Chapter 17, Problem 119MP

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is

$\text{Ni}\left(s\right)+\text{4CO}\left(g\right)\rightleftharpoons \text{Ni}{\left(\text{CO}\right)}_4\left(g\right)$

a. Without referring to Appendix 4, predict the sign of ∆S° for the above reaction. Explain.

b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of ∆S_surr, for this reaction. Explain

c. For Ni(CO)₄(g), $\Delta H_{\text{f}}^{\text{o}}=-607\text{KJ/mol}$ and S° = 417 J/K ·mol at 298 K. Using these values and data in Appendix 4, calculate ∆H° and ∆S° for the above reaction.

d. Calculate the temperature at which ∆G° = 0 (K = 1) for the above reaction, assuming that ∆H° and ∆S° do not depend on temperature.

e. The first step of the Mood process involves equilibrating impure nickel with CO(g) and Ni(CO)₄(g) at about 50°C. The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at 50.°C.

f. In the second step of the Mood process, the gaseous Ni(CO)₄ is isolated and heated to 227°C. The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at 227°C.

g. Why is temperature increased for the second step of the Mood process?

h. The Mond process relies on the volatility of Ni(CO)₄ for its success. Only pressures and temperatures at which Ni(CO)₄ is a gas are useful. A recently developed variation of the Mood process carries out the first step at higher pressures and a temperature of l52°C. Estimate the maximum pressure of Ni(CO)₄(g) that can be attained before the gas will liquefy at 152°C. The boiling point for Ni(CO)₄ is 42°C and the enthalpy of vaporization is 29.0 kJ/mol.

[Hint: The phase change reaction and the corresponding equilibrium expression are

$\text{Ni}{\left(\text{CO}\right)}_4\left(l\right)\rightleftharpoons \text{Ni}{\left(\text{CO}\right)}_4\left(g\right)K=P_{\text{Ni}{\left(\text{CO}\right)}_4}$

Ni(CO)₄(g) will liquefy when the pressure of Ni(CO)₄ is greater than the K value.]

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The sign of $\Delta S^{\text{ο}}$ for the reaction.

Answer to Problem 119MP

Answer

The sign of $\Delta S^{\text{ο}}$ for the reaction is negative.

Explanation of Solution

Explanation

The given reaction is,

$\text{Ni}\left(s\right)+4\text{CO}\left(g\right)\rightleftharpoons \text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$

There are four gaseous molecules present on the reactant side and one gaseous molecule present at the product side. Since the number of gaseous molecules decreases in the reaction, hence the standard entropy change $\Delta S^{\text{ο}}$ for the reaction becomes negative.

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The sign of $\Delta S_{\text{surr}}$ for the reaction.

Answer to Problem 119MP

Answer

The sign of $\Delta S_{\text{surr}}$ for the reaction is positive.

Explanation of Solution

Explanation

Since the entropy for the system is decreasing, therefore the sign of $\Delta S_{\text{surr}}$ should be positive so that the entropy of universe always increases.

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The value of $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ for the reaction.

Answer to Problem 119MP

Answer

The value of $\Delta H^{\text{ο}}$ is $\underset{\_}{-565kJ/mol}$ and the value of $\Delta S^{\text{ο}}$ is $\underset{\_}{-405J/mol}$.

Explanation of Solution

Explanation

The given reaction is,

$\text{Ni}\left(s\right)+4\text{CO}\left(g\right)\rightleftharpoons \text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$

The standard enthalpy change ( $\Delta H^{\text{ο}}$) is calculated by the formula,

$\Delta H^{\text{ο}}=\left(\begin{array}{l}\sum \text{Number of moles of products}\times \Delta H_f^{\text{ο}}\text{Products }-\\ \sum \text{Number of moles of reactants}\times \Delta H_f^{\text{ο}}\text{Reactants}\end{array}\right)$

$\Delta H^{\text{ο}}=\left(\begin{array}{l}\text{Number of moles of Ni}{\left(\text{Co}\right)}_4\left(g\right)\times \Delta H_f^{\text{ο}}\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)-\\ \left(\begin{array}{l}\text{Number of moles of Ni}\left(s\right)\times \Delta H_f^{\text{ο}}\text{Ni}\left(s\right)+\\ \text{Number of moles of}\text{CO}\left(g\right)\times \Delta H_f^{\text{ο}}\text{CO}\left(g\right)\end{array}\right)\end{array}\right)$

The number of moles of $\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$ is $1$.

The number of moles of $\text{Ni}\left(s\right)$ is $1$.

The number of moles of $\text{CO}\left(g\right)$ is $4$.

