This is “Spontaneity and Equilibrium”, section 18.6 from the book Principles of General Chemistry (v. 1.0M).
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We have identified three criteria for whether a given reaction will occur spontaneously: ΔS_{univ} > 0, ΔG_{sys} < 0, and the relative magnitude of the reaction quotient Q versus the equilibrium constant K. (For more information on the reaction quotient and the equilibrium constant, see Chapter 15 "Chemical Equilibrium".) Recall that if Q < K, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if Q > K, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If Q = K, then the system is at equilibrium, and no net reaction occurs. Table 18.3 "Criteria for the Spontaneity of a Process as Written" summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes. Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. The relationship between ΔS_{univ} and ΔG_{sys} was described in Section 18.5 "Free Energy". In this section, we explore the relationship between the standard free energy of reaction (ΔG°) and the equilibrium constant (K).
Table 18.3 Criteria for the Spontaneity of a Process as Written
Spontaneous | Equilibrium | Nonspontaneous* |
---|---|---|
ΔS_{univ} > 0 | ΔS_{univ} = 0 | ΔS_{univ} < 0 |
ΔG_{sys} < 0 | ΔG_{sys} = 0 | ΔG_{sys} > 0 |
Q < K | Q = K | Q > K |
*Spontaneous in the reverse direction. |
Because ΔH° and ΔS° determine the magnitude of ΔG° (Equation 18.26), and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. As you learned in Section 18.5 "Free Energy", ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. Using higher math, the general relationship can be shown as follows:
Equation 18.29
ΔG = VΔP − SΔTIf a reaction is carried out at constant temperature (ΔT = 0), then Equation 18.29 simplifies to
Equation 18.30
ΔG = VΔPUnder normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the V in Equation 18.30 by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express ΔG in terms of the initial and final pressures (P_{i} and P_{f}, respectively) as in Equation 18.20:
Equation 18.31
$$\Delta G=\left(\frac{nRT}{P}\right)\Delta P=nRT\frac{\Delta P}{P}=nRT\mathrm{ln}\left(\frac{{P}_{\text{f}}}{{P}_{\text{i}}}\right)$$If the initial state is the standard state with P_{i} = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
G − G° = nRTln PThis can be rearranged as follows:
Equation 18.32
G = G° + nRTln PAs you will soon discover, Equation 18.32 allows us to relate ΔG° and K_{p}. Any relationship that is true for K_{p} must also be true for K because K_{p} and K are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
Equation 18.33
$$a\text{A}+bB\rightleftharpoons c\text{C}+d\text{D}$$Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:
Equation 18.34
$$\Delta G=\sum m{G}_{\text{products}}-\sum n{G}_{\text{reactants}}=(c{G}_{\text{C}}+d{G}_{\text{D}})-(a{G}_{\text{A}}+b{G}_{\text{B}})$$Substituting Equation 18.32 for each term into Equation 18.34,
$$\Delta G=[(cG{\xb0}_{\text{C}}+cRT\text{ln}{P}_{\text{C}})+(dG{\xb0}_{\text{D}}+dRT\text{ln}{P}_{\text{D}})]-[(aG{\xb0}_{\text{A}}+aRT\text{ln}{P}_{\text{A}})+(bG{\xb0}_{\text{B}}+bRT\text{ln}{P}_{\text{B}})]$$Combining terms gives the following relationship between ΔG and the reaction quotient Q:
Equation 18.35
$$\Delta G=\Delta G\xb0+RT\mathrm{ln}\left(\frac{{P}_{\text{C}}^{c}{P}_{\text{D}}^{d}}{{P}_{\text{A}}^{a}{P}_{\text{B}}^{b}}\right)=\Delta G\xb0+RT\text{ln}Q$$where ΔG° indicates that all reactants and products are in their standard states. In Chapter 15 "Chemical Equilibrium", you learned that for gases Q = K_{p} at equilibrium, and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and K_{p} for gases as follows:
Equation 18.36
$$\begin{array}{cc}\hfill 0& =\Delta G\xb0+RT\text{ln}{K}_{\text{p}}\hfill \\ \hfill \Delta G\xb0& =-RT\text{ln}{K}_{\text{p}}\hfill \end{array}$$If the products and reactants are in their standard states and ΔG° < 0, then K_{p} > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then K_{p} < 1, and reactants are favored over products. If ΔG° = 0, then K_{p} = 1, and neither reactants nor products are favored: the system is at equilibrium.
