Chapter 22, Problem 76AP

Interpretation Introduction

Interpretation:

The standard Gibbs free energy and the equilibrium constant for the given reaction are to be calculated with the formation constant for the reaction.

Concept introduction:

The change in free energy is called Gibb’s free energy and is denoted as $\Delta G^o$.

If $\Delta G^o$ is negative, it indicates that the electrical work has been done by the system on the surroundings and if it is positive then it means the electrical work has been done on the system by the surroundings.

The value of $K_1$ for the reverse of reaction is given as:

$K_1=\frac{1}{K}$, $K$ is the formation constant for the reaction.

The expression of equilibrium constant $\left(K_c\right)$ is given as:

$K_c=K_1\times K_2$

The relation between free energy change and standard free energy change is as:

$\Delta G=\Delta G^{\text{o}}+\text{R}T\ln Q$

Here, $\Delta G$ is free energy change, $\Delta G^o$ is standard free energy change, $Q$ is the reaction quotient, $\text{R}$ is the gas constant, and $T$ is the temperature.

At equilibrium the above equation is reduced to the expression: $\Delta G^{\text{o}}\text{=}-\text{R }T\text{lnK }$

Answer to Problem 76AP

Solution: $K_c = 6.7\times 10^{13}\text{ and }\Delta G°= -7.89\times 10^{4}J/mol$

Explanation of Solution

The overall reaction is obtained by adding the two half reactions as follows: $\begin{array}{l}\text{ anode}\left(oxidation\right){\text{: Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}^{\text{+}}\left(aq\right)\rightleftarrows {\text{ Ag}}^{\text{+}}\left(aq\right){\text{ + 2NH}}_{\text{3}}\text{(}aq\text{) }\\ \text{cathode}\left(oxidation\right){\text{:Ag}}^{\text{+}}\left(aq\right)+2C{N}^{-}\rightleftarrows {\text{ Ag(CN)}}_2^{-}\left(aq\right)\text{] }\\ \text{ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{ overall: Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}^{\text{+}}\left(aq\right) +2C{N}^{-}\left(aq\right)\rightleftarrows {\text{ Ag(CN)}}_2^{-}{\text{ + 2NH}}_{\text{3}}\left(aq\right)\end{array}$

The value of $K_1$ for the reverse of reaction is given as follows:

$\begin{array}{c}{K}_1=\frac{\text{1}}{\text{1}\text{.5}\text{×}{\text{10}}^{\text{7}}}\\ \text{=}\text{6}\text{.7}\text{×}{\text{10}}^{-8}\end{array}$

The expression of equilibrium constant $\left(K_c\right)$ is given as follows:

$K_c=K_1\times K_2$

Substitute $\text{6}\text{.7}\text{×}{\text{10}}^{-8}$ for $K_1$ and $1.0\times {10}^{21}$ for $K_2$ in the above equation as follows:

$\begin{array}{c}{K}_c=\text{(6}\text{.7 }\times {10}^{-8}\text{)}\left(1.0\times {10}^{21}\right) \\ \text{=} 6.7\times 10^{13}\end{array}$

The change in free energy or Gibbs free energy is calculated as follows:

$\Delta G^{\text{o}}=-RT\text{ ln }{K}_c$

Now, substitute $6.7\times 10^{13}$ for $K_c$, $\text{8}\text{.314}\text{J/Kmol}$ for $R$, and $298\text{ K}$ for $T$ in the above equation as follows:

$\begin{array}{c}\Delta G^{\text{o}}=-\text{(8}\text{.314 J/mol}\text{. K)}\left(\text{298 K}\right)\text{ln}\left(\text{6}{\text{.7 × 10}}^{\text{13}}\right) \\ = -7.89\times 10^{4}J/mol\end{array}$

Conclusion

The standard Gibbs free energy and equilibrium constant for the given reaction are $-7.89\times 10^{4}J/mol$ and $6.7\times 10^{13}$.