conditions, such as temperature or pressure. As the following Sections will discuss, there are three main forms of radioactive emissions. The first is called an alpha particle,which is symbolized by the Greek letter α. An alpha particle is composed of two protons and two neutrons, and so it is the same as a helium nucleus. (We often use $\ce{^{4}_{2}He}$ to represent an alpha particle.) It has a 2+ charge. When a radioactive atom emits an alpha particle, the original atom’s atomic number decreases by two (because of the loss of two protons), and its mass number decreases by four (because of the loss of four nuclear particles). We can represent the emission of an alpha particle with a nuclear equation—for example, the alpha-particle emission of uranium-235 is as follows: $\ce{^{235}_{92}U \rightarrow \,_2^4He + \, _{90}^{231}Th} \label{Eq2}$ Chemists often use the names parent isotope and daughter isotope to represent the original atom and the product other than the alpha particle. In the previous example, $\ce{^{235}_{92}U}$ is the parent isotope, and $\ce{^{231}_{90}Th}$ is the daughter isotope. When one element changes into another in this manner, it undergoes radioactive decay. Major Differences between Nuclear and Chemical Reactions 1. Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei. 2. Different isotopes of an element normally behave similarly in chemical reactions. The nuclear chemistry of different isotopes vary greatly from each other. 3. Rates of chemical reactions are influenced by temperature and catalysts. Rates of nuclear reactions are unaffected by such factors. 4. Nuclear reactions are independent of the chemical form of the element. 5. Energy changes accompanying nuclear reactions are much larger. This energy comes from destruction of mass. 6. In a nuclear reaction, mass is not strictly conserved. Some of the mass is converted into energy.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.01%3A_Nuclear_Reactions.txt
Learning Objectives • To define and give examples of the major types of radioactivity. Atomic theory in the 19th century presumed that nuclei had fixed compositions. But in 1896, the French scientist Henri Becquerel found that a uranium compound placed near a photographic plate made an image on the plate, even if the compound was wrapped in black cloth. He reasoned that the uranium compound was emitting some kind of radiation that passed through the cloth to expose the photographic plate. Further investigations showed that the radiation was a combination of particles and electromagnetic rays, with its ultimate source as the atomic nucleus. These emanations were ultimately called, collectively, radioactivity. There are three main forms of radioactive emissions. The first is called an alpha particle,which is symbolized by the Greek letter $α$. An alpha particle is composed of two protons and two neutrons, and so it is the same as a helium nucleus. (We often use $\ce{^{4}_{2}He}$ to represent an alpha particle.) It has a 2+ charge. When a radioactive atom emits an alpha particle, the original atom’s atomic number decreases by two (because of the loss of two protons), and its mass number decreases by four (because of the loss of four nuclear particles). We can represent the emission of an alpha particle with a nuclear equation—for example, the alpha-particle emission of uranium-235 is as follows: $\ce{^{235}_{92}U \rightarrow \,_2^4He + \, _{90}^{231}Th} \label{Eq2}$ Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure $1$) helped him determine that one type of radiation consisted of positively charged and relatively massive $α$ particles; a second type was made up of negatively charged and much less massive $β$ particles; and a third was uncharged electromagnetic waves, $γ$ rays. We now know that $α$ particles are high-energy helium nuclei, $β$ particles are high-energy electrons, and $γ$ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced. Alpha, beta, and gamma emissions have different abilities to penetrate matter (Figure $2$). The relatively large alpha particle is easily stopped by matter (although it may impart a significant amount of energy to the matter it contacts). Beta particles penetrate slightly into matter, perhaps a few centimeters at most. Gamma rays can penetrate deeply into matter and can impart a large amount of energy into the surrounding matter. Table $1$ summarizes the properties of the three main types of radioactive emissions. Table $1$: The Three Main Forms of Radioactive Emissions Characteristic Alpha Particles Beta Particles Gamma Rays symbols α, $\mathrm{_{2}^{4}He}$ β, $\ce{^{0}_{-1} e}$ γ identity helium nucleus electron electromagnetic radiation charge 2+ 1− none mass number 4 0 0 penetrating power minimal (will not penetrate skin) short (will penetrate skin and some tissues slightly) deep (will penetrate tissues deeply) Key Takeaway The major types of radioactivity include alpha particles, beta particles, and gamma rays. 11.03: Stable and Unstable Isotopes Learning Outcomes • Recognize that radioactivity is a signature of unstable nuclide - radioisotopes. • Describe a radioisotope. • Explain how the stability of isotopes depends on the composition of its nucleus. • Use the "band of stability" to identify stable isotopes. The discovery of radioactivity and its effects on the nuclei of elements disproved Dalton's assumption that atoms are indivisible. A nuclide is a term for an atom with a specific number of protons and neutrons in its nucleus. When nuclides of one type emit radiation, they are changed into different nuclides. Radioactive decay is spontaneous and does not required an input of energy to occur. The stability of a particular nuclide depends on the composition of its nucleus, including the number of protons, the number of neutrons, and the proton-to-neutron ratio. In nuclear reactions, it is the nucleus of the atom that gains stability by undergoing a change of some kind. Some elements have no stable isotopes, which means that any atom of that element is radioactive. For some other elements, only certain isotopes are radioactive. A radioisotope is an isotope of an element that is unstable and undergoes radioactive decay. The energies that are released in nuclear reactions are many orders of magnitude greater than the energies involved in chemical reactions. Unlike chemical reactions, nuclear reactions are not noticeably affected by changes in environmental conditions, such as temperature or pressure. Carbon-12, with six protons and six neutrons, is a stable nucleus, meaning that it does not spontaneously emit radioactivity. Carbon-14, with six protons and eight neutrons, is unstable and naturally radioactive. Among atoms with lower atomic numbers, the ideal ratio of neutrons to protons is approximately 1:1. As the atomic number increases, the stable neutron-proton ratio gradually increases to about 1.5:1 for the heaviest known elements. For example, lead-206 is a stable nucleus that contains 124 neutrons and 82 protons, a ratio of 1.51 to 1. This observation is shown in Figure \(1\). The band of stability is the range of stable nuclei on a graph that plots the number of neutrons in a nuclide against the number of protons. Known stable nuclides are shown with individual blue dots, while the 1:1 and 1.5:1 ratios are shown with a solid red line and a green line, respectively. It should be noted that just because a nucleus is "unstable" (able to undergo spontaneous radioactive decay) does not mean that it will rapidly decompose. For example, uranium-238 is unstable because it spontaneously decays over time, but if a sample of uranium-238 is allowed to sit for 1000 years, only \(0.0000155\%\) of the sample will have decayed. However, other unstable nuclei, such as berkelium-243, will be almost completely gone (>\(99.9999\%\) decayed) in less than a day. Contributors and Attributions • Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.02%3A_The_Discovery_and_Nature_of_Radioactivity.txt
Learning Objectives • Write and balance nuclear equations. Unstable nuclei spontaneously emit radiation in the form of particles and energy. This generally changes the number of protons and/or neutrons in the nucleus, resulting in a more stable nuclide. One type of a nuclear reaction is radioactive decay, a reaction in which a nucleus spontaneously disintegrates into a slightly lighter nucleus, accompanied by the emission of particles, energy, or both. An example is shown below, in which the nucleus of a polonium atom radioactively decays into a lead nucleus. $\ce{^{235}_{92}U \rightarrow \, _2^4He + \, _{90}^{231}Th} \label{Eq2}$ Note that in a balanced nuclear equation, the sum of the atomic numbers (subscripts) and the sum of the mass numbers (superscripts) must be equal on both sides of the equation. How do we know that a product of the reaction is $\ce{_{90}^{231}Th}$? We use a modified type of the law of conservation of matter, which says that we must have the same number of protons and neutrons on both sides of the chemical equation. If our uranium nucleus loses 2 protons from the alpha particle, then there are 90 protons remaining, identifying the element as thorium. Moreover, if we lose 4 nuclear particles of the original 235, there are 231 remaining. Thus, we use subtraction to identify the isotope of the thorium atom—in this case, $\ce{^{231}_{90}Th}$. Because the number of protons changes as a result of this nuclear reaction, the identity of the element changes. Transmutation is a change in the identity of a nucleus as a result of a change in the number of protons. There are several different types of naturally occurring radioactive decay, and we will examine each separately. Alpha Emission An alpha particle $\left( \alpha \right)$ is a helium nucleus with two protons and two neutrons. Alpha particles are emitted during some types of radioactive decay. The net charge of an alpha particle is $2+$, and its mass is approximately $4 \: \text{amu}$. The symbol for an alpha particle in a nuclear equation is usually $\ce{^4_2He}$, though sometimes $\alpha$ is used. Alpha emission typically occurs for very heavy nuclei in which the nuclei are unstable due to large numbers of nucleons. For nuclei that undergo alpha decay, their stability is increased by the subtraction of two protons and two neutrons. For example, uranium-238 decays into thorium-234 by the emission of an alpha particle (see Figure $1$). Example $1$: Radon-222 Write the nuclear equation that represents the radioactive decay of radon-222 by alpha particle emission and identify the daughter isotope. Solution Radon has an atomic number of 86, so the parent isotope is represented as $\ce{^{222}_{86}Rn}$. We represent the alpha particle as $\ce{^{4}_{2}He}$ and use subtraction (222 − 4 = 218 and 86 − 2 = 84) to identify the daughter isotope as an isotope of polonium, $\mathrm{^{218}_{84}Po}$: $\ce{_{86}^{222}Rn\rightarrow \, _2^4He + \, _{84}^{218}Po} \nonumber$ Exercise $1$: Polonium-209 Write the nuclear equation that represents the radioactive decay of polonium-209 by alpha particle emission and identify the daughter isotope. Answer $\ce{_{84}^{209}Po\rightarrow \, _2^4He + \, _{82}^{205}Pb} \nonumber$ Beta Emission Nuclei above the band of stability are unstable because their neutron to proton ratio is too high. To decrease that ratio, a neutron in the nucleus is capable of turning into a proton and an electron. The electron is immediately ejected at a high speed from the nucleus. A beta particle $\left( \beta \right)$ is a high-speed electron emitted from the nucleus of an atom during some kinds of radioactive decay (see Figure $2$). The symbol for a beta particle in an equation is either $\beta$ or $\ce{^0_{-1}e}$. Carbon-14 undergoes beta decay, transmutating into a nitrogen-14 nucleus. $\ce{^{14}_6C} \rightarrow \ce{^{14}_7N} + \ce{^0_{-1}e}$ Note that beta decay increases the atomic number by one, but the mass number remains the same. Example $2$: Boron-12 Write the nuclear equation that represents the radioactive decay of boron-12 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle. Solution The parent isotope is $\ce{^{12}_{5}B}$ while one of the products is an electron, $\ce{^{0}_{-1}e}$. So that the mass and atomic numbers have the same value on both sides, the mass number of the daughter isotope must be 12, and its atomic number must be 6. The element having an atomic number of 6 is carbon. Thus, the complete nuclear equation is as follows: $\ce{_5^{12}B\rightarrow \, _6^{12}C + \, _{-1}^0e + \gamma} \nonumber$ The daughter isotope is $\ce{^{12}_6 C}$. Exercise $2$: Iodine-131 Write the nuclear equation that represents the radioactive decay of iodine-131 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle. Answer $\ce{_53^{131}I\rightarrow \, _54^{131}Xe + \, _{-1}^0e + \gamma} \nonumber$ Gamma Emission Gamma rays $\left( \gamma \right)$ are very high energy electromagnetic waves emitted from a nucleus. Gamma rays are emitted by a nucleus when nuclear particles undergo transitions between nuclear energy levels. This is analogous to the electromagnetic radiation emitted when excited electrons drop from higher to lower energy levels; the only difference is that nuclear transitions release much more energetic radiation. Gamma ray emission often accompanies the decay of a nuclide by other means. $\ce{^{230}_{90}Th} \rightarrow \ce{^{226}_{88}Ra} + \ce{^4_2He} + \gamma$ The emission of gamma radiation has no effect on the atomic number or mass number of the products, but it reduces their energy. Positron Emission Nuclei below the band of stability are unstable because their neutron to proton ratio is too low. One way to increase that ratio is for a proton in the nucleus to turn into a neutron and another particle called a positron. A positron is a particle with the same mass as an electron, but with a positive charge. Like the beta particle, a positron is immediately ejected from the nucleus upon its formation. The symbol for a positron in an equation is $\ce{^0_{+1}e}$. For example, potassium-38 emits a positron, becoming argon-38. $\ce{^{38}_{19}K} \rightarrow \ce{^{38}_{18}Ar} + \ce{^0_1e}$ Positron emission decreases the atomic number by one, but the mass number remains the same. Electron Capture An alternate way for a nuclide to increase its neutron to proton ratio is by a phenomenon called electron capture, sympolized E.C. In electron capture, an electron from an inner orbital is captured by the nucleus of the atom and combined with a proton to form a neutron. For example, silver-106 undergoes electron capture to become palladium-106. $\ce{^{106}_{47}Ag} + \ce{^0_{-1}e} \rightarrow \ce{^{106}_{46}Pd}$ Note that the overall result of electron capture is identical to positron emission. The atomic number decreases by one while the mass number remains the same. Summary of Nuclear Radiation Table $1$ lists the characteristics of the different types of radioactive decay. Table $1$ Summary of types of radioactive decay. Type Symbol Change in Atomic Number Change in Mass Number Change in Number of Neutrons Alpha emission $\ce{^4_2He}$ or $\alpha$ –2 –4 –2 Beta emission $\ce{^0_{-1}e}$ or $\beta$ +1 0 –1 Gamma emission $\gamma$ or $^0_0\gamma$ 0 0 0 Positron emission $\ce{^0_1e}$ or $\beta^+$ –1 0 +1 electron capture E.C. –1 0 +1 Example $3$ Write a balanced nuclear equation to describe each reaction. 1. the beta decay of $^{35}_{16}\textrm{S}$ 2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture 3. the decay of $^{30}_{15}\textrm{P}$ by positron emission Given: radioactive nuclide and mode of decay Asked for: balanced nuclear equation Strategy: A Identify the reactants and the products from the information given. B Use the values of A and Z to identify any missing components needed to balance the equation. Solution a. A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$: $^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber$ B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows: $^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber$ b. A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows: $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber$ B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber$ c. A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore $^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber$ B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: $^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber$ Exercise $3$ Write a balanced nuclear equation to describe each reaction. 1. $^{11}_{6}\textrm{C}$ by positron emission 2. the beta decay of molybdenum-99 3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$ Answer a $^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$ Answer d $^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$ Answer c $^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$ Example $4$ Predict the kind of nuclear change each unstable nuclide undergoes when it decays. 1. $^{45}_{22}\textrm{Ti}$ 2. $^{242}_{94}\textrm{Pu}$ 3. $^{12}_{5}\textrm{B}$ 4. $^{256}_{100}\textrm{Fm}$ Given: nuclide Asked for: type of nuclear decay Strategy: Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide. Solution 1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time. 2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission. 3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay. 4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio. Exercise $4$ Predict the kind of nuclear change each unstable nuclide undergoes when it decays. 1. $^{32}_{14}\textrm{Si}$ 2. $^{43}_{21}\textrm{Sc}$ 3. $^{231}_{91}\textrm{Pa}$ Answer a beta decay Answer d positron emission or electron capture Answer c alpha decay Contributors and Attributions • Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.04%3A_Nuclear_Decay.txt
Learning Objectives • Define half-life. • Determine the amount of radioactive substance remaining after a given number of half-lives. • Describe common radiometric carbon-14 dating technique. Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is half life (t1/2), which is the amount of time it takes for one-half of a radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope. Each radioactive nuclide has a characteristic, constant half-life (t1/2), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem. For example, cobalt-60 source, since half of the $\ce{^{60}_{27}Co}$ nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective. We can determine the amount of a radioactive isotope remaining after a given number half-lives by using the following expression: $\text{amount remaining} = \text{initial amount} \times \left ( \frac{1}{2} \right )^{n}$ where $n$ is the number of half-lives. This expression works even if the number of half-lives is not a whole number. Example $1$: Fluorine-20 The half-life of fluorine-20 is 11.0 s. If a sample initially contains 5.00 g of fluorine-20, how much remains after 44.0 s? Solution If we compare the time that has passed to the isotope's half-life, we note that 44.0 s is exactly 4 half-lives, so using the previous expression, n = 4. Substituting and solving results in the following: \begin{align*} \text{amount remaining} &= 5.00\,g \times \left ( \frac{1}{2} \right )^{4} \[4pt] & =\: 5.00\,g\times \left ( \frac{1}{16} \right ) \[4pt] &= 0.313\,g \end{align*} Less than one-third of a gram of fluorine-20 remains. Exercise $1$: Titanium-44 The half-life of titanium-44 is 60.0 y. A sample of titanium contains 0.600 g of titanium-44. How much remains after 240.0 y? Answer 0.0375 g Half-lives of isotopes range from fractions of a microsecond to billions of years. Table $1$ - Half-Lives of Various Isotopes, lists the half-lives of some isotopes. Table $1$ Half-Lives of Various Isotopes Isotope Half-Life 3H 12.3 y 14C 5730 y 40K 1.26 × 109 y 51Cr 27.70 d 90Sr 29.1 y 131I 8.04 d 222Rn 3.823 d 235U 7.04 × 108 y 238U 4.47 × 109 y 241Am 432.7 y 248Bk 23.7 h 260Sg 4 ms Chemistry Is Everywhere: Radioactive Elements in the Body You may not think of yourself as radioactive, but you are. A small portion of certain elements in the human body are radioactive and constantly undergo decay. Most of the radioactivity in the human body comes from potassium-40 and carbon-14. Potassium and carbon are two elements that we absolutely cannot live without, so unless we can remove all the radioactive isotopes of these elements, there is no way to escape at least some radioactivity. There is debate about which radioactive element is more problematic. There is more potassium-40 in the body than carbon-14, and it has a much longer half-life. Potassium-40 also decays with about 10 times more energy than carbon-14, making each decay potentially more problematic. However, carbon is the element that makes up the backbone of most living molecules, making carbon-14 more likely to be present around important molecules, such as proteins and DNA molecules. Most experts agree that while it is foolhardy to expect absolutely no exposure to radioactivity, we can and should minimize exposure to excess radioactivity. Radiometric Dating Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old. Naturally occurring carbon consists of three isotopes: $\ce{^{12}_6C}$, which constitutes about 99% of the carbon on earth; $\ce{^{13}_6C}$, about 1% of the total; and trace amounts of $\ce{^{14}_6C}$. Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space: $\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C + ^1_1H}\nonumber$ All isotopes of carbon react with oxygen to produce CO2 molecules. The ratio of $\ce{^{14}_6CO2}$ to $\ce{^{12}_6CO2}$ depends on the ratio of $\ce{^{14}_6CO}$ to $\ce{^{12}_6CO}$ in the atmosphere. The natural abundance of $\ce{^{14}_6CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of $\ce{^{14}_6C ^{14}_6CO2}$ and $\ce{^{12}_6CO2}$ into plants is a regular part of the photosynthesis process, which means that the $\ce{^{14}_6C: ^{12}_6C}$ ratio found in a living plant is the same as the $\ce{^{14}_6C: ^{12}_6C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because $\ce{^{12}_6C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years: $\ce{^{14}_6C⟶ ^{14}_7N + ^0_{-1}e}\nonumber$ Thus, the $\ce{^{14}_6C: ^{12}_6C}$ ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure $2$ visually depicts this process. For example, with the half-life of $\ce{^{14}_6C}$ being 5730 years, if the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of $\ce{^{14}_6C : ^{12}_6C}$ ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer. Key Takeaways • Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively. • The amount of material left over after a certain number of half-lives can be easily calculated.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.05%3A_Radioactive_Half-Life.txt
Learning Objectives • To know the differences between ionizing and nonionizing radiation and their effects on matter. • To identify natural and artificial sources of radiation. Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an $α$ particle can act as a powerful oxidant. Ionizing versus Nonionizing Radiation The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling. In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions: $\mathrm{atom + ionizing\: radiation \rightarrow ion^+ + \, {e^-}\label{Eq1}}$ Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (Figure $1$). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle: $\text{1 MeV/particle} = \text{96 billion J/mol}.$ The Effects of Ionizing Radiation on Matter The effects of ionizing radiation depend on four factors: 1. The type of radiation, which dictates how far it can penetrate into matter 2. The energy of the individual particles or photons 3. The number of particles or photons that strike a given area per unit time 4. The chemical nature of the substance exposed to the radiation Because of its high charge and mass, $α$ radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, $γ$ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop $γ$ rays. Because $β$ particles are intermediate in mass and charge between $α$ particles and $γ$ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal. Because of their great penetrating ability, $γ$ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Table $1$. Table $1$: Some Properties of Ionizing Radiation Type Energy Range (MeV) Penetration Distance in Water* Penetration Distance in Air* *Distance at which half of the radiation has been absorbed. $\alpha$ particles 3–9 < 0.05 mm < 10 cm $\beta$ particles ≤ 3 < 4 mm 1 m X rays <10−2 < 1 cm < 3 m $\gamma$ rays 10−2 –101 < 20 cm > 3 m There are many different ways to measure radiation exposure, or the dose. The roentgen (R), which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure. Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered X rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose); the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram: $\mathrm{1\: rad = 0.010\: J/kg \hspace{25 pt} 1\: Gy = 1\: J/kg \label{Eq2}}$ Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle. Because $α$ particles have a much higher mass and charge than $β$ particles or $γ$ rays, the difference in mass between $α$ and $β$ particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of $α$ particles is much greater than the damage caused by 1 rad of $β$ particles or $γ$ rays. Wilhelm Röntgen Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics. Summary Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive. The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas $γ$ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.06%3A_Ionizing_Radiation.txt
Learning Objectives • Define units for measuring radiation exposure Radioactivity is determined by measuring the number of decay processes per unit time. Perhaps the easiest way is simply to determine the number of counts/minute, with each count measuring a single decay process, such as the emission of an $\alpha$-particle. A particular isotope may have an activity of 5,000 counts/minute $\left( \text{cpm} \right)$ while another isotope might only have $250 \: \text{cpm}$. The amount of activity gives a rough indication of the amount of the radioisotope present - the higher the activity, the more radioactivity isotope in the sample. Measurement of exposure to radioactivity is important for anyone who deals with radioactive materials on a regular basis. Perhaps the simplest device is a personal dosimeter - a film badge that will fog up when exposed to radiation (Figure $1$). The amount of fogging is proportional to the amount of radiation present. These devices are not very sensitive to low levels of radiation. More sensitive systems use crystals that respond in some way to radioactivity by registering the number of emissions in a given time. These systems tend to be more sensitive and more reliable than film badges. When alpha, beta or gamma particles collides with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization. A Geiger counter (or Geiger-Müller counter) takes advantage of this to detect these particles (Figure $2$). In a Geiger tube, the electron produced by ionization of a captive gas travels to the anode and the change in voltage is detected by the attached circuitry. Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter. We previously used mass to indicate the amount of radioactive substance present. However, this is only one of several units used to express amounts of radiation. Some units describe the number of radioactive events occurring per unit time, while others express the amount of a person's exposure to radiation. A variety of units are used to measure various aspects of radiation (Table $1$). Table $1$: Units Used for Measuring Radiation Measurement Purpose Unit Quantity Measured Description activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second curie (Ci) amount of sample that undergoes $\mathrm{3.7 \times 10^{10}\; decays/second}$ absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue biologically effective dose sievert (Sv) tissue damage Sv = RBE × Gy roentgen equivalent for humans (rem) Rem = RBE × rad The roentgen equivalent for humans (rem) is the unit for radiation damage that is used most frequently in medicine (1 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy) along with a biological factor referred to as the RBE (for relative biological effectiveness) that is an approximate measure of the relative damage done by the radiation. These are related by: $\text{number of rems}=\text{RBE} \times \text{number of rads} \label{Eq2}$ with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation. The Becquerel Unit Perhaps the direct way of reporting radioactivity is the number of radioactive decays per second. One decay per second is called one becquerel (Bq). Even in a small mass of radioactive material, however, there are thousands upon thousands of decays or disintegrations per second. The becquerel is named after Henri Becquerel, who discovered radioactivity in 1896. The Curie Unit The curie $\left( \text{Ci} \right)$ is one measure of the rate of decay (named after Pierre and Marie Curie). One curie is equivalent to $3.7 \times 10^{10}$ disintegrations per second. Since this is obviously a large and unwieldy number, radiation is often expressed in millicuries or microcuries (still very large numbers). The curie is named after Polish scientist Marie Curie, who performed some of the initial investigations into radioactive phenomena in the early 1900s. The curie can be used in place of grams to describe quantities of radioactive material. As an example, the amount of americium in an average smoke detector has an activity of 0.9 µCi. The Roentgen Unit There are many different ways to measure radiation exposure, or the dose. The roentgen (R), which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The Rad Unit The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose); the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram: $1\: rad = 0.010 \: J/kg \quad 1\: Gy = 1\: J/kg \label{Eq3}$ Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle. The Gray Unit Another unit of radiation absorption is the gray (Gy): 1 Gy = 100 rad The rad is more common. To get an idea of the amount of energy this represents, consider that the absorption of 1 rad by 70,000 g of H2O (approximately the same mass as a 150 lb person) would increase its temperature by only 0.002°C. This may not seem like a lot, but it is enough energy to break about 1 × 1021 molecular C–C bonds in a person’s body. That amount of damage would not be desirable. The Rem Unit Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in human) was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor: $em = rad × RBE$ where RBE is the relative biological effectiveness factor is a number greater than or equal to 1 that takes into account the type of radioactive emission and sometimes the type of tissue being exposed. For beta particles, RBE factor equals 1. For alpha particles striking most tissues, the factor is 10, but for eye tissue, the factor is 30. Most radioactive emissions that people are exposed to are on the order of a few dozen millirems (mrem) or less; a medical X ray is about 20 mrem. The Sievert Unit A sievert (Sv) is related to the rem unit and is defined as 100 rem. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem). Assessing the Impact of Radiation Exposure One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in Table $2$. Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid. Table $2$: The Effects of a Single Radiation Dose on a 70 kg Human Dose (rem) Symptoms/Effects < 5 no observable effect 5–20 possible chromosomal damage 20–100 temporary reduction in white blood cell count 50–100 temporary sterility in men (up to a year) 100–200 mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded > 300 permanent sterility in women > 500 fatal to 50% within 30 days; destruction of bone marrow and intestine > 3000 fatal within hours Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess. The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure. Summary The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = $3.7 \times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.07%3A_Detecting_and_Measuring_Radiation.txt