For a spontaneous process under standard conditions, K_{eq} and K_{p} are greater than 1.
In Example 10, we calculated that ΔG° = −32.7 kJ/mol of N_{2} for the reaction ${\text{N}}_{2}\text{(g)}+3{\text{H}}_{2}\text{(g)}\rightleftharpoons 2{\text{NH}}_{3}\text{(g)}\text{.}$ This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions: ${P}_{{\text{N}}_{2}}$ = 2.00 atm, ${P}_{{\text{H}}_{2}}$ = 7.00 atm, ${P}_{{\text{NH}}_{3}}$ = 0.021 atm, and T = 100°C. Does the reaction favor products or reactants?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether products or reactants are favored
Strategy:
A Using the values given and Equation 18.35, calculate Q.
B Substitute the values of ΔG° and Q into Equation 18.35 to obtain ΔG for the reaction under nonstandard conditions.
Solution:
A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation 18.35. Substituting the partial pressures given, we can calculate Q:
$$Q=\frac{{P}_{{\text{NH}}_{3}}^{2}}{{P}_{{\text{N}}_{2}}{P}_{{\text{H}}_{2}}^{3}}=\frac{{(0.021)}^{2}}{(2.00){(7.00)}^{3}}=6.4\times {10}^{-7}$$B Substituting the values of ΔG° and Q into Equation 18.35,
$$\begin{array}{cc}\hfill \Delta G=\Delta G\xb0+RT\text{ln}Q& =-32.7\text{kJ}+\left[(8.314\overline{)\text{J}}\text{/}\overline{)\text{K}})(373\text{}\overline{)\text{K}})\left(\frac{1\text{kJ}}{1000\text{}\overline{)\text{J}}}\right)\mathrm{ln}(6.4\times {10}^{-7})\right]\hfill \\ \hfill & \begin{array}{l}\hfill =-32.7\text{kJ}+(-44\text{kJ})\\ \hfill =-77{\text{kJ/molofN}}_{2}\end{array}\end{array}$$Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.
Exercise
Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P_{NO} = 0.0100 atm, ${P}_{{\text{O}}_{2}}$ = 0.200 atm, and ${P}_{{\text{NO}}_{2}}$ = 1.00 × 10^{−4} atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O_{2}. Are products or reactants favored?
Answer: −92.9 kJ/mol of O_{2}; the reaction is spontaneous to the right as written, so products are favored.
Calculate K_{p} for the reaction of H_{2} with N_{2} to give NH_{3} at 25°C. As calculated in Example 10, ΔG° for this reaction is −32.7 kJ/mol of N_{2}.
Given: balanced chemical equation from Example 10, ΔG°, and temperature
Asked for: K _{p}
Strategy:
Substitute values for ΔG° and T (in kelvins) into Equation 18.36 to calculate K_{p}, the equilibrium constant for the formation of ammonia.
Solution:
In Example 10, we used tabulated values of $\Delta {G}_{\text{f}}^{\xb0}$ to calculate ΔG° for this reaction (−32.7 kJ/mol of N_{2}). For equilibrium conditions, rearranging Equation 18.36,
$$\begin{array}{cc}\hfill \Delta G\xb0& =-RT\text{ln}{K}_{\text{p}}\hfill \\ \hfill \frac{-\Delta G\xb0}{RT}& =\text{ln}{K}_{\text{p}}\hfill \end{array}$$Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,
$$\begin{array}{cc}\hfill \text{ln}{K}_{\text{p}}& =-\frac{(-32.7\overline{)\text{kJ}})(1000\text{}\overline{)\text{J}}\text{/}\overline{)\text{kJ}})}{(8.314\text{}\overline{)\text{J}}\text{/}\overline{)\text{K}})(298\text{}\overline{)\text{K}})}=13.2\hfill \\ \hfill {K}_{\text{p}}& =5.4\times {10}^{5}\hfill \end{array}$$Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. As we saw in Chapter 15 "Chemical Equilibrium", however, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise
Calculate K_{p} for the reaction of NO with O_{2} to give NO_{2} at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O_{2}.
