Chemistry: Atoms First 2e

# 13.4Equilibrium Calculations

Chemistry: Atoms First 2e13.4 Equilibrium Calculations

### Learning Objectives

By the end of this section, you will be able to:

• Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
• Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches
• Explain how temperature affects the spontaneity of some proceses
• Relate standard free energy changes to equilibrium constants

Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.

Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:

$2NH3(g)⇌N2(g)+3H2(g)2NH3(g)⇌N2(g)+3H2(g)$

As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x:

$Δ[N2]=+x,Δ[N2]=+x,$

the corresponding changes in the other species concentrations are

$Δ[H2]=Δ[N2](3molH21molN2)=+3xΔ[NH3]=−Δ[N2](2molNH31molN2)=−2x,Δ[H2]=Δ[N2](3molH21molN2)=+3xΔ[NH3]=−Δ[N2](2molNH31molN2)=−2x,$

### Example 13.5

#### Determining Relative Changes in Concentration

Derive the missing terms representing concentration changes for each of the following reactions.

(a) $C2H2(g)+2Br2(g)⇌C2H2Br4(g)x__________C2H2(g)+2Br2(g)⇌C2H2Br4(g)x__________$

(b) $I2(aq)+I−(aq)⇌I3−(aq)__________xI2(aq)+I−(aq)⇌I3−(aq)__________x$

(c) $C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)x_______________C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)x_______________$

#### Solution

(a) $C2H2(g)+2Br2(g)⇌C2H2Br4(g)x2x−xC2H2(g)+2Br2(g)⇌C2H2Br4(g)x2x−x$

(b) $I2(aq)+I−(aq)⇌I3−(aq)−x−xxI2(aq)+I−(aq)⇌I3−(aq)−x−xx$

(c) $C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)x5x−3x−4xC3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)x5x−3x−4x$

Complete the changes in concentrations for each of the following reactions:

(a) $2SO2(g)+O2(g)⇌2SO3(g)_____x_____2SO2(g)+O2(g)⇌2SO3(g)_____x_____$

(b) $C4H8(g)⇌2C2H4(g)_____−2xC4H8(g)⇌2C2H4(g)_____−2x$

(c) $4NH3(g)+7H2O(g)⇌4NO2(g)+6H2O(g)____________________4NH3(g)+7H2O(g)⇌4NO2(g)+6H2O(g)____________________$

(a) 2x, x, −2x; (b) x, −2x; (c) 4x, 7x, −4x, −6x or −4x, −7x, 4x, 6x

### Calculation of an Equilibrium Constant

The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example 13.2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.

### Example 13.6

#### Calculation of an Equilibrium Constant

Iodine molecules react reversibly with iodide ions to produce triiodide ions.
$I2(aq)+I−(aq)⇌I3−(aq)I2(aq)+I−(aq)⇌I3−(aq)$

If a solution with the concentrations of I2 and I both equal to 1.000 $××$ 10−3 M before reaction gives an equilibrium concentration of I2 of 6.61 $××$ 10−4 M, what is the equilibrium constant for the reaction?

#### Solution

To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:
$KC=[I3−][I2][I−]KC=[I3−][I2][I−]$

Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table. At equilibrium the concentration of I2 is 6.61 $××$ 10−4 M so that

$1.000×10−3−x=6.61×10−41.000×10−3−x=6.61×10−4$
$x=1.000×10−3−6.61×10−4x=1.000×10−3−6.61×10−4$
$=3.39×10−4M=3.39×10−4M$

The ICE table may now be updated with numerical values for all its concentrations: Finally, substitute the equilibrium concentrations into the K expression and solve:

$Kc=[I3−][I2][I−]Kc=[I3−][I2][I−]$
$=3.39×10−4M(6.61×10−4M)(6.61×10−4M)=776=3.39×10−4M(6.61×10−4M)(6.61×10−4M)=776$

Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.
$C2H5OH+CH3CO2H⇌CH3CO2C2H5+H2OC2H5OH+CH3CO2H⇌CH3CO2C2H5+H2O$

When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $1313$ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)

Kc = 4

### Calculation of a Missing Equilibrium Concentration

When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.