Learning Objectives • Describe transmutation. • Write and balance transmutation equations. Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another artificially. The conversion of one element to another is the process of transmutation. Between 1921 and 1924, Patrick Blackett conducted experiments in which he converted a stable isotope of nitrogen to a stable isotope of oxygen. By bombarding $\ce{^{14}N}$ with $\alpha$ particles he created $\ce{^{17}O}$. Transmutation may also be accomplished by bombardment with neutrons. $\ce{^{14}_7N + ^4_2He \rightarrow ^{17}_8O + ^1_1H} \nonumber$ The $\ce{^{17}_8O}$ and $\ce{^1_1H}$ nuclei that are produced are stable, so no further (nuclear) changes occur. To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. CERN Particle Accelerator Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure $1$). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers. Figure $1$: A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere) In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2103 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously. Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are: \begin{align*} \ce{^{238}_{92}U + ^1_0n} &⟶ \ce{^{239}_{92}U} && \ \ce{^{239}_{92}U} &⟶ \ce{^{239}_{93}Np + ^0_{−1}e} &&\textrm{half-life}=\mathrm{23.5\: min} \ \ce{^{239}_{93}Np } &⟶ \ce{^{239}_{94}Pu + ^0_{−1}e} &&\textrm{half-life}=\mathrm{2.36\: days} \end{align*} Plutonium is now mostly formed in nuclear reactors as a byproduct during the decay of uranium. Some of the neutrons that are released during U-235 decay combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. It is possible to summarize these equations as: $\mathrm{\ce{^{238}_{92}U} + {^1_0n}⟶ \ce{^{239}_{92}U} \xrightarrow{β^-} \ce{^{239}_{93}Np} \xrightarrow{β^-} \ce{^{239}_{94}Pu}}$ Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years. Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses. The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table $1$. Table $1$: Preparation of Some of the Transuranium Elements Name Symbol Atomic Number Reaction americium Am 95 $\ce{^{239}_{94}Pu + ^1_0n ⟶ ^{240}_{95}Am + ^0_{−1}e}$ curium Cm 96 $\ce{^{239}_{94}Pu + ^4_2He ⟶ ^{242}_{96}Cm + ^1_0n}$ californium Cf 98 $\ce{^{242}_{96}Cm + ^4_2He⟶ ^{243}_{97}Bk + 2^1_0n}$ einsteinium Es 99 $\ce{^{238}_{92}U + 15^1_0n⟶ ^{253}_{99}Es + 7^0_{−1}e}$ mendelevium Md 101 $\ce{^{253}_{99}Es + ^4_2He ⟶ ^{256}_{101}Md + ^1_0n}$ nobelium No 102 $\ce{^{246}_{96}Cm + ^{12}_6C ⟶ ^{254}_{102}No + 4 ^1_0n}$ rutherfordium Rf 104 $\ce{^{249}_{98}Cf + ^{12}_6C⟶ ^{257}_{104}Rf + 4 ^1_0n}$ seaborgium Sg 106 $\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}$ $\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}$ meitnerium Mt 107 $\ce{^{209}_{83}Bi + ^{58}_{26}Fe ⟶ ^{266}_{109}Mt + ^1_0n}$ Example $1$ Write the balanced nuclear equation for the production of the following transuranium elements: 1. berkelium-244, made by the reaction of Am-241 and He-4 2. fermium-254, made by the reaction of Pu-239 with a large number of neutrons 3. lawrencium-257, made by the reaction of Cf-250 and B-11 4. dubnium-260, made by the reaction of Cf-249 and N-15 Solution a From the given information we can write the nuclear equation $^{241}_{95}\text{Am} +^4_2\text{He}\rightarrow ^{244}_{97}\text{Bk} \nonumber$ On the left side the total mass number is $241 + 4 = 245 \nonumber$ and the total atomic number is $95 + 2 = 97 \nonumber$ On the right side the total mass number is 244 and the total atomic number is 97. This shows that one neutron needs to be added which would increase the total mass number needs to by one while keeping the total atomic number the same. The balanced nuclear equation would be $^{241}_{95}\text{Am} +^4_2\text{He}\rightarrow ^{244}_{97}\text{Bk}+^1_0\text{n} \nonumber$ b From the given information we can write the nuclear equation $^{239}_{94}\text{Pu}+\text{x }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm} \nonumber$ On the left side we see that the total mass number is the sum of $239+(1)\text{x}$. on the right side we see that the total mass number is 254. since the total mass number of the reactants must equal that of the products we can write $239+\text{x}=254 \nonumber$ showing 15 neutrons need to be added to balance the mass number. $\ce{^{239}_{94}Pu + 15 ^1_0n \rightarrow ^{254}_{100}Fm} \nonumber$ To balance the total atomic number of the equation, 6 electrons need to be added to the right side. Therefore the balanced equation reads: $^{239}_{94}\text{Pu}+15\text{ }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm}+\ce{6 ^0_{-1}e} \nonumber$ c From the given information we can write the nuclear equation $^{250}_{98}\text{Cf}+\ce{^{11}_5B}\rightarrow^{257}_{103}\text{Lr} \nonumber$ On the left side the total mass number is $250+11=261 \nonumber$ and the total atomic number is $98+5=103 \nonumber$ On the right side the total mass number is 257 and the total atomic number is 103. This means that 4 neutrons need to be added to the right side to balance the equation. The balanced nuclear equation is $^{250}_{98}\text{Cf}+\ce{^{11}_5B}\rightarrow^{257}_{103}\text{Lr}+4\text{ }^1_0\text{n} \nonumber$ d From the given information we can write the nuclear equation $^{249}_{98}\text{Cf}+\ce{^{15}_7N}\rightarrow^{260}_{105}\text{Db} \nonumber$ On the left side the total mass number is $249+15=264 \nonumber$ and the total atomic number is $98+7=105 \nonumber$ On the right side the total mass number is 260 and the total atomic number is 105. This means that 4 neutrons need to be added to the right side t obalance the equation. The balanced nuclear equation is $^{249}_{98}\text{Cf}+\ce{^{15}_7N}\rightarrow^{260}_{105}\text{Lr}+4\text{ }^1_0\text{n} \nonumber$
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.08%3A_Artificial_Transmutation.txt
Learning Objectives • Explain nuclear fission and fusion processes. • Write and balance nuclear fission and fusion equations. • Relate the concepts of critical mass and nuclear chain reactions. Nuclear fusion and nuclear fission are two different types of energy-releasing reactions in which energy is released from high-powered atomic bonds between the particles within the nucleus. The main difference between these two processes is that fission is the splitting of an atom into two or more smaller ones while fusion is the fusing of two or more smaller atoms into a larger one. Protons and neutrons make up a nucleus, which is the foundation of nuclear science. Fission and fusion involves the dispersal and combination of elemental nucleus and isotopes, and part of nuclear science is to understand the process behind this phenomenon. Adding up the individual masses of each of these subatomic particles of any given element will always give you a greater mass than the mass of the nucleus as a whole. The missing idea in this observation is the concept called nuclear binding energy. Nuclear binding energy is the energy required to keep the protons and neutrons of a nucleus intact, and the energy that is released during a nuclear fission or fusion is nuclear power. There are some things to consider however. The mass of an element's nucleus as a whole is less than the total mass of its individual protons and neutrons. To calculate the energy released during mass destruction in both nuclear fission and fusion, we use Einstein’s equation that equates energy and mass: $E=mc^2 \label{1}$ with $m$ is mass (kilograms), $c$ is speed of light (meters/sec) and $E$ is energy (Joules). Nuclear Fission Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure $1$. Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure $2$. Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium. A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal. When undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure $3$). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur. Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents. Nuclear Accidents The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima). In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen: $\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g) \nonumber$ The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process. Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure $4$). Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland. In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events. An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure $5$). The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate. Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. Nuclear Fusion The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events: $\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}}$ A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ). It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1$ and a triton, $^3_1$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron: $\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}$ This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide. The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure $7$). Example $1$ Calculate the energy released in each of the following hypothetical processes. 1. $\ce{3 ^4_2He \rightarrow ^{12}_6C}$ 2. $\ce{6 ^1_1H + 6 ^1_0n \rightarrow ^{12}_6C}$ 3. $\ce{6 ^2_1D \rightarrow ^{12}_6C}$ Solution 1. $Q_a = 3 \times 4.0026 - 12.000) \,amu \times (1.4924\times 10^{-10} \,J/amu) = 1.17 \times 10^{-12} \,J$ 2. $Q_b = (6 \times (1.007825 + 1.008665) - 12.00000)\, amu \times (1.4924\times 10^{1-0} J/amu) = 1.476\times 10^{-11} \,J$ 3. $Q_c = 6 \times 2.014102 - 12.00000 \, amu \times (1.4924\times 10^{-10} \, J/amu) = 1.263\times 10^{-11}\, J$ Fusion of $\ce{He}$ to give $\ce{C}$ releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy. Nuclear Reactors Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur. Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $8$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/11%3A_Nuclear_Chemistry/11.09%3A_Nuclear_Fission_and_Nuclear_Fusion.txt
Learning Objectives • Describe the basic structural properties of simple organic molecules. Organic chemistry is the study of the chemistry of carbon-containing compounds. Carbon is singled out because it has a chemical diversity unrivaled by any other chemical element. Its diversity is based on the following: • Carbon atoms bond reasonably strongly with other carbon atoms. • Carbon atoms bond reasonably strongly with atoms of other elements. • Carbon atoms make a large number of covalent bonds (four). Curiously, elemental carbon is not particularly abundant. It does not even appear in the list of the most common elements in Earth’s crust. Nevertheless, all living things consist of organic compounds. Most organic chemicals are covalent compounds, which is why we introduce organic chemistry here. By convention, compounds containing carbonate ions and bicarbonate ions, as well as carbon dioxide and carbon monoxide, are not considered part of organic chemistry, even though they contain carbon. Structural Properties of Carbon Compounds A carbon atom has four valence electrons, it is tetravalent. Carbon can form four covalent bonds, or share electrons with up to four atoms in order to gain a complete octet. The simplest carbon compounds contain only carbon and hydrogen and are called hydrocarbons. Methane, the simplest hydrocarbon, contains a single carbon with four covalently bonded hydrogen atoms. Recalling what you have learned about molecular structures and VSEPR, we know that methane is tetrahedral (four electron groups and no lone pairs). Carbon can also form double bonds by sharing four electrons with a neighboring carbon atom or triple bonds by sharing six electrons with a neighboring carbon atom. As shown in Figure $2$ below, carbon with three electron groups attached will be trigonal planar, and carbon with two electron groups attached will be linear. Simple hydrocarbon compounds are nonpolar due to the shape and the small electronegativity difference between carbon and hydrogen atoms. When carbon is bonded to a halogen or oxygen atom, the resulting bond is polar. It may be useful to review the section on electronegativity and polarity of bonds and molecules to be able to describe the properties of different organic compounds, specifically how they react and interact with other molecules. Comparing Organic and Inorganic Compounds Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$. Keep in mind, however, that there are exceptions to every category in this table. Table $1$: Contrasting Properties and Examples of Organic and Inorganic Compounds Organic Properties Example: Hexane Inorganic Properties Example: NaCl low melting points −95°C high melting points 801°C low boiling points 69°C high boiling points 1,413°C low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline flammable highly flammable nonflammable nonflammable aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds 12.02: Families of Organic Molecules - Functional Groups Learning Objectives • Identify and describe functional groups in organic molecules. Organic molecules can be classified into families based on structural similarities. Within a family, molecules have similar physical behavior and often have predictable chemical reactivity. The structural components differentiating different organic families involve specific arrangements of atoms or bonds, called functional groups. If you understand the behavior of a particular functional group, you can describe the general properties of that class of compounds. The simplest organic compounds are in the alkane family and contain only carbon–carbon and carbon–hydrogen single bonds but do not have any specific functional group. Hydrocarbons containing at least one carbon–carbon double bond, (denoted C=C), are in the alkene family. Alkynes have at least one carbon–carbon triple bond (C≡C). Both carbon–carbon double bonds and triple bonds chemically react in specific ways that differ from reactions of alkanes and each other, making these specific functional groups. In the next few chapters, we will learn more about additional functional groups that are made up of atoms or groups of atoms attached to hydrocarbons. Being able to recognize different functional groups will help to understand and describe common medications and biomolecules such as amino acids, carbohydrates, and fats. Table \(1\) and Figure \(1\) below list several of the functional groups to become familiar with as you learn about organic chemistry. Table \(1\): Organic Families and Functional Groups Family Name Functional Group Structure Simple Example Structure Simple Example Name Name Suffix alkane none CH3CH2CH3 propane -ane alkene H2C=CH2 ethene (ethylene) -ene alkyne HC≡CH ethyne (acetylene) -yne aromatic benzene none alkyl halide (X = F, Cl, Br, I) CH3CH2Cl chloroethane none alcohol CH3CH2OH ethanol -ol ether CH3CH2–O–CH2CH3 diethyl ether none* amine CH3CH2NH2 ethylamine -amine aldehyde ethanal -al ketone propanone (acetone) -one carboxylic acid ethanoic acid (acetic acid) -oic acid anhydride acetic anhydride none ester methyl ethanoate (methyl acetate) -ate amide acetamide -amide thiol CH3CH2SH ethanethiol -thiol disulfide CH3S–SCH3 dimethyl disulfide none sulfide CH3CH2SCH3 ethyl methyl sulfide none Atoms and bonds in red indicate the functional group. Bonds not specified are attached to R groups (carbons and hydrogens). *Ethers do not have a suffix in their common name; all ethers end with the word ether. Figure \(1\): Functional groups in organic chemistry. (CC BY-NC-ND, CompoundChem.com).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.01%3A_The_Nature_of_Organic_Molecules.txt
Learning Objectives • To identify simple alkanes as straight-chain or branched-chain. • Describe and recognize structural and functional group isomers. As you just learned, there is a wide variety of organic compounds containing different functional groups. However, all organic compounds are hydrocarbons, they contain hydrogen and carbon. The general rule for hydrocarbons is that any carbon must be bonded to at least one other carbon atom, except in the case of methane which only contains one carbon. The bonded carbons form the backbone of the molecule to which the hydrogen atoms (or other functional groups) are attached. Hydrocarbons with only carbon-to-carbon single bonds (C–C) are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in these molecules. Saturated fats and oils are organic molecules that do not have carbon-to-carbon double bonds (C=C). The three simplest alkanes—methane (CH4), ethane (C2H6), and propane (C3H8) shown in Figure \(1\), are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit (called methylene). Alkanes follow the general formula: CnH2n+2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18. Isomers Alkanes that contain one continuous chain of linked carbons are called straight-chain alkanes. As the number of carbons in a chain increases beyond three, the arrangement of atoms can expand to include branched-chain alkanes. In a branched chain, one or more hydrogen atoms along the chain is replaced by a carbon atom (or a separate chain of carbon atoms). It is important to note that while the structural arrangement of these chains are different, continuous versus branched, they both still follow the same general formula for alkanes as introduced above, CnH2n+2. In fact, alkane chains that have the same molecular formula (same number of carbon and hydrogen), but a different arrangement of atoms, are called isomers. Let's look at an example below: The structure of butane (C4H10) is written by stringing four carbon atoms in a row, and then adding enough hydrogen atoms to give each carbon atom four bonds: Butane is a straight-chain alkane, but there is another way to put 4 carbon atoms and 10 hydrogen atoms together. Place 3 of the carbon atoms in a row and then branch the fourth carbon off the middle carbon atom: Now we add enough hydrogen atoms to give each carbon four bonds: The result is the isomer 2-methylpropane (also called isobutane), which is a branched-chain alkane with the same formula as butane, (C4H10). However, it is a different molecule with a different name and different chemical properties. A side-by-side comparison of these two molecules is shown in the below figure. The four-carbon straight chain butane may be drawn with different bends or kinks in the backbone (Figure \(2\)) because the groups can rotate freely about the C–C bonds. This rotation or bending of the carbon chain does not change the identity of the compound; all of the following structures represent the same compound, butane, with different bends in the chain: When identifying isomers, it is useful to trace the carbon backbone with your finger or a pencil and count carbons until you need to lift your hand or pencil to get the another carbon. Try this with each of the above arrangements of four carbons, then do the same with 2-methylpropane. Butane has a continuous chain of four carbons no matter how the bonds are rotated – you can connect the carbons in a line without lifting your finger from the page. How many continuous carbons are in the 2-methylpropane backbone? You should be able to count a continuous chain of three carbon atoms only, with the fourth carbon attached as a branch, (compare the two structures in Figure \(1\)). In a later chapter, you will learn how to systematically name compounds by counting the number of carbons in the longest continuous chain and identifying any functional groups present. Adding one more carbon to the butane chain gives pentane, which has the formula, C5H12. Pentane and its two branched-chain isomers are shown below. The compound at the far left is pentane because it has all five carbon atoms in a continuous chain. The compound in the middle is isopentane; like isobutane, it has a one CH3 branch off the second carbon atom of the continuous chain. The compound at the far right, discovered after the other two, was named neopentane (from the Greek neos, meaning “new”). Although all three have the same molecular formula, they have different properties, including boiling points: pentane, 36.1°C; isopentane, 27.7°C; and neopentane, 9.5°C. The names isopentane and neopentane are common names for these molecules. As mentioned above, we will learn the systematic rules for naming compounds in later chapters. A continuous (unbranched) chain of carbon atoms is often called a straight chain even though the tetrahedral arrangement about each carbon gives it a zigzag shape. Straight-chain alkanes are sometimes called normal alkanes, and their names are given the prefix n-. For example, butane is called n-butane. We will not use that prefix here because it is not a part of the system established by the International Union of Pure and Applied Chemistry. 12.04: Drawing Organic Structures Learning Objectives • Draw condensed structures and line structures for simple compounds from the given molecular formulas. • Convert between expanded, condensed and line structures. We use several kinds of formulas to describe organic compounds. A molecular formula shows only the number and type of atoms in a molecule. For example, the molecular formula C4H10 tells us there are 4 carbon atoms and 10 hydrogen atoms in a molecule, but it doesn’t distinguish between butane and 2-methylpropane. A structural formula shows all the carbon and hydrogen atoms and the bonds attaching them (expanded structure). This type of structure allows for easy identification of specific isomers by showing the order of attachment of the various atoms. Unfortunately, structural formulas that show the bonds between all atoms are sometimes difficult to type/write and take up a lot of space, especially when the number of atoms greatly increases. Chemists often use condensed structures, that show hydrogen atoms right next to the carbon atoms to which they are attached, to alleviate these problems. The ultimate condensed formula is a line (or line-angle) structure, in which carbon atoms are implied at the corners and ends of lines rather than written out, and each carbon atom is understood to be attached to the appropriate amount of hydrogen atoms to give each carbon atom four bonds. Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom. All three of these structure types are illustrated for butane and its isomer, 2-methylpropane in Figure \(1\) below.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.03%3A_The_Structure_of_Organic_Molecules_-_Alkanes_and_Their_Isom.txt
Learning Objectives • To name alkanes by the IUPAC system and write formulas for alkanes given IUPAC names As noted in previously, the number of isomers increases rapidly as the number of carbon atoms increases: there are 3 pentanes, 5 hexanes, 9 heptanes, and 18 octanes, etc. It would be difficult to assign each compound unique individual names that we could remember easily. A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. (Some of the names mentioned earlier, such as isobutane, isopentane, and neopentane, do not follow these rules and are called common names.) In the IUPAC system, a compound is named according to the number of carbons in the longest continuous chain (LCC) or parent chain and the family it belongs to. Atoms or groups attached to this carbon chain, called substituents, are then named, with their positions indicated by a numerical prefix at the beginning of the name: Prefix (substituent) Parent (# carbons) Suffix (family name) 2-methylpropane (Table $1$) below lists the IUPAC parent names that are used for charbon chains containing 1 to 10 carbons, along with straight-chain alkane examples for each. Notice that the suffix for each example in this table is -ane, which indicates these are members of the alkane family. Table $1$: Parent name for 1-10 carbons and Example Alkanes Number of Carbons Parent Chain (LCC) Name Example Alkane Name Example Condensed Structural Formula 1 meth- methane CH4 2 eth- ethane CH3CH3 3 prop- propane CH3CH2CH3 4 but- butane CH3CH2CH2CH3 5 pent- pentane CH3CH2CH2CH2CH3 6 hex- hexane CH3CH2CH2CH2CH2CH3 7 hept- heptane CH3CH2CH2CH2CH2CH2CH3 8 oct- octane CH3CH2CH2CH2CH2CH2CH2CH3 9 non- nonane CH3CH2CH2CH2CH2CH2CH2CH2CH3 10 dec- decane CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 Atoms or groups of atoms that branch off the parent chain are called substituents. When the substituent is a carbon or group of carbons, such as –CH3 or –CH2CH3, it is called an alkyl group. Alkyl groups are alkanes that have had one hydrogen removed to allow for binding to a main chain carbon and are named by replacing the -ane suffix of the parent hydrocarbon with -yl. For example, the –CH3 group derived from methane (CH4) results from subtracting one hydrogen atom and is called a methyl group. Removing a hydrogen from ethane, CH3CH3, gives –CH2CH3, or the ethyl group. The alkyl groups we will use most frequently are listed in Table $2$. Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically. Table $2$: Common Alkyl Groups Parent Alkane Alkyl Group Condensed Structural Formula methane methyl CH3 ethane ethyl CH3CH2 propane propyl CH3CH2CH2 isopropyl (CH3)2CH– butane butyl CH3CH2CH2CH2 sec-butyl isobutyl tert-butyl (tBu) Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example $1$). Step 1: Name the parent chain. Find the longest continuous chain, (it may not always be the most obvious chain written in one line), and name according to the number of carbon atoms it contains. Add the suffix -ane to indicate that the molecule is an alkane. Use Table $1$ as a reference to start, but it is a good idea to commit these to memory. Step 2: Number the carbon atoms in the parent chain, giving carbons with any substituents attached the lowest number possible. These numbers are used to locate where substituents are attached to a main chain. Step 3: Name any substituents (including the location number). If the same alkyl group appears more than once, the numbers of all the carbon atoms to which it is attached are expressed. If the same group appears more than once on the same carbon atom, the number of that carbon atom is repeated as many times as the group appears. Moreover, the number of identical groups is indicated by the Greek prefixes di-, tri-, tetra-, and so on. These prefixes are not considered in determining the alphabetical order of the substituents. For example, ethyl is listed before dimethyl; the di- is simply ignored. The last alkyl group named is prefixed to the name of the parent alkane to form one word. Step 4: Write the name of the compound as a single word placing the substituent groups first (in alphabetical order), then the parent name, then the family name. Hyphens are used to separate numbers from the names of substituents; commas separate numbers from each other. When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to practice it, not just memorize the rules. It’s easier than it looks. Example $1$ Name each compound. Solution 1. Step 1: The LCC has five carbon atoms, and so the parent compound name is pentane. Step 2: Number the carbons in the LCC from left to right. Step 3: There is a methyl group attached to carbon #2 of the pentane chain. Step 4: The name is 2-methylpentane. 2. Step 1: The LCC has six carbon atoms, so the parent compound is hexane. Step 2: Number the carbons in the LCC from left to right (or right to left, either way will be identical numbering). Step 3: There are two methyl groups attached to the second and fifth carbon atoms. Step 4: The name is 2,5-dimethylhexane. 3. Step 1: The LCC has eight carbon atoms, so the parent compound is octane. Step 2: Number the carbons in the LCC from left to right to give the lower number. Step 3: There are methyl and ethyl groups, both attached to the fourth carbon atom. Step 4: The correct name is thus 4-ethyl-4-methyloctane. Exercise $1$ Name each compound. Example $2$ Draw the structure for each compound. 1. 2,3-dimethylbutane 2. 4-ethyl-2-methylheptane Solution In drawing structures, always start with the parent chain. 1. The parent chain is butane, indicating four carbon atoms in the LCC. Then add the substituents at their proper positions. You can number the parent chain from either direction as long as you are consistent; just don’t change directions before the structure is done. The name indicates two methyl (–CH3) groups, one on the second carbon atom and one on the third. Finally, fill in all the hydrogen atoms, keeping in mind that each carbon atom must have four bonds total. • Adding the substituents at their proper positions gives Filling in all the hydrogen atoms gives the following condensed structural formulas (both are correct): Note that the bonds (dashes) can be shown or not; sometimes they are needed for spacing. Exercise $2$ Draw the structure for each compound. 1. 4-ethyloctane 2. 3-ethyl-2-methylpentane 3. 3,3,5-trimethylheptane