$\Delta H_f^{\text{ο}}\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$=Standard enthalpy of formation of $\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$= $\text{-607}\text{kJ/mol}$.

$\Delta H_f^{\text{ο}}\text{Ni}\left(s\right)$=Standard enthalpy of formation of $\text{Ni}\left(s\right)=0$

$\Delta H_f^{\text{ο}}\text{CO}\left(g\right)$= Standard enthalpy of formation of $\text{CO}\left(g\right)=\text{-10}\text{.5}\text{kJ/mol}$

Substitute the number of moles of reactants and products and their standard enthalpy of formation in the above equation.

$\begin{array}{c}\Delta H^{\text{ο}}=\left(\left(1\times -607\text{ kJ/mol}\right)-\left(1\times 0+4\times -10.5\text{ kJ/mol}\right)\right)\\ =\underset{\_}{-565kJ/mol}\end{array}$

The conversion of $\text{kJ/mol}$ to $\text{J/mol}$ is done as,

$\text{1 kJ/mol}={10}^3\text{ J/mol}$

Therefore, the conversion of $\text{-565 kJ/mol}$ to $\text{J/mol}$ is done as,

$\text{-565 kJ/mol}=-\text{565}\times {10}^3\text{ J/mol}$

Therefore, the value of standard enthalpy change ( $\Delta H^{\text{ο}}$) is $-\text{565}\times {10}^3\text{ J/mol}$.

The standard entropy change ( $\Delta S^{\text{ο}}$) is calculated by the formula,

$\Delta S^{\text{ο}}=\left(\begin{array}{l}\sum \text{Number of moles of products}\times S^{\text{ο}}\text{Products }-\\ \sum \text{Number of moles of reactants}\times S^{\text{ο}}\text{Reactants}\end{array}\right)$

$\Delta S^{\text{ο}}=\left(\begin{array}{l}\text{Number of moles of Ni}{\left(\text{Co}\right)}_4\left(g\right)\times S^{\text{ο}}\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)-\\ \left(\begin{array}{l}\text{Number of moles of Ni}\left(s\right)\times S^{\text{ο}}\text{Ni}\left(s\right)+\\ \text{Number of moles of}\text{CO}\left(g\right)\times S^{\text{ο}}\text{CO}\left(g\right)\end{array}\right)\end{array}\right)$

The number of moles of $\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$ is $1$.

The number of moles of $\text{Ni}\left(s\right)$ is $1$.

The number of moles of $\text{CO}\left(g\right)$ is $4$.

$S^{\text{ο}}\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$=Standard entropy of $\text{Ni}{\left(\text{Co}\right)}_4\left(g\right)$= $417\text{J/K}\cdot \text{mol}$.

$S^{\text{ο}}\text{Ni}\left(s\right)$=Standard entropy of $\text{Ni}\left(s\right)=30\text{J/K}\cdot \text{mol}$

$S^{\text{ο}}\text{CO}\left(g\right)$= Standard entropy of $\text{CO}\left(g\right)=198\text{J/K}\cdot \text{mol}$

Substitute the number of moles of reactants and products and their standard entropy in the above equation.

$\begin{array}{c}\Delta S^{\text{ο}}=\left(\left(1\text{mol}\times 417\text{ J/K}\cdot \text{mol}\right)-\left(1\text{mol}\times 30\text{ J/K}\cdot \text{mol}+4\text{mol}\times 198\text{J/K}\cdot \text{mol}\right)\right)\\ =\underset{\_}{-405J/K}\end{array}$

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The temperature at which $\Delta G^{\text{ο}}$ for the reaction is zero.

Answer to Problem 119MP

Answer

The temperature at which $\Delta G^{\text{ο}}=0$ is $\underset{\_}{1395K}$.

Explanation of Solution

Explanation

The standard free energy change formula is,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

It is given that $\Delta G^{\text{ο}}=0$. Therefore, the equation becomes,

$T=\frac{\Delta H^{\text{ο}}}{\Delta S^{\text{ο}}}$

Substitute the value of $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ in the above equation.

$\begin{array}{c}T=\frac{-\text{565}\times {10}^3\text{ J/mol}}{\text{-405}\text{J/K}}\\ =\underset{\_}{1395K}\end{array}$

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The equilibrium constant for the reaction at $\text{50}{}^{\text{ο}}\text{C}$.

Answer to Problem 119MP

Answer

The equilibrium constant for the reaction at $\text{50}{}^{\text{ο}}\text{C}$ is $\underset{\_}{1.95\times 10^{70}}$.

Explanation of Solution

Explanation

The standard free energy change formula is,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

The temperature ( $T$) is $\text{50}{}^{\text{ο}}\text{C}=323\text{K}$.