### Example 13.7

#### Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the Kc for the reaction, $N2(g)+O2(g)⇌2NO(g),N2(g)+O2(g)⇌2NO(g),$ is 4.1 $××$ 10−4. Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of N2 and O2 at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

#### Solution

Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:
$Kc=[NO]2[N2][O2]Kc=[NO]2[N2][O2]$
$[NO]2=Kc[N2][O2][NO]2=Kc[N2][O2]$
$[NO]=Kc[N2][O2][NO]=Kc[N2][O2]$
$=(4.1×10−4)(0.036)(0.0089)=(4.1×10−4)(0.036)(0.0089)$
$=1.31×10−7=1.31×10−7$
$=3.6×10−4=3.6×10−4$

Thus [NO] is 3.6 $××$ 10−4 mol/L at equilibrium under these conditions.

To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for K:

$Kc=[NO]2[N2][O2] =(3.6×10−4)2(0.036)(0.0089) =4.0×10−4Kc=[NO]2[N2][O2] =(3.6×10−4)2(0.036)(0.0089) =4.0×10−4$

This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.

The equilibrium constant Kc for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 $××$ 10−2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.

1.53 mol/L

### Calculation of Equilibrium Concentrations from Initial Concentrations

Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

1. Identify the direction in which the reaction will proceed to reach equilibrium.
2. Develop an ICE table.
3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
4. Confirm the calculated equilibrium concentrations.

The last two example exercises of this chapter demonstrate the application of this strategy.

### Example 13.8

#### Calculation of Equilibrium Concentrations

Under certain conditions, the equilibrium constant Kc for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of PCl5, PCl3, and Cl2 in a mixture that initially contained only PCl5 at a concentration of 1.00 M?

#### Solution

Use the stepwise process described earlier.
1. Step 1.

Determine the direction the reaction proceeds.

The balanced equation for the decomposition of PCl5 is

$PCl5(g)⇌PCl3(g)+Cl2(g)PCl5(g)⇌PCl3(g)+Cl2(g)$

Because only the reactant is present initially Qc = 0 and the reaction will proceed to the right.

2. Step 2.

Develop an ICE table. 3. Step 3.

Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives

$Kc=[PCl3][Cl2][PCl5]=0.0211Kc=[PCl3][Cl2][PCl5]=0.0211$
$=(x)(x)(1.00−x)=(x)(x)(1.00−x)$
$0.0211=(x)(x)(1.00−x)0.0211=(x)(x)(1.00−x)$
$0.0211(1.00−x)=x20.0211(1.00−x)=x2$
$x2+0.0211x−0.0211=0x2+0.0211x−0.0211=0$

Appendix B shows an equation of the form ax2 + bx + c = 0 can be rearranged to solve for x:

$x=−b±b2−4ac2ax=−b±b2−4ac2a$

In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:

$x=−0.0211±(0.0211)2−4(1)(−0.0211)2(1)x=−0.0211±(0.0211)2−4(1)(−0.0211)2(1)$
$=−0.0211±(4.45×10−4)+(8.44×10−2)2=−0.0211±(4.45×10−4)+(8.44×10−2)2$
$=−0.0211±0.2912=−0.0211±0.2912$

The two roots of the quadratic are, therefore,

$x=−0.0211+0.2912=0.135x=−0.0211+0.2912=0.135$

and

$x=−0.0211−0.2912=−0.156x=−0.0211−0.2912=−0.156$

For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M.

The equilibrium concentrations are

$[PCl5]=1.00−0.135=0.87M[PCl5]=1.00−0.135=0.87M$
$[PCl3]=x=0.135M[PCl3]=x=0.135M$
$[Cl2]=x=0.135M[Cl2]=x=0.135M$
4. Step 4.

Confirm the calculated equilibrium concentrations.

Substitution into the expression for Kc (to check the calculation) gives

$Kc=[PCl3][Cl2][PCl5]=(0.135)(0.135)0.87=0.021Kc=[PCl3][Cl2][PCl5]=(0.135)(0.135)0.87=0.021$

The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures).

Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5.
$CH3CO2H+C2H5OH⇌CH3CO2C2H5+H2OCH3CO2H+C2H5OH⇌CH3CO2C2H5+H2O$

The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O?

[CH3CO2H] = 0.37 M, [C2H5OH] = 0.37 M, [CH3CO2C2H5] = 0.18 M, [H2O] = 0.18 M

A 1.00-L flask is filled with 1.00 moles of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2, I2, and HI in moles/L?
$H2(g)+I2(g)⇌2HI(g)H2(g)+I2(g)⇌2HI(g)$

[H2] = 0.06 M, [I2] = 1.06 M, [HI] = 1.88 M

### Example 13.9

#### Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption

What are the concentrations at equilibrium of a 0.15 M solution of HCN?
$HCN(aq)⇌H+(aq)+CN−(aq)Kc=4.9×10−10HCN(aq)⇌H+(aq)+CN−(aq)Kc=4.9×10−10$

#### Solution

Using “x” to represent the concentration of each product at equilibrium gives this ICE table. Substitute the equilibrium concentration terms into the Kc expression

$Kc=(x)(x)0.15−xKc=(x)(x)0.15−x$

Rearrange to the quadratic form and solve for x

$x2+4.9×10−10−7.35×10−11=0x2+4.9×10−10−7.35×10−11=0$
$x=8.56×10−6M(3 sig. figs.)=8.6×10−6M(2 sig. figs.)x=8.56×10−6M(3 sig. figs.)=8.6×10−6M(2 sig. figs.)$

Thus [H+] = [CN] = x = 8.6 $××$ 10–6 M and [HCN] = 0.15 – x = 0.15 M.

Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change:

$ifx≪0.15M,then(0.15−x)≈0.15ifx≪0.15M,then(0.15−x)≈0.15$

This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:

$Kc=(x)(x)0.15−x≈x20.15Kc=(x)(x)0.15−x≈x20.15$
$4.9×10−10=x20.154.9×10−10=x20.15$
$x2=(0.15)(4.9×10−10)=7.4×10−11x2=(0.15)(4.9×10−10)=7.4×10−11$
$x=7.4×10−11=8.6×10−6Mx=7.4×10−11=8.6×10−6M$

The value of x calculated is, indeed, much less than the initial concentration

$8.6×10−6≪0.158.6×10−6≪0.15$

and so the approximation was justified. If this simplified approach were to yield a value for x that did not justify the approximation, the calculation would need to be repeated without making the approximation.

What are the equilibrium concentrations in a 0.25 M NH3 solution?
$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kc=1.8×10−5NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kc=1.8×10−5$

$[OH−]=[NH4+]=0.0021M;[OH−]=[NH4+]=0.0021M;$ [NH3] = 0.25 M

### Temperature Dependence of Spontaneity

As was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 13.8.

Figure 13.8 There are four possibilities regarding the signs of enthalpy and entropy changes.

### Example 13.10

#### Predicting the Temperature Dependence of Spontaneity

The incomplete combustion of carbon is described by the following equation:
$2C(s)+O2(g)⟶2CO(g)2C(s)+O2(g)⟶2CO(g)$

How does the spontaneity of this process depend upon temperature?

#### Solution

Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.

Popular chemical hand warmers generate heat by the air-oxidation of iron:
$4Fe(s)+3O2(g)⟶2Fe2O3(s)4Fe(s)+3O2(g)⟶2Fe2O3(s)$

How does the spontaneity of this process depend upon temperature?

ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$
$y=b+mxy=b+mx$

Such a plot is shown in Figure 13.9. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:

$ΔG=0=ΔH−TΔSΔG=0=ΔH−TΔS$
$T=ΔHΔST=ΔHΔS$

So, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

Figure 13.9 These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

### Example 13.11

#### Equilibrium Temperature for a Phase Transition

As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.