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.06%3A_Naming_Alkanes.txt
Learning Objectives • To identify the physical properties of alkanes and describe trends in these properties. Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first. Table \(1\) describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H2O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface. Table \(1\): Physical Properties of Some Alkanes Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C) methane CH4 –182 –164 0.668 g/L gas ethane C2H6 –183 –89 1.265 g/L gas propane C3H8 –190 –42 1.867 g/L gas butane C4H10 –138 –1 2.493 g/L gas pentane C5H12 –130 36 0.626 g/mL liquid hexane C6H14 –95 69 0.659 g/mL liquid octane C8H18 –57 125 0.703 g/mL liquid decane C10H22 –30 174 0.730 g/mL liquid *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. Looking Closer: Gas Densities and Fire Hazards Table \(1\) indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas. Also shown in Table \(1\) are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state. Looking Closer: An Alkane Basis for Properties of Other Compounds An understanding of the physical properties of the alkanes is important in that petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fatlike compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes. Key Takeaway • Alkanes are nonpolar compounds that are low boiling and insoluble in water.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.07%3A_Properties_of_Alkanes.txt
Learning Objectives • Understand the reactions of alkanes: combustion and halogenation. Alkanes are relatively stable, nonpolar molecules, that will not react with acids, bases, or oxidizing or reducing reagents. Alkanes undergo so few reactions that they are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.” However, heat or light can initiate the breaking of C–H or C–C single bonds in reactions called combustion and halogenation. Combustion Nothing happens when alkanes are merely mixed with oxygen ($O_2$) at room temperature, but when a flame or spark provides the activation energy, a highly exothermic combustion reaction proceeds vigorously. For methane (CH4), the combustion reaction is as follows: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat} \label{1}$ As a consequence, alkanes are excellent fuels. For example, methane, CH4, is the principal component of natural gas. Butane, C4H10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of straight- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (Figure $1$). You may recall that boiling point is a function of intermolecular interactions, which was discussed in an earlier chapter. If the reactants of combustion reactions are adequately mixed, and there is sufficient oxygen, the only products are carbon dioxide ($CO_2$), water ($H_2O$), and energy—heat for cooking foods, heating homes, and drying clothes. Because conditions are rarely ideal, other unwanted by-products are frequently formed. When the oxygen supply is limited, carbon monoxide ($CO$) is a by-product: $2CH_4 + 3O_2 \rightarrow​ 2CO + 4H_2O\label{2}$ This reaction is responsible for dozens of deaths each year from unventilated or improperly adjusted gas heaters. (Similar reactions with similar results occur with kerosene heaters.) Halogenation In halogenation reactions, alkanes react with the halogens chlorine ($Cl_2$) and bromine ($Br_2$) in the presence of ultraviolet light or at high temperatures to yield chlorinated and brominated alkanes. For example, chlorine reacts with excess methane ($CH_4$) to give methyl chloride ($CH_3Cl$). $CH_4 + Cl_2 \rightarrow​ CH_3Cl + HCl\label{12.7.3}$ With more chlorine, a mixture of products is obtained: CH3Cl, CH2Cl2, CHCl3, and CCl4. Fluorine ($F_2$), the lightest halogen, combines explosively with most hydrocarbons. Iodine ($I_2$) is relatively unreactive. Fluorinated and iodinated alkanes are produced by indirect methods. A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH4) can react with chlorine (Cl2), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH3CH3) are listed in Table $1$, along with some of their uses. Table $1$: Some Halogenated Hydrocarbons Formula Common Name IUPAC Name Some Important Uses Derived from CH4 CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent CHCl3 chloroform trichloromethane industrial solvent CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use) CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant Derived from CH3CH3 CH3CH2Cl ethyl chloride chloroethane local anesthetic ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics Note To Your Health: Halogenated Hydrocarbons Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl4), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl2) gas, which makes the use of CCl4 in fire extinguishers particularly dangerous. Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic. Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds. Note To Your Health: Chlorofluorocarbons and The Ozone Layer Alkanes substituted with both fluorine (F) and chlorine (Cl) atoms have been used as the dispersing gases in aerosol cans, as foaming agents for plastics, and as refrigerants. Two of the best known of these chlorofluorocarbons (CFCs) are listed in Table $2$. Chlorofluorocarbons contribute to the greenhouse effect in the lower atmosphere. They also diffuse into the stratosphere, where they are broken down by ultraviolet (UV) radiation to release Cl atoms. These in turn break down the ozone (O3) molecules that protect Earth from harmful UV radiation. Worldwide action has reduced the use of CFCs and related compounds. The CFCs and other Cl- or bromine (Br)-containing ozone-destroying compounds are being replaced with more benign substances. Hydrofluorocarbons (HFCs), such as CH2FCF3, which have no Cl or Br to form radicals, are one alternative. Another is hydrochlorofluorocarbons (HCFCs), such as CHCl2CF3. HCFC molecules break down more readily in the troposphere, and fewer ozone-destroying molecules reach the stratosphere. Figure $2$: Ozone in the upper atmosphere shields Earth’s surface from UV radiation from the sun, which can cause skin cancer in humans and is also harmful to other animals and to some plants. Ozone “holes” in the upper atmosphere (the gray, pink, and purple areas at the center) are large areas of substantial ozone depletion. They occur mainly over Antarctica from late August through early October and fill in about mid-November. Ozone depletion has also been noted over the Arctic regions. The largest ozone hole ever observed occurred on 24 September 2006. Source: Image courtesy of NASA, http://ozonewatch.gsfc.nasa.gov/daily.php?date=2006-09-24.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.08%3A_Reactions_of_Alkanes.txt
Learning Objectives • Identify the structures of cycloalkanes. A cyclic hydrocarbon is a hydrocarbon in which the carbon chain joins to itself in a ring. A cycloalkane is a cyclic hydrocarbon in which all of the carbon-carbon bonds are single bonds and each carbon is bonded to two hydrogen atoms, they are saturated compounds. Cycloalkanes have the general formula $\ce{C_{n}H_{2n}}$. The simplest of these cyclic hydrocarbons, cyclopropane, has the formula C3H6, which makes a three-carbon ring. The structural formulas of cyclic hydrocarbons can be represented in multiple ways, two of which are shown above. Each atom can be shown as in the structure on the left from the figure above. A convenient shorthand is to omit the element symbols and only show the shape, as in the triangle on the right. Carbon atoms are understood to be the vertices of the triangle. The carbon atoms in cycloalkanes have a bond angle of $109.5^\text{o}$. However, an examination of the cyclopropane structure shows that the triangular structure results in a $\ce{C-C-C}$ bond angle of $60^\text{o}$. This deviation from the ideal angle is called ring strain and makes cyclopropane a fairly unstable and reactive molecule. Ring strain is decreased for cyclobutane, with a bond angle of $90^\text{o}$, but is still significant. Cyclopentane has a bond angle of about $108^\text{o}$. This minimal ring strain for cyclopentane makes it a more stable compound. Cyclohexane is a six-carbon cycloalkane, shown below. All three of the depictions of cyclohexane above are somewhat misleading, because the molecule is not planar. In order to reduce the ring strain and attain a bond angle of approximately $109.5^\text{o}$, the molecule is actually puckered. The ring structure in cycloalkanes also prevents rotation around the carbon–carbon bonds without breaking open the ring, thus they are more rigid and less flexible than acyclic alkanes. This property is called restricted rotation. Note To Your Health: Cyclopropane as an Anesthetic With its boiling point of −33°C, cyclopropane is a gas at room temperature. It is also a potent, quick-acting anesthetic with few undesirable side effects in the body. It is no longer used in surgery, however, because it forms explosive mixtures with air at nearly all concentrations. 12.10: Drawing and Naming Cycloalkanes Learning Objectives • To name cycloalkanes given their formulas and write formulas for these compounds given their names. The cycloalkanes—cyclic hydrocarbons with only single bonds—are named by adding the prefix cyclo- to the name of the open-chain compound having the same number of carbon atoms as there are in the ring. Thus the name for the cyclic compound C4H8 is cyclobutane. The carbon atoms in cyclic compounds can be represented by line-angle formulas that result in regular geometric figures. Keep in mind, however, that each corner of the geometric figure represents a carbon atom plus as many hydrogen atoms as needed to give each carbon atom four bonds. Some cyclic compounds have substituent groups attached. Example \(1\) interprets the name of a cycloalkane with a single substituent group. Example \(1\) Draw the structure for each compound. 1. cyclopentane 2. methylcyclobutane Solution 1. The name cyclopentane indicates a cyclic (cyclo) alkane with five (pent-) carbon atoms. It can be represented as a pentagon. • The name methylcyclobutane indicates a cyclic alkane with four (but-) carbon atoms in the cyclic part. It can be represented as a square with a CH3 group attached. Exercise \(1\) Draw the structure for each compound. 1. cycloheptane 2. ethylcyclohexane
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/12%3A_Introduction_to_Organic_Chemistry_-_Alkanes/12.09%3A_Cycloalkanes.txt
Learning Objectives • Identify the difference between saturated and unsaturated hydrocarbons. • Describe the functional groups, alkenes and alkynes. As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas: You have likely heard of unsaturated fats. These are complex organic molecules with long chains of carbon atoms, which contain at least one double bond between carbon atoms. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats. Alkenes Ethene, C2H4, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure \(1\)); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism. Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules. Alkynes Hydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond have bond angles of 180°, giving these types of bonds a linear shape. The simplest member of the alkyne series is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is: Acetylene is used in oxyacetylene torches for cutting and welding metals. The flame from such a torch can be very hot. Most acetylene, however, is converted to chemical intermediates that are used to make vinyl and acrylic plastics, fibers, resins, and a variety of other products. Alkynes are similar to alkenes in both physical and chemical properties. For example, alkynes undergo many of the typical addition reactions of alkenes. The International Union of Pure and Applied Chemistry (IUPAC) names for alkynes parallel those of alkenes, except that the family ending is -yne rather than -ene. 13.02: Naming Alkenes and Alkynes Learning Objectives • Objective 1 • Objective 2 Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC): 1. The longest chain of carbon atoms containing the double bond is considered the parent chain. It is named using the same stem as the alkane having the same number of carbon atoms but ends in -ene to identify it as an alkene. Thus the compound CH2=CHCH3 is propene. 2. If there are four or more carbon atoms in a chain, we must indicate the position of the double bond. The carbons atoms are numbered so that the first of the two that are doubly bonded is given the lower of the two possible numbers.The compound CH3CH=CHCH2CH3, for example, has the double bond between the second and third carbon atoms. Its name is 2-pentene (not 3-pentene). 3. Substituent groups are named as with alkanes, and their position is indicated by a number. Thus, the structure below is 5-methyl-2-hexene. Note that the numbering of the parent chain is always done in such a way as to give the double bond the lowest number, even if that causes a substituent to have a higher number. The double bond always has priority in numbering. Example $1$ Name each compound. Solution 1. The longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene. 2. The longest chain containing the double bond has five carbon atoms, so the parent compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the third carbon atom (rule 3), so the compound’s name is 3-methyl-2-pentene. Exercise $1$ Name each compound. 1. CH3CH2CH2CH2CH2CH=CHCH3 Just as there are cycloalkanes, there are cycloalkenes. These compounds are named like alkenes, but with the prefix cyclo- attached to the beginning of the parent alkene name. Example $2$ Draw the structure for each compound. 1. 3-methyl-2-pentene 2. cyclohexene Solution 1. First write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms: Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds. • First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and -ene means a double bond. Exercise $2$ Draw the structure for each compound. 1. 2-ethyl-1-hexene 2. cyclopentene The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, $\mathrm{CH_3CH_2C≡CH}$ is called 1-butyne. Example $6$: Structure of Alkynes Describe the geometry and hybridization of the carbon atoms in the following molecule: Solution Carbon atoms 1 and 4 have four single bonds and are thus tetrahedral with sp3 hybridization. Carbon atoms 2 and 3 are involved in the triple bond, so they have linear geometries and would be classified as sp hybrids. Exercise $6$ Identify the hybridization and bond angles at the carbon atoms in the molecule shown: Answer carbon 1: sp, 180°; carbon 2: sp, 180°; carbon 3: sp2, 120°; carbon 4: sp2, 120°; carbon 5: sp3, 109.5°
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/13%3A_Alkenes_Alkynes_and_Aromatic_Compounds/13.01%3A_Alkenes_and_Alkynes.txt
Learning Objectives • Recognize that alkenes that can exist as cis-trans isomers. • Classify isomers as cis or trans. • Draw structures for cis-trans isomers given their names. There is free rotation about the carbon-to-carbon single bonds (C–C) in alkanes. In contrast, the structure of alkenes requires that the carbon atoms of a double bond and the two atoms bonded to each carbon atom all lie in a single plane, and that each doubly bonded carbon atom lies in the center of a triangle. This part of the molecule’s structure is rigid; rotation about doubly bonded carbon atoms is not possible without rupturing the bond. Look at the two chlorinated hydrocarbons in Figure \(1\). Figure \(1\): Rotation about Bonds. In 1,2-dichloroethane (a), free rotation about the C–C bond allows the two structures to be interconverted by a twist of one end relative to the other. In 1,2-dichloroethene (b), restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond are significant. In 1,2-dichloroethane (part (a) of Figure \(1\)), there is free rotation about the C–C bond. The two models shown represent exactly the same molecule; they are not isomers. You can draw structural formulas that look different, but if you bear in mind the possibility of this free rotation about single bonds, you should recognize that these two structures represent the same molecule: In 1,2-dichloroethene (Figure \(\PageIndex{1b}\)), however, restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism. The isomer in which the two chlorine (Cl) atoms lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-1,2-dichloroethene. The isomer with the two Cl atoms on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-1,2-dichloroethene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. Consider the alkene with the condensed structural formula CH3CH=CHCH3. We could name it 2-butene, but there are actually two such compounds; the double bond results in cis-trans isomerism (Figure \(2\)). Cis-2-butene has both methyl groups on the same side of the molecule. Trans-2-butene has the methyl groups on opposite sides of the molecule. Their structural formulas are as follows: Figure \(3\): Models of (left) Cis-2-Butene and (right) Trans-2-Butene. Note, however, that the presence of a double bond does not necessarily lead to cis-trans isomerism (Figure \(4\)). We can draw two seemingly different propenes: Figure \(4\): Different views of the propene molecule (flip vertically). These are not isomers. However, these two structures are not really different from each other. If you could pick up either molecule from the page and flip it over top to bottom, you would see that the two formulas are identical. Thus there are two requirements for cis-trans isomerism: 1. Rotation must be restricted in the molecule. 2. There must be two nonidentical groups on each doubly bonded carbon atom. In these propene structures, the second requirement for cis-trans isomerism is not fulfilled. One of the doubly bonded carbon atoms does have two different groups attached, but the rules require that both carbon atoms have two different groups. In general, the following statements hold true in cis-trans isomerism: • Alkenes with a C=CH2 unit do not exist as cis-trans isomers. • Alkenes with a C=CR2 unit, where the two R groups are the same, do not exist as cis-trans isomers. • Alkenes of the type R–CH=CH–R can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond. Advanced Note: E/Z Isomerization If a molecule has a C=C bond with one non-hydrogen group attached to each of the carbons, cis/trans nomenclature descried above is enough to describe it. However, if you have three different groups (or four), then the cis/trans approach is insufficient to describe the different isomers, since we do not know which two of the three groups are being described. For example, if you have a C=C bond, with a methyl group and a bromine on one carbon , and an ethyl group on the other, it is neither trans nor cis, since it is not clear whether the ethyl group is trans to the bromine or the methyl. This is addressed with a more advanced E/Z nomenclature discussed elsewhere. Cis-trans isomerism also occurs in cyclic compounds. In ring structures, groups are unable to rotate about any of the ring carbon–carbon bonds. Therefore, groups can be either on the same side of the ring (cis) or on opposite sides of the ring (trans). For our purposes here, we represent all cycloalkanes as planar structures, and we indicate the positions of the groups, either above or below the plane of the ring. Example \(1\) Which compounds can exist as cis-trans (geometric) isomers? Draw them. 1. CHCl=CHBr 2. CH2=CBrCH3 3. (CH3)2C=CHCH2CH3 4. CH3CH=CHCH2CH3 Solution All four structures have a double bond and thus meet rule 1 for cis-trans isomerism. 1. This compound meets rule 2; it has two nonidentical groups on each carbon atom (H and Cl on one and H and Br on the other). It exists as both cis and trans isomers: • This compound has two hydrogen atoms on one of its doubly bonded carbon atoms; it fails rule 2 and does not exist as cis and trans isomers. • This compound has two methyl (CH3) groups on one of its doubly bonded carbon atoms. It fails rule 2 and does not exist as cis and trans isomers. • This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers: Exercise \(1\) Which compounds can exist as cis-trans isomers? Draw them. 1. CH2=CHCH2CH2CH3 2. CH3CH=CHCH2CH3 3. CH3CH2CH=CHCH2CH3 Concept Review Exercises 1. What are cis-trans (geometric) isomers? What two types of compounds can exhibit cis-trans isomerism? 2. Classify each compound as a cis isomer, a trans isomer, or neither. Answers 1. Cis-trans isomers are compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. Alkenes and cyclic compounds can exhibit cis-trans isomerism. 1. trans (the two hydrogen atoms are on opposite sides) 2. cis (the two hydrogen atoms are on the same side, as are the two ethyl groups) 3. cis (the two ethyl groups are on the same side) 4. neither (fliping the bond does not change the molecule. There are no isomers for this molecule)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/13%3A_Alkenes_Alkynes_and_Aromatic_Compounds/13.03%3A_The_Structure_of_Alkenes-_Cis-Trans_Isomerism.txt
Learning Objectives • To identify the physical properties of alkenes and describe trends in these properties. The physical properties of alkenes are similar to those of the alkanes. The table at the start of the chapter shows that the boiling points of straight-chain alkenes increase with increasing molar mass, just as with alkanes. For molecules with the same number of carbon atoms and the same general shape, the boiling points usually differ only slightly, just as we would expect for substances whose molar mass differs by only 2 u (equivalent to two hydrogen atoms). Like other hydrocarbons, the alkenes are insoluble in water but soluble in organic solvents. Looking Closer: Environmental Note Alkenes occur widely in nature. Ripening fruits and vegetables give off ethylene, which triggers further ripening. Fruit processors artificially introduce ethylene to hasten the ripening process; exposure to as little as 0.1 mg of ethylene for 24 h can ripen 1 kg of tomatoes. Unfortunately, this process does not exactly duplicate the ripening process, and tomatoes picked green and treated this way don’t taste much like vine-ripened tomatoes fresh from the garden. The bright red color of tomatoes is due to lycopene—a polyene. Other alkenes that occur in nature include 1-octene, a constituent of lemon oil, and octadecene (C18H36) found in fish liver. Dienes (two double bonds) and polyenes (three or more double bonds) are also common. Butadiene (CH2=CHCH=CH2) is found in coffee. Lycopene and the carotenes are isomeric polyenes (C40H56) that give the attractive red, orange, and yellow colors to watermelons, tomatoes, carrots, and other fruits and vegetables. Vitamin A, essential to good vision, is derived from a carotene. The world would be a much less colorful place without alkenes. 13.06: Addition Reactions of Alkenes Learning Objectives • To write equations for the addition reactions of alkenes with hydrogen, halogens, and water Alkenes are valued mainly for addition reactions, in which one of the bonds in the double bond is broken. Each of the carbon atoms in the bond can then attach another atom or group while remaining joined to each other by a single bond. Perhaps the simplest addition reaction is hydrogenation—a reaction with hydrogen (H2) in the presence of a catalyst such as nickel (Ni) or platinum (Pt). The product is an alkane having the same carbon skeleton as the alkene. Alkenes also readily undergo halogenation—the addition of halogens. Indeed, the reaction with bromine (Br2) can be used to test for alkenes. Bromine solutions are brownish red. When we add a Br2 solution to an alkene, the color of the solution disappears because the alkene reacts with the bromine: Another important addition reaction is that between an alkene and water to form an alcohol. This reaction, called hydration, requires a catalyst—usually a strong acid, such as sulfuric acid (H2SO4): The hydration reaction is discussed later, where we deal with this reaction in the synthesis of alcohols. Example \(1\) Write the equation for the reaction between CH3CH=CHCH3 and each substance. 1. H2 (Ni catalyst) 2. Br2 3. H2O (H2SO4 catalyst) Solution In each reaction, the reagent adds across the double bond. Exercise \(1\) Write the equation for each reaction. 1. CH3CH2CH=CH2 with H2 (Ni catalyst) 2. CH3CH=CH2 with Cl2 3. CH3CH2CH=CHCH2CH3 with H2O (H2SO4 catalyst) 13.07: Alkene Polymers Learning Objectives • To draw structures for monomers that can undergo addition polymerization and for four-monomer-unit sections of an addition polymer. The most important commercial reactions of alkenes are polymerizations, reactions in which small molecules, referred to in general as monomers (from the Greek monos, meaning “one,” and meros, meaning “parts”), are assembled into giant molecules referred to as polymers (from the Greek poly, meaning “many,” and meros, meaning “parts”). A polymer is as different from its monomer as a long strand of spaghetti is from a tiny speck of flour. For example, polyethylene, the familiar waxy material used to make plastic bags, is made from the monomer ethylene—a gas. There are two general types of polymerization reactions: addition polymerization and condensation polymerization. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers. Ethylene molecules are joined together in long chains. The polymerization can be represented by the reaction of a few monomer units: The bond lines extending at the ends in the formula of the product indicate that the structure extends for many units in each direction. Notice that all the atoms—two carbon atoms and four hydrogen atoms—of each monomer molecule are incorporated into the polymer structure. Because displays such as the one above are cumbersome, the polymerization is often abbreviated as follows: nCH2=CH2 [ CH2CH2 ] n Many natural materials—such as proteins, cellulose and starch, and complex silicate minerals—are polymers. Artificial fibers, films, plastics, semisolid resins, and rubbers are also polymers. More than half the compounds produced by the chemical industry are synthetic polymers. Some common addition polymers are listed in Table \(1\). Note that all the monomers have carbon-to-carbon double bonds. Many polymers are mundane (e.g., plastic bags, food wrap, toys, and tableware), but there are also polymers that conduct electricity, have amazing adhesive properties, or are stronger than steel but much lighter in weight. Table \(1\): Some Addition Polymers Monomer Polymer Polymer Name Some Uses CH2=CH2 ~CH2CH2CH2CH2CH2CH2~ polyethylene plastic bags, bottles, toys, electrical insulation CH2=CHCH3 polypropylene carpeting, bottles, luggage, exercise clothing CH2=CHCl polyvinyl chloride bags for intravenous solutions, pipes, tubing, floor coverings CF2=CF2 ~CF2CF2CF2CF2CF2CF2~ polytetrafluoroethylene nonstick coatings, electrical insulation
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/13%3A_Alkenes_Alkynes_and_Aromatic_Compounds/13.04%3A_Properties_of_Alkenes_and_Alkynes.txt
Learning Objectives • To describe the bonding in benzene and the way typical reactions of benzene differ from those of the alkenes. Next we consider a class of hydrocarbons with molecular formulas like those of unsaturated hydrocarbons, but which, unlike the alkenes, do not readily undergo addition reactions. These compounds comprise a distinct class, called aromatic hydrocarbons, with unique structures and properties. We start with the simplest of these compounds. Benzene (C6H6) is of great commercial importance, but it also has noteworthy health effects. The formula C6H6 seems to indicate that benzene has a high degree of unsaturation. (Hexane, the saturated hydrocarbon with six carbon atoms has the formula C6H14—eight more hydrogen atoms than benzene.) However, despite the seeming low level of saturation, benzene is rather unreactive. It does not, for example, react readily with bromine, which, is a test for unsaturation. Note Benzene is a liquid that smells like gasoline, boils at 80°C, and freezes at 5.5°C. It is the aromatic hydrocarbon produced in the largest volume. It was formerly used to decaffeinate coffee and was a significant component of many consumer products, such as paint strippers, rubber cements, and home dry-cleaning spot removers. It was removed from many product formulations in the 1950s, but others continued to use benzene in products until the 1970s when it was associated with leukemia deaths. Benzene is still important in industry as a precursor in the production of plastics (such as Styrofoam and nylon), drugs, detergents, synthetic rubber, pesticides, and dyes. It is used as a solvent for such things as cleaning and maintaining printing equipment and for adhesives such as those used to attach soles to shoes. Benzene is a natural constituent of petroleum products, but because it is a known carcinogen, its use as an additive in gasoline is now limited. To explain the surprising properties of benzene, chemists suppose the molecule has a cyclic, hexagonal, planar structure of six carbon atoms with one hydrogen atom bonded to each. We can write a structure with alternate single and double bonds, either as a full structural formula or as a line-angle formula: However, these structures do not explain the unique properties of benzene. Furthermore, experimental evidence indicates that all the carbon-to-carbon bonds in benzene are equivalent, and the molecule is unusually stable. Chemists often represent benzene as a hexagon with an inscribed circle: The inner circle indicates that the valence electrons are shared equally by all six carbon atoms (that is, the electrons are delocalized, or spread out, over all the carbon atoms). It is understood that each corner of the hexagon is occupied by one carbon atom, and each carbon atom has one hydrogen atom attached to it. Any other atom or groups of atoms substituted for a hydrogen atom must be shown bonded to a particular corner of the hexagon. We use this modern symbolism, but many scientists still use the earlier structure with alternate double and single bonds. To Your Health: Benzene and Us Most of the benzene used commercially comes from petroleum. It is employed as a starting material for the production of detergents, drugs, dyes, insecticides, and plastics. Once widely used as an organic solvent, benzene is now known to have both short- and long-term toxic effects. The inhalation of large concentrations can cause nausea and even death due to respiratory or heart failure, while repeated exposure leads to a progressive disease in which the ability of the bone marrow to make new blood cells is eventually destroyed. This results in a condition called aplastic anemia, in which there is a decrease in the numbers of both the red and white blood cells.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/13%3A_Alkenes_Alkynes_and_Aromatic_Compounds/13.08%3A_Aromatic_Compounds_and_the_Structure_of_Benzene.txt
