Question

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 $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ a. Without referring to Appendix 4 , predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). mol at \(298 \mathrm{~K}\). Using these values and data in Appendix 4 , calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). 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 \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). 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 \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\). [Hint: The phase change reaction and the corresponding equilibrium expression are $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

Short answer

The signs of \(\Delta S^{\circ}\) and \(\Delta S_{\text{surr}}\) for the reaction are negative and positive, respectively. The calculated values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are -165 kJ/mol and -404.1 J/Kmol, respectively. The temperature at which \(\Delta G^{\circ} = 0\) is approximately 408 K. The equilibrium constants at \(50^{\circ}C\) and \(227^{\circ}C\) are approximately 3.61 and 0.27, respectively. The increase in temperature for the second step of the Mond process drives the reverse reaction to deposit pure solid nickel. Lastly, the maximum pressure of Ni(CO)\(_4\) before it liquefies at \(152^{\circ}C\) is approximately 88,000 atm.

Step-by-step solution

a. Predicting the sign of \(\Delta S^{\circ}\) for the reaction.

The reaction involves one mole of solid nickel combining with four moles of carbon monoxide gas to form one mole of nickel tetracarbonyl gas. As the number of gas molecules is decreasing from 4 to 1, the total entropy of the system will decrease. Therefore, the sign of \(\Delta S^{\circ}\) for this reaction is negative.

b. Predicting the sign of \(\Delta S_{\text {surr }}\) for the reaction.

As the reaction is exothermic (releases heat to the surroundings), the surrounding temperature will increase. Therefore, the entropy of the surroundings will increase. The sign of \(\Delta S_{\text {surr }}\) for this reaction is positive.

c. Calculating \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction.

We are given that for Ni(CO)\(_4\), \(\Delta H_{\mathrm{f}}^{\circ}=-607~\text{kJ/mol}\) and \(S^{\circ}=417~\text{J/Kmol}\). From Appendix 4, we have the following data for Ni(s) and CO(g): Ni(s): \(\Delta H_{\mathrm{f}}^{\circ}=0~\text{kJ/mol}\) and \(S^{\circ}=29.9~\text{J/Kmol}\) CO(g): \(\Delta H_{\mathrm{f}}^{\circ}=-110.5~\text{kJ/mol}\)l and \(S^{\circ}=197.9~\text{J/Kmol}\) Using Hess's law, we can calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction: $$ \Delta H^{\circ}_{\text{reaction}} = \Delta H^{\circ}_{\text{products}} - \Delta H^{\circ}_{\text{reactants}} $$ $$ \Delta H^{\circ}_{\text{reaction}} = (-607) - [0 + 4(-110.5)] = -607 + 442 = -165~\text{kJ/mol} $$ Similarly, we can calculate \(\Delta S^{\circ}_{\text{reaction}}\): $$ \Delta S^{\circ}_{\text{reaction}} = S^{\circ}_{\text{products}} - S^{\circ}_{\text{reactants}} $$ $$ \Delta S^{\circ}_{\text{reaction}} = (417) - [29.9 + 4(197.9)] = 417 - 821.1 = -404.1~\text{J/Kmol} $$ Therefore, \(\Delta H^{\circ} = -165~\text{kJ/mol}\) and \(\Delta S^{\circ} = -404.1~\text{J/Kmol}\) for the reaction.

d. Calculating the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the reaction.

With \(\Delta G^{\circ}=0\), we can use the following equation: $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = 0 $$ We are given that \(\Delta H^{\circ} = -165~\text{kJ/mol}\) and \(\Delta S^{\circ} = -404.1~\text{J/Kmol}\). Solving for T: $$ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{-165~\text{kJ/mol}}{-404.1~\text{J/Kmol}} = \frac{-165 \cdot 1000}{-404.1} \approx 408~K $$ Therefore, the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the reaction is about 408K.

e. Calculating the equilibrium constant at \(50^{\circ}C\).

Using the data from part (c) and the Van 't Hoff equation, we can calculate the equilibrium constant at \(50^{\circ}C\): $$ \Delta G^{\circ} = -RT\,\ln K $$ where R = 8.314 J/Kmol, T = 273 + 50 = 323K As \(\Delta G^{\circ}=0\) when T = 408K, we have \(K=1\) at T = 408K. Now we can use the Van 't Hoff equation: $$ \ln \frac{K_2}{K_1} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$ Plugging in the values, we get: $$ \ln \frac{K_2}{1} = -\frac{-165 \cdot 1000}{8.314} \left(\frac{1}{323} - \frac{1}{408}\right) $$ $$ \ln K_2 = 19.86(0.0030953 - 0.00245098) \approx 1.2831 $$ $$ K_2 \approx e^{1.2831} \approx 3.61 $$ The equilibrium constant at \(50^{\circ}C\) is approximately 3.61.

f. Calculating the equilibrium constant at \(227^{\circ}C\).