The value of $\Delta H^{\text{ο}}$ is $\text{-565}\text{kJ}$.

The value of $\Delta S^{\text{ο}}$ is $\text{-405}\text{J/K}$.

Substitute the value of $T$, $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=-\text{565}\text{kJ}-323\text{K}\times -\text{405}\text{J/K}\\ \text{=}-\text{565}\text{kJ}+131\times {10}^3\text{J}\\ \text{=}-\text{565}\text{kJ}+131\text{kJ}\\ \text{=}-434\text{kJ}\end{array}$

The conversion of $\text{kJ/mol}$ to $\text{J/mol}$ is done as,

$\text{1 kJ/mol}={10}^3\text{ J/mol}$

Therefore, the conversion of $-434\text{kJ}$ to $\text{J/mol}$ is done as,

$-434\text{kJ}=-434\times {10}^3\text{ J/mol}$

Therefore, the value of standard free energy change ( $\Delta G^{\text{ο}}$) is $-434\times {10}^3\text{ J/mol}$.

Since,

$\Delta G=\Delta G^{\text{ο}}+RT\ln Q$.

Where,

• $\Delta G$ is the change in gibbs free energy.
• $Q$ is the reaction quotient.
• $K$ is the equilibrium constant.
• $R$ is the gas constant ( $8.3145\text{J/K}\cdot \text{mol}$).

At equilibrium, $\Delta G=0$ and $Q=K$.

Substitute the value of $\Delta G$ and $Q$ in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=-RT\ln K\\ \ln K=-\frac{\Delta G^{\text{ο}}}{RT}\end{array}$

Substitute the value of $\Delta G^{\text{ο}}$, $R$ and $T$ in the above equation.

$\begin{array}{c}\ln K=-\frac{-434\times {10}^3\text{ J/mol}}{8.3145\text{J/K}\cdot \text{mol}\times 323\text{K}}\\ \log K=\frac{162}{2.303}\\ \log K=70.3\\ K=1.95\times {10}^{70}\end{array}$

Thus, the value of equilibrium constant for the reaction is $\underset{\_}{1.95\times 10^{70}}$.

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The equilibrium constant for the reaction at $227{}^{\text{ο}}\text{C}$.

Answer to Problem 119MP

Answer

The equilibrium constant for the reaction at $227{}^{\text{ο}}\text{C}$ is $\underset{\_}{7.94\times 10^{37}}$.

Explanation of Solution

Explanation

The standard free energy change formula is,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

The temperature ( $T$) is $227{}^{\text{ο}}\text{C}=500\text{K}$.

The value of $\Delta H^{\text{ο}}$ is $\text{-565}\text{kJ}$.

The value of $\Delta S^{\text{ο}}$ is $\text{-405}\text{J/K}$.

Substitute the value of $T$, $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=-\text{565}\text{kJ}-500\text{K}\times -\text{405}\text{J/K}\\ \text{=}-\text{565}\text{kJ}+202\times {10}^3\text{J}\\ \text{=}-\text{565}\text{kJ}+204.5\text{kJ}\\ \text{=}-363\text{kJ}\end{array}$

The conversion of $\text{kJ/mol}$ to $\text{J/mol}$ is done as,

$\text{1 kJ/mol}={10}^3\text{ J/mol}$

Therefore, the conversion of $-363\text{kJ}$ to $\text{J/mol}$ is done as,

$-363\text{kJ}=-363\times {10}^3\text{ J/mol}$

Therefore, the value of standard free energy change ( $\Delta G^{\text{ο}}$) is $-363\times {10}^3\text{ J/mol}$.

Since,

$\Delta G=\Delta G^{\text{ο}}+RT\ln Q$.

Where,

• $\Delta G$ is the change in gibbs free energy.
• $Q$ is the reaction quotient.
• $K$ is the equilibrium constant.
• $R$ is the gas constant ( $8.3145\text{J/K}\cdot \text{mol}$).

At equilibrium, $\Delta G=0$ and $Q=K$.

Substitute the value of $\Delta G$ and $Q$ in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=-RT\ln K\\ \ln K=-\frac{\Delta G^{\text{ο}}}{RT}\end{array}$

Substitute the value of $\Delta G^{\text{ο}}$, $R$ and $T$ in the above equation.

$\begin{array}{c}\ln K=-\frac{-363\times {10}^3\text{ J/mol}}{8.3145\text{J/K}\cdot \text{mol}\times 500\text{K}}\\ \log K=\frac{87.2}{2.303}\\ \log K=37.9\\ K=7.94\times {10}^{37}\end{array}$

Thus, the value of equilibrium constant for the reaction is $\underset{\_}{7.94\times 10^{37}}$.