Learning Objectives • Recognize aromatic compounds from structural formulas. • Name aromatic compounds given formulas. • Write formulas for aromatic compounds given their names. Historically, benzene-like substances were called aromatic hydrocarbons because they had distinctive aromas. Today, an aromatic compound is any compound that contains a benzene ring or has certain benzene-like properties (but not necessarily a strong aroma). You can recognize the aromatic compounds in this text by the presence of one or more benzene rings in their structure. Some representative aromatic compounds and their uses are listed in Table \(1\), where the benzene ring is represented as C6H5. Table \(1\): Some Representative Aromatic Compounds Name Structure Typical Uses aniline C6H5–NH2 starting material for the synthesis of dyes, drugs, resins, varnishes, perfumes; solvent; vulcanizing rubber benzoic acid C6H5–COOH food preservative; starting material for the synthesis of dyes and other organic compounds; curing of tobacco bromobenzene C6H5–Br starting material for the synthesis of many other aromatic compounds; solvent; motor oil additive nitrobenzene C6H5–NO2 starting material for the synthesis of aniline; solvent for cellulose nitrate; in soaps and shoe polish phenol C6H5–OH disinfectant; starting material for the synthesis of resins, drugs, and other organic compounds toluene C6H5–CH3 solvent; gasoline octane booster; starting material for the synthesis of benzoic acid, benzaldehyde, and many other organic compounds Example \(1\) Which compounds are aromatic? Solution 1. The compound has a benzene ring (with a chlorine atom substituted for one of the hydrogen atoms); it is aromatic. 2. The compound is cyclic, but it does not have a benzene ring; it is not aromatic. 3. The compound has a benzene ring (with a propyl group substituted for one of the hydrogen atoms); it is aromatic. 4. The compound is cyclic, but it does not have a benzene ring; it is not aromatic. Exercise \(1\) Which compounds are aromatic? In the International Union of Pure and Applied Chemistry (IUPAC) system, aromatic hydrocarbons are named as derivatives of benzene. Figure \(1\) shows four examples. In these structures, it is immaterial whether the single substituent is written at the top, side, or bottom of the ring: a hexagon is symmetrical, and therefore all positions are equivalent. Although some compounds are referred to exclusively by IUPAC names, some are more frequently denoted by common names, as is indicated in Table \(1\). When there is more than one substituent, the corners of the hexagon are no longer equivalent, so we must designate the relative positions. There are three possible disubstituted benzenes, and we can use numbers to distinguish them (Figure \(2\)). We start numbering at the carbon atom to which one of the groups is attached and count toward the carbon atom that bears the other substituent group by the shortest path. In Figure \(2\), common names are also used: the prefix ortho (o-) for 1,2-disubstitution, meta (m-) for 1,3-disubstitution, and para (p-) for 1,4-disubstitution. The substituent names are listed in alphabetical order. The first substituent is given the lowest number. When a common name is used, the carbon atom that bears the group responsible for the name is given the number 1: Example \(2\) Name each compound using both the common name and the IUPAC name. Solution 1. The benzene ring has two chlorine atoms (dichloro) in the first and second positions. The compound is o-dichlorobenzene or 1,2-dichlorobenzene. 2. The benzene ring has a methyl (CH3) group. The compound is therefore named as a derivative of toluene. The bromine atom is on the fourth carbon atom, counting from the methyl group. The compound is p-bromotoluene or 4-bromotoluene. 3. The benzene ring has two nitro (NO2) groups in the first and third positions. It is m-dinitrobenzene or 1,3-dinitrobenzene. Note: The nitro (NO2) group is a common substituent in aromatic compounds. Many nitro compounds are explosive, most notably 2,4,6-trinitrotoluene (TNT). Exercise \(2\) Name each compound using both the common name and the IUPAC name. Sometimes an aromatic group is found as a substituent bonded to a nonaromatic entity or to another aromatic ring. The group of atoms remaining when a hydrogen atom is removed from an aromatic compound is called an aryl group. The most common aryl group is derived from benzene (C6H6) by removing one hydrogen atom (C6H5) and is called a phenyl group, from pheno, an old name for benzene. Polycyclic Aromatic Hydrocarbons Some common aromatic hydrocarbons consist of fused benzene rings—rings that share a common side. These compounds are called polycyclic aromatic hydrocarbons (PAHs). The three examples shown here are colorless, crystalline solids generally obtained from coal tar. Naphthalene has a pungent odor and is used in mothballs. Anthracene is used in the manufacture of certain dyes. Steroids, a large group of naturally occurring substances, contain the phenanthrene structure. To Your Health: Polycyclic Aromatic Hydrocarbons and Cancer The intense heating required for distilling coal tar results in the formation of PAHs. For many years, it has been known that workers in coal-tar refineries are susceptible to a type of skin cancer known as tar cancer. Investigations have shown that a number of PAHs are carcinogens. One of the most active carcinogenic compounds, benzopyrene, occurs in coal tar and has also been isolated from cigarette smoke, automobile exhaust gases, and charcoal-broiled steaks. It is estimated that more than 1,000 t of benzopyrene are emitted into the air over the United States each year. Only a few milligrams of benzopyrene per kilogram of body weight are required to induce cancer in experimental animals. Biologically Important Compounds with Benzene Rings Substances containing the benzene ring are common in both animals and plants, although they are more abundant in the latter. Plants can synthesize the benzene ring from carbon dioxide, water, and inorganic materials. Animals cannot synthesize it, but they are dependent on certain aromatic compounds for survival and therefore must obtain them from food. Phenylalanine, tyrosine, and tryptophan (essential amino acids) and vitamins K, B2 (riboflavin), and B9 (folic acid) all contain the benzene ring. Many important drugs, a few of which are shown in Table \(2\), also feature a benzene ring. So far we have studied only aromatic compounds with carbon-containing rings. However, many cyclic compounds have an element other than carbon atoms in the ring. These compounds, called heterocyclic compounds, are discussed later. Some of these are heterocyclic aromatic compounds. Table \(2\): Some Drugs That Contain a Benzene Ring Name Structure aspirin acetaminophen ibuprofen amphetamine sulfanilamide
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/13%3A_Alkenes_Alkynes_and_Aromatic_Compounds/13.09%3A_Naming_Aromatic_Compounds.txt
Learning Objectives 1. Give two major types of reactions of alcohols. 2. Describe the result of the oxidation of a primary alcohol. 3. Describe the result of the oxidation of a secondary alcohol. Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH-bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure \(1\), two—dehydration and oxidation—are considered here. The third reaction type—esterification—is covered elsewhere. Solution The first step is to recognize the class of each alcohol as primary, secondary, or tertiary. Exercise \(1\) Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow. Summary Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized. • Dehydration As noted in Figure \(1\), an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule: Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule. (Ethers are discussed in elsewhere) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols. Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds. Oxidation Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows: We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones. The oxidation of isopropyl alcohol by potassium dichromate (K2Cr2O7) gives acetone, the simplest ketone: Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH2OH) and secondary (R2CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group. These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows: Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group: The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone. Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized. Example \(1\) Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow. 1. CH3CH2CH2CH2CH2OH 1. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid. 2. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs. 3. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/14%3A_Some_Compounds_with_Oxygen_Sulfur_or_a_Halogen/14.04%3A_Reactions_of_Alcohols.txt
Learning Objectives • To describe the structure and uses of some phenols Compounds in which an OH group is attached directly to an aromatic ring are designated ArOH and called phenols. Phenols differ from alcohols in that they are slightly acidic in water. They react with aqueous sodium hydroxide (NaOH) to form salts. $\ce{ArOH (aq) + NaOH (aq) \rightarrow ArONa (aq) + H_2O} \nonumber$ The parent compound, C6H5OH, is itself called phenol. (An old name, emphasizing its slight acidity, was carbolic acid.) Phenol is a white crystalline compound that has a distinctive (“hospital smell”) odor. Figure $1$: (Left) Structure of Phenol (right) Approximately two grams of phenol in glass vial. Image used with permisison from Wikipedia To Your Health: Phenols and Us Phenols are widely used as antiseptics (substances that kill microorganisms on living tissue) and as disinfectants (substances intended to kill microorganisms on inanimate objects such as furniture or floors). The first widely used antiseptic was phenol. Joseph Lister used it for antiseptic surgery in 1867. Phenol is toxic to humans, however, and can cause severe burns when applied to the skin. In the bloodstream, it is a systemic poison—that is, one that is carried to and affects all parts of the body. Its severe side effects led to searches for safer antiseptics, a number of which have been found. An operation in 1753, painted by Gaspare Traversi, of a surgery before antiseptics were used. One safer phenolic antiseptic is 4-hexylresorcinol (4-hexyl-1,3-dihydroxybenzene; resorcinol is the common name for 1,3-dihydroxybenzene, and 4-hexylresorcinol has a hexyl group on the fourth carbon atom of the resorcinol ring). It is much more powerful than phenol as a germicide and has fewer undesirable side effects. Indeed, it is safe enough to be used as the active ingredient in some mouthwashes and throat lozenges. The compound 4-hexylresorcinol is mild enough to be used as the active ingredient in antiseptic preparations for use on the skin. Summary Phenols are compounds in which an OH group is attached directly to an aromatic ring. Many phenols are used as antiseptics. 14.07: Ethers Learning Objectives • Describe the structural difference between an alcohol and an ether that affects physical characteristics and reactivity of each. • Name simple ethers. • Describe the structure and uses of some ethers. With the general formula ROR′, an ether may be considered a derivative of water in which both hydrogen atoms are replaced by alkyl or aryl groups. It may also be considered a derivative of an alcohol (ROH) in which the hydrogen atom of the OH group is been replaced by a second alkyl or aryl group: $\mathrm{HOH\underset{H\: atoms}{\xrightarrow{replace\: both}}ROR'\underset{of\: OH\: group}{\xleftarrow{replace\: H\: atom}}ROH} \nonumber$ Simple ethers have simple common names, formed from the names of the groups attached to oxygen atom, followed by the generic name ether. For example, CH3–O–CH2CH2CH3 is methyl propyl ether. If both groups are the same, the group name should be preceded by the prefix di-, as in dimethyl ether (CH3–O–CH3) and diethyl ether CH3CH2–O–CH2CH3. Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols (Table $1$). Table $1$: Comparison of Boiling Points of Alkanes, Alcohols, and Ethers Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid? CH3CH2CH3 propane 44 –42 no CH3OCH3 dimethyl ether 46 –25 no CH3CH2OH ethyl alcohol 46 78 yes CH3CH2CH2CH2CH3 pentane 72 36 no CH3CH2OCH2CH3 diethyl ether 74 35 no CH3CH2CH2CH2OH butyl alcohol 74 117 yes Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water). Example $1$ What is the common name for each ether? 1. CH3CH2CH2OCH2CH2CH3 Solution 1. The carbon groups on either side of the oxygen atom are propyl (CH3CH2CH2) groups, so the compound is dipropyl ether. 2. The three-carbon group is attached by the middle carbon atom, so it is an isopropyl group. The one-carbon group is a methyl group. The compound is isopropyl methyl ether. Exercise $1$ What is the common name for each ether? 1. CH3CH2CH2CH2OCH2CH2CH2CH3 To Your Health: Ethers as General Anesthetics A general anesthetic acts on the brain to produce unconsciousness and a general insensitivity to feeling or pain. Diethyl ether (CH3CH2OCH2CH3) was the first general anesthetic to be used. Diethyl ether is relatively safe because there is a fairly wide gap between the dose that produces an effective level of anesthesia and the lethal dose. However, because it is highly flammable and has the added disadvantage of causing nausea, it has been replaced by newer inhalant anesthetics, including the fluorine-containing compounds halothane, enflurane, and isoflurane. Unfortunately, the safety of these compounds for operating room personnel has been questioned. For example, female operating room workers exposed to halothane suffer a higher rate of miscarriages than women in the general population. These three modern, inhalant, halogen-containing, anesthetic compounds are less flammable than diethyl ether. Summary To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as the alcohol that is isomeric with it.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/14%3A_Some_Compounds_with_Oxygen_Sulfur_or_a_Halogen/14.05%3A_Phenols.txt
Learning Objectives • Identify thiols (mercaptans) by the presence of an SH group. • The mild oxidation of thiols gives disulfides. Because sulfur is in the same group (6A) of the periodic table as oxygen, the two elements have some similar properties. We might expect sulfur to form organic compounds related to those of oxygen, and indeed it does. Thiols (also called mercaptans), which are sulfur analogs of alcohols, have the general formula RSH. Methanethiol (also called methyl mercaptan), has the formula CH3SH. Ethanethiol (ethyl mercaptan) is the most common odorant for liquid propane (LP) gas. The mild oxidation of thiols gives compounds called disulfides. $\mathrm{2 RSH \xrightarrow{\,[O]\,} RSSR} \nonumber$ The amino acids cysteine [HSCH2CH(NH2)COOH] and methionine [CH3SCH2CH2CH(NH2)COOH] contain sulfur atoms, as do all proteins that contain these amino acids. Disulfide linkages (–S–S–) between protein chains are extremely important in protein structure. Thioethers, which are sulfur analogs of ethers, have the form general formula RSR′. An example is dimethylsulfide (CH3SCH3), which is responsible for the sometimes unpleasant odor of cooking cabbage and related vegetables. Note that methionine has a thioether functional group. Career Focus: Paramedic Paramedics are highly trained experts at providing emergency medical treatment. Their critical duties often include rescue work and emergency medical procedures in a wide variety of settings, sometimes under extremely harsh and difficult conditions. Like other science-based professions, their work requires knowledge, ingenuity, and complex thinking, as well as a great deal of technical skill. The recommended courses for preparation in this field include anatomy, physiology, medical terminology, and—not surprisingly—chemistry. An understanding of basic principles of organic chemistry, for example, is useful when paramedics have to deal with such traumas as burns from fuel (hydrocarbons) or solvent (alcohols, ethers, esters, and so on) fires and alcohol and drug overdoses. To become a paramedic requires 2–4 y of training and usually includes a stint as an emergency medical technician (EMT). An EMT provides basic care, can administer certain medications and treatments, such as oxygen for respiratory problems and epinephrine (adrenalin) for allergic reactions, and has some knowledge of common medical conditions. A paramedic, in contrast, must have extensive knowledge of common medical problems and be trained to administer a wide variety of emergency drugs. Paramedics usually work under the direction of a medical doctor with a title such as “medical director.” Some paramedics are employed by fire departments and may work from a fire engine that carries medical equipment as well as fire-fighting gear. Some work from hospital-sponsored ambulances and continue to care for their patients after reaching the hospital emergency room. Still other paramedics work for a government department responsible for emergency health care in a specific geographical area. Finally, some work for private companies that contract to provide service for a government body. An experienced paramedic has a broad range of employment options, including training for mountain or ocean rescue, working with police department special weapons and tactics (SWAT) teams, or working in isolated settings such as on oil rigs. With their expertise at treating and stabilizing patients before quickly moving them to a hospital, paramedics often provide the first critical steps in saving an endangered life. The following quotation, inscribed on the Arlington National Cemetery headstone of Army Lieutenant R. Adams Cowley, who is often called the “father” of shock trauma medicine, serves as the motto for many paramedic units: “Next to creating a life the finest thing a man can do is save one.” —Abraham Lincoln Summary Thiols, thioethers, and disulfides are common in biological compounds. Concept Review Exercises 1. What is the functional group of a thiol? Write the condensed structural formula for ethanethiol (ethyl mercaptan). 2. What is the functional group of a disulfide? Write the condensed structural formula for dipropyl disulfide. Answers 1. SH; CH3CH2SH 2. –S–S–; CH3CH2CH2SSCH2CH2CH3 Exercises 1. A common natural gas odorant is tert-butyl mercaptan. What is its condensed structural formula? 2. Write the equation for the oxidation of ethanethiol to diethyl disulfide. 1. (CH3)3CSH
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/14%3A_Some_Compounds_with_Oxygen_Sulfur_or_a_Halogen/14.08%3A_Thiols_and_Disulfides.txt
Learning Objectives • Identify and name a simple alkyl halide. The presence of a halogen atom (F, Cl, Br, or I; also, X is used to represent any halogen atom) is one of the simplest functional groups. Organic compounds that contain a halogen atom are called alkyl halides. We have already seen some examples of alkyl halides when the addition of halogens across double and triple bonds was introduced in Section 16.3 - "Branched Hydrocarbons;" the products of these reactions were alkyl halides. A simple alkyl halide can be named like an ionic salt, first by stating the name of the parent alkane as a substituent group (with the -yl suffix) and then the name of the halogen as if it were the anion. So CH3Cl has the common name of methyl chloride, while CH3CH2Br is ethyl bromide and CH3CH2CH2I is propyl iodide. However, this system is not ideal for more complicated alkyl halides. The systematic way of naming alkyl halides is to name the halogen as a substituent, just like an alkyl group, and use numbers to indicate the position of the halogen atom on the main chain. The name of the halogen as a substituent comes from the stem of the element's name plus the ending -o, so the substituent names are fluoro-, chloro-, bromo- and iodo-. If there is more than one of a certain halogen, we use numerical prefixes to indicate the number of each kind, just as with alkyl groups. For example, this molecule is 2-bromobutane, while this molecule is 2,3-dichloropentane. If alkyl groups are present, the substituents are listed alphabetically. Numerical prefixes are ignored when determining the alphabetical ordering of substituent groups. Example \(1\) Name this molecule. Solution The longest carbon chain has five C atoms, so the molecule is a pentane. There are two chlorine substituents located on the second and third C atoms, with a one-carbon methyl group on the third C atom as well. The correct name for this molecule is 2,3-dichloro-3-methylpentane. Exercise \(1\) Name this molecule. Answer 1,1,2-tribromopropane Most alkyl halides are insoluble in H2O. Smaller alcohols, however, are very soluble in H2O because these molecules can engage in hydrogen bonding with H2O molecules. For larger molecules, however, the polar OH group is overwhelmed by the nonpolar alkyl part of the molecule. While methanol is soluble in H2O in all proportions, only about 2.6 g of pentanol will dissolve in 100 g of H2O. Larger alcohols have an even lower solubility in H2O. One reaction common to alcohols and alkyl halides is elimination, the removal of the functional group (either X or OH) and an H atom from an adjacent carbon. The general reaction can be written as follows: where Z represents either the X or the OH group. The biggest difference between elimination in alkyl halides and elimination in alcohols is the identity of the catalyst: for alkyl halides, the catalyst is a strong base; for alcohols, the catalyst is a strong acid. For compounds in which there are H atoms on more than one adjacent carbon, a mixture of products results. Example \(3\) Predict the organic product(s) of this reaction. Solution Under these conditions, an HOH (otherwise known as H2O) molecule will be eliminated, and an alkene will be formed. It does not matter which adjacent carbon loses the H atom; in either case the product will be which is propene. Exercise \(3\) Predict the organic product(s) of this reaction. Answer 1-butene and 2-butene Key Takeaways • Alkyl halides have a halogen atom as a functional group. • Alcohols have an OH group as a functional group. • Nomenclature rules allow us to name alkyl halides and alcohols. • In an elimination reaction, a double bond is formed as an HX or an HOH molecule is removed.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/14%3A_Some_Compounds_with_Oxygen_Sulfur_or_a_Halogen/14.09%3A_Halogen-Containing_Compounds.txt
Learning Objective • Identify the aldehyde and ketone functional groups. There are other functional groups that contain O atoms. Before we introduce them, we define the carbonyl group, which is formed when an O atom and a C atom are joined by a double bond: The other two bonds on the C atom are attached to other atoms. It is the identities of these other atoms that define what specific type of compound an organic molecule is. If one bond of the carbonyl group is made to an H atom, then the molecule is classified as an aldehyde (If there are two H atoms, there is only 1 C atom). When naming aldehydes, the main chain of C atoms must include the carbon in the carbonyl group, which is numbered as position 1 in the carbon chain. The parent name of the hydrocarbon is used, but the suffix -al is appended. (Do not confuse -al with -ol, which is the suffix used for alcohols.) So we have Methanal has a common name with which you may be familiar: formaldehyde. The main thing to note about aldehydes is that the carbonyl group is at the end of a carbon chain. A carbonyl group in the middle of a carbon chain implies that both remaining bonds of the carbonyl group are made to C atoms. This type of molecule is called a ketone. Despite the fact that aldehydes and ketones have the same carbonyl group, they have different chemical and physical properties and are properly grouped as two different types of compounds. The smallest ketone has three C atoms in it. When naming a ketone, we take the name of the parent hydrocarbon and change the suffix to -one: The common name for propanone is acetone. With larger ketones, we must use a number to indicate the position of the carbonyl group, much like a number is used with alkenes and alkynes: There is another way to name ketones: name the alkyl groups that are attached to the carbonyl group and add the word ketone to the name. So propanone can also be called dimethyl ketone, while 2-butanone is called methyl ethyl ketone. Example \(1\) Draw the structure of 2-pentanone. Solution This molecule has five C atoms in a chain, with the carbonyl group on the second C atom. Its structure is as follows: 15.02: Naming Aldehydes and Ketones Learning Objectives • Use the IUPAC system to name and draw aldehydes and ketones. Both common and International Union of Pure and Applied Chemistry (IUPAC) names are frequently used for aldehydes and ketones, with common names predominating for the lower homologs. The common names of aldehydes are taken from the names of the acids into which the aldehydes can be converted by oxidation. The stems for the common names of the first four aldehydes are as follows: • 1 carbon atom: form- • 2 carbon atoms: acet- • 3 carbon atoms: propion- • 4 carbon atoms: butyr- Because the carbonyl group in a ketone must be attached to two carbon groups, the simplest ketone has three carbon atoms. It is widely known as acetone, a unique name unrelated to other common names for ketones. Generally, the common names of ketones consist of the names of the groups attached to the carbonyl group, followed by the word ketone. (Note the similarity to the naming of ethers.) Another name for acetone, then, is dimethyl ketone. The ketone with four carbon atoms is ethyl methyl ketone. Example \(1\) Classify each compound as an aldehyde or a ketone. Give the common name for each ketone. Solution 1. This compound has the carbonyl group on an end carbon atom, so it is an aldehyde. 2. This compound has the carbonyl group on an interior carbon atom, so it is a ketone. Both alkyl groups are propyl groups. The name is therefore dipropyl ketone. 3. This compound has the carbonyl group between two alkyl groups, so it is a ketone. One alkyl group has three carbon atoms and is attached by the middle carbon atom; it is an isopropyl group. A group with one carbon atom is a methyl group. The name is therefore isopropyl methyl ketone. Exercise \(1\) Classify each compound as an aldehyde or a ketone. Give the common name for each ketone. Here are some simple IUPAC rules for naming aldehydes and ketones: Example \(2\) Give the IUPAC name for each compound. Solution 1. There are five carbon atoms in the LCC. The methyl group (CH3) is a substituent on the second carbon atom of the chain; the aldehyde carbon atom is always C1. The name is derived from pentane. Dropping the -e and adding the ending -al gives pentanal. The methyl group on the second carbon atom makes the name 2-methylpentanal. 2. There are five carbon atoms in the LCC. The carbonyl carbon atom is C3, and there are methyl groups on C2 and C4. The IUPAC name is 2,4-dimethyl-3-pentanone. 3. There are six carbon atoms in the ring. The compound is cyclohexanone. No number is needed to indicate the position of the carbonyl group because all six carbon atoms are equivalent. Exercise Give the IUPAC name for each compound. Example \(3\) Draw the structure for each compound. 1. 7-chlorooctanal 2. 4-methyl–3-hexanone Solution 1. The octan- part of the name tells us that the LCC has eight carbon atoms. There is a chlorine (Cl) atom on the seventh carbon atom; numbering from the carbonyl group and counting the carbonyl carbon atom as C1, we place the Cl atom on the seventh carbon atom. • The hexan- part of the name tells us that the LCC has six carbon atoms. The 3 means that the carbonyl carbon atom is C3 in this chain, and the 4 tells us that there is a methyl (CH3) group at C4: Exercise Draw the structure for each compound. 1. 5-bromo-3-iodoheptanal 2. 5-bromo-4-ethyl-2-heptanone Summary The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone. Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone. 1. The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group. 2. For an aldehyde, drop the -e from the alkane name and add the ending -al. Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde. 3. For a ketone, drop the -e from the alkane name and add the ending -one. Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone. 4. To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number. 5. To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1. 6. Give the structure and IUPAC name for the compound that has the common name m-bromobenzaldehyde. 7. Give the IUPAC name for glyceraldehyde, (HOCH2CHOHCHO). (Hint: as a substituent, the OH group is named hydroxy.)