#### Solution

The process of interest is the following phase change:
$H2O(l)⟶H2O(g)H2O(l)⟶H2O(g)$

When this process is at equilibrium, ΔG = 0, so the following is true:

$0=ΔH°−TΔS°orT=ΔH°ΔS°0=ΔH°−TΔS°orT=ΔH°ΔS°$

Using the standard thermodynamic data from Appendix G,

$ΔH°=1 mol×ΔHf°(H2O(g))−1 mol×ΔHf°(H2O(l))=(1 mol)−241.82 kJ/mol−(1 mol)(−241.82 kJ/mol)=44.01 kJ ΔS°=1 mol×ΔS°(H2O(g))−1 mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol−(1 mol)70.0 J/K·mol=118.8 J/K T=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°CΔH°=1 mol×ΔHf°(H2O(g))−1 mol×ΔHf°(H2O(l))=(1 mol)−241.82 kJ/mol−(1 mol)(−241.82 kJ/mol)=44.01 kJ ΔS°=1 mol×ΔS°(H2O(g))−1 mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol−(1 mol)70.0 J/K·mol=118.8 J/K T=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°C$

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

Use the information in Appendix G to estimate the boiling point of CS2.

313 K (accepted value 319 K)

### Free Energy and Equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).

In the section on equilibrium, the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.

The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change, according to this equation:

$ΔG=ΔG°+RTlnQΔG=ΔG°+RTlnQ$

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.12.

### Example 13.12

#### Calculating ΔG under Nonstandard Conditions

What is the free energy change for the process shown here under the specified conditions?

T = 25 °C, $PN2=0.870 atm,PN2=0.870 atm,$ $PH2=0.250 atm,PH2=0.250 atm,$ and $PNH3=12.9 atmPNH3=12.9 atm$

$2NH3(g)⟶3H2(g)+N2(g)ΔG°=33.0 kJ/mol2NH3(g)⟶3H2(g)+N2(g)ΔG°=33.0 kJ/mol$

#### Solution

The equation relating free energy change to standard free energy change and reaction quotient may be used directly:
$ΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/molΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/mol$

Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.

Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

ΔG = −47 kJ/mol; yes

For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as

$0=ΔG°+RTlnK(at equilibrium)0=ΔG°+RTlnK(at equilibrium)$
$ΔG°=−RTlnKorK=e−ΔG°RTΔG°=−RTlnKorK=e−ΔG°RT$

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 13.1.

Relations between Standard Free Energy Changes and Equilibrium Constants
K ΔG° Composition of an Equilibrium Mixture
> 1 < 0 Products are more abundant
< 1 > 0 Reactants are more abundant
= 1 = 0 Reactants and products are comparably abundant
Table 13.1

### Example 13.13

#### Calculating an Equilibrium Constant using Standard Free Energy Change

Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.

#### Solution

The reaction of interest is the following:
$AgCl(s)⇌Ag+(aq)+Cl−(aq)Ksp=[Ag+][Cl−]AgCl(s)⇌Ag+(aq)+Cl−(aq)Ksp=[Ag+][Cl−]$

The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

$ΔG°=[ΔGf°(Ag+(aq))+ΔGf°(Cl−(aq))]−[ΔGf°(AgCl(s))]=[77.1 kJ/mol−131.2 kJ/mol]−[−109.8 kJ/mol]=55.7 kJ/molΔG°=[ΔGf°(Ag+(aq))+ΔGf°(Cl−(aq))]−[ΔGf°(AgCl(s))]=[77.1 kJ/mol−131.2 kJ/mol]−[−109.8 kJ/mol]=55.7 kJ/mol$

The equilibrium constant for the reaction may then be derived from its standard free energy change:

$Ksp=e−ΔG°RT=exp(−ΔG°RT)=exp(−55.7×103J/mol8.314J/mol·K×298.15K)=exp(−22.470)=e−22.470=1.74×10−10Ksp=e−ΔG°RT=exp(−ΔG°RT)=exp(−55.7×103J/mol8.314J/mol·K×298.15K)=exp(−22.470)=e−22.470=1.74×10−10$

This result is in reasonable agreement with the value provided in Appendix J.

Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
$N2O4 (g)⇌ 2NO2(g)N2O4 (g)⇌ 2NO2(g)$

K = 0.32

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure 13.10). If a system consists of reactants and products in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Figure 13.10 These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.