Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ${\text{Δ}}_{\text{r}}{G}^{\text{°}}$ for the reaction of silver with $\text{HX}$ where $\text{X}$ is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\Delta }_{\text{r}}{G}^{\circ }$. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

${\text{Δ}}_{\text{r}}{G}^{\text{°}}=\sum n{\Delta }_f{G}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{G}^{\text{°}}\left(\text{reactants}\right)$

It is related to entropy and entropy by the following expression,

${\Delta }_{\text{r}}{G}^{\circ }={\Delta }_{\text{r}}{H}^{\circ }-T{\Delta }_{\text{r}}{S}^{\circ }$

Here, ${\Delta }_{\text{r}}{H}^{\circ }$ is the change in enthalpy and ${\Delta }_{\text{r}}{S}^{\circ }$ is the change in entropy.

Answer to Problem 92GQ

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen fluoride is $\text{+}79.4kJ\text{/}mol\text{-}rxn$.

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen chloride is $-14.67kJ\text{/}mol\text{-}rxn$.

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen bromide is $-43.45kJ\text{/}mol\text{-}rxn$.

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen iodide is $-64.63kJ\text{/}mol\text{-}rxn$.

Explanation of Solution

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For $\text{X}=\text{F}$,

The given reaction is,

  $\text{Ag}\left(\text{s}\right)+\text{HF}\left(\text{g}\right)\to \text{AgF}\left(\text{s}\right)+\frac{1}{2}{\text{H}}_{\text{2}}\left(\text{g}\right)$

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{AgF}\left(\text{s}\right)$ is $-193.8\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{HF}\left(\text{g}\right)$ is $-273.2\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for ${\text{H}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{Ag}\left(\text{s}\right)$ is $\text{0 kJ/mol}$.

The expression for free energy change is,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\sum n{\Delta }_f{G}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{G}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgF}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{AgF}\left(\text{s}\right)\right]+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{Ag}\left(\text{s}\right)\right]\text{+}\\ \left(\text{1 mol HF}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{HF}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgF}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-193.8\text{ kJ/mol}\right)+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1 mol HF}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-273.2\text{ kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =\text{+79}\text{.4 kJ/mol-rxn}\end{array}$

For $\text{X}=\text{Cl}$,

The given reaction is,

  $\text{Ag}\left(\text{s}\right)+\text{HCl}\left(\text{g}\right)\to \text{AgCl}\left(\text{s}\right)+\frac{1}{2}{\text{H}}_{\text{2}}\left(\text{g}\right)$

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{AgCl}\left(\text{s}\right)$ is $-109.76\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{HCl}\left(\text{g}\right)$ is $-95.09\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for ${\text{H}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{Ag}\left(\text{s}\right)$ is $\text{0 kJ/mol}$.

The expression for free energy change is,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\sum n{\Delta }_f{G}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{G}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgCl}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{AgCl}\left(\text{s}\right)\right]+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{Ag}\left(\text{s}\right)\right]\text{+}\\ \left(\text{1 mol HCl}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{HCl}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgCl}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-109.76\text{ kJ/mol}\right)+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1 mol HCl}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-95.09\text{ kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =-14.67\text{ kJ/mol-rxn}\end{array}$

For $\text{X}=\text{Br}$,

The given reaction is,

  $\text{Ag}\left(\text{s}\right)+\text{HBr}\left(\text{g}\right)\to \text{AgBr}\left(\text{s}\right)+\frac{1}{2}{\text{H}}_{\text{2}}\left(\text{g}\right)$

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{AgBr}\left(\text{s}\right)$ is $-96.90\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{HBr}\left(\text{g}\right)$ is $-53.45\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for ${\text{H}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{Ag}\left(\text{s}\right)$ is $\text{0 kJ/mol}$.

The expression for free energy change is,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\sum n{\Delta }_f{G}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{G}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgBr}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{AgBr}\left(\text{s}\right)\right]+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{Ag}\left(\text{s}\right)\right]\text{+}\\ \left(\text{1 mol HBr}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{HBr}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgBr}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-96.90\text{ kJ/mol}\right)+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1 mol HBr}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-53.45\text{ kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =-43.45\text{ kJ/mol-rxn}\end{array}$

For $\text{X}=\text{I}$,

The given reaction is,

  $\text{Ag}\left(\text{s}\right)+\text{HI}\left(\text{g}\right)\to \text{AgI}\left(\text{s}\right)+\frac{1}{2}{\text{H}}_{\text{2}}\left(\text{g}\right)$

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{AgI}\left(\text{s}\right)$ is $-66.19\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{HI}\left(\text{g}\right)$ is $-1.56\text{ kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for ${\text{H}}_2\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The ${\text{Δ}}_f{G}^{\text{°}}$ for $\text{Ag}\left(\text{s}\right)$ is $\text{0 kJ/mol}$.

The expression for free energy change is,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\sum n{\Delta }_f{G}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{G}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgI}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{AgI}\left(\text{s}\right)\right]+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[{\text{H}}_{\text{2}}\left(\text{g}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{Ag}\left(\text{s}\right)\right]\text{+}\\ \left(\text{1 mol HI}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{G}^{°}\left[\text{HI}\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the values,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{G}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{1 mol AgI}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-66.19\text{ kJ/mol}\right)+\\ \left(\text{0}{\text{.5 mol H}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left(\text{1 mol Ag}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{0 kJ/mol}\right)\text{+}\\ \left(\text{1 mol HI}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-1.56\text{ kJ/mol}\right)\end{array}\right]\end{array}\right]\\ =-64.63\text{ kJ/mol-rxn}\end{array}$

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen fluoride is $\text{+}79.4kJ\text{/}mol\text{-}rxn$.

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen chloride is $-14.67kJ\text{/}mol\text{-}rxn$.

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen bromide is $-43.45kJ\text{/}mol\text{-}rxn$.

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the reaction of silver with hydrogen iodide is $-64.63kJ\text{/}mol\text{-}rxn$.