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/15%3A_Aldehydes_and_Ketones/15.01%3A_The_Carbonyl_Group.txt
Learning Objectives • Explain why the boiling points of aldehydes and ketones are higher than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols. • Compare the solubilities in water of aldehydes and ketones of four or fewer carbon atoms with the solubilities of comparable alkanes and alcohols. • Name the typical reactions take place with aldehydes and ketones. • Describe some of the uses of common aldehydes and ketones. The carbon-to-oxygen double bond is quite polar, more polar than a carbon-to-oxygen single bond. The electronegative oxygen atom has a much greater attraction for the bonding electron pairs than does the carbon atom. The carbon atom has a partial positive charge, and the oxygen atom has a partial negative charge: In aldehydes and ketones, this charge separation leads to dipole-dipole interactions that are great enough to significantly affect the boiling points. Table $1$ shows that the polar single bonds in ethers have little such effect, whereas hydrogen bonding between alcohol molecules is even stronger. Table $1$: Boiling Points of Compounds Having Similar Molar Masses but Different Types of Intermolecular Forces Compound Family Molar Mass Type of Intermolecular Forces Boiling Point (°C) CH3CH2CH2CH3 alkane 58 dispersion only –1 CH3OCH2CH3 ether 60 weak dipole 6 CH3CH2CHO aldehyde 58 strong dipole 49 CH3CH2CH2OH alcohol 60 hydrogen bonding 97 Formaldehyde is a gas at room temperature. Acetaldehyde boils at 20°C; in an open vessel, it boils away in a warm room. Most other common aldehydes are liquids at room temperature. Although the lower members of the homologous series have pungent odors, many higher aldehydes have pleasant odors and are used in perfumes and artificial flavorings. As for the ketones, acetone has a pleasant odor, but most of the higher homologs have rather bland odors. The oxygen atom of the carbonyl group engages in hydrogen bonding with a water molecule. The solubility of aldehydes is therefore about the same as that of alcohols and ethers. Formaldehyde, acetaldehyde, and acetone are soluble in water. As the carbon chain increases in length, solubility in water decreases. The borderline of solubility occurs at about four carbon atoms per oxygen atom. All aldehydes and ketones are soluble in organic solvents and, in general, are less dense than water. Oxidation of Aldehydes and Ketones Aldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen (O2) in air to carboxylic acids. $2RCHO + O_2 \rightarrow 2RCOOH \label{14.10.1}$ The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent, for example, is an alkaline solution of silver (Ag+) ion complexed with ammonia (NH3), which keeps the Ag+ ion in solution. $H_3N—Ag^+—NH_3 \label{14.10.2}$ When Tollens’ reagent oxidizes an aldehyde, the Ag+ ion is reduced to free silver (Ag). Deposited on a clean glass surface, the silver produces a mirror (Figure $1$). Ordinary ketones do not react with Tollens’ reagent. Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes. Summary The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/15%3A_Aldehydes_and_Ketones/15.03%3A_Properties_of_Aldehydes_and_Ketones.txt
Learning Objectives • Objective 1 • Objective 2 Formaldehyde has an irritating odor. Because of its reactivity, it is difficult to handle in the gaseous state. For many uses, it is therefore dissolved in water and sold as a 37% to 40% aqueous solution called formalin. Formaldehyde denatures proteins, rendering them insoluble in water and resistant to bacterial decay. For this reason, formalin is used in embalming solutions and in preserving biological specimens. Aldehydes are the active components in many other familiar substances. Large quantities of formaldehyde are used to make phenol-formaldehyde resins for gluing the wood sheets in plywood and as adhesives in other building materials. Sometimes the formaldehyde escapes from the materials and causes health problems in some people. While some people seem unaffected, others experience coughing, wheezing, eye irritation, and other symptoms. The odor of green leaves is due in part to a carbonyl compound, cis-3-hexenal, which with related compounds is used to impart a “green” herbal odor to shampoos and other products. Acetaldehyde is an extremely volatile, colorless liquid. It is a starting material for the preparation of many other organic compounds. Acetaldehyde is formed as a metabolite in the fermentation of sugars and in the detoxification of alcohol in the liver. Aldehydes are the active components of many other familiar materials (Figure \(2\)). Acetone is the simplest and most important ketone. Because it is miscible with water as well as with most organic solvents, its chief use is as an industrial solvent (for example, for paints and lacquers). It is also the chief ingredient in some brands of nail polish remover. To Your Health: Acetone in Blood, Urine, and Breath Acetone is formed in the human body as a by-product of lipid metabolism. Normally, acetone does not accumulate to an appreciable extent because it is oxidized to carbon dioxide and water. The normal concentration of acetone in the human body is less than 1 mg/100 mL of blood. In certain disease states, such as uncontrolled diabetes mellitus, the acetone concentration rises to higher levels. It is then excreted in the urine, where it is easily detected. In severe cases, its odor can be noted on the breath. Ketones are also the active components of other familiar substances, some of which are noted in the accompanying figure. Some ketones have interesting properties: (a) Butter flavoring comes from 2,3-butanedione; (b) β-ionone is responsible for the odor of violets; (c) muscone is musk oil, an ingredient in perfumes; and (d) camphor is used in some insect repellents. Certain steroid hormones have the ketone functional group as a part of their structure. Two examples are progesterone, a hormone secreted by the ovaries that stimulates the growth of cells in the uterine wall and prepares it for attachment of a fertilized egg, and testosterone, the main male sex hormone. These and other sex hormones affect our development and our lives in fundamental ways. 15.05: Oxidation of Aldehydes Learning Objectives • Objective 1 • Objective 2 ldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen (O2) in air to carboxylic acids. $2RCHO + O_2 \rightarrow 2RCOOH \label{14.10.1}$ The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent, for example, is an alkaline solution of silver (Ag+) ion complexed with ammonia (NH3), which keeps the Ag+ ion in solution. $H_3N—Ag^+—NH_3 \label{14.10.2}$ When Tollens’ reagent oxidizes an aldehyde, the Ag+ ion is reduced to free silver (Ag). Deposited on a clean glass surface, the silver produces a mirror (Figure $1$). Ordinary ketones do not react with Tollens’ reagent. Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/15%3A_Aldehydes_and_Ketones/15.04%3A_Some_Common_Aldehydes_and_Ketones.txt
Learning Objectives • Determine the structural feature that classifies amines as primary, secondary, or tertiary. Amines are classified according to the number of carbon atoms bonded directly to the nitrogen atom. A primary (1°) amine has one alkyl (or aryl) group on the nitrogen atom, a secondary (2°) amine has two, and a tertiary (3°) amine has three (Figure \(1\)). To classify alcohols, we look at the number of carbon atoms bonded to the carbon atom bearing the OH group, not the oxygen atom itself. Thus, although isopropylamine looks similar to isopropyl alcohol, the former is a primary amine, while the latter is a secondary alcohol. The common names for simple aliphatic amines consist of an alphabetic list of alkyl groups attached to the nitrogen atom, followed by the suffix -amine. (Systematic names are often used by some chemists.) The amino group (NH2) is named as a substituent in more complicated amines, such as those that incorporate other functional groups or in which the alkyl groups cannot be simply named. Example \(1\) Name and classify each compound. 1. CH3CH2CH2NH2 • CH3CH2NHCH2CH3 • CH3CH2CH2NHCH3 Solution 1. There is only one alkyl group attached to the nitrogen atom, so the amine is primary. A group of three carbon atoms (a propyl group) is attached to the NH2 group through an end carbon atom, so the name is propylamine. 2. There are two methyl groups and one ethyl group on the nitrogen atom. The compound is ethyldimethylamine, a tertiary amine. 3. There are two ethyl groups attached to the nitrogen atom; the amine is secondary, so the compound is diethylamine. 4. The nitrogen atom has a methyl group and a propyl group, so the compound is methylpropylamine, a secondary amine. 16.02: Naming and Drawing Amines Learning Objectives • Objective 1 • Objective 2 The common names for simple aliphatic amines consist of an alphabetic list of alkyl groups attached to the nitrogen atom, followed by the suffix -amine. (Systematic names are often used by some chemists.) The amino group (NH2) is named as a substituent in more complicated amines, such as those that incorporate other functional groups or in which the alkyl groups cannot be simply named. Example \(1\) Name and classify each compound. 1. CH3CH2CH2NH2 • CH3CH2NHCH2CH3 • CH3CH2CH2NHCH3 Solution 1. There is only one alkyl group attached to the nitrogen atom, so the amine is primary. A group of three carbon atoms (a propyl group) is attached to the NH2 group through an end carbon atom, so the name is propylamine. 2. There are two methyl groups and one ethyl group on the nitrogen atom. The compound is ethyldimethylamine, a tertiary amine. 3. There are two ethyl groups attached to the nitrogen atom; the amine is secondary, so the compound is diethylamine. 4. The nitrogen atom has a methyl group and a propyl group, so the compound is methylpropylamine, a secondary amine. Exercise \(1\) Name and classify each compound. • CH3CH2CH2CH2NH2 • CH3CH2CH2NHCH2CH2 CH3 Example \(2\) Draw the structure for each compound and classify. 1. isopropyldimethylamine 2. dipropylamine Solution 1. The name indicates that there are an isopropyl group (in red) and two methyl groups (in green) attached to the nitrogen atom; the amine is tertiary. • The name indicates that there are two propyl groups attached to the nitrogen atom; the amine is secondary. (The third bond on the nitrogen atom goes to a hydrogen atom.) CH3CH2CH2NHCH2CH2CH3 Exercise \(2\) Draw the structure for each compound and classify. 1. ethylisopropylamine 2. diethylpropylamine The primary amine in which the nitrogen atom is attached directly to a benzene ring has a special name—aniline. Aryl amines are named as derivatives of aniline. Example \(3\) Name this compound. Solution The benzene ring with an amino (NH2) group is aniline. The compound is named as a derivative of aniline: 3-bromoaniline or m-bromoaniline. Exercise \(3\) Name this compound. Example \(4\) Draw the structure for p-ethylaniline and classify. Solution The compound is a derivative of aniline. It is a primary amine having an ethyl group located para to the amino (NH2) group. Exercise \(4\) Draw the structure for p-isopropylaniline and classify. Example \(5\) Draw the structure for 2-amino-3-methylpentane. Solution Always start with the parent compound: draw the pentane chain. Then attach a methyl group at the third carbon atom and an amino group at the second carbon atom. Exercise \(5\) Draw the structure for 2-amino-3-ethyl-1-chloroheptane. Ammonium (NH4+) ions, in which one or more hydrogen atoms are replaced with alkyl groups, are named in a manner analogous to that used for simple amines. The alkyl groups are named as substituents, and the parent species is regarded as the NH4+ ion. For example, CH3NH3+ is the methylammonium ion. The ion formed from aniline (C6H5NH3+) is called the anilinium ion. Example \(6\) Name each ion. 1. CH3NH3+ 2. (CH3)2NH2+ 3. (CH3)3NH+ 4. (CH3)4N+ Solution The ions have one, two, three, and four methyl (CH3) groups attached to a nitrogen atom. Their names are as follows: 1. methylammonium ion 2. dimethylammonium ion 3. trimethylammonium ion 4. tetramethylammonium ion Exercise \(6\) Name each ion. 1. CH3CH2NH3+ 2. (CH3CH2)3NH+ 3. (CH3CH2CH2)2NH2+ 4. (CH3CH2CH2CH2)4N+ Summary An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. The amine functional group is as follows: Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. Amines are named by naming the alkyl groups attached to the nitrogen atom, followed by the suffix -amine.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/16%3A_Amines/16.01%3A_Classifying_Amines.txt
Learning Objectives • Explain why the boiling points of primary and secondary amines are higher than those of alkanes or ethers of similar molar mass but are lower than those of alcohols. • Compare the boiling points of tertiary amines with alcohols, alkanes, and ethers of similar molar mass. • Compare the solubilities in water of amines of five or fewer carbon atoms with the solubilities of comparable alkanes and alcohols in water. Primary and secondary amines have hydrogen atoms bonded to an nitrogen atom and are therefore capable of hydrogen bonding (part (a) of Figure \(1\)), although not as strongly as alcohol molecules (which have hydrogen atoms bonded to an oxygen atom, which is more electronegative than nitrogen). These amines boil at higher temperatures than alkanes but at lower temperatures than alcohols of comparable molar mass. For example, compare the boiling point of methylamine (CH3NH2; −6°C) with those of ethane (CH3CH3; −89°C) and methanol (CH3OH; 65°C). Tertiary amines have no hydrogen atom bonded to the nitrogen atom and so cannot participate in intermolecular hydrogen bonding. They have boiling points comparable to those of ethers (Table \(1\)). Table \(1\): Physical Properties of Some Amines and Comparable Oxygen-Containing Compounds Name Condensed Structural Formula Class Molar Mass Boiling Point (°C) Solubility at 25°C (g/100 g Water) butylamine CH3CH2CH2CH2NH2 73 78 miscible diethylamine (CH3CH2)2NH 73 55 miscible butyl alcohol CH3CH2CH2CH2OH 74 118 8 dipropylamine (CH3CH2CH2)2NH 101 111 4 triethylamine (CH3CH2)3N 101 90 14 dipropyl ether (CH3CH2CH2)2O 102 91 0.25 All three classes of amines can engage in hydrogen bonding with water (Figure \(\PageIndex{1b}\)). Amines of low molar mass are quite soluble in water; the borderline of solubility in water is at five or six carbon atoms. To Your Health: Amines in Death and Life Amines have “interesting” odors. The simple ones smell very much like ammonia. Higher aliphatic amines smell like decaying fish. Or perhaps we should put it the other way around: Decaying fish give off odorous amines. The stench of rotting fish is due in part to two diamines: putrescine and cadaverine. They arise from the decarboxylation of ornithine and lysine, respectively, amino acids that are found in animal cells. Aromatic amines generally are quite toxic. They are readily absorbed through the skin, and workers must exercise caution when handling these compounds. Several aromatic amines, including β-naphthylamine, are potent carcinogens. Key Takeaways • Primary and secondary amines have higher boiling points than those of alkanes or ethers of similar molar mass because they can engage in intermolecular hydrogen bonding. Their boiling points are lower than those of alcohols because alcohol molecules have hydrogen atoms bonded to an oxygen atom, which is more electronegative. • The boiling points of tertiary amines, which cannot engage in hydrogen bonding because they have no hydrogen atom on the nitrogen atom, are comparable to those of alkanes and ethers of similar molar mass. • Because all three classes of amines can engage in hydrogen bonding with water, amines of low molar mass are quite soluble in water. Concept Review Exercises 1. Which compound has the higher boiling point, CH3CH2CH2CH2CH2NH2 or CH3CH2CH2CH2CH2CH3? Explain. 2. Which compound is more soluble in water, CH3CH2CH2CH2CH3 or CH3CH2NHCH2CH3? Explain. Answers 1. CH3CH2CH2CH2CH2NH2 because the nitrogen-to-hydrogen (N–H) bonds can engage in hydrogen bonding; CH3CH2CH2CH2CH2CH3 cannot engage in hydrogen bonding 2. CH3CH2NHCH2CH3 because amines can engage in hydrogen bonding with water; alkanes cannot engage in hydrogen bonding Exercises 1. Which compound of each pair has the higher boiling point? Explain. 1. butylamine or pentane 2. CH3NH2 or CH3CH2CH2CH2CH2NH2 2. Which compound of each pair has the higher boiling point? Explain. 1. butylamine or butyl alcohol 2. trimethylamine or propylamine 3. Which compound is more soluble in water—CH3CH2CH3 or CH3CH2NH2? Explain. 4. Which compound is more soluble in water—CH3CH2CH2NH2 or CH3CH2CH2CH2CH2CH2NH2? Explain. Answers 1. butylamine because the N–H bonds can engage in hydrogen bonding; pentane cannot engage in hydrogen bonding 2. CH3CH2CH2CH2CH2NH2 because it has a greater molar mass than CH3NH2 1. CH3CH2NH2 because amines can engage in hydrogen bonding with water; alkanes cannot engage in hydrogen bonding
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/16%3A_Amines/16.03%3A_Properties_of_Amines.txt
• Objective 1 • Objective 2 Heterocyclic Amines Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros, meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis. Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid, a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine. To Your Health: Three Well-Known Alkaloids Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea. Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide. Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine. Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine. Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s. Summary Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring. 16.05: Basicity of Amines Learning Objectives • Name the typical reactions that take place with amines. • Describe heterocyclic amines. Recall that ammonia (NH3) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH4+) ions and hydroxide (OH) ions: As a specific example, methylamine reacts with water to form the methylammonium ion and the OH ion. Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water. Amine salts are named like other salts: the name of the cation is followed by the name of the anion. Example \(1\) What are the formulas of the acid and base that react to form [CH3NH2CH2CH3]+CH3COO? Solution The cation has two groups—methyl and ethyl—attached to the nitrogen atom. It comes from ethylmethylamine (CH3NHCH2CH3). The anion is the acetate ion. It comes from acetic acid (CH3COOH). Exercise \(1\) What are the formulas of the acid and base that react to form (CH3CH2CH2)3NH+I? To Your Health: Amine Salts as Drugs Salts of aniline are properly named as anilinium compounds, but an older system, still in use for naming drugs, identifies the salt of aniline and hydrochloric acid as “aniline hydrochloride.” These compounds are ionic—they are salts—and the properties of the compounds (solubility, for example) are those characteristic of salts. Many drugs that are amines are converted to hydrochloride salts to increase their solubility in aqueous solution. 16.07: Amines in Plants- Alkaloids • Objective 1 • Objective 2 Addictive Alkaloids Since ancient times, plants have been used for medicinal purposes. One class of substances, called alkaloids, found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with H3O+ in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant: $\ce{R3N + H3O+ + Cl- ⟶[R3NH+]Cl- + H2O} \nonumber$ The name alkaloid means “like an alkali.” Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base: $\ce{[R3NH+]Cl- + OH- ⟶R3N + H2O + Cl-} \nonumber$ The structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant:
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/16%3A_Amines/16.04%3A_Heterocyclic_Nitrogen_Compounds.txt
• Objective 1 • Objective 2 17.02: Acidity of Carboxylic Acids Learning Objectives • Define \(pK_a\) and use it to determine acidity of different carboxylic acids. • Describe the reactions between carboxylic acids and strong bases. Ionization of Carboxylic Acids Carboxylic acids are named such because they tend to be more acidic than other functional groups in organic chemistry. In dilute aqueous solutions, they act as weak acids that partially dissociate to produce the corresponding carboxylate anion and hydronium cation (H3O+). Carboxylate anions are named by replacing the -ic acid ending from the carboxylic acid with -ate, see examples below. The extent of dissociation of these weak acids in water is described by \(K_a\) values. Remember that a compound with a smaller \(K_a\) value will be a weaker acid. \({RCOOH+H_2O\rightleftharpoons RCOO^-+H_3O^+}\) \(K_a = \dfrac{[\ce{RCOO^{-}}][\ce{H3O^{+}}]}{[\ce{RCOOH}]}\) When comparing the acidity of organic and biomolecules, it is useful (and more preferable) to use \(pK_a\) values instead of \(K_a\) values, which are calculated by taking the negative log of \(K_a\): \(pK_a =\ –\log K_a\). When using the \(pK_a\) scale, it is important to know that weaker acids have larger and more positive \(pK_a\) values, this is opposite of \(K_a\) values. The \(pK_a\) values of some typical carboxylic acids are listed in Table \(1\). (Remember that \(pK_a\) is a log expression, which means that every 1 \(pK_a\) unit represents a 10-fold change in acidity.) Table \(1\): Comparisons of Carboxylic Acid \(K_a\) and \(pK_a\) Values Name Compound \(K_a\) \(pK_a\) formic acid HCOOH 1.8 X 10–4 3.74 acetic acid CH3COOH 1.8 X 10–5 4.74 propanoic acid CH3CH2COOH 1.3 X 10–5 4.89 butanoic acid CH3CH2CH2COOH 1.5 X 10–5 4.82 chloroacetic acid ClCH2COOH 1.4 X 10–3 2.85 trichloroacetic acid Cl3CCOOH 2.3 X 10–1 0.64 hexanoic acid CH3(CH2)4COOH 1.3 X 10–5 4.89 benzoic acid C6H5COOH 6.5 X 10–5 4.19 oxalic acid HOOCCOOH 5.4 X 10–2 1.27 OOCCOOH 5.2 X 10–5 4.28 glutaric acid HOOC(CH2)3COOH 4.5 X 10–5 4.35 OOC(CH2)3COOH 3.8 X 10–6 5.42 Neutralization of Carboxylic Acids Carboxylic acids will react with bases such as sodium hydroxide (NaOH), sodium carbonate (Na2CO3), and sodium bicarbonate (NaHCO3) to form water and a carboxylic acid salt: RCOOH + NaOH(aq) → RCOONa+(aq) + H2O 2RCOOH + Na2CO3(aq) → 2RCOONa+(aq) + H2O + CO2(g) RCOOH + NaHCO3(aq) → RCOONa+(aq) + H2O + CO2(g) In these reactions, the carboxylic acids act like inorganic acids: they neutralize basic compounds. With solutions of carbonate (\(CO_3^{2–}\)) and bicarbonate (\(HCO_3^{–}\)) ions, they also form carbon dioxide gas. Carboxylic acid salts are named in the same manner as inorganic salts: the name of the cation is followed by the name of the organic anion. The name of the anion is obtained by dropping the -ic ending of the acid name and replacing it with the suffix -ate. This rule applies whether we are using common names or International Union of Pure and Applied Chemistry (IUPAC) names: Note The salts of long-chain carboxylic acids are called soaps. Example \(1\) Write an equation for each reaction. 1. the ionization of propionic acid in water (H2O) 2. the neutralization of propionic acid with aqueous sodium hydroxide (NaOH) Solution Propionic acid has three carbon atoms, so its formula is CH2CH2COOH. 1. Propionic acid ionizes in water to form a propionate ion and a hydronium (H3O+) ion. CH3CH2COOH(aq) + H2O(ℓ) → CH3CH2COO(aq) + H3O+(aq) 2. Propionic acid reacts with NaOH(aq) to form sodium propionate and water. CH3CH2COOH(aq) + NaOH(aq) → CH3CH2COONa+(aq) + H2O(ℓ) Exercise \(1\) Write an equation for each reaction. 1. the ionization of formic acid in water 2. the ionization of p-chlorobenzoic acid in water Example \(2\) Write an equation for the reaction of decanoic acid with each compound. 1. aqueous sodium hydoxide (NaOH) 2. aqueous sodium bicarbonate (NaHCO3) Solution 1. Decanoic acid has 10 carbon atoms. It reacts with NaOH to form a salt and water (H2O). CH3(CH2)8COOH + NaOH(aq) → CH3(CH2)8COONa+(aq) + H2O(ℓ) 2. With NaHCO3, the products are a salt, H2O, and carbon dioxide (CO2). CH3(CH2)8COOH + NaHCO3(aq) → CH3(CH2)8COONa+(aq) + H2O(ℓ) + CO2(g) Exercise \(3\) Write an equation for the reaction of benzoic acid with each compound. 1. aqueous sodium hydroxide (NaOH) 2. aqueous sodium bicarbonate (NaHCO3) Note To Your Health: Organic Salts as Preservatives Some organic salts are used as preservatives in food products. They prevent spoilage by inhibiting the growth of bacteria and fungi. Calcium and sodium propionate, for example, are added to processed cheese and bakery goods; sodium benzoate is added to cider, jellies, pickles, and syrups; and sodium sorbate and potassium sorbate are added to fruit juices, sauerkraut, soft drinks, and wine. Look for them on ingredient labels the next time you shop for groceries.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/17%3A_Carboxylic_Acids_and_their_Derivatives/17.01%3A_Carboxylic_Acids_and_Their_Derivatives-_Properties_and_Names.txt
Learning Objectives • To identify and describe the substances from which most esters are prepared. • To identify and describe the substances from which most amides are prepared. Carboxylic acids will react with alcohols and amines following a similar pattern. In both cases, the –OH group of the carboxylic acid will be replaced by a different group to form either an ester or an amide, with water formed as a by-product. When the reaction involves an alcohol, the –OH of the acid is replaced by the –OR' of the alcohol. When the reaction involves an amine, the –OH of the acid is replaced by the –NH2, or –NHR', or –NR'2 of the amine. Ester formation Amide formation Esterification Esters are prepared by esterification, a reaction in which a carboxylic acid and an alcohol are heated in the presence of an acid catalyst: The reaction is reversible and will reach equilibrium with approximately equivalent amounts of reactants and products. Using excess amounts of alcohol and continuously removing a product, can drive the reaction towards the product side as per LeChatelier's principle. A Closer Look: Condensation Polymers A commercially important esterification reaction is condensation polymerization, in which a reaction occurs between a dicarboxylic acid and a dihydric alcohol (diol), with the elimination of water. Such a reaction yields an ester that contains a free (unreacted) carboxyl group at one end and a free alcohol group at the other end. Further condensation reactions then occur, producing polyester polymers. The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers: Polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda pop and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes. Synthetic arteries can be made from PET, polytetrafluoroethylene, and other polymers. Amide Formation When a carboxylic acid reacts with ammonia (NH3) a primary amide is formed: When a carboxylic acid reacts with primary or secondary amines, secondary or tertiary amides are produced, respectively. Need another figure here Tertiary amines do not have a hydrogen attached to the nitrogen and therefore do not form amides when mixed with carboxylic acids. However, an acid-base reaction does occur with the amine accepting a proton (acts as a base) and the carboxylic acid donating a proton. In this case the ammonium and carboxylate salts are formed: Need another figure here Note Polyamides Just as the reaction of a diol and a diacid forms a polyester, the reaction of a diacid and a diamine yields a polyamide. The two difunctional monomers often employed are adipic acid and 1,6-hexanediamine. The monomers condense by splitting out water to form a new product, which is still difunctional and thus can react further to yield a polyamide polymer. Some polyamides are known as nylons. Nylons are among the most widely used synthetic fibers—for example, they are used in ropes, sails, carpets, clothing, tires, brushes, and parachutes. They also can be molded into blocks for use in electrical equipment, gears, bearings, and valves.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/17%3A_Carboxylic_Acids_and_their_Derivatives/17.03%3A_Reactions_of_Carboxylic_Acids_-_Ester_and_Amide_Formation.txt
Learning Objectives • Describe the typical reaction that takes place with esters. • Identify the products of an acidic hydrolysis of an ester. • Identify the products of a basic hydrolysis of an ester. Esters are neutral compounds, unlike the acids from which they are formed. In typical reactions, the alkoxy (OR′) group of an ester is replaced by another group. One such reaction is hydrolysis, literally “splitting with water.” The hydrolysis of esters is catalyzed by either an acid or a base. Acidic hydrolysis is simply the reverse of esterification. The ester is heated with a large excess of water containing a strong-acid catalyst. Like esterification, the reaction is reversible and does not go to completion. As a specific example, butyl acetate and water react to form acetic acid and 1-butanol. The reaction is reversible and does not go to completion. Example \(1\) Write an equation for the acidic hydrolysis of ethyl butyrate (CH3CH2CH2COOCH2CH3) and name the products. Solution Remember that in acidic hydrolysis, water (HOH) splits the ester bond. The H of HOH joins to the oxygen atom in the OR part of the original ester, and the OH of HOH joins to the carbonyl carbon atom: The products are butyric acid (butanoic acid) and ethanol. Exercise \(1\) Write an equation for the acidic hydrolysis of methyl butanoate and name the products. When a base (such as sodium hydroxide [NaOH] or potassium hydroxide [KOH]) is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils, alkaline hydrolysis of esters is called saponification (Latin sapon, meaning “soap,” and facere, meaning “to make”). In a saponification reaction, the base is a reactant, not simply a catalyst. The reaction goes to completion: As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol: Example \(2\) Write an equation for the hydrolysis of methyl benzoate in a potassium hydroxide solution. Solution In basic hydrolysis, the molecule of the base splits the ester linkage. The acid portion of the ester ends up as the salt of the acid (in this case, the potassium salt). The alcohol portion of the ester ends up as the free alcohol. Exercise \(2\) Write the equation for the hydrolysis of ethyl propanoate in a sodium hydroxide solution. Hydrolysis of Amides Generally, amides resist hydrolysis in plain water, even after prolonged heating. In the presence of added acid or base, however, hydrolysis proceeds at a moderate rate. In living cells, amide hydrolysis is catalyzed by enzymes. Amide hydrolysis is illustrated in the following example: Hydrolysis of an amide in acid solution actually gives a carboxylic acid and the salt of ammonia or an amine (the ammonia or amine initially formed is neutralized by the acid). Basic hydrolysis gives a salt of the carboxylic acid and ammonia or an amine. Example \(1\) Write the equation for the hydrolysis of each compound. 1. butyramide 2. benzamide Solution 1. The hydrolysis of a simple amide produces an organic acid and ammonia. Butyramide thus yields butyric acid and ammonia. • The hydrolysis of an amide produces an organic acid and ammonia. Benzamide thus yields benzoic acid and ammonia. Exercise \(1\) Write the equation for the hydrolysis of each compound. 1. propionamide (propanamide) 2. hexanamide Career Focus: Athletic Trainer Athletic training is an allied health-care profession recognized by the American Medical Association. The athletic trainer’s role is to recognize, evaluate, and provide immediate care for athletic injuries; prevent athletic injuries by taping, bandaging, and bracing vulnerable body parts; make referrals to medical doctors when necessary; and rehabilitate injured athletes. Athletic trainers work in high schools, colleges, and other organizations where athletics programs are found. Athletic trainers usually have a degree from an accredited athletic training program whose curriculum includes such basic science courses as biology, chemistry, and physics. These studies provide the necessary background for more applied courses, such as anatomy and physiology, exercise physiology, kinesiology, and nutrition. Knowledge of chemistry is necessary for understanding pharmacological and medical terminology. For example, athletic trainers must understand the action of numerous drugs, many of which are esters, amines, or amides like those mentioned in this chapter. Athletic trainers may have administrative duties, such as the responsibility for ordering supplies. They also need to be able to evaluate nutritional supplements because providing the wrong one can get an athlete banned from competition and may bring sanctions against a school. In short, the athletic trainer is responsible for the overall health and well-being of the athletes in his or her charge. Summary Hydrolysis is a most important reaction of esters. Acidic hydrolysis of an ester gives a carboxylic acid and an alcohol. Basic hydrolysis of an ester gives a carboxylate salt and an alcohol.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/17%3A_Carboxylic_Acids_and_their_Derivatives/17.04%3A_Hydrolysis_of_Esters_and_Amides.txt