Answer: 2.2 × 10^{12}
Although K_{p} is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of K_{p} and K in Chapter 15 "Chemical Equilibrium" and showed that they are related:
Equation 18.37
K_{p} = K(RT)^{Δ}^{n}where Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so K_{p} = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation 18.36 can be written in a more general form:
Equation 18.38
ΔG° = −RT ln KOnly when a reaction results in a net production or consumption of gases is it necessary to correct Equation 18.38 for the difference between K_{p} and K.Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equation 18.26 and Equation 18.38 provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:
Equation 18.39
ΔG° = ΔH° − TΔS° = −RT ln KNotice that K becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible.
The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation 18.39, which can be rearranged as follows:
Equation 18.40
$$\text{ln}K=-\frac{\Delta H\xb0}{RT}+\frac{\Delta S\xb0}{R}$$Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation 18.40 agrees with the qualitative predictions made by applying Le Châtelier’s principle, which we discussed in Chapter 15 "Chemical Equilibrium". Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation 18.40 also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.
If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K_{1} and K_{2} are the equilibrium constants for a reaction at temperatures T_{1} and T_{2}, respectively. Applying Equation 18.40 gives the following relationship at each temperature:
$$\begin{array}{cc}\hfill \text{ln}{K}_{1}& =\frac{-\Delta H\xb0}{R{T}_{1}}+\frac{\Delta S\xb0}{R}\hfill \\ \hfill \text{ln}{K}_{2}& =\frac{-\Delta H\xb0}{R{T}_{2}}+\frac{\Delta S\xb0}{R}\hfill \end{array}$$Subtracting ln K_{1} from ln K_{2},
Equation 18.41
$$\text{ln}{K}_{2}-\text{ln}{K}_{1}=\text{ln}\frac{{K}_{2}}{{K}_{1}}=\frac{\Delta H\xb0}{R}\left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)$$Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K_{1}) allow us to calculate the value of the equilibrium constant at any other temperature (K_{2}), assuming that ΔH° and ΔS° are independent of temperature.
The equilibrium constant for the formation of NH_{3} from H_{2} and N_{2} at 25°C was calculated to be K_{p} = 5.4 × 10^{5} in Example 13. What is K_{p} at 500°C? (Use the data from Example 10.)
Given: balanced chemical equation, ΔH°, initial and final T, and K_{p} at 25°C
Asked for: K_{p} at 500°C
Strategy:
Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation 18.41 to obtain K_{2}, the equilibrium constant at the final temperature.
Solution:
The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N_{2}. If we set T_{1} = 25°C = 298.K and T_{2} = 500°C = 773 K, then from Equation 18.41 we obtain the following:
$$\begin{array}{cc}\hfill \text{ln}\frac{{K}_{2}}{{K}_{1}}& =\frac{\Delta H\xb0}{R}\left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)\hfill \\ \hfill & =\frac{(-91.8\text{}\overline{)\text{kJ}})(1000\text{}\overline{)\text{J}}\text{/}\overline{)\text{kJ}})}{8.314\text{}\overline{)\text{J}}\text{/}\overline{)\text{K}}}\left(\frac{1}{298\text{}\overline{)\text{K}}}-\frac{1}{773\text{}\overline{)\text{K}}}\right)=-22.8\hfill \\ \hfill \frac{{K}_{2}}{{K}_{1}}& =1.3\times {10}^{-10}\hfill \\ \hfill {K}_{2}& =(5.4\times {10}^{5})(1.3\times {10}^{-10})=7.0\times {10}^{-5}\hfill \end{array}$$Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise
In the exercise in Example 13, you calculated K_{p} = 2.2 × 10^{12} for the reaction of NO with O_{2} to give NO_{2} at 25°C. Use the $\Delta {H}_{\text{f}}^{\xb0}$ values in the exercise in Example 10 to calculate K_{p} for this reaction at 1000°C.
Answer: 5.6 × 10^{−4}
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and K_{p}, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or K_{p} > 1, and products are favored over reactants. If ΔG° > 0, then K or K_{p} < 1, and reactants are favored over products. If ΔG° = 0, then K or K_{p} = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Relationship between standard free-energy change and equilibrium constant
Equation 18.38: ΔG° = −RT ln K
Temperature dependence of equilibrium constant
Equation 18.40: $\text{ln}K=\frac{-\Delta H\xb0}{RT}+\frac{\Delta S\xb0}{R}$
Calculation of K at second temperature
Equation 18.41: $\text{ln}\frac{{K}_{2}}{{K}_{1}}=\frac{\Delta H\xb0}{R}\left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)$
Do you expect products or reactants to dominate at equilibrium in a reaction for which ΔG° is equal to
The change in free energy enables us to determine whether a reaction will proceed spontaneously. How is this related to the extent to which a reaction proceeds?