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The reason for increase in temperature in second step.

Answer to Problem 119MP

Answer

The purpose of increase in temperature is to enhance the yield of pure nickel.

Explanation of Solution

Explanation

In the second step of Mond process, the temperature is increased. By doing this, the equilibrium will shift in reverse direction. Hence, deposition of as much as nickel as pure solid will take place.

Interpretation Introduction

Interpretation: The given questions based upon Mond process are to be answered.

Concept introduction: The Mond process relies on the volatility of $\text{Ni}{\left(\text{CO}\right)}_4$ for its success. The temperatures and pressures at which $\text{Ni}{\left(\text{CO}\right)}_4$ is gas are useful.

To determine: The maximum pressure of ${\text{NiCO}}_{\text{4}}\left(g\right)$ that can be attained before the gas liquefy at $152{}^{\text{ο}}\text{C}$.

Answer to Problem 119MP

Answer

The maximum pressure of ${\text{NiCO}}_{\text{4}}\left(g\right)$ that can be attained before the gas will liquefy at $152{}^{\text{ο}}\text{C}$ is $3.38\times {10}^{40}$.

Explanation of Solution

Explanation

The standard free energy change formula is,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

The temperature ( $T$) is $152{}^{\text{ο}}\text{C}=425\text{K}$.

The value of $\Delta H^{\text{ο}}$ is $\text{-565}\text{kJ}$.

The value of $\Delta S^{\text{ο}}$ is $\text{-405}\text{J/K}$.

Substitute the value of $T$, $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=-\text{565}\text{kJ}-425\text{K}\times -\text{405}\text{J/K}\\ \text{=}-\text{565}\text{kJ}+172.1\times {10}^3\text{J}\\ \text{=}-\text{565}\text{kJ}+172.1\text{kJ}\\ \text{=}-392.9\text{kJ}\end{array}$

The conversion of $\text{kJ/mol}$ to $\text{J/mol}$ is done as,

$\text{1 kJ/mol}={10}^3\text{ J/mol}$

Therefore, the conversion of $-392.9\text{kJ}$ to $\text{J/mol}$ is done as,

$-392.9\text{kJ}=-392.9\times {10}^3\text{ J/mol}$

Therefore, the value of standard free energy change ( $\Delta G^{\text{ο}}$) is $-392.9\times {10}^3\text{ J/mol}$.

Since,

$\Delta G=\Delta G^{\text{ο}}+RT\ln Q$.

Where,

• $\Delta G$ is the change in Gibbs free energy.
• $Q$ is the reaction quotient.
• $K$ is the equilibrium constant.
• $R$ is the gas constant ( $8.3145\text{J/K}\cdot \text{mol}$).

At equilibrium, $\Delta G=0$ and $Q=K$.

Substitute the value of $\Delta G$ and $Q$ in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=-RT\ln K\\ \ln K=-\frac{\Delta G^{\text{ο}}}{RT}\end{array}$

Substitute the value of $\Delta G^{\text{ο}}$, $R$ and $T$ in the above equation.

$\begin{array}{c}\ln K=-\frac{-392.9\times {10}^3\text{ J/mol}}{8.3145\text{J/K}\cdot \text{mol}\times 425\text{K}}\\ \log K=\frac{93.35}{2.303}\\ \log K=40.53\\ K=3.38\times {10}^{40}\end{array}$

Thus, the value of equilibrium constant for the reaction is $3.38\times {10}^{40}$.

The phase change reaction and the corresponding equilibrium expression is,

${\text{NiCO}}_{\text{4}}\left(l\right)\underset{}{\rightleftharpoons }{\text{NiCO}}_{\text{4}}\left(g\right)$

The equilibrium constant ( $K$) expression for the above reaction is,

$K=P_{\text{Ni}{\left(\text{CO}\right)}_{\text{4}}}$

Where,

• $P_{\text{Ni}{\left(\text{CO}\right)}_{\text{4}}}$ is the pressure of ${\text{NiCO}}_{\text{4}}\left(g\right)$.

Substitute the value of $K$ in the above equation.

$P_{\text{Ni}{\left(\text{CO}\right)}_{\text{4}}}=3.38\times {10}^{40}$

The ${\text{NiCO}}_{\text{4}}\left(g\right)$ will liquefy when pressure of ${\text{NiCO}}_{\text{4}}\left(g\right)$ is greater than $K$ value. Therefore, the maximum pressure of ${\text{NiCO}}_{\text{4}}\left(g\right)$ that can be attained before the gas will liquefy at $152{}^{\text{ο}}\text{C}$ is $3.38\times {10}^{40}$