Learning Objectives • Describe how polyamides and polyesters are formed. Animal intestines and silk were used for all guitar strings for centuries, until modern technology and changes in musical taste brought about significant changes. There are two major types of guitar strings in use today. Steel strings (first developed around 1900) are found on acoustic and electric guitars. They have a bright, crisp sound that lends itself well to diverse music such as jazz, rock 'n' roll, and bluegrass. Nylon strings are a more recent development. During World War II, the silk and animal products needed to manufacture steel guitar strings were not available. Nylon quickly proved to be a more-than-adequate substitute. Now nylon strings are found on all classical guitars. Their sound is somewhat softer than the steel strings, making the tone quality well-suited for the classical genre of music. Polymerization - Condensation Polymers A condensation polymer is a polymer formed by condensation reactions. Monomers of condensation polymers must contain two functional groups so that each monomer can link up with two other monomers. One type of condensation polymer is called a polyamide. One pair of monomers that can form a polyamide is that of adipic acid and hexanediamine. Adipic acid is a carboxylic acid with two carboxyl groups on either end of the molecule. Hexanediamine has amino groups on either end of a six-carbon chain. When these molecules react with each other, a molecule of water is eliminated, classifying it as a condensation reaction (see figure below). The polymer that results from the repetition of the condensation reaction is a polyamide called nylon-66. Nylon-66 was first invented in 1935 and has been used in all sorts of products. Polyamides, including Nylon-66, are commonly found in fibers and clothing, cooking utensils, fishing line, and carpeting—among many other applications. Polyester is another common type of condensation polymer. Recall that esters are formed from the reaction of an alcohol with a carboxylic acid. When both the acid and alcohol have two functional groups, the ester is capable of being polymerized. One such polyester is called polyethylene terephthalate (PET) and is formed from the reaction of ethylene glycol with terephthalic acid. The structure of PET is shown below. PET is used in tires, photographic film, food packaging, and clothing. Polyester fabric is used in permanent-press clothing. Its resistance to wrinkling comes from the cross-linking of the polymer strands. Summary • A condensation polymer is a polymer formed by condensation reactions. • Polyamides and polyesters are common types of condensation polymers. 17.06: Phosphoric Acid Derivatives Learning Objectives • Objective 1 • Objective 2 Just as carboxylic acids do, inorganic acids such as nitric acid (HNO3), sulfuric acid (H2SO4), and phosphoric acid (H3PO4) also form esters. The esters of phosphoric acid are especially important in biochemistry. A phosphoric acid molecule can form a monoalkyl, a dialkyl, or a trialkyl ester by reaction with one, two, or three molecules of an alcohol. Esters of pyrophosphoric acid and triphosphoric acid are also important in biochemistry. Esters of these acids are present in every plant and animal cell. They are biochemical intermediates in the transformation of food into usable energy. The bonds between phosphate units in adenosine triphosphate (ATP) are called phosphoanhydride bonds. These are high-energy bonds that store energy from the metabolism of foods. Hydrolysis of ATP releases energy as it is needed for biochemical processes (for instance, for muscle contraction). Phosphate esters are also important structural constituents of phospholipids and nucleic acids. The explosive nitroglycerin (glyceryl trinitrate) is an ester formed from glycerol and nitric acid. It is used in medicine to relieve chest pain in heart disease. Summary Inorganic acids such as H3PO4 form esters. The esters of phosphoric acid are especially important in biochemistry.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/17%3A_Carboxylic_Acids_and_their_Derivatives/17.05%3A_Polyamides_and_Polyesters.txt
Learning Objectives • Explain what a biomolecule is and list the four main types. Biochemistry is the study of the molecules of life, (biomolecules); those that structurally make up living organisms and function to keep them alive. Although the complexity of biomolecules ranges from individual small molecules, such as glycine, to very large complexes made up of multiple molecules linked together, like ATP synthase, most biomolecules can be categorized into for main groups based on their structural similarities: carbohydrates, proteins, lipids, or nucleic acids. There are also many other small molecules and ions that play a wide variety of roles in the cell, ranging chemical messengers (or signals), to toxins produced as a means of defense against invaders, to coordination complexes that play important rolls in protein function. Now that you have learned the basics of general and organic chemistry, you are ready to apply your knowledge to understanding the chemistry of these biomolecules. You will recognize some of the organic functional groups in these biomolecules, which will help you predict the physical and chemical properties of these new molecules. Living cells are very complex, however, the same principles that govern chemistry apply to all of biochemistry. 18.02: Proteins and Their Functions- An Overview Learning Objectives • Describe the different structural classes of proteins. • Understand the different functional roles of proteins. The ability to serve a variety of functions is characteristic of most biomolecules. Nowhere is this versatility better exemplified than by the proteins. Perhaps because of their many functions, proteins are the most abundant organic molecules in living cells, constituting more than 50 percent of the mass once water is removed. It is estimated that the human body contains well over a million different kinds of protein, and even a single-cell organism contains thousands. Each of these is a polymer of amino acids which has a highly specific composition, a unique molecular weight (usually in the range from 6000 to 1 000 000) and its own sequence of different amino acids along the polymer chain. Proteins may be subdivided on the basis of their molecular shape or conformation. In the fibrous proteins, long polymer chains are arranged parallel or nearly parallel to one another to give long fibers or sheets. This arrangement results in physically tough materials which do not dissolve in water. The fibrous proteins are fundamental components of structural tissues such as tendons, bone, hair, horn, leather, claws, and feathers. By contrast, polymer chains of the globular proteins fold back on themselves to produce compact, nearly spherical shapes. Most globular proteins are water soluble and hence are relatively mobile within a cell. Some examples are enzymes, antibodies, hormones, toxins, and substances such as hemoglobin whose function is to transport simple molecules or even electrons from one place to another. The enzyme trypsin, is a typical globular protein. Another class of proteins are the membrane proteins, which, as the name would suggest, reside in a cell's lipid bilayer membrane. Such proteins can act as channels for ions or other molecules unable to pass through the lipid bilayer; as signal transducers, able to respond to signal molecules on one side of a membrane to begin a molecular response on the other side of the membrane; or as anchors of other molecules to the cell membrane, to name a few exemplars of membrane protein function. Because these proteins interface with non-polar portions of the lipid bilayer, they do no maintain function and structure in an aqueous solution, making them far more difficult to study than globular proteins or fibrous proteins. Proteins are also classified based on their function as listed in Table \(1\) below. As you will see in the following chapters, the structure and function of a protein are directly related, meaning that if a protein adopts a certain structure, the general function of that protein can be predicted with a good amount of certainty. Table \(1\): Classes of Proteins by Function Type Function Example Enzymes Catalyze biochemical reactions Amylase – helps digest carbohydrates Hormone proteins Regulate and coordinate cell functions Insulin – controls the amount of sugar (glucose) in the blood Storage proteins Storage and release of essential substances Myoglobin – stores oxygen in muscle tissue Transport proteins Carries substances through the body Hemoglobin – transports oxygen between the lungs and other tissues Structural proteins Provides structural shape and support Keratin – main structural component of hair, nails, feathers, hooves, etc. Defense proteins Protects the body against foreign invaders Immunoglobulin – recognizes and binds to foreign matter, aiding in destruction Contractile proteins Mediate contractile processes, i.e., movement Actin and Myosin – control the movement of muscles
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/18%3A_Amino_Acids_and_Proteins/18.01%3A_An_Introduction_to_Biochemistry.txt
Learning Objectives • To recognize amino acids and classify them based on the characteristics of their side chains. • Identify which amino acids are chiral. The proteins in all living species, from bacteria to humans, are polymers constructed from the same set of 20 amino acids. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002. Amino Acid Structure Every amino acid contains an amino group, (–NH2), a carboxyl group, (–COOH), and a side chain or R group, which are all attached to the alpha ($\alpha$-) carbon (the one directly bonded to the carboxyl functional group). Therefore, amino acids are commonly called alpha-amino ($\alpha$-amino) acids. Figure $1$ below shows the structure of a generic $\alpha$-amino acid. Amino Acid Side Chains Amino acid side chains or R groups can range from a single hydrogen atom (as in glycine), to a simple hydrocarbon chain, to a hydrocarbon containing a functional group. Each R group has differences in size, shape, solubility, and ionization properties, which contributes to the unique properties of an individual amino acid, and can have an effect on the overall structure and function of a protein. Table $1$ below lists the 20 common amino acids along with their names, their three- and one-letter codes, structures, and distinctive features. The three-letter codes are generally the first three letters of the amino acid name except in a few cases, such as isoleucine (Ile) and tryptophan (Trp). Similarly, the one-letter code is usually the first letter in the amino acid name, but where the letter is not unique, a letter that is phonetically similar the amino acid name is used: F for Fenylalanine, R for aRginine, and W for tWiptophan. This table also groups the amino acids according to whether the side chain at neutral pH is nonpolar, polar uncharged, positively charged, or negatively charged. Table $1$: Common Amino Acids Found in Proteins. (Isoelectric points are explained in a later section.) Common Name Three-letter (one-letter) Code Systematic (IUPAC) Name Structural Formula (at pH 6) Isoelectric Point (pI) Distinctive Feature Amino acids with a nonpolar R group glycine Gly (G) aminoethanoic acid 6.0 the only amino acid lacking a chiral carbon alanine Ala (A) 2-aminopropanoic acid 6.0 a methyl group, it is the second smallest side chain valine Val (V) 2-amino-3-methylbutanoic acid 6.0 a branched-chain amino acid leucine Leu (L) 2-amino-4-methylpentanoic acid 6.0 a branched-chain amino acid isoleucine Ile (I) 2-amino-3-methylpentanoic acid 6.0 an essential amino acid because most animals cannot synthesize branched-chain amino acids phenylalanine Phe (F) 2-amino-3-phenylpropanoic acid 5.5 also classified as an aromatic amino acid tryptophan Trp (W) 2-Amino-3-(1H-indol-3-yl)-propanoic acid 5.9 also classified as an aromatic amino acid methionine Met (M) 2-amino-4-(methylthio)butanoic acid 5.7 side chain functions as a methyl group donor proline Pro (P) pyrrolidine-2-carboxylic acid 6.3 contains a secondary amine group; referred to as an α-imino acid Amino acids with a polar but neutral R group serine Ser (S) 2-amino-3-hydroxypropanoic acid 5.7 found at the active site of many enzymes threonine Thr (T) 2-amino-3-hydroxybutanoic acid 5.6 named for its similarity to the sugar threose cysteine Cys (C) 2-amino-3-mercaptopropanoic acid 5.0 oxidation of two cysteine molecules yields cystine tyrosine Tyr (Y) 2-amino-3-(4-hydroxyphenyl)-propanoic acid 5.7 also classified as an aromatic amino acid asparagine Asn (N) 2-amino-3-carbamoylpropanoic acid 5.4 the amide of aspartic acid glutamine Gln (Q) 2-amino-4-carbamoylbutanoic acid 5.7 the amide of glutamic acid Amino acids with a negatively charged R group aspartic acid Asp (D) 2-aminobutanedioic acid 3.0 carboxyl groups are ionized at physiological pH; also known as aspartate glutamic acid Glu (E) 2-aminopentanedioic acid 3.2 carboxyl groups are ionized at physiological pH; also known as glutamate Amino acids with a positively charged R group histidine His (H) 2-Amino-3-(1H-imidazol-4-yl)-propanoic acid 7.6 the only amino acid whose R group has a pKa (6.0) near physiological pH lysine Lys (K) 2,6-diaminohexanoic acid 9.7 is somewhat amphipathic due to the long hydrocarbon tail and positively charged amino group on the $\epsilon$ carbon arginine Arg (R) 2-amino-5-guanidinopentanoic acid 10.8 almost as strong a base as sodium hydroxide Note: Interesting Facts About Amino Acids The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek glykys, meaning “sweet”). Glutamic acid is named as such because it was first isolated from gluten. The crystalline salt of glutamic acid is called monosodium glutamate (MSG), which is naturally occuring in some foods but is also added to as a savory or "umami" flavor enhancer. In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification of proline occurs after the amino acid has been assembled into a protein. Chirality of Amino Acids Notice in Table $1$ that glycine is the only amino acid whose (\alpha\)-carbon is not chiral, in other words the molecule and the mirror-image of glycine are identical. All other amino acids have two forms that are mirror images of each other, they are enantiomers. As you can see in the figure below, the "left-handed" form of the molecule is known as the L-amino acid and the "right-handed" form is the D-amino acid. Summary Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/18%3A_Amino_Acids_and_Proteins/18.03%3A_Amino_Acids.txt
Learning Objectives • To explain how an amino acid can act as both an acid and a base. The structure of an amino acid allows it to act as both an acid and a base. An amino acid has this ability because at a certain pH value (different for each amino acid) nearly all the amino acid molecules exist as zwitterions. If acid is added to a solution containing the zwitterion, the carboxylate group captures a hydrogen (H+) ion, and the amino acid becomes positively charged. If base is added, ion removal of the H+ ion from the amino group of the zwitterion produces a negatively charged amino acid. In both circumstances, the amino acid acts to maintain the pH of the system—that is, to remove the added acid (H+) or base (OH) from solution. Example \(1\) 1. Draw the structure for the anion formed when glycine (at neutral pH) reacts with a base. 2. Draw the structure for the cation formed when glycine (at neutral pH) reacts with an acid. Solution 1. The base removes H+ from the protonated amine group. • The acid adds H+ to the carboxylate group. Exercise \(1\) 1. Draw the structure for the cation formed when valine (at neutral pH) reacts with an acid. 2. Draw the structure for the anion formed when valine (at neutral pH) reacts with a base. The particular pH at which a given amino acid exists in solution as a zwitterion is called the isoelectric point (pI). At its pI, the positive and negative charges on the amino acid balance, and the molecule as a whole is electrically neutral. The amino acids whose side chains are always neutral have isoelectric points ranging from 5.0 to 6.5. The basic amino acids (which have positively charged side chains at neutral pH) have relatively high examples. Acidic amino acids (which have negatively charged side chains at neutral pH) have quite low examples (Table \(1\)). Table \(1\): ExampIes of Some Representative Amino Acids Amino Acid Classification pI alanine nonpolar 6.0 valine nonpolar 6.0 serine polar, uncharged 5.7 threonine polar, uncharged 6.5 arginine positively charged (basic) 10.8 histidine positively charged (basic) 7.6 lysine positively charged (basic) 9.8 aspartic acid negatively charged (acidic) 3.0 glutamic acid negatively charged (acidic) 3.2 Amino acids undergo reactions characteristic of carboxylic acids and amines. The reactivity of these functional groups is particularly important in linking amino acids together to form peptides and proteins, as you will see later in this chapter. Simple chemical tests that are used to detect amino acids take advantage of the reactivity of these functional groups. An example is the ninhydrin test in which the amine functional group of α-amino acids reacts with ninhydrin to form purple-colored compounds. Ninhydrin is used to detect fingerprints because it reacts with amino acids from the proteins in skin cells transferred to the surface by the individual leaving the fingerprint. Summary Amino acids can act as both an acid and a base due to the presence of the amino and carboxyl functional groups. The pH at which a given amino acid exists in solution as a zwitterion is called the isoelectric point (pI).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/18%3A_Amino_Acids_and_Proteins/18.04%3A_Acid-Base_Properties_of_Amino_Acids.txt
Learning Objectives • Explain how a peptide is formed from individual amino acids. • Explain why the sequence of amino acids in a protein is important. Two or more amino acids can join together into chains called peptides. In an earlier chapter, we discussed the reaction between ammonia and a carboxylic acid to form an amide. In a similar reaction, the amino group on one amino acid molecule reacts with the carboxyl group on another, releasing a molecule of water and forming an amide linkage: An amide bond joining two amino acid units is called a peptide bond. Note that the product molecule still has a reactive amino group on the left and a reactive carboxyl group on the right. These can react with additional amino acids to lengthen the peptide. The process can continue until thousands of units have joined, resulting in large proteins. A chain consisting of only two amino acid units is called a dipeptide; a chain consisting of three is a tripeptide. By convention, peptide and protein structures are depicted with the amino acid whose amino group is free (the amino-terminal or N-terminal end) on the left and the amino acid with a free carboxyl group (the carboxyl-terminal or C-terminal end) to the right. Individual amino acids joined in a chain are called residues. The general term peptide refers to an amino acid chain of unspecified length. However, chains of about 50 amino acids or more are usually called proteins or polypeptides. In its physiologically active form, a protein may be composed of one or more polypeptide chains. Note: Order is Important For peptides and proteins to be physiologically active, it is not enough that they incorporate certain amounts of specific amino acids. The order, or sequence, in which the amino acids are connected is also of critical importance. Bradykinin is a nine-amino acid peptide (Figure \(1\)) produced in the blood that has the following amino acid sequence: arg-pro-pro-gly-phe-ser-pro-phe-arg This peptide lowers blood pressure, stimulates smooth muscle tissue, increases capillary permeability, and causes pain. When the order of amino acids in bradykinin is reversed, arg-phe-pro-ser-phe-gly-pro-pro-arg the peptide resulting from this synthesis shows none of the activity of bradykinin. Just as millions of different words are spelled with our 26-letter English alphabet, millions of different proteins are made with the 20 common amino acids. However, just as the English alphabet can be used to write gibberish, amino acids can be put together in the wrong sequence to produce nonfunctional proteins. Although the correct sequence is ordinarily of utmost importance, it is not always absolutely required. Just as you can sometimes make sense of incorrectly spelled English words, a protein with a small percentage of “incorrect” amino acids may continue to function. However, it rarely functions as well as a protein having the correct sequence. There are also instances in which seemingly minor errors of sequence have disastrous effects. For example, in some people, every molecule of hemoglobin (a protein in the blood that transports oxygen) has a single incorrect amino acid unit out of about 300 (a single valine replaces a glutamic acid). That “minor” error is responsible for sickle cell anemia, an inherited condition that usually is fatal. Summary The amino group of one amino acid can react with the carboxyl group on another amino acid to form a peptide bond that links the two amino acids together. Additional amino acids can be added on through the formation of addition peptide (amide) bonds. A sequence of amino acids in a peptide or protein is written with the N-terminal amino acid first and the C-terminal amino acid at the end (writing left to right). Concept Review Exercises 1. Distinguish between the N-terminal amino acid and the C-terminal amino acid of a peptide or protein. 2. Describe the difference between an amino acid and a peptide. 3. Amino acid units in a protein are connected by peptide bonds. What is another name for the functional group linking the amino acids? Answers 1. The N-terminal end is the end of a peptide or protein whose amino group is free (not involved in the formation of a peptide bond), while the C-terminal end has a free carboxyl group. 2. A peptide is composed of two or more amino acids. Amino acids are the building blocks of peptides. 3. amide bond Exercises 1. Draw the structure for each peptide. 1. gly-val 2. val-gly 2. Draw the structure for cys-val-ala. 3. Identify the C- and N-terminal amino acids for the peptide lys-val-phe-gly-arg-cys. 4. Identify the C- and N-terminal amino acids for the peptide asp-arg-val-tyr-ile-his-pro-phe. Answers 1. C-terminal amino acid: cys; N-terminal amino acid: lys 18.06: Protein Structure- An Overview and Primary Protein Structure Learning Objectives • Describe primary protein structure. • Explain how sequence can affect function. Each of the thousands of naturally occurring proteins has its own characteristic amino acid composition and sequence that result in a unique three-dimensional shape. Since the 1950s, scientists have determined the amino acid sequences and three-dimensional conformation of numerous proteins and thus obtained important clues on how each protein performs its specific function in the body. Levels of Protein Structure The structure of proteins is generally described as having four organizational levels. The first of these is the primary structure, which is the number and sequence of amino acids in a protein’s polypeptide chain or chains, beginning with the free amino group and maintained by the peptide bonds connecting each amino acid to the next. The primary structure of insulin, composed of 51 amino acids, is shown in Figure \(1\). 18.07: Secondary Protein Structure Learning Objectives • Objective 1 • Objective 2 A protein molecule is not a random tangle of polypeptide chains. Instead, the chains are arranged in unique but specific conformations. The term secondary structure refers to the fixed arrangement of the polypeptide backbone. On the basis of X ray studies, Linus Pauling and Robert Corey postulated that certain proteins or portions of proteins twist into a spiral or a helix. This helix is stabilized by intrachain hydrogen bonding between the carbonyl oxygen atom of one amino acid and the amide hydrogen atom four amino acids up the chain (located on the next turn of the helix) and is known as a right-handed α-helix. X ray data indicate that this helix makes one turn for every 3.6 amino acids, and the side chains of these amino acids project outward from the coiled backbone (Figure \(1\)). The α-keratins, found in hair and wool, are exclusively α-helical in conformation. Some proteins, such as gamma globulin, chymotrypsin, and cytochrome c, have little or no helical structure. Others, such as hemoglobin and myoglobin, are helical in certain regions but not in others. Another common type of secondary structure, called the β-pleated sheet conformation, is a sheetlike arrangement in which two or more extended polypeptide chains (or separate regions on the same chain) are aligned side by side. The aligned segments can run either parallel or antiparallel—that is, the N-terminals can face in the same direction on adjacent chains or in different directions—and are connected by interchain hydrogen bonding (Figure \(3\)). The β-pleated sheet is particularly important in structural proteins, such as silk fibroin. It is also seen in portions of many enzymes, such as carboxypeptidase A and lysozyme.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/18%3A_Amino_Acids_and_Proteins/18.05%3A_Peptides.txt
Learning Objectives • Objective 1 • Objective 2 Tertiary structure refers to the unique three-dimensional shape of the protein as a whole, which results from the folding and bending of the protein backbone. The tertiary structure is intimately tied to the proper biochemical functioning of the protein. Figure \(1\) shows a depiction of the three-dimensional structure of insulin. Four major types of attractive interactions determine the shape and stability of the tertiary structure of proteins. You studied several of them previously. 1. Ionic bonding. Ionic bonds result from electrostatic attractions between positively and negatively charged side chains of amino acids. For example, the mutual attraction between an aspartic acid carboxylate ion and a lysine ammonium ion helps to maintain a particular folded area of a protein (part (a) of Figure \(5\)). 2. Hydrogen bonding. Hydrogen bonding forms between a highly electronegative oxygen atom or a nitrogen atom and a hydrogen atom attached to another oxygen atom or a nitrogen atom, such as those found in polar amino acid side chains. Hydrogen bonding (as well as ionic attractions) is extremely important in both the intra- and intermolecular interactions of proteins (part (b) of Figure \(5\)). 3. Disulfide linkages. Two cysteine amino acid units may be brought close together as the protein molecule folds. Subsequent oxidation and linkage of the sulfur atoms in the highly reactive sulfhydryl (SH) groups leads to the formation of cystine (part (c) of Figure \(2\)). Intrachain disulfide linkages are found in many proteins, including insulin (yellow bars in Figure \(1\)) and have a strong stabilizing effect on the tertiary structure. 1. Dispersion forces. Dispersion forces arise when a normally nonpolar atom becomes momentarily polar due to an uneven distribution of electrons, leading to an instantaneous dipole that induces a shift of electrons in a neighboring nonpolar atom. Dispersion forces are weak but can be important when other types of interactions are either missing or minimal (part (d) of Figure \(2\)). This is the case with fibroin, the major protein in silk, in which a high proportion of amino acids in the protein have nonpolar side chains. The term hydrophobic interaction is often misused as a synonym for dispersion forces. Hydrophobic interactions arise because water molecules engage in hydrogen bonding with other water molecules (or groups in proteins capable of hydrogen bonding). Because nonpolar groups cannot engage in hydrogen bonding, the protein folds in such a way that these groups are buried in the interior part of the protein structure, minimizing their contact with water. 18.09: Quaternary Protein Structure Learning Objectives • Objective 1 • Objective 2 When a protein contains more than one polypeptide chain, each chain is called a subunit. The arrangement of multiple subunits represents a fourth level of structure, the quaternary structure of a protein. Hemoglobin, with four polypeptide chains or subunits, is the most frequently cited example of a protein having quaternary structure (Figure \(1\)). The quaternary structure of a protein is produced and stabilized by the same kinds of interactions that produce and maintain the tertiary structure. A schematic representation of the four levels of protein structure is in Figure \(2\). The primary structure consists of the specific amino acid sequence. The resulting peptide chain can twist into an α-helix, which is one type of secondary structure. This helical segment is incorporated into the tertiary structure of the folded polypeptide chain. The single polypeptide chain is a subunit that constitutes the quaternary structure of a protein, such as hemoglobin that has four polypeptide chains. 18.10: Chemical Properties of Proteins • Objective 1 • Objective 2 Protein Denaturation The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job. (Sometimes denaturation is equated with the precipitation or coagulation of a protein; our definition is a bit broader.) A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation (Figure \(1\)). Figure \(1\): Protein Denaturation Methods Method Effect on Protein Structure Heat above 50°C or ultraviolet (UV) radiation Heat or UV radiation supplies kinetic energy to protein molecules, causing their atoms to vibrate more rapidly and disrupting relatively weak hydrogen bonding and dispersion forces. Use of organic compounds, such as ethyl alcohol These compounds are capable of engaging in intermolecular hydrogen bonding with protein molecules, disrupting intramolecular hydrogen bonding within the protein. Salts of heavy metal ions, such as mercury, silver, and lead These ions form strong bonds with the carboxylate anions of the acidic amino acids or SH groups of cysteine, disrupting ionic bonds and disulfide linkages. Alkaloid reagents, such as tannic acid (used in tanning leather) These reagents combine with positively charged amino groups in proteins to disrupt ionic bonds. Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. No one has yet reversed that process. However, given the proper circumstances and enough time, a protein that has unfolded under sufficiently gentle conditions can refold and may again exhibit biological activity (Figure \(8\)). Such evidence suggests that, at least for these proteins, the primary structure determines the secondary and tertiary structure. A given sequence of amino acids seems to adopt its particular three-dimensional arrangement naturally if conditions are right. The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/18%3A_Amino_Acids_and_Proteins/18.08%3A_Tertiary_Protein_Structure.txt
Thumbnail: Ball-and-stick model of the pyridoxal phosphate molecule, the active form of vitamin B6. (Public Domain) 19: Enzymes and Vitamins Learning Objectives • Describe how enzymes catalyze biochemical reactions. 19.02: Enzyme Cofactors Learning Objectives • To explain why vitamins are necessary in the diet. Many enzymes are simple proteins consisting entirely of one or more amino acid chains. Other enzymes contain a nonprotein component called a cofactor that is necessary for the enzyme’s proper functioning. There are two types of cofactors: inorganic ions [e.g., zinc or Cu(I) ions] and organic molecules known as coenzymes. Most coenzymes are vitamins or are derived from vitamins. Vitamins are organic compounds that are essential in very small (trace) amounts for the maintenance of normal metabolism. They generally cannot be synthesized at adequate levels by the body and must be obtained from the diet. The absence or shortage of a vitamin may result in a vitamin-deficiency disease. In the first half of the 20th century, a major focus of biochemistry was the identification, isolation, and characterization of vitamins. Despite accumulating evidence that people needed more than just carbohydrates, fats, and proteins in their diets for normal growth and health, it was not until the early 1900s that research established the need for trace nutrients in the diet. Vitamin Physiological Function Effect of Deficiency Table \(1\): Fat-Soluble Vitamins and Physiological Functions vitamin A (retinol) formation of vision pigments; differentiation of epithelial cells night blindness; continued deficiency leads to total blindness vitamin D (cholecalciferol) increases the body’s ability to absorb calcium and phosphorus osteomalacia (softening of the bones); known as rickets in children vitamin E (tocopherol) fat-soluble antioxidant damage to cell membranes vitamin K (phylloquinone) formation of prothrombin, a key enzyme in the blood-clotting process increases the time required for blood to clot Because organisms differ in their synthetic abilities, a substance that is a vitamin for one species may not be so for another. Over the past 100 years, scientists have identified and isolated 13 vitamins required in the human diet and have divided them into two broad categories: the fat-soluble vitamins, which include vitamins A, D, E, and K, and the water-soluble vitamins, which are the B complex vitamins and vitamin C. All fat-soluble vitamins contain a high proportion of hydrocarbon structural components. There are one or two oxygen atoms present, but the compounds as a whole are nonpolar. In contrast, water-soluble vitamins contain large numbers of electronegative oxygen and nitrogen atoms, which can engage in hydrogen bonding with water. Most water-soluble vitamins act as coenzymes or are required for the synthesis of coenzymes. The fat-soluble vitamins are important for a variety of physiological functions. The key vitamins and their functions are found in Tables \(1\) and \(2\). Table \(2\): Water-Soluble Vitamins and Physiological Functions Vitamin Coenzyme Coenzyme Function Deficiency Disease vitamin B1 (thiamine) thiamine pyrophosphate decarboxylation reactions beri-beri vitamin B2 (riboflavin) flavin mononucleotide or flavin adenine dinucleotide oxidation-reduction reactions involving two hydrogen atoms vitamin B3 (niacin) nicotinamide adenine dinucleotide or nicotinamide adenine dinucleotide phosphate oxidation-reduction reactions involving the hydride ion (H) pellagra vitamin B6 (pyridoxine) pyridoxal phosphate variety of reactions including the transfer of amino groups vitamin B12 (cyanocobalamin) methylcobalamin or deoxyadenoxylcobalamin intramolecular rearrangement reactions pernicious anemia biotin biotin carboxylation reactions folic acid tetrahydrofolate carrier of one-carbon units such as the formyl group anemia pantothenic Acid coenzyme A carrier of acyl groups vitamin C (ascorbic acid) none antioxidant; formation of collagen, a protein found in tendons, ligaments, and bone scurvy Vitamins C and E, as well as the provitamin β-carotene can act as antioxidants in the body. Antioxidants prevent damage from free radicals, which are molecules that are highly reactive because they have unpaired electrons. Free radicals are formed not only through metabolic reactions involving oxygen but also by such environmental factors as radiation and pollution. β-carotene is known as a provitamin because it can be converted to vitamin A in the body. Free radicals react most commonly react with lipoproteins and unsaturated fatty acids in cell membranes, removing an electron from those molecules and thus generating a new free radical. The process becomes a chain reaction that finally leads to the oxidative degradation of the affected compounds. Antioxidants react with free radicals to stop these chain reactions by forming a more stable molecule or, in the case of vitamin E, a free radical that is much less reactive (vitamin E is converted back to its original form through interaction with vitamin C). Summary Vitamins are organic compounds that are essential in very small amounts for the maintenance of normal metabolism. Vitamins are divided into two broad categories: fat-soluble vitamins and water-soluble vitamins. Most water-soluble vitamins are needed for the formation of coenzymes, which are organic molecules needed by some enzymes for catalytic activity.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/19%3A_Enzymes_and_Vitamins/19.01%3A_Catalysis_by_Enzymes.txt
