Chapter 8, Problem 15P

Interpretation Introduction

Interpretation:

The matching of the given $K_{\text{eq}}^{\text{'}}$ values with the appropriate $\Delta G°\text{'}$ values is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K_{\text{eq}}$. The equilibrium constant is independent of the initial amount of the reactant and product.

Answer to Problem 15P

The correct matching of the given $K_{\text{eq}}^{\text{'}}$ values with the appropriate $\Delta G°\text{'}$ values is given in the table below.

$K_{\text{eq}}^{\text{'}}$	$\Delta G°\text{'}$$\left(\text{kJ}{\text{mol}}^{-1}\right)$
(a) $1$	$0$
(b) ${10}^{-5}$	$28.53\text{kJ}{\text{mol}}^{-1}$
(c) ${10}^4$	$-22.84\text{kJ}{\text{mol}}^{-1}$
(d) ${10}^2$	$-11.42\text{kJ}{\text{mol}}^{-1}$
(e) ${10}^{-1}$	$5.69\text{kJ}{\text{mol}}^{-1}$

Explanation of Solution

(a) The given value of $K_{\text{eq}}^{\text{'}}$ is $1$.

The standard temperature is $25°\text{C}$.

The conversion of degrees Celsius into Kelvin is done as,

$0°\text{C}=\text{273}\text{K}$

Thus, the given temperature becomes,

$\begin{array}{c}25°\text{C}=25+\text{273}\text{K}\\ =\text{298}\text{K}\end{array}$

The value of universal gas constant is $8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}$.

The value of $\Delta G^{°\text{'}}$ is calculated by the expression,

$\Delta G^{°\text{'}}=-RT\ln K_{\text{eq}}$  (1)

Where,

• $R$ is the universal gas constant.
• $T$ is the temperature.
• $K_{\text{eq}}$ is the equilibrium constant.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

$\begin{array}{c}\Delta G^{°\text{'}}=-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \left(1\right)\\ =-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times 0\\ =0\end{array}$

Thus, the value of Gibbs free energy for $K_{\text{eq}}^{\text{'}}$ equals to $1$ is $0$.

(b) The given value of $K_{\text{eq}}^{\text{'}}$ is ${10}^{-5}$.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

$\begin{array}{c}\Delta G^{°\text{'}}=-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \left({10}^{-5}\right)\\ =-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times -11.5129\\ =28.53\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the value of Gibbs free energy for $K_{\text{eq}}^{\text{'}}$ equals to ${10}^{-5}$ is $28.53\text{kJ}{\text{mol}}^{-1}$.

(c) The given value of $K_{\text{eq}}^{\text{'}}$ is ${10}^4$.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

$\begin{array}{c}\Delta G^{°\text{'}}=-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \left({10}^4\right)\\ =-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times 9.21034\\ =-22.84\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the value of Gibbs free energy for $K_{\text{eq}}^{\text{'}}$ equals to ${10}^4$ is $-22.84\text{kJ}{\text{mol}}^{-1}$.

(d) The given value of $K_{\text{eq}}^{\text{'}}$ is ${10}^2$.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

$\begin{array}{c}\Delta G^{°\text{'}}=-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \left({10}^2\right)\\ =-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times 4.6052\\ =-11.42\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the value of Gibbs free energy for $K_{\text{eq}}^{\text{'}}$ equals to ${10}^2$ is $-11.42\text{kJ}{\text{mol}}^{-1}$.

(e) The given value of $K_{\text{eq}}^{\text{'}}$ is ${10}^{-1}$.

Substitute the values of universal gas constant, temperature and equilibrium constant in equation (1).

$\begin{array}{c}\Delta G^{°\text{'}}=-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times \ln \left({10}^{-1}\right)\\ =-8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}\times -2.303\\ =5.69\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the value of Gibbs free energy for $K_{\text{eq}}^{\text{'}}$ equals to ${10}^{-1}$ is $5.69\text{kJ}{\text{mol}}^{-1}$.

Conclusion

Therefore, the correct matching of the given $K_{\text{eq}}^{\text{'}}$ values with the appropriate $\Delta G°\text{'}$ values is done using the following relation:

$\Delta G^{°\text{'}}=-RT\ln K_{\text{eq}}$