What happens to the change in free energy of the reaction N_{2}(g) + 3F_{2}(g) → 2NF_{3}(g) if the pressure is increased while the temperature remains constant? if the temperature is increased at constant pressure? Why are these effects not so important for reactions that involve liquids and solids?
Compare the expressions for the relationship between the change in free energy of a reaction and its equilibrium constant where the reactants are gases versus liquids. What are the differences between these expressions?
Carbon monoxide, a toxic product from the incomplete combustion of fossil fuels, reacts with water to form CO_{2} and H_{2}, as shown in the equation $\text{CO(g)}+{\text{H}}_{2}\text{O(g)}\rightleftharpoons {\text{CO}}_{2}\text{(g)}+{\text{H}}_{2}\text{(g),}$ for which ΔH° = −41.0 kJ/mol and ΔS° = −42.3 J cal/(mol·K) at 25°C and 1 atm.
Methane and water react to form carbon monoxide and hydrogen according to the equation ${\text{CH}}_{4}\text{(g)}+{\text{H}}_{2}\text{O(g)}\rightleftharpoons \text{CO(g)}+3{\text{H}}_{2}\text{(g)}\text{.}$
Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
The gas-phase decomposition of N_{2}O_{4} to NO_{2} is an equilibrium reaction with K_{p} = 4.66 × 10^{−3}. Calculate the standard free-energy change for the equilibrium reaction between N_{2}O_{4} and NO_{2}.
The standard free-energy change for the dissolution ${\text{K}}_{4}{\text{Fe(CN)}}_{6}\text{\u22c5}{\text{H}}_{2}\text{O(s)}\rightleftharpoons 4{\text{K}}^{+}\text{(aq)}+{\text{Fe(CN)}}_{6}{}^{4-}\text{(aq)}+{\text{H}}_{2}\text{O(l)}$ is 26.1 kJ/mol. What is the equilibrium constant for this process at 25°C?
Ammonia reacts with water in liquid ammonia solution (am) according to the equation ${\text{NH}}_{3}\text{(g)}+{\text{H}}_{2}\text{O(am)}\rightleftharpoons {\text{NH}}_{4}{}^{+}\text{(am)}+{\text{OH}}^{-}\text{(am)}\text{.}$ The change in enthalpy for this reaction is 21 kJ/mol, and ΔS° = −303 J/(mol·K). What is the equilibrium constant for the reaction at the boiling point of liquid ammonia (−31°C)?
At 25°C, a saturated solution of barium carbonate is found to have a concentration of [Ba^{2+}] = [CO_{3}^{2−}] = 5.08 × 10^{−5} M. Determine ΔG° for the dissolution of BaCO_{3}.
Lead phosphates are believed to play a major role in controlling the overall solubility of lead in acidic soils. One of the dissolution reactions is ${\text{Pb}}_{3}{({\text{PO}}_{4})}_{2}\text{(s)}+4{\text{H}}^{+}\text{(aq)}\rightleftharpoons 3{\text{Pb}}^{2+}\text{(aq)}+2{\text{H}}_{2}{\text{PO}}_{4}{}^{-}\text{(aq),}$ for which log K = −1.80. What is ΔG° for this reaction?
The conversion of butane to 2-methylpropane is an equilibrium process with ΔH° = −2.05 kcal/mol and ΔG° = −0.89 kcal/mol.
The reaction of CaCO_{3}(s) to produce CaO(s) and CO_{2}(g) has an equilibrium constant at 25°C of 2 × 10^{−23}. Values of $\Delta {H}_{\text{f}}^{\xb0}$ are as follows: CaCO_{3}, −1207.6 kJ/mol; CaO, −634.9 kJ/mol; and CO_{2}, −393.5 kJ/mol.
In acidic soils, dissolved Al^{3+} undergoes a complex formation reaction with SO_{4}^{2−} to form [AlSO_{4}^{+}]. The equilibrium constant at 25°C for the reaction ${\text{Al}}^{3+}\text{(aq)}+{\text{SO}}_{4}{}^{2-}\text{(aq)}\rightleftharpoons {\text{AlSO}}_{4}{}^{+}\text{(aq)}$ is 1585.
13.3 kJ/mol
5.1 × 10^{−21}
10.3 kJ/mol