Learning Objectives • Objective 1 • Objective 2 Hundreds of enzymes have been purified and studied in an effort to understand how they work so effectively and with such specificity. The resulting knowledge has been used to design drugs that inhibit or activate particular enzymes. An example is the intensive research to improve the treatment of or find a cure for acquired immunodeficiency syndrome (AIDS). AIDS is caused by the human immunodeficiency virus (HIV). Researchers are studying the enzymes produced by this virus and are developing drugs intended to block the action of those enzymes without interfering with enzymes produced by the human body. Several of these drugs have now been approved for use by AIDS patients. Enzyme Nomenclature Most enzymes can be recognized because they have the family name ending –ase. However, the first enzymes to be discovered were named according to their source or method of discovery. The enzyme pepsin, which aids in the hydrolysis of proteins, is found in the digestive juices of the stomach (Greek pepsis, meaning “digestion”). Papain, another enzyme that hydrolyzes protein (in fact, it is used in meat tenderizers), is isolated from papayas. In addition to the family name, more systematic enzyme names will give two sepcific pieces of information: the first part is the substrate upon which the enzyme acts, and the second part is the type of reaction it catalyzes. For example, alcohol dehydrogenase (Figure \(1\)) catalyzes the oxidation of an alcohol to an aldehyde. Enzyme Classification As more enzymes were discovered, chemists recognized the need for a more systematic and chemically informative identification scheme. In the current numbering and naming scheme, under the oversight of the Nomenclature Commission of the International Union of Biochemistry, enzymes are arranged into six groups according to the general type of reaction they catalyze (Table \(1\)), with subgroups and secondary subgroups that specify the reaction more precisely. Each enzyme is assigned a four-digit number, preceded by the prefix EC—for enzyme classification—that indicates its group, subgroup, and so forth. This is demonstrated in Table \(2\) for alcohol dehydrogenase. Table \(1\): Classes of Enzymes Main Class Type of Reaction Catalyzed Subclasses Examples Oxidoreductases oxidation-reduction reactions Dehydrogenases catalyze oxidation-reduction reactions involving hydrogen. Alcohol dehydrogenase Oxidases catalyze oxidation by addition of O2 to a substrate. Reductases catalyze reactions in which a substrate is reduced. Transferases transfer reactions of functional groups Transaminases catalyze the transfer of amino group. Kinases catalyze the transfer of a phosphate group. Phosphofructokinase Hydrolases reactions that use water to break a chemical bond Lipases catalyze the hydrolysis of lipids Proteases catalyze the hydrolysis of proteins Amylases catalyze the hydrolysis of carbohydrates Nucleases catalyze the hydrolysis of DNA and RNA Lyases reactions in which functional groups are added or removed without hydrolysis Decarboxylases catalyze the removal of carboxyl groups. Deaminases catalyze the removal of amino groups. Dehydratases catalyze the removal of water. Hydratases catalyze the addition of water. Fumarase Isomerases reactions in which a compound is converted to its isomer Isomerases may catalyze the conversion of an aldose to a ketose. Triose Phosphate Isomerase Mutases catalyze reactions in which a functional group is transferred from one atom in a substrate to another. Ligases reactions in which new bonds are formed between carbon and another atom; energy is required Synthetases catalyze reactions in which two smaller molecules are linked to form a larger one. Carboxylases catalyze the addition of CO2 using ATP Pyruvate Carboxylase Table \(2\): Assignment of an Enzyme Classification Number Alcohol Dehydrogenase: EC 1.1.1.1 The first digit indicates that this enzyme is an oxidoreductase; that is, an enzyme that catalyzes an oxidation-reduction reaction. The second digit indicates that this oxidoreductase catalyzes a reaction involving a primary or secondary alcohol. The third digit indicates that either the coenzyme NAD+ or NADP+ is required for this reaction. The fourth digit indicates that this was the first enzyme isolated, characterized, and named using this system of nomenclature. The systematic name for this enzyme is alcohol:NAD+ oxidoreductase, while the recommended or common name is alcohol dehydrogenase. Reaction catalyzed: Figure \(1\): Structure of the alcohol dehydrogenase protein (E.C.1.1.1.1) (EE ISOZYME) complexed wtih nicotinamide adenini dinulceotide (NAD) and zinc (PDB: 1CDO). Summary An enzyme is a biological catalyst, a substance that increases the rate of a chemical reaction without being changed or consumed in the reaction. A systematic process is used to name and classify enzymes.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/19%3A_Enzymes_and_Vitamins/19.03%3A_Enzyme_Classification.txt
Learning Objectives • To describe the interaction between an enzyme and its substrate. Enzyme-catalyzed reactions occur in at least two steps. In the first step, an enzyme (E) and the substrate molecule or molecules (S) collide and react to form an intermediate compound called the enzyme-substrate (ES) complex. (This step is reversible because the complex can break apart into the original substrate or substrates and the free enzyme.) Once the ES complex forms, the enzyme is able to catalyze the formation of product (P), which is then released from the enzyme surface: $E + S \leftrightarrow ES \tag{$1$}$ $ES \rightarrow E + P \tag{$2$}$ Hydrogen bonding and other electrostatic interactions hold the enzyme and substrate together in the complex. The structural features or functional groups on the enzyme that participate in these interactions are located in a cleft or pocket on the enzyme surface. This pocket, where the enzyme combines with the substrate and transforms the substrate to product is called the active site of the enzyme (Figure $1$). Models of Enzyme-Substrate Interaction The active site of an enzyme possesses a unique conformation (including correctly positioned bonding groups) that is complementary to the structure of the substrate, so that the enzyme and substrate molecules fit together in much the same manner as a key fits into a tumbler lock. In fact, an early model describing the formation of the enzyme-substrate complex was called the lock-and-key model (Figure $2$). This model portrayed the enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. Working out the precise three-dimensional structures of numerous enzymes has enabled chemists to refine the original lock-and-key model of enzyme actions. They discovered that the binding of a substrate often leads to a large conformational change in the enzyme, as well as to changes in the structure of the substrate or substrates. The current theory, known as the induced-fit model, says that enzymes can undergo a change in conformation when they bind substrate molecules, and the active site has a shape complementary to that of the substrate only after the substrate is bound, as shown for hexokinase in Figure $3$. After catalysis, the enzyme resumes its original structure. The structural changes that occur when an enzyme and a substrate join together bring specific parts of a substrate into alignment with specific parts of the enzyme’s active site. Amino acid side chains in or near the binding site can then act as acid or base catalysts, provide binding sites for the transfer of functional groups from one substrate to another or aid in the rearrangement of a substrate. The participating amino acids, which are usually widely separated in the primary sequence of the protein, are brought close together in the active site as a result of the folding and bending of the polypeptide chain or chains when the protein acquires its tertiary and quaternary structure. Binding to enzymes brings reactants close to each other and aligns them properly, which has the same effect as increasing the concentration of the reacting compounds. Example $1$ 1. What type of interaction would occur between an OH group present on a substrate molecule and a functional group in the active site of an enzyme? 2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified. Solution 1. An OH group would most likely engage in hydrogen bonding with an appropriate functional group present in the active site of an enzyme. 2. Several amino acid side chains would be able to engage in hydrogen bonding with an OH group. One example would be asparagine, which has an amide functional group. Exercise $1$ 1. What type of interaction would occur between an COO group present on a substrate molecule and a functional group in the active site of an enzyme? 2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified. One characteristic that distinguishes an enzyme from all other types of catalysts is its substrate specificity. An inorganic acid such as sulfuric acid can be used to increase the reaction rates of many different reactions, such as the hydrolysis of disaccharides, polysaccharides, lipids, and proteins, with complete impartiality. In contrast, enzymes are much more specific. Some enzymes act on a single substrate, while other enzymes act on any of a group of related molecules containing a similar functional group or chemical bond. Some enzymes even distinguish between D- and L-stereoisomers, binding one stereoisomer but not the other. Urease, for example, is an enzyme that catalyzes the hydrolysis of a single substrate—urea—but not the closely related compounds methyl urea, thiourea, or biuret. The enzyme carboxypeptidase, on the other hand, is far less specific. It catalyzes the removal of nearly any amino acid from the carboxyl end of any peptide or protein. Enzyme specificity results from the uniqueness of the active site in each different enzyme because of the identity, charge, and spatial orientation of the functional groups located there. It regulates cell chemistry so that the proper reactions occur in the proper place at the proper time. Clearly, it is crucial to the proper functioning of the living cell. Summary A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product. The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions. The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate. Enzymes exhibit varying degrees of substrate specificity.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/19%3A_Enzymes_and_Vitamins/19.04%3A_How_Enzymes_Work.txt
Learning Objectives • To describe how pH, temperature, and the concentration of an enzyme and its substrate influence enzyme activity. The single most important property of enzymes is the ability to increase the rates of reactions occurring in living organisms, a property known as catalytic activity. Because most enzymes are proteins, their activity is affected by factors that disrupt protein structure, as well as by factors that affect catalysts in general. Factors that disrupt protein structure include temperature and pH; factors that affect catalysts in general include reactant or substrate concentration and catalyst or enzyme concentration. The activity of an enzyme can be measured by monitoring either the rate at which a substrate disappears or the rate at which a product forms. Substrate Concentration In the presence of a given amount of enzyme, the rate of an enzymatic reaction increases as the substrate concentration increases until a limiting rate is reached, after which further increase in the substrate concentration produces no significant change in the reaction rate (part (a) of Figure \(1\)). At this point, so much substrate is present that essentially all of the enzyme active sites have substrate bound to them. In other words, the enzyme molecules are saturated with substrate. The excess substrate molecules cannot react until the substrate already bound to the enzymes has reacted and been released (or been released without reacting). Let’s consider an analogy. Ten taxis (enzyme molecules) are waiting at a taxi stand to take people (substrate) on a 10-minute trip to a concert hall, one passenger at a time. If only 5 people are present at the stand, the rate of their arrival at the concert hall is 5 people in 10 minutes. If the number of people at the stand is increased to 10, the rate increases to 10 arrivals in 10 minutes. With 20 people at the stand, the rate would still be 10 arrivals in 10 minutes. The taxis have been “saturated.” If the taxis could carry 2 or 3 passengers each, the same principle would apply. The rate would simply be higher (20 or 30 people in 10 minutes) before it leveled off. Enzyme Concentration When the concentration of the enzyme is significantly lower than the concentration of the substrate (as when the number of taxis is far lower than the number of waiting passengers), the rate of an enzyme-catalyzed reaction is directly dependent on the enzyme concentration (part (b) of Figure \(1\)). This is true for any catalyst; the reaction rate increases as the concentration of the catalyst is increased. Effect of Temperature on Activity A general rule of thumb for most chemical reactions is that a temperature rise of 10°C approximately doubles the reaction rate. To some extent, this rule holds for all enzymatic reactions. After a certain point, however, an increase in temperature causes a decrease in the enzyme reaction rate, due to denaturation of the protein structure and disruption of the active site (part (a) of Figure \(2\)). For many proteins, denaturation occurs between 45°C and 55°C. Furthermore, even though an enzyme may appear to have a maximum reaction rate between 40°C and 50°C, most biochemical reactions are carried out at lower temperatures because enzymes are not stable at these higher temperatures and will denature after a few minutes. At 0°C and 100°C, the rate of enzyme-catalyzed reactions is nearly zero. This fact has several practical applications. We sterilize objects by placing them in boiling water, which denatures the enzymes of any bacteria that may be in or on them. We preserve our food by refrigerating or freezing it, which slows enzyme activity. When animals go into hibernation in winter, their body temperature drops, decreasing the rates of their metabolic processes to levels that can be maintained by the amount of energy stored in the fat reserves in the animals’ tissues. Effect of Hydrogen Ion Concentration (pH) on Activity Because most enzymes are proteins, they are sensitive to changes in the hydrogen ion concentration or pH. Enzymes may be denatured by extreme levels of hydrogen ions (whether high or low); any change in pH, even a small one, alters the degree of ionization of an enzyme’s acidic and basic side groups and the substrate components as well. Ionizable side groups located in the active site must have a certain charge for the enzyme to bind its substrate. Neutralization of even one of these charges alters an enzyme’s catalytic activity. An enzyme exhibits maximum activity over the narrow pH range in which a molecule exists in its properly charged form. The median value of this pH range is called the optimum pH of the enzyme (part (b) of Figure \(2\)). With the notable exception of gastric juice (the fluids secreted in the stomach), most body fluids have pH values between 6 and 8. Not surprisingly, most enzymes exhibit optimal activity in this pH range. However, a few enzymes have optimum pH values outside this range. For example, the optimum pH for pepsin, an enzyme that is active in the stomach, is 2.0. Summary Initially, an increase in substrate concentration leads to an increase in the rate of an enzyme-catalyzed reaction. As the enzyme molecules become saturated with substrate, this increase in reaction rate levels off. The rate of an enzyme-catalyzed reaction increases with an increase in the concentration of an enzyme. At low temperatures, an increase in temperature increases the rate of an enzyme-catalyzed reaction. At higher temperatures, the protein is denatured, and the rate of the reaction dramatically decreases. An enzyme has an optimum pH range in which it exhibits maximum activity. Concept Review Exercises 1. The concentration of substrate X is low. What happens to the rate of the enzyme-catalyzed reaction if the concentration of X is doubled? 2. What effect does an increase in the enzyme concentration have on the rate of an enzyme-catalyzed reaction? Answers 1. If the concentration of the substrate is low, increasing its concentration will increase the rate of the reaction. 2. An increase in the amount of enzyme will increase the rate of the reaction (provided sufficient substrate is present). Exercises 1. In non-enzyme-catalyzed reactions, the reaction rate increases as the concentration of reactant is increased. In an enzyme-catalyzed reaction, the reaction rate initially increases as the substrate concentration is increased but then begins to level off, so that the increase in reaction rate becomes less and less as the substrate concentration increases. Explain this difference. 2. Why do enzymes become inactive at very high temperatures? 3. An enzyme has an optimum pH of 7.4. What is most likely to happen to the activity of the enzyme if the pH drops to 6.3? Explain. 4. An enzyme has an optimum pH of 7.2. What is most likely to happen to the activity of the enzyme if the pH increases to 8.5? Explain. Answers 1. In an enzyme-catalyzed reaction, the substrate binds to the enzyme to form an enzyme-substrate complex. If more substrate is present than enzyme, all of the enzyme binding sites will have substrate bound, and further increases in substrate concentration cannot increase the rate. 1. The activity will decrease; a pH of 6.3 is more acidic than 7.4, and one or more key groups in the active site may bind a hydrogen ion, changing the charge on that group.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/19%3A_Enzymes_and_Vitamins/19.05%3A_Factors_Affecting_Enzyme_Activity.txt
Learning Objectives • Explain what an enzyme inhibitor is. • Distinguish between reversible and irreversible inhibitors. • Distinguish between competitive, noncompetitive, and uncompetitive inhibitors. Previously, we noted that enzymes can be inactivated at high temperatures and by changes in pH. These are nonspecific factors that would inactivate any enzyme. The activity of enzymes can also be regulated by more specific inhibitors that slow or stop catalysis. Enzyme inhibition can either be reversible or irreversible. In reversible inhibition, the inhibitor can bind (usually non-covalently) and dissociate, allowing enzyme activity to return back to its original, uninhibited level. Irreversible inhibitors bind to the enzyme permanently and thus permanently inhibit enzyme activity. Reversible Inhibition Reversible enzyme inhibition can be competitive, noncompetitive, or uncompetitive, depending on where the inhibitor binds to the enzyme, substrate, or enzyme-substrate complex. Competitive inhibition is when an inhibitor reversibly binds to an enzyme at the enzyme active site; competing with the substrate for binding. A competitive inhibitor must be a molecule that is structurally similar to the substrate molecule, allowing it to interact with the enzyme active site through similar non-covalent interactions, but it does not, or cannot, undergo the same chemical reaction. When the inhibitor is bound to the active site, it blocks the correct substrate from binding and catalysis from occurring. However, as a reversible inhibitor, it can disassociate from the enzyme eventually allowing for the correct substrate to bind and the catalysis to occur. Because the inhibitor and substrate are in competition for the same active site, inhibition is concentration-dependent. As shown in the below plot of rate of reaction vs. substrate concentration (Figure \(1\)), the competitive inhibitor slows the rate of reaction, but at higher substrate concentrations, the normal maximum rate can be reached. Studies of competitive inhibition have provided helpful information about certain enzyme-substrate complexes and the interactions of specific groups at the active sites. As a result, pharmaceutical companies have synthesized drugs that competitively inhibit metabolic processes in bacteria and certain cancer cells. Many drugs are competitive inhibitors of specific enzymes. A classic example of competitive inhibition is the effect of malonate on the enzyme activity of succinate dehydrogenase (Figure \(2\)). Malonate and succinate are the anions of dicarboxylic acids and contain three and four carbon atoms, respectively. The malonate molecule binds to the active site because the spacing of its carboxyl groups is not greatly different from that of succinate. However, no catalytic reaction occurs because malonate does not have a CH2CH2 group to convert to CH=CH. This reaction will also be discussed in connection with the Krebs cycle and energy production in a later chapter. In uncompetitive inhibition, the inhibitor can only bind the enzyme when the substrate is already bound, in other words it binds the enzyme-substrate complex but not the enzyme alone. The maximum reaction rate in the presence of an uncompetitive inhibitor is lowered, however, unlike with competitive inhibition, the rate cannot be increased by adding more substrate. This type of inhibition is most commonly seen when the enzyme reaction involves two substrates and as long as the concentration of inhibitor remains constant, the maximum reaction rate does not change. A noncompetitive inhibitor can bind to either the free enzyme or the enzyme-substrate complex because its binding site on the enzyme is distinct from the active site. Binding of this kind of inhibitor alters the three-dimensional conformation of the enzyme, changing the configuration of the active site with one of two results. Either the enzyme-substrate complex does not form at its normal rate, or, once formed, it does not yield products at the normal rate (see Figure \(1\)). Because the inhibitor does not structurally resemble the substrate, nor is it competing with the substrate for the active site, the addition of excess substrate does not reverse the inhibitory effect. Chemotherapy is the strategic use of chemicals (that is, drugs) to destroy infectious microorganisms or cancer cells without causing excessive damage to the other, healthy cells of the host. From bacteria to humans, the metabolic pathways of all living organisms are quite similar, so the search for safe and effective chemotherapeutic agents is a formidable task. Many well-established chemotherapeutic drugs function by inhibiting a critical enzyme in the cells of the invading organism. An antibiotic is a compound that kills bacteria; it may come from a natural source such as molds or be synthesized with a structure analogous to a naturally occurring antibacterial compound. Antibiotics constitute no well-defined class of chemically related substances, but many of them work by effectively inhibiting a variety of enzymes essential to bacterial growth. To Your Health: Penicillin Penicillin, one of the most widely used antibiotics in the world, was fortuitously discovered by Alexander Fleming in 1928, when he noticed antibacterial properties in a mold growing on a bacterial culture plate. In 1938, Ernst Chain and Howard Florey began an intensive effort to isolate penicillin from the mold and study its properties. The large quantities of penicillin needed for this research became available through development of a corn-based nutrient medium that the mold loved and through the discovery of a higher-yielding strain of mold at a United States Department of Agriculture research center near Peoria, Illinois. Even so, it was not until 1944 that large quantities of penicillin were being produced and made available for the treatment of bacterial infections. Penicillin functions by interfering with the synthesis of cell walls of reproducing bacteria. It does so by inhibiting an enzyme—transpeptidase—that catalyzes the last step in bacterial cell-wall biosynthesis. The defective walls cause bacterial cells to burst. Human cells are not affected because they have cell membranes, not cell walls. Several naturally occurring penicillins have been isolated. They are distinguished by different R groups connected to a common structure: a four-member cyclic amide (called a lactam ring) fused to a five-member ring. The addition of appropriate organic compounds to the culture medium leads to the production of the different kinds of penicillin. The penicillins are effective against gram-positive bacteria (bacteria capable of being stained by Gram’s stain) and a few gram-negative bacteria (including the intestinal bacterium Escherichia coli). They are effective in the treatment of diphtheria, gonorrhea, pneumonia, syphilis, many pus infections, and certain types of boils. Penicillin G was the earliest penicillin to be used on a wide scale. However, it cannot be administered orally because it is quite unstable; the acidic pH of the stomach converts it to an inactive derivative. The major oral penicillins—penicillin V, ampicillin, and amoxicillin—on the other hand, are acid stable. Some strains of bacteria become resistant to penicillin through a mutation that allows them to synthesize an enzyme—penicillinase—that breaks the antibiotic down (by cleavage of the amide linkage in the lactam ring). To combat these strains, scientists have synthesized penicillin analogs (such as methicillin) that are not inactivated by penicillinase. Some people (perhaps 5% of the population) are allergic to penicillin and therefore must be treated with other antibiotics. Their allergic reaction can be so severe that a fatal coma may occur if penicillin is inadvertently administered to them. Fortunately, several other antibiotics have been discovered. Most, including aureomycin and streptomycin, are the products of microbial synthesis. Others, such as the semisynthetic penicillins and tetracyclines, are made by chemical modifications of antibiotics; and some, like chloramphenicol, are manufactured entirely by chemical synthesis. They are as effective as penicillin in destroying infectious microorganisms. Many of these antibiotics exert their effects by blocking protein synthesis in microorganisms. Initially, antibiotics were considered miracle drugs, substantially reducing the number of deaths from blood poisoning, pneumonia, and other infectious diseases. Some seven decades ago, a person with a major infection almost always died. Today, such deaths are rare. Seven decades ago, pneumonia was a dreaded killer of people of all ages. Today, it kills only the very old or those ill from other causes. Antibiotics have indeed worked miracles in our time, but even miracle drugs have limitations. Not long after the drugs were first used, disease organisms began to develop strains resistant to them. In a race to stay ahead of resistant bacterial strains, scientists continue to seek new antibiotics. The penicillins have now been partially displaced by related compounds, such as the cephalosporins and vancomycin. Unfortunately, some strains of bacteria have already shown resistance to these antibiotics. Irreversible Inhibition An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. When the inhibitor is bound, the enzyme active site is blocked, the substrate does not bind, and catalysis cannot occur, similar to competitive inhibition. The difference here is that the inhibition is irreversible, meaning that the inhibitor remains bound and does not dissociate from the enzyme because the enzyme-inhibitor covalent bonds are not easily broken. In the presence of an irreversible inhibitor, the substrate cannot bind the active site at all, nor can high substrate concentrations outcompete the inhibitor, hence the enzyme is completely inactivated. Many of the known irreversible inhibitors are poisons because they inactivate an enzyme completely. Some examples are provided in Table \(1\) below. Table \(1\): Poisons as Enzyme Inhibitors Poison Formula Example of Enzyme Inhibited Action arsenate \(\ce{AsO4^{3−}}\) glyceraldehyde 3-phosphate dehydrogenase substitutes for phosphate iodoacetate \(\ce{ICH2COO^{−}}\) triose phosphate dehydrogenase binds to cysteine \(\ce{SH}\) group diisopropylfluoro-phosphate (DIFP; a nerve poison) acetylcholinesterase binds to serine \(\ce{OH}\) group Summary An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. A reversible inhibitor inactivates an enzyme through noncovalent, reversible interactions. A competitive inhibitor competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site. Concept Review Exercises 1. What are the characteristics of an irreversible inhibitor? 2. In what ways does a competitive inhibitor differ from a noncompetitive inhibitor? Answers 1. It inactivates an enzyme by bonding covalently to a particular group at the active site. 2. A competitive inhibitor structurally resembles the substrate for a given enzyme and competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site and can bind to either the free enzyme or the enzyme-substrate complex. Exercises 1. What amino acid is present in the active site of all enzymes that are irreversibly inhibited by nerve gases such as DIFP? 2. Oxaloacetate (OOCCH2COCOO) inhibits succinate dehydrogenase. Would you expect oxaloacetate to be a competitive or noncompetitive inhibitor? Explain. 1. serine
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/19%3A_Enzymes_and_Vitamins/19.06%3A_Enzyme_Regulation_-_Inhibition.txt
Learning Objectives • Objective 1 • Objective 2 In the previous section you learned about the different types of enzyme inhibitors and how they can be used to slow or stop enzyme activity by binding to an enzyme or enzyme-substrate complex. All of these inhibitor types, except noncompetitive inhibitors, work by binding to enzyme active sites. Noncompetitive inhibitors, however, work by binding to an enzyme at a location other than the active site, an allosteric site. Inhibitors and other molecules, called activators, that bind to enzymes at allosteric sites are considered an important part of enzyme regulation called allosteric control. In this section, we will take a look at allosteric control and feedback control, two ways in which enzyme activity is regulated differently. Allosteric Control Allosteric enzymes have both a binding site, for substrate binding and catalysis, and an allosteric site, for regulation of enzyme activity. When a regulator molecule binds to the allosteric site of an enzyme, usually by noncovalent interactions, a conformational change occurs in the enzyme active site, which affects substrate binding and reaction rates. Allosteric regulation of enzyme activity can be either positive, increasing reaction rates, or negative, decreasing reaction rates. When an enzyme binds a negative regulator (or inhibitor), it will undergo a change in the active site in a way that prevents substrate binding, thereby lowering the reaction rate. As illustrated in the left-hand panel in Figure \(1\), the active site changes (becomes smaller in this case) and the substrate can no longer bind. Positive regulators (activators) bind to allosteric sites and cause conformational changes that open up an active site to promote substrate binding, allowing catalysis or increasing the reaction rate. The right panel of Figure \(1\) shows an enzyme that will only bind substrate when the active site is formed after the allosteric activator binds. Some enzymes will have more than one allosteric site that can interact with one another, which allows for highly-controlled or finely-tuned reaction rates. Feedback Control Many biological processes involve the sequential action of multiple enzymes, a reaction pathway, in which the product of one reaction is the substrate for the next enzyme and so on until the final product is formed. Positive or negative regulation of these pathways often occurs by feedback control, where a product from one of the steps in the path feedback to an earlier step in the process to increase or decrease production. It may help to visualize a factory assembly line with each person responsible for one step (catalytic reaction) in making a perfect box of 12 donuts. If the last person in the line, who is responsible for putting 12 donuts in the box, falls behind, donuts will start to pile up. In order to not waste donuts or have less than full boxes at the end, it would be beneficial to signal to the other people to slow down or take a break. The process is similar in biochemical pathways: if too much product is being formed, the pathway needs to be turned off so energy and resources are not wasted. Consider the pathway shown below in which substrate A is converted to product D through three enzymes and two intermediate products (B and C): \(A\ \xrightarrow{Enzyme\ 1}\ B\ \xrightarrow{Enzyme\ 2}\ C\ \xrightarrow{Enzyme\ 3}\ D\) If there is a lot of product D formed, there would be enough to bind to Enzyme 1, which would inhibit formation of products B, and subsequently product C, and D. This type of feedback control is useful to prevent waste of substrate A and any energy that is needed for the activity of Enzymes 1-3. As you will see in later chapters, there are many different types of feedback control that can both negatively and positively regulate pathways. Typically, feedback control occurs at points in pathways where it would be energetically unfavorable to proceede if the final product is not needed. Feedback inhibition is used to regulate the synthesis of many amino acids. For example, bacteria synthesize isoleucine from threonine in a series of five enzyme-catalyzed steps. As the concentration of isoleucine increases, some of it binds as a noncompetitive inhibitor to the first enzyme of the series (threonine deaminase), thus bringing about a decrease in the amount of isoleucine being formed (Figure \(2\)).
