Chapter 18, Problem 56AP

Interpretation Introduction

Interpretation:

The standard free energy change for the given equilibrium reaction and initial concentrations at ${\text{25}}^{\circ }\text{C}$ is to be calculated.

Concept introduction:

For a general chemical reaction, $\text{aA + bB }\underset{}{\overset{}{\rightleftarrows }}\text{cC+dD}$, the reaction quotient is represented as:

$Q=\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

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

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

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

The value of $\Delta G$ for a reaction to be spontaneous must be negative.

The value of $\Delta G$ for a reaction at equilibrium must be zero.

The value of $\Delta G$ for a reaction to be non-spontaneous must be positive.

Answer to Problem 56AP

Solution:

a)

$80\text{ kJ/mol}$

b)

$4.0\times {10}^4\text{ J/mol}$

c)

$-3.2\times {10}^4\text{ J/mol}$

d)

$6.4\times {10}^4\text{ J/mol}$

Explanation of Solution

a) $\left[{\text{H}}^{+}\right]=1.0\times {10}^{-7}\text{ M},\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-7}\text{M}$

The given reaction for which $\Delta \text{G}$ is to be calculated is as follows:

${\text{H}}_{\text{2}}\text{O}(l)\rightleftarrows {\text{ H}}^{\text{+}}(aq)+{\text{OH}}^{-}(aq)$

Theconcentration of pure liquids and pure solids istaken as unity. Therefore, the reaction quotient for the above reaction will be represented as follows:

$Q={\text{[H}}^{+}{\text{] [OH}}^{-}]$

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

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

In this case, the given concentrations are equilibrium concentrations at ${\text{25}}^{\circ }\text{C}$.

Thevalue of $\Delta G$ for a reaction at equilibrium is taken to be zero.

Thus, the above expression is reduced to the expression, which is as follows:

$\begin{array}{c}0=\Delta G^{\circ }+\text{R }T\ln Q\\ \Delta G^{\circ }=-\text{R }T\ln Q\end{array}$

Substitute all the values in the above expression as follows:

$\begin{array}{l}\Delta G^o=\left(-\text{0}\text{.008314}\text{kJ/mol}\text{.K}\text{×}\text{298}\text{.15}\text{K}\text{×}{\text{ln[H}}^{+}{\text{] [OH}}^{-}]\right)\\ \Delta G^o=\left(-\text{0}\text{.008314}\text{kJ/mol}\text{.K}\text{×}\text{298}\text{.15}\text{K}\text{×}\ln {\text{K}}_w\right)\\ \Delta G^o=\left(-\text{0}\text{.008314}\text{kJ/mol}\text{.K}\text{×}\text{298}\text{.15}\text{K}\text{×}\ln (1.0\times {10}^{-14})\right)\\ \Delta G^o=\left(-\text{0}\text{.008314}\text{kJ/mol}\text{.K}\text{×}\text{298}\text{.15}\text{K × }(-32.2361)\right)\end{array}$

On further calculation, the standard free energy change will be calculated as follows:

$\begin{array}{l}\Delta G^o=79.9074\text{ kJ/mol}\\ \Delta G^o\cong 80\text{ kJ/mol}\end{array}$

Thus, $\Delta G^o$ for the given equilibrium reaction at ${\text{25}}^{\circ }\text{C}$ is $80\text{ kJ/mol}$.

b) $\left[{\text{H}}^{+}\right]=1.0\times {10}^{-3}\text{ M},\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-4}\text{ M}$

The given reaction for which $\Delta \text{G}$ is to be calculated is as follows:

${\text{H}}_{\text{2}}\text{O}(l)\rightleftarrows {\text{ H}}^{\text{+}}(aq)+{\text{OH}}^{-}(aq)$

The concentration of pure liquids and pure solids istaken as unity. Therefore, the reaction quotient for the above reaction will be represented as follows:

$Q={\text{[H}}^{+}{\text{] [OH}}^{-}]$

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

$\begin{array}{l}\Delta G=\Delta G^{\circ }+\text{R }T\ln Q\\ \Delta G=\Delta G^{\circ }+\text{R }T\ln \left({\text{[H}}^{+}{\text{] [OH}}^{-}]\right)\\ \Delta G=(8.0\times {10}^4\text{J/mol})+(8.314\text{ J/K×mol})\left(\text{298 K}\right)\text{ln}[(1.0\times {10}^{-3})(1.0\times {10}^{-4})] \\ \Delta G=4.0\times {10}^4\text{J/mol}\end{array}$

Thus, $\Delta G$ for the given equilibrium reaction at ${\text{25}}^{\circ }\text{C}$ is $4.0\times {10}^4\text{ J/mol}$.

c) $\left[{\text{H}}^{+}\right]=1.0\times {10}^{-12}\text{ M},\left[{\text{OH}}^{-}\right]=2.0\times {10}^{-8}\text{ M}$

The given reaction for which $\Delta G$ is to be calculated is as follows:

${\text{H}}_{\text{2}}\text{O}(l)\rightleftarrows {\text{ H}}^{\text{+}}(aq)+{\text{OH}}^{-}(aq)$

The concentration of pure liquids and pure solids istaken as unity. Therefore, the reaction quotient for the above reaction will be represented as follows:

$Q={\text{[H}}^{+}{\text{] [OH}}^{-}]$

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

$\begin{array}{l}\Delta G=\Delta G^{\circ }+\text{R }T\ln Q\\ \Delta G=\Delta G^{\circ }+\text{R }T\ln \left({\text{[H}}^{+}{\text{] [OH}}^{-}]\right)\\ \Delta G=(8.0\times {10}^4\text{J/mol})+(8.314\text{ J/K×mol})\left(\text{298 K}\right)\text{ln}[(1.0\times {10}^{-12})(2.0\times {10}^{-8})] \\ \Delta G=-3.2\times {10}^4\text{J/mol}\end{array}$

Thus, $\Delta G$ for the given equilibrium reaction at ${\text{25}}^{\circ }\text{C}$ is $-3.2\times {10}^4\text{J/mol}$.

d) $\left[{\text{H}}^{+}\right]=3.5\text{ M},\left[{\text{OH}}^{-}\right]=4.8\times {10}^{-4}\text{ M}$

The given reaction for which $\Delta G$ is to be calculated is as follows:

${\text{H}}_{\text{2}}\text{O}(l)\rightleftarrows {\text{H}}^{\text{+}}(aq)+{\text{OH}}^{-}(aq)$

The concentration of pure liquids and pure solids istaken as unity. Therefore, the reaction quotient for the above reaction will be represented as follows:

$Q={\text{[H}}^{+}{\text{] [OH}}^{-}]$

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

$\begin{array}{l}\Delta G=\Delta G^{\circ }+\text{R }T\ln Q\\ \Delta G=\Delta G^{\circ }+\text{R }T\ln \left({\text{[H}}^{+}{\text{] [OH}}^{-}]\right)\\ \Delta G=(8.0\times {10}^4\text{J/mol})+(8.314\text{ J/K×mol})\left(\text{298 K}\right)\text{ln}[\left(3.5\right)(4.8\times {10}^{-4})] \\ \Delta G=6.4\times {10}^4\text{J/mol}\end{array}$

Thus, $\Delta G$ for the given equilibrium reaction at ${\text{25}}^{\circ }\text{C}$ is $6.4\times {10}^4\text{J/mol}$.