Chapter 10, Problem 71E

(a)

Interpretation Introduction

Interpretation:

The values of $\Delta H^0,\Delta S^0\text{ and }K\text{ at 298 K}$ for the synthesis of ammonia by Haber process should be calculated. The value of $\Delta G$ for this reaction under $T=298{\text{ K, P}}_{N_2}={\text{P}}_{H_2}=200{\text{ atm, P}}_{N{H}_3}=50\text{ atm}$ should be calculated.

Concept Introduction :

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

  $\Delta G^0=-RT\ln K$

  $\Delta G=\Delta G^0+RT\ln K$

  $\begin{array}{l}\Delta G\text{ - Gibbs free energy change}\\ \Delta G^0\text{ - Gibbs free energy change at standard conditions}\\ \text{R - universal gas constant}\\ \text{T - temperature}\\ \text{K - equilibrium constant}\end{array}$

Answer to Problem 71E

  $\Delta {\text{H}}^0=-92\text{ kJ/mol}$

  $\Delta S^0=-199\text{ J/K}$

  $\text{K = 9}\text{.1}\times {\text{10}}^5$

  $\Delta G\text{ = }-67\text{ kJ/mol}$

Explanation of Solution

  ${\text{N}}_2{\text{(g) + 3 H}}_2\text{(g) }\rightleftharpoons {\text{ 2 NH}}_3\text{(g)}$

  $\begin{array}{l}\Delta {\text{H}}^0=\Delta {\text{H}}_f^0{\text{(NH}}_3\text{) = 2}\times \left(-46\text{ kJ/mol}\right)\\ \Delta {\text{H}}^0=-92\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta S^0=\Delta S_{}^0{\text{(NH}}_3\text{)}-\left\{\Delta S^0\left({\text{N}}_2\right)+3\Delta S^0\left({\text{H}}_2\right)\right\}\\ \Delta S^0\text{= 2}\times 193\text{ J/K }-\left(192\text{ J/K + }\left(3\times 131\text{ J/K}\right)\right)\\ \Delta S^0=-199\text{ J/K}\end{array}$

  $\begin{array}{l}\Delta G^0=\Delta G_f^0{\text{(NH}}_3\text{) = 2}\times \left(-17\text{ kJ/mol}\right)\\ \Delta G^0=-34\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K\\ -34000\text{ J = }-\text{8}\text{.314 J/mol}\text{.K }\times \text{ 298 K }\times \text{ ln K}\\ \text{ln K = 13}\text{.72}\\ \text{K = 9}\text{.1}\times {\text{10}}^5\end{array}$

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{P_{N{H}_3}^2}{P_{N_2}\times P_{H_2}^3}\\ \text{ =}-34000\text{ J/mol + 8}\text{.314 J/mol}\text{.K }\times \text{298 K ln}\frac{{\left(50\right)}^2}{200{\left(200\right)}^3}\\ \text{ = }-34000-33000\\ \text{ = }-67000\text{ J/mol}\\ \Delta G\text{ = }-67\text{ kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta G$ for this reaction under $T=298{\text{ K, P}}_{N_2}=200{\text{ atm, P}}_{H_2}=600{\text{ atm, P}}_{N{H}_3}=200\text{ atm}$ should be calculated.

Concept Introduction :

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

  $\Delta G=\Delta G^0+RT\ln K$

  $\begin{array}{l}\Delta G\text{ - Gibbs free energy change}\\ \Delta G^0\text{ - Gibbs free energy change at standard conditions}\\ \text{R - universal gas constant}\\ \text{T - temperature}\\ \text{K - equilibrium constant}\end{array}$

Answer to Problem 71E

  $\Delta G\text{ = }-68.4\text{ kJ/mol}$

Explanation of Solution

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{P_{N{H}_3}^2}{P_{N_2}\times P_{H_2}^3}\\ \text{ =}-34000\text{ J/mol + 8}\text{.314 J/mol}\text{.K }\times \text{298 K ln}\frac{{\left(200\right)}^2}{200{\left(600\right)}^3}\\ \text{ = }-34000-34400\\ \text{ = }-68400\text{ J/mol}\\ \Delta G\text{ = }-68.4\text{ kJ/mol}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $\Delta G$ for this reaction under $T=100{\text{ K, P}}_{N_2}=50{\text{ atm, P}}_{H_2}=200{\text{ atm, P}}_{N{H}_3}=10\text{ atm}$ should be calculated.

Concept Introduction :

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

  $\Delta G=\Delta G^0+RT\ln K$

  $\begin{array}{l}\Delta G\text{ - Gibbs free energy change}\\ \Delta G^0\text{ - Gibbs free energy change at standard conditions}\\ \text{R - universal gas constant}\\ \text{T - temperature}\\ \text{K - equilibrium constant}\end{array}$

Answer to Problem 71E

  $\Delta G\text{ = }-85\text{ kJ/mol}$

Explanation of Solution

Assuming $\Delta H^0\text{ and }\Delta S^0$ are temperature independent.

  $\begin{array}{l}\Delta G_{100}^0=\Delta H^0-T.\Delta S^0\\ \text{ = }-92\text{ kJ/mol }-100\text{ K }\times \left(-0.199\text{ kJ/K}\right)\\ \text{ = }-72\text{ kJ}\end{array}$

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{P_{N{H}_3}^2}{P_{N_2}\times P_{H_2}^3}\\ \text{ =}-72000\text{ J/mol + 8}\text{.314 J/mol}\text{.K }\times 100\text{ K ln}\frac{{\left(10\right)}^2}{50{\left(200\right)}^3}\\ \text{ = }-72000-13000\\ \text{ = }-85000\text{ J/mol}\\ \Delta G\text{ = }-85\text{ kJ/mol}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The value of $\Delta G$ for this reaction under $T=700{\text{ K, P}}_{N_2}=50{\text{ atm, P}}_{H_2}=200{\text{ atm, P}}_{N{H}_3}=10\text{ atm}$ should be calculated.

Concept Introduction :

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

  $\Delta G=\Delta G^0+RT\ln K$

  $\begin{array}{l}\Delta G\text{ - Gibbs free energy change}\\ \Delta G^0\text{ - Gibbs free energy change at standard conditions}\\ \text{R - universal gas constant}\\ \text{T - temperature}\\ \text{K - equilibrium constant}\end{array}$

Answer to Problem 71E

  $\Delta G\text{ = }-41\text{ kJ/mol}$

Explanation of Solution

Assuming $\Delta H^0\text{ and }\Delta S^0$ are temperature independent.

  $\begin{array}{l}\Delta G_{700}^0=\Delta H^0-T.\Delta S^0\\ \text{ = }-92\text{ kJ/mol }-700\text{ K }\times \left(-0.199\text{ kJ/K}\right)\\ \text{ = }47\text{ kJ}\end{array}$

  $\begin{array}{l}\Delta G=\Delta G^0+RT\ln \frac{P_{N{H}_3}^2}{P_{N_2}\times P_{H_2}^3}\\ \text{ =}47000\text{ J/mol + 8}\text{.314 J/mol}\text{.K }\times 700\text{ K ln}\frac{{\left(10\right)}^2}{50{\left(200\right)}^3}\\ \text{ = }47000-88000\\ \text{ = }-41000\text{ J/mol}\\ \Delta G\text{ = }-41\text{ kJ/mol}\end{array}$