Using the same method as in part (e), we can calculate the equilibrium constant at \(227^{\circ}C\): T = 273 + 227 = 500K $$ \ln \frac{K_3}{1} = -\frac{-165 \cdot 1000}{8.314} \left(\frac{1}{500} - \frac{1}{408}\right) $$ $$ \ln K_3 = 19.86(0.002 - 0.00245098) \approx -1.3084 $$ $$ K_3 \approx e^{-1.3084} \approx 0.27 $$ The equilibrium constant at \(227^{\circ}C\) is approximately 0.27.

g. Explaining the increase in temperature for the second step of the Mond process.

By increasing the temperature for the second step of the Mond process, the reaction is driven in the reverse direction which favors the deposition of the nickel in its solid form as a pure metal. In other words, a higher temperature forces the reverse reaction to occur, thus converting nickel tetracarbonyl back to solid nickel and carbon monoxide gas.

h. Estimating the maximum pressure of Ni(CO)\(_4\) before it liquefies at \(152^{\circ}C\).

We are given the boiling point of Ni(CO)\(_4\) as \(42^{\circ}C\) with an enthalpy of vaporization of \(29.0~\text{kJ/mol}\). We can use the Clausius-Clapeyron equation to estimate the maximum pressure at which Ni(CO)\(_4\) remains in the gas phase at \(152^{\circ}C\): $$ \ln \frac{P_2}{P_1} = \frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$ We are looking for \(P_2\), the pressure at \(T_2 = 152^{\circ}C = 425K\). The boiling point corresponds to the pressure at \(T_1 = 42^{\circ}C = 315K\), with \(P_1 = 1\) atm (assuming standard conditions): $$ \ln \frac{P_2}{1} = \frac{-29.0 \cdot 1000}{8.314} \left(\frac{1}{315} - \frac{1}{425}\right) $$ $$ \ln P_2 = -328.54(-0.00079365) \approx 11.39 $$ $$ P_2 \approx e^{11.39} \approx 88,000~\text{atm} $$ The maximum pressure of Ni(CO)\(_4\) before it liquefies at \(152^{\circ}C\) is approximately 88,000 atm.

Key concepts

Nickel refining

The Mond process is a fascinating method to refine nickel, achieving remarkably high purity levels from 99.90% to 99.99%. This process takes advantage of nickel's reaction with carbon monoxide to form nickel tetracarbonyl, a volatile compound, in a gas phase.

The resulting gas is then decomposed back into pure nickel upon heating.

• It begins with impure nickel reacting in the presence of carbon monoxide gas.
• The gas phase facilitates separation since impurities do not form similar volatile compounds.
• Finally, heating the gas decomposes it back into pure nickel and carbon monoxide.

This method cleverly exploits the physical properties of nickel tetracarbonyl, enabling a cyclic purification process. Understanding the conditions and reactions is key to maximizing effectiveness.

Entropy change

Entropy, often associated with disorder, plays a critical role in determining reaction spontaneity. In the Mond process, we see a change in entropy, denoted as \(\Delta S^{\circ}\), due to the reformation of volatile nickel tetracarbonyl from nickel and carbon monoxide.

The process turns a solid and multiple gas molecules into a single gaseous product.

This results in decreased disorder, indicating a negative value for \(\Delta S^{\circ}\) for the system.

However, as the Mond reaction is exothermic and releases heat into the surroundings, the surrounding's entropy (\(\Delta S_{\text{surr}}\)) increases, suggesting a positive shift.

• Exothermic nature releases energy, increasing site disorder.
  
• The overall entropy change depends significantly on both system and surroundings contributions.

Calculating and predicting entropy changes help us understand when this process proceeds spontaneously at given conditions.

Equilibrium constant

The equilibrium constant \(K\) is crucial for understanding the balance between reactants and products at a given temperature. This constant changes with temperature, as seen in different steps of the Mond process involving Nickle:

• At lower temperatures (around 50°C), a higher equilibrium constant indicates a greater formation of nickel tetracarbonyl from nickel and carbon monoxide.
  
• As temperature rises to 227°C, the equilibrium constant decreases, favoring the decomposition of nickel tetracarbonyl back to solid nickel.

This change aligns with Le Chatelier's principle, where increasing temperature shifts the equilibrium towards the endothermic direction, affecting the equilibrium constant values crucially.

Clausius-Clapeyron equation

The Clausius-Clapeyron equation is fundamental for predicting phase changes in gases like nickel tetracarbonyl. It describes how vapor pressure changes with temperature, enabling calculations needed in the Mond process modification for maximum efficiency:

• The equation connects pressure, temperature, and enthalpy of vaporization.
• It helps estimate pressure conditions where nickel tetracarbonyl remains gaseous and doesn't liquefy.
• Assists in ensuring the process stays in the desired phase at specific temperatures.
  

By using the Clausius-Clapeyron equation, one can ensure that the conditions favor the desired outcomes in phase stability, aiding in industrial applications.