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 temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) 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} \cdot \mathrm{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 Nic CO) 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 $\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}$ greater than the \(K\) value. \(]\)

Step by step solution

01

Predict the sign of ∆S° for the given reaction

In the given reaction, 5 moles of reactants (solid and gaseous) are combined to form 1 mole of gaseous product: $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ Here, there is a decrease in the number of moles from the reactants to the product side; hence, there will be a decrease in the disorder of the system. Therefore, the entropy change (∆S°) will have a negative value.

02

Predict the sign of ∆S_sum for the given reaction

The spontaneity of the reaction is temperature-dependent. The total entropy change (∆S_sum) can be defined as the difference in entropy of the surroundings and the entropy of the system. Since the reaction is exothermic, heat is released, which increases the entropy of the surroundings. The sign of ∆S_sum will depend on the magnitude of the entropy changes in the system and the surroundings. We cannot predict its sign without more information.

03

Calculate ∆H° and ∆S° for the reaction

Given for \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\), \(\Delta H_{\mathrm{f}}^{\circ}=-607\mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 K. Using the standard enthalpy and entropy of CO(g) and Ni(s) from the Appendix 4, we can calculate the ∆H° and ∆S° for the given reaction. For CO(g), \(\Delta H_{\mathrm{f}}^{\circ}=-110.5 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=197.7 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) . For Ni(s), \(\Delta H_{\mathrm{f}}^{\circ}=0 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=29.87 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) . To calculate ∆H° for the reaction: $$\Delta H^{\circ}=\Delta H_{\mathrm{f}}^{\circ}(\mathrm{Ni}(\mathrm{CO})_{4}(g)) - 4\Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}(g)) - \Delta H_{\mathrm{f}}^{\circ}(\mathrm{Ni}(s))$$ $$\Delta H^{\circ}= (-607) - 4(-110.5) - 0 =-164.0 \mathrm{kJ/mol}$$ To calculate ∆S° for the reaction: $$\Delta S^{\circ} = S^{\circ}(\mathrm{Ni}(\mathrm{CO})_{4}(g)) - 4S^{\circ}(\mathrm{CO}(g)) - S^{\circ}(\mathrm{Ni}(s))$$ $$\Delta S^{\circ} = 417 - 4(197.7) - 29.87 = -426.67 \mathrm{J/K \cdot mol}$$

04

Calculate the temperature at which ∆G°=0 (K=1) for the reaction

At equilibrium, ∆G°=0, and according to the Gibbs-Helmholtz equation $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = 0$$ Solving for T: $$T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{-164000 \mathrm{J/mol}}{-426.67 \mathrm{J/K \cdot mol}} = 384.2 \mathrm{K}$$ The temperature at which ∆G°=0 (K=1) for the given reaction is 384.2 K.

05

Calculate the equilibrium constant for the reaction at 50°C

To calculate the equilibrium constant (K) at 50°C (323.15 K), we can use the Van't Hoff equation: $$\ln K = -\frac{\Delta H^{\circ}}{R} \cdot \frac{1}{T} + \frac{\Delta S^{\circ}}{R}$$ where R is the gas constant (8.314 J/K∙mol) $$\ln K = -\frac{-164000}{8.314} \cdot \frac{1}{323.15} + \frac{-426.67}{8.314}$$ $$\ln K = 4.75224$$ The equilibrium constant at 50°C: $$K = e^{4.75224} = 115.93$$

06

Calculate the equilibrium constant for the reaction at 227°C

Now, let's calculate the equilibrium constant (K) at 227°C (500.15 K): $$\ln K = -\frac{-164000}{8.314} \cdot \frac{1}{500.15} + \frac{-426.67}{8.314}$$ $$\ln K = 1.60117$$ The equilibrium constant at 227°C: $$K = e^{1.60117} = 4.96$$

07

Explain why the temperature is increased for the second step of the Mond process

The temperature is increased for the second step of the Mond process because, at higher temperatures, the reaction shifts towards the left, favoring the formation of nickel as a solid and CO(g). In the second step of the process, the goal is to deposit as much nickel as possible in its pure solid form, so increasing the temperature helps achieve this goal.

08

Estimate the maximum pressure of Ni(CO)4(g) that can be attained before the gas will liquefy at 152°C

Given, the boiling point for Ni(CO)₄ is 42°C and the enthalpy of vaporization is 29.0 kJ/mol. At the boiling point, the vapor pressure is equal to the external atmospheric pressure. We can use the Clausius-Clapeyron equation to determine the max pressure at 152°C: $$\ln \frac{P_2}{P_1} = -\frac{\Delta H_\mathrm{vap}}{R} \cdot \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ where \(P_1\) is the atmospheric pressure (assume 1 atm), \(T_1\) is the boiling point temperature (42+273.15=315.15 K), and \(T_2\) is the temperature for the first step (152+273.15=425.15 K). $$\ln \frac{P_2}{1} = -\frac{29000}{8.314} \cdot \left(\frac{1}{425.15} - \frac{1}{315.15}\right)$$ $$P_2 = e^{4.72113} = 112.45 \mathrm{atm}$$ The maximum pressure of Ni(CO)₄(g) that can be attained before the gas will liquefy at 152°C is 112.45 atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!