Chapter 10, Problem 78E

(a)

Interpretation Introduction

Interpretation:

The value of $\Delta H^0,\Delta S^0,\Delta G^0\text{ and }K\text{ (at 298 K)}$ should be calculated for each step in the Ostwald process.

Concept Introduction :

The relation between standard Gibbs free energy change, standard enthalpy change and standard entropy change is as follows:

  $\Delta G^0=\Delta H^0-T.\Delta S^0$

Here,

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy}\\ \Delta H^0\text{ - enthalpy change}\\ \Delta S^0\text{ - entropy change}\\ T\text{ - temperature}\end{array}$

The Gibbs free energy change and equilibrium constant is related to each other as follows:

  $\Delta G^0=-RT\ln K$

Here,

  $\begin{array}{l}R\text{ - universal gas constant}\\ T\text{ - temperature}\\ K\text{ - equilibrium constant}\end{array}$

Answer to Problem 78E

	ΔH0 (kJ/mol)	ΔS0 (J/K.mol)	ΔG0 (kJ/mol)	K
Step 1	-908	181	-962	4.33 x 10168
Step 2	-112	-147	-68	8.3 x 1011
Step 3	-74	-267	6	9.0 x 10-2

Explanation of Solution

  $4{\text{ NH}}_3{\text{(g) + 5 O}}_2\text{(g) }\underset{{82.5}^{0}C}{\overset{Pt}{\to }}{\text{ 4 NO(g) + 6 H}}_2\text{O(g)}$

  $\begin{array}{l}\Delta H^0=\left[\left(4\times 90\text{ kJ/mol}\right)+\left(6\times \left(-242\right)\text{ kJ/mol}\right)\right]-\left(4\times \left(-46\right)\text{ kJ/mol}\right)\\ \text{ = }-908\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta S^0=\left[\left(4\times 211\text{ J/K}\text{.mol}\right)+\left(6\times 189\text{ J/K}\text{.mol}\right)\right]-\left[\left(4\times 193\text{ J/K}\text{.mol}\right)+\left(5\times 205\text{ J/K}\text{.mol}\right)\right]\\ \text{ = }-181\text{ J/K}\text{.mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-908\text{ kJ/mol}-\left(298\text{ K }\times 0.181\text{ kJ/K}\text{.mol}\right)\\ \text{ = }-962\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ -962\text{000 J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{ 298 K }\times \text{ ln K}\\ \text{ln K = }\frac{962\text{000 J/mol}}{8.314\text{ J/mol}\text{.K }\times \text{ 298 K}}=388.3\\ \text{K = 4}\text{.33 }\times {\text{ 10}}^{168}\end{array}$

  ${\text{2 NO(g) + O}}_2\text{(g) }\to {\text{ 2 NO}}_2\text{(g)}$

  $\begin{array}{l}\Delta H^0=\left(2\times 34\text{ kJ/mol}\right)-\left(2\times 90\text{ kJ/mol}\right)\\ \text{ = }-112\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta S^0=\left(2\times 240\text{ J/K}\text{.mol}\right)-\left[\left(2\times 211\text{ J/K}\text{.mol}\right)+\left(205\text{ J/K}\text{.mol}\right)\right]\\ \text{ = }-147\text{ J/K}\text{.mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-112\text{ kJ/mol}-\left(298\text{ K }\times 0.147\text{ kJ/K}\text{.mol}\right)\\ \text{ = }-68\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ -68000\text{ J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{ 298 K }\times \text{ ln K}\\ \text{ln K = }\frac{\text{68000 J/mol}}{8.314\text{ J/mol}\text{.K }\times \text{ 298 K}}=27.45\\ \text{K = 8}\text{.3 }\times {\text{ 10}}^{11}\end{array}$

  $3{\text{ NO}}_2{\text{(g) + H}}_2\text{O(l) }\to {\text{ 2 HNO}}_3\text{(l) + NO(g)}$

  $\begin{array}{l}\Delta H^0=\left[\left(2\times \left(-174\right)\text{ kJ/mol}\right)+90\text{ kJ/mol}\right]-\left[\left(3\times 34\text{ kJ/mol}\right)+\left(-286\text{ kJ/mol}\right)\right]\\ \text{ = }-74\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta S^0=\left[\left(2\times 156\text{ J/K}\text{.mol}\right)+211\text{ J/K}\text{.mol}\right]-\left[\left(3\times 240\text{ J/K}\text{.mol}\right)+\left(\text{70 J/K}\text{.mol}\right)\right]\\ \text{ = }-267\text{ J/K}\text{.mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-74\text{ kJ/mol}-\left(298\text{ K }\times 0.267\text{ kJ/K}\text{.mol}\right)\\ \text{ = }-6\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ -6000\text{ J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{ 298 K }\times \text{ ln K}\\ \text{ln K = }\frac{\text{6000 J/mol}}{8.314\text{ J/mol}\text{.K }\times \text{ 298 K}}=2.4\\ \text{K = 9}\text{.0 }\times {\text{ 10}}^{-2}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant for the first step at 825 ⁰C should be calculated, assuming standard enthalpy change and entropy change are temperature independent.

Concept Introduction :

The relation between standard Gibbs free energy change, standard enthalpy change and standard entropy change is as follows:

  $\Delta G^0=\Delta H^0-T.\Delta S^0$

Here,

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy}\\ \Delta H^0\text{ - enthalpy change}\\ \Delta S^0\text{ - entropy change}\\ T\text{ - temperature}\end{array}$

The Gibbs free energy change and equilibrium constant is related to each other as follows:

  $\Delta G^0=-RT\ln K$

Here,

  $\begin{array}{l}R\text{ - universal gas constant}\\ T\text{ - temperature}\\ K\text{ - equilibrium constant}\end{array}$

Answer to Problem 78E

  $\text{K = 4}\text{.6 }\times {\text{10}}^{52}$

Explanation of Solution

  $\begin{array}{l}\Delta G^0=\Delta H^0-T.\Delta S^0\\ \text{ = }-908\text{ kJ/mol - }\left(\text{1098 K }\times \text{ 0}\text{.0181 kJ/K}\text{.mol}\right)\\ \text{ = }-1107\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ -1107000\text{ J/mol = }-8.314\text{ J/mol}\text{.K }\times \text{ 1098 K }\times \text{ ln K}\\ \text{ln K = }\frac{-1107000\text{ J/mol}}{-8.314\text{ J/mol}\text{.K }\times \text{ 1098 K}}=121.26\\ \text{K = 4}\text{.6 }\times {\text{10}}^{52}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Whether there is a thermodynamic reason for the high temperature in the first step assuming standard conditions should be explained.

Concept Introduction :

Ostwald process is a chemical process that converts ammonia into nitric acid.

Answer to Problem 78E

No thermodynamic reason. The purpose of using higher temperature is to increase the rate of the reaction.

Explanation of Solution

There is no thermodynamic reason for using higher temperature in step 1 of the Ostwald process. Since enthalpy change is negative and entropy change is positive, temperature should be lowered to shift the equilibrium to right side. But here, the purpose of using higher temperature is to increase the rate of the reaction.