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/19%3A_Enzymes_and_Vitamins/19.07%3A_Enzyme_Regulation-_Allosteric_Control_and_Feedback_Inhibition.txt
Learning Objectives • Define what is meant by anomers and describe how they are formed. • Explain what is meant by mutarotation. So far we have represented monosaccharides as linear molecules, but many of them also adopt cyclic structures. This conversion occurs because of the ability of aldehydes and ketones to react with alcohols: You might wonder why the aldehyde reacts with the OH group on the fifth carbon atom rather than the OH group on the second carbon atom next to it. Recall that cyclic alkanes containing five or six carbon atoms in the ring are the most stable. The same is true for monosaccharides that form cyclic structures: rings consisting of five or six carbon atoms are the most stable. When a straight-chain monosaccharide, such as any of the structures shown in Figure \(1\), forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers, as shown in Figure \(2\). The structure shown on the left side of Figure \(2\), with the OH group on the first carbon atom projected downward, represent what is called the alpha (α) form. The structures on the right side, with the OH group on the first carbon atom pointed upward, is the beta (β) form. These two stereoisomers of a cyclic monosaccharide are known as anomers; they differ in structure around the anomeric carbon—that is, the carbon atom that was the carbonyl carbon atom in the straight-chain form. It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium (part (a) of Figure \(2\)). You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare, meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°. Even though only a small percentage of the molecules are in the open-chain aldehyde form at any time, the solution will nevertheless exhibit the characteristic reactions of an aldehyde. As the small amount of free aldehyde is used up in a reaction, there is a shift in the equilibrium to yield more aldehyde. Thus, all the molecules may eventually react, even though very little free aldehyde is present at a time. Commonly, (e.g., in Figures \(1\) and \(2\)) the cyclic forms of sugars are depicted using a convention first suggested by Walter N. Haworth, an English chemist. The molecules are drawn as planar hexagons with a darkened edge representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above the plane in a Haworth projection. The difference between the α and the β forms of sugars may seem trivial, but such structural differences are often crucial in biochemical reactions. This explains why we can get energy from the starch in potatoes and other plants but not from cellulose, even though both starch and cellulose are polysaccharides composed of glucose molecules linked together. Summary Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/20%3A_Carbohydrates/20.03%3A_Structure_of_Glucose_and_Other_Monosaccharides.txt
Learning Objectives • To identify the structures of D-glucose, D-galactose, and D-fructose and describe how they differ from each other. Although a variety of monosaccharides are found in living organisms, three hexoses are particularly abundant: D-glucose, D-galactose, and D-fructose (Figure \(1\)). Glucose and galactose are both aldohexoses, while fructose is a ketohexose. Glucose D-Glucose, generally referred to as simply glucose, is the most abundant sugar found in nature; most of the carbohydrates we eat are eventually converted to it in a series of biochemical reactions that produce energy for our cells. It is also known by three other names: dextrose, from the fact that it rotates plane-polarized light in a clockwise (dextrorotatory) direction; corn sugar because in the United States cornstarch is used in the commercial process that produces glucose from the hydrolysis of starch; and blood sugar because it is the carbohydrate found in the circulatory system of animals. Normal blood sugar values range from 70 to 105 mg glucose/dL plasma, and normal urine may contain anywhere from a trace to 20 mg glucose/dL urine. The Fischer projection of D-glucose is given in Figure \(2\). Glucose is a D sugar because the OH group on the fifth carbon atom (the chiral center farthest from the carbonyl group) is on the right. In fact, all the OH groups except the one on the third carbon atom are to the right. 20.05: Reactions of Monosaccharides Learning Objectives • Identify the structures of sucrose, lactose, and maltose. • Identify the monosaccharides that are needed to form sucrose, lactose, and maltose Disaccharide Formation Previously, you learned that monosaccharides can form cyclic structures by the reaction of the carbonyl group with an OH group. These cyclic molecules can in turn react with another alcohol. Disaccharides (C12H22O11) are sugars composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a glycosidic linkage. This linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide. The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them. There are three common disaccharides: maltose, lactose, and sucrose. All three are white crystalline solids at room temperature and are soluble in water. We’ll consider each sugar in more detail. 20.06: Common Disaccharides • Objective 1 • Objective 2 Maltose Maltose occurs to a limited extent in sprouting grain. It is formed most often by the partial hydrolysis of starch and glycogen. In the manufacture of beer, maltose is liberated by the action of malt (germinating barley) on starch; for this reason, it is often referred to as malt sugar. Maltose is about 30% as sweet as sucrose. The human body is unable to metabolize maltose or any other disaccharide directly from the diet because the molecules are too large to pass through the cell membranes of the intestinal wall. Therefore, an ingested disaccharide must first be broken down by hydrolysis into its two constituent monosaccharide units. In the body, such hydrolysis reactions are catalyzed by enzymes such as maltase. The same reactions can be carried out in the laboratory with dilute acid as a catalyst, although in that case the rate is much slower, and high temperatures are required. Whether it occurs in the body or a glass beaker, the hydrolysis of maltose produces two molecules of D-glucose. $\mathrm{maltose \xrightarrow{H^+\: or\: maltase} \textrm{2 D-glucose}} \tag{16.6.2}$ Maltose is a reducing sugar. Thus, its two glucose molecules must be linked in such a way as to leave one anomeric carbon that can open to form an aldehyde group. The glucose units in maltose are joined in a head-to-tail fashion through an α-linkage from the first carbon atom of one glucose molecule to the fourth carbon atom of the second glucose molecule (that is, an α-1,4-glycosidic linkage; see Figure $1$). The bond from the anomeric carbon of the first monosaccharide unit is directed downward, which is why this is known as an α-glycosidic linkage. The OH group on the anomeric carbon of the second glucose can be in either the α or the β position, as shown in Figure $1$.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/20%3A_Carbohydrates/20.04%3A_Some_Important_Monosaccharides.txt
Learning Objectives • To compare and contrast the structures and uses of starch, glycogen, and cellulose. The polysaccharides are the most abundant carbohydrates in nature and serve a variety of functions, such as energy storage or as components of plant cell walls. Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages. The three most abundant polysaccharides are starch, glycogen, and cellulose. These three are referred to as homopolymers because each yields only one type of monosaccharide (glucose) after complete hydrolysis. Heteropolymers may contain sugar acids, amino sugars, or noncarbohydrate substances in addition to monosaccharides. Heteropolymers are common in nature (gums, pectins, and other substances) but will not be discussed further in this textbook. The polysaccharides are nonreducing carbohydrates, are not sweet tasting, and do not undergo mutarotation. Starch Starch is the most important source of carbohydrates in the human diet and accounts for more than 50% of our carbohydrate intake. It occurs in plants in the form of granules, and these are particularly abundant in seeds (especially the cereal grains) and tubers, where they serve as a storage form of carbohydrates. The breakdown of starch to glucose nourishes the plant during periods of reduced photosynthetic activity. We often think of potatoes as a “starchy” food, yet other plants contain a much greater percentage of starch (potatoes 15%, wheat 55%, corn 65%, and rice 75%). Commercial starch is a white powder. Starch is a mixture of two polymers: amylose and amylopectin. Natural starches consist of about 10%–30% amylose and 70%–90% amylopectin. Amylose is a linear polysaccharide composed entirely of D-glucose units joined by the α-1,4-glycosidic linkages we saw in maltose (part (a) of Figure \(1\)). Experimental evidence indicates that amylose is not a straight chain of glucose units but instead is coiled like a spring, with six glucose monomers per turn (part (b) of Figure \(1\)). When coiled in this fashion, amylose has just enough room in its core to accommodate an iodine molecule. The characteristic blue-violet color that appears when starch is treated with iodine is due to the formation of the amylose-iodine complex. This color test is sensitive enough to detect even minute amounts of starch in solution. Amylopectin is a branched-chain polysaccharide composed of glucose units linked primarily by α-1,4-glycosidic bonds but with occasional α-1,6-glycosidic bonds, which are responsible for the branching. A molecule of amylopectin may contain many thousands of glucose units with branch points occurring about every 25–30 units (Figure \(2\)). The helical structure of amylopectin is disrupted by the branching of the chain, so instead of the deep blue-violet color amylose gives with iodine, amylopectin produces a less intense reddish brown. Dextrins are glucose polysaccharides of intermediate size. The shine and stiffness imparted to clothing by starch are due to the presence of dextrins formed when clothing is ironed. Because of their characteristic stickiness with wetting, dextrins are used as adhesives on stamps, envelopes, and labels; as binders to hold pills and tablets together; and as pastes. Dextrins are more easily digested than starch and are therefore used extensively in the commercial preparation of infant foods. The complete hydrolysis of starch yields, in successive stages, glucose: starch → dextrins → maltose → glucose In the human body, several enzymes known collectively as amylases degrade starch sequentially into usable glucose units. • Glycogen Glycogen is the energy reserve carbohydrate of animals. Practically all mammalian cells contain some stored carbohydrates in the form of glycogen, but it is especially abundant in the liver (4%–8% by weight of tissue) and in skeletal muscle cells (0.5%–1.0%). Like starch in plants, glycogen is found as granules in liver and muscle cells. When fasting, animals draw on these glycogen reserves during the first day without food to obtain the glucose needed to maintain metabolic balance. Glycogen is structurally quite similar to amylopectin, although glycogen is more highly branched (8–12 glucose units between branches) and the branches are shorter. When treated with iodine, glycogen gives a reddish brown color. Glycogen can be broken down into its D-glucose subunits by acid hydrolysis or by the same enzymes that catalyze the breakdown of starch. In animals, the enzyme phosphorylase catalyzes the breakdown of glycogen to phosphate esters of glucose. About 70% of the total glycogen in the body is stored in muscle cells. Although the percentage of glycogen (by weight) is higher in the liver, the much greater mass of skeletal muscle stores a greater total amount of glycogen. Cellulose Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%–20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of noncellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production. Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by β-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure \(3\)). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure \(3\)). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product. Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the β-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes. Career Focus: Certified Diabetes Educator Certified diabetes educators come from a variety of health professions, such as nursing and dietetics, and specialize in the education and treatment of patients with diabetes. A diabetes educator will work with patients to manage their diabetes. This involves teaching the patient to monitor blood sugar levels, make good food choices, develop and maintain an exercise program, and take medication, if required. Summary Starch is a storage form of energy in plants. It contains two polymers composed of glucose units: amylose (linear) and amylopectin (branched). Glycogen is a storage form of energy in animals. It is a branched polymer composed of glucose units. It is more highly branched than amylopectin. Cellulose is a structural polymer of glucose units found in plants. It is a linear polymer with the glucose units linked through β-1,4-glycosidic bonds. Answers 1. What purposes do starch and cellulose serve in plants? 2. What purpose does glycogen serve in animals? 3. Starch is the storage form of glucose (energy) in plants, while cellulose is a structural component of the plant cell wall. 4. Glycogen is the storage form of glucose (energy) in animals. 5. What monosaccharide is obtained from the hydrolysis of each carbohydrate? 1. starch 2. cellulose 3. glycogen 6. For each carbohydrate listed in Exercise 1, indicate whether it is found in plants or mammals. 7. Describe the similarities and differences between amylose and cellulose. 8. Describe the similarities and differences between amylopectin and glycogen. 1. glucose 2. glucose 3. glucose 9. Amylose and cellulose are both linear polymers of glucose units, but the glycosidic linkages between the glucose units differ. The linkages in amylose are α-1,4-glycosidic linkages, while the linkages in cellulose they are β-1,4-glycosidic linkages.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/20%3A_Carbohydrates/20.07%3A_Some_Important_Polysaccharides_Based_on_Glucose.txt
Thumbnail: Ball-and-stick model of adenosine triphosphate (ATP) based on x-ray diffraction data. (Public Domain; Ben Mills). 21: The Generation of Biochemical Energy Learning Objectives • Objective 1 • Objective 2 Life requires energy. Animals, for example, require heat energy to maintain body temperature, mechanical energy to move their limbs, and chemical energy to synthesize the compounds needed by their cells. Living cells remain organized and functioning properly only through a continual supply of energy. But only specific forms of energy can be used. Supplying a plant with energy by holding it in a flame will not prolong its life. On the other hand, a green plant is able to absorb radiant energy from the sun, the most abundant source of energy for life on the earth. Plants use this energy first to form glucose and then to make other carbohydrates, as well as lipids and proteins. Unlike plants, animals cannot directly use the sun’s energy to synthesize new compounds. They must eat plants or other animals to get carbohydrates, fats, and proteins and the chemical energy stored in them. Once digested and transported to the cells, the nutrient molecules can be used in either of two ways: as building blocks for making new cell parts or repairing old ones or “burned” for energy. The thousands of coordinated chemical reactions that keep cells alive are referred to collectively as metabolism. In general, metabolic reactions are divided into two classes: the breaking down of molecules to obtain energy is catabolism, and the building of new molecules needed by living systems is anabolism. Definition: Metabolite Any chemical compound that participates in a metabolic reaction is a metabolite. Most of the energy required by animals is generated from lipids and carbohydrates. These fuels must be oxidized, or “burned,” for the energy to be released. The oxidation process ultimately converts the lipid or carbohydrate to carbon dioxide (CO2) and water (H2O). Carbohydrate: $\ce{C6H12O6 + 6O2 → 6CO2 + 6H2O + 670 kcal} \nonumber$ Lipid: $\ce{C16H32O2 + 23O2 → 16CO2 + 16H2O + 2,385 kcal} \nonumber$ These two equations summarize the biological combustion of a carbohydrate and a lipid by the cell through respiration. Respiration is the collective name for all metabolic processes in which gaseous oxygen is used to oxidize organic matter to carbon dioxide, water, and energy. Like the combustion of the common fuels we burn in our homes and cars (wood, coal, gasoline), respiration uses oxygen from the air to break down complex organic substances to carbon dioxide and water. But the energy released in the burning of wood is manifested entirely in the form of heat, and excess heat energy is not only useless but also injurious to the living cell. Living organisms instead conserve much of the energy respiration releases by channeling it into a series of stepwise reactions that produce adenosine triphosphate (ATP) or other compounds that ultimately lead to the synthesis of ATP. The remainder of the energy is released as heat and manifested as body temperature. 21.02: Cells and Their Structure Learning Objectives • Describe the two types of cell: prokaryotic and eukaryotic. • Explain the function of each cellular component. Cells fall into one of two broad categories: prokaryotic and eukaryotic. Only the predominantly single-celled organisms of the domains Bacteria and Archaea are classified as prokaryotes (pro- = “before”; -kary- = “nucleus”). Cells of animals, plants, fungi, and protists are all eukaryotes (eu- = “true”) and are made up of eukaryotic cells. Components of Prokaryotic Cells All cells share four common components: 1) a plasma membrane, an outer covering that separates the cell’s interior from its surrounding environment; 2) cytoplasm, consisting of a jelly-like cytosol within the cell in which other cellular components are found; 3) DNA, the genetic material of the cell; and 4) ribosomes, which synthesize proteins. However, prokaryotes differ from eukaryotic cells in several ways. A prokaryote is a simple, mostly single-celled (unicellular) organism that lacks a nucleus, or any other membrane-bound organelle. We will shortly come to see that this is significantly different in eukaryotes. Prokaryotic DNA is found in a central part of the cell: the nucleoid (Figure \(1\)). Unlike prokaryotic cells, eukaryotic cells have: 1) a membrane-bound nucleus; 2) numerous membrane-bound organelles such as the endoplasmic reticulum, Golgi apparatus, chloroplasts, mitochondria, and others; and 3) several, rod-shaped chromosomes. Because a eukaryotic cell’s nucleus is surrounded by a membrane, it is often said to have a “true nucleus.” The word “organelle” means “little organ,” and, as already mentioned, organelles have specialized cellular functions, just as the organs of your body have specialized functions. At this point, it should be clear to you that eukaryotic cells have a more complex structure than prokaryotic cells. Organelles allow different functions to be compartmentalized in different areas of the cell. Before turning to organelles, let’s first examine two important components of the cell: the plasma membrane and the cytoplasm. Mitochondria (singular = mitochondrion) are often called the “powerhouses” or “energy factories” of a cell because they are responsible for making adenosine triphosphate (ATP), the cell’s main energy-carrying molecule. ATP represents the short-term stored energy of the cell. Cellular respiration is the process of making ATP using the chemical energy found in glucose and other nutrients. In mitochondria, this process uses oxygen and produces carbon dioxide as a waste product. In fact, the carbon dioxide that you exhale with every breath comes from the cellular reactions that produce carbon dioxide as a byproduct. Mitochondria are oval-shaped, double membrane organelles (Figure \(3\)) that have their own ribosomes and DNA. Each membrane is a phospholipid bilayer embedded with proteins. The inner layer has folds called cristae. The area surrounded by the folds is called the mitochondrial matrix. The cristae and the matrix have different roles in cellular respiration.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/21%3A_The_Generation_of_Biochemical_Energy/21.01%3A_Energy_Life_and_Biochemical_Reactions.txt
Learning Objectives • Describe the stages of catabolism from food to ATP. Metabolism describes all of the chemical reactions that take place in an organism. A metabolic pathway is a series of interconnected biochemical reactions that convert a substrate molecule or molecules, step-by-step, through a series of metabolic intermediates, eventually yielding a final product or products. In the case of sugar metabolism, the first metabolic pathway synthesized sugar from smaller molecules, and the other pathway broke sugar down into smaller molecules. These two opposite processes—the first requiring energy and the second producing energy—are referred to as anabolic (building) and catabolic (breaking down) pathways, respectively. Consequently, metabolism is composed of building (anabolism) and degradation (catabolism). It is important to know that the chemical reactions of metabolic pathways don’t take place spontaneously. Each reaction step is facilitated, or catalyzed, by enzymes. Enzymes are important for catalyzing all types of biological reactions—those that require energy as well as those that release energy. See a simple graphic below, (Figure \(1\)). We can think of catabolism as occurring in three stages (Figure \(1\)). In stage I, carbohydrates, fats, and proteins are broken down into their individual monomer units: carbohydrates into simple sugars, fats into fatty acids and glycerol, and proteins into amino acids. One part of stage I of catabolism is the breakdown of food molecules by hydrolysis reactions into the individual monomer units—which occurs in the mouth, stomach, and small intestine—and is referred to as digestion. In stage II, these monomer units (or building blocks) are further broken down through different reaction pathways, one of which produces ATP, to form a common end product, Acetyl-coenzyme A, that can then be used in stage III to produce even more ATP. In this chapter, we will look at each stage of catabolism—as an overview and in detail. 21.04: Strategies of Metabolism - ATP and Energy Transfer Learning Objectives • To describe the importance of ATP as a source of energy in living organisms. Adenosine triphosphate (ATP), a nucleotide composed of adenine, ribose, and three phosphate groups, is perhaps the most important of the so-called energy-rich compounds in a cell. Its concentration in the cell varies from 0.5 to 2.5 mg/mL of cell fluid. Energy-rich compounds are substances having particular structural features that lead to a release of energy after hydrolysis. As a result, these compounds are able to supply energy for biochemical processes that require energy. The structural feature important in ATP is the phosphoric acid anhydride, or pyrophosphate, linkage: The pyrophosphate bond, symbolized by a squiggle (~), is hydrolyzed when ATP is converted to adenosine diphosphate (ADP). In this hydrolysis reaction, the products contain less energy than the reactants; there is a release of energy (> 7 kcal/mol). One reason for the amount of energy released is that hydrolysis relieves the electron-electron repulsions experienced by the negatively charged phosphate groups when they are bonded to each other (Figure 20.1.1). Energy is released because the products (ADP and phosphate ion) have less energy than the reactants [ATP and water (H2O)]. The general equation for ATP hydrolysis is as follows: $ATP + H_2O → ADP + P_i + 7.4\; kcal/mol \nonumber$ If the hydrolysis of ATP releases energy, its synthesis (from ADP) requires energy. In the cell, ATP is produced by those processes that supply energy to the organism (absorption of radiant energy from the sun in green plants and breakdown of food in animals), and it is hydrolyzed by those processes that require energy (the syntheses of carbohydrates, lipids, proteins; the transmission of nerve impulses; muscle contractions). In fact, ATP is the principal medium of energy exchange in biological systems. Many scientists call it the energy currency of cells. $P_i$ is the symbol for the inorganic phosphate anions $H_2PO_4^−$ and $HPO_4^{2−}$. ATP is not the only high-energy compound needed for metabolism. Several others are listed in Table $1$. Notice, however, that the energy released when ATP is hydrolyzed is approximately midway between those of the high-energy and the low-energy phosphate compounds. This means that the hydrolysis of ATP can provide energy for the phosphorylation of the compounds below it in the table. For example, the hydrolysis of ATP provides sufficient energy for the phosphorylation of glucose to form glucose 1-phosphate. By the same token, the hydrolysis of compounds, such as creatine phosphate, that appear above ATP in the table can provide the energy needed to resynthesize ATP from ADP. Table $1$: Energy Released by Hydrolysis of Some Phosphate Compounds Type Example Energy Released (kcal/mol) acyl phosphate 1,3-bisphosphoglycerate (BPG) −11.8 acetyl phosphate −11.3 guanidine phosphates creatine phosphate −10.3 arginine phosphate −9.1 pyrophosphates PPi* → 2Pi −7.8 ATP → AMP + PPi −7.7 ATP → ADP + Pi −7.5 ADP → AMP + Pi −7.5 sugar phosphates glucose 1-phosphate −5.0 fructose 6-phosphate −3.8 AMP → adenosine + Pi −3.4 glucose 6-phosphate −3.3 glycerol 3-phosphate −2.2 *PPi is the pyrophosphate ion. Summary The hydrolysis of ATP releases energy that can be used for cellular processes that require energy.
textbooks/chem/Introductory_Chemistry/Map%3A_Fundamentals_of_General_Organic_and_Biological_Chemistry_(McMurry_et_al.)/21%3A_The_Generation_of_Biochemical_Energy/21.03%3A_An_Overview_of_Metabolism_and_Energy_Production.txt