Chapter 6, Problem 6.7P

a) Which $K_{\text{eq}}$ corresponds to a negative value of $\Delta \text{G}°$, $K_{\text{eq}}=1000$ or $K_{\text{eq}}=.001$?

b) Which $K_{\text{eq}}$ corresponds to a lower value of $\Delta \text{G}°$, $K_{\text{eq}}={10}^{-2}$ or $K_{\text{eq}}={10}^{-5}$?

Interpretation Introduction

(a)

Interpretation: The $K_{\text{eq}}$ value which corresponds to a negative value of $\Delta G°$ is to be predicted.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T{\mathrm{logK}}_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.7P

The $K_{\text{eq}}$ value which corresponds to a negative value of $\Delta G°$ is $1000$.

Explanation of Solution

Given

The values of $K_{\text{eq}}$ are $1000$ or $0.001$.

The value of $\Delta G°$ becomes negative if the value of equilibrium constant is greater than $1$.

The free energy change is calculated as,

$\Delta G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$ ……(1)

Substitute the values of $\text{R}$, $T$ and $K_{\text{eq}}$ for $K_{\text{eq}}=1000$ in the equation (1).

$\begin{array}{rll}\text{ΔG°}& =& -2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 298\text{K}\times \text{log}\left(1000\right)\\ & =& -17117.5\text{J}/\text{mol}\end{array}$

The conversion of units of $\text{ΔG°}$ into $\text{k}\text{J}/\text{mol}$ is done as,

$\begin{array}{rll}\text{1}\text{k}\text{J}/\text{mol}& =& \text{1000}\text{k}\text{J}/\text{mol}\\ -17117.5\text{J}/\text{mol}& =& -17117.5\text{J}/\text{mol}\times \frac{1\text{kJ}}{1000\text{J}}\\ & =& -17.1\text{k}\text{J}/\text{mol}\end{array}$

The value of $\text{ΔG°}$ is $-17.1\text{k}\text{J}/\text{mol}$ for $K_{\text{eq}}=1000$.

Similarly, for $K_{\text{eq}}=0.001$,

Substitute the values of $\text{R}$, $T$ and $K_{\text{eq}}$ in the equation (1).

$\begin{array}{rll}\text{Δ}G\text{°}& =& -2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 298\text{K}\times \text{log}\left(.001\right)\\ & =& 17117.5\text{J}/\text{mol}\end{array}$

The conversion of units of $\text{Δ}G\text{°}$ into $\text{k}\text{J}/\text{mol}$ is done as,

$\begin{array}{rll}\text{1}\text{k}\text{J}/\text{mol}& =& \text{1000}\text{k}\text{J}/\text{mol}\\ 17117.5\text{J}/\text{mol}& =& 17117.5\text{J}/\text{mol}\times \frac{1\text{kJ}}{1000\text{J}}\\ & =& 17.1\text{k}\text{J}/\text{mol}\end{array}$

The value of $\text{Δ}G\text{°}$ is $17.1\text{k}\text{J}/\text{mol}$ for $K_{\text{eq}}=.001$. Therefore, the $K_{\text{eq}}$ value which corresponds to a negative value of $\Delta G°$ is $1000$.

Conclusion

The $K_{\text{eq}}$ value which corresponds to a negative value of $\Delta G°$ is $1000$.

Interpretation Introduction

(b)

Interpretation: The $K_{\text{eq}}$ value which corresponds to a lower value of $\Delta \text{G}°$ is to be stated.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.7P

The $K_{\text{eq}}$ value corresponds to a lower value of $\Delta G°$ is ${10}^{-2}$.

Explanation of Solution

The value of $\Delta G°$ becomes negative if the value of equilibrium constant is greater than $1$.

The free energy change is calculated as,

$\Delta G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$ ……(2)

Substitute the values of $\text{R}$, $T$ and $K_{\text{eq}}$ for $K_{\text{eq}}={10}^{-2}$ in the equation (2).

$\begin{array}{rll}\text{Δ}G\text{°}& =& -2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 298\text{K}\times \text{log}\left({10}^{-2}\right)\\ & =& 11411.7\text{J}/\text{mol}\end{array}$

The conversion of units of $\text{Δ}G\text{°}$ into $\text{k}\text{J}/\text{mol}$ is done as,

$\begin{array}{rll}\text{1}\text{k}\text{J}/\text{mol}& =& \text{1000}\text{k}\text{J}/\text{mol}\\ 11411.7\text{J}/\text{mol}& =& 11411.7\text{J}/\text{mol}\times \frac{1\text{kJ}}{1000\text{J}}\\ & =& +11.4\text{k}\text{J}/\text{mol}\end{array}$

The value of $\Delta G°$ is $+11.4\text{k}\text{J}/\text{mol}$ for $K_{\text{eq}}={10}^{-2}$.

Similarly, for $K_{\text{eq}}={10}^{-5}$,

Substitute the values of $\text{R}$, $T$ and $K_{\text{eq}}$ for $K_{\text{eq}}={10}^{-5}$ in the equation (2).

$\begin{array}{rll}\text{Δ}G\text{°}& =& -2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 298\text{K}\times \text{log}\left({10}^{-5}\right)\\ & =& 28529.2\text{J}/\text{mol}\end{array}$

The conversion of units of $\Delta G\text{°}$ into $\text{k}\text{J}/\text{mol}$ is done as,

$\begin{array}{rll}\text{1}\text{k}\text{J}/\text{mol}& =& \text{1000}\text{k}\text{J}/\text{mol}\\ 28529.2\text{J}/\text{mol}& =& 28529.2\text{J}/\text{mol}\times \frac{1\text{kJ}}{1000\text{J}}\\ & =& +28.5\text{k}\text{J}/\text{mol}\end{array}$

The value of $\Delta G\text{°}$ is $+28.5\text{k}\text{J}/\text{mol}$ for $K_{\text{eq}}={10}^{-5}$.

Therefore, the $K_{\text{eq}}$ value corresponds to a lower value of $\Delta G°$ is ${10}^{-2}$.

Conclusion

The $K_{\text{eq}}$ which value corresponds to a lower value of $\Delta G°$ is ${10}^{-2}$.