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Chapter 17: Problem 66

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce \(\mathrm{Ni}(\mathrm{CO})_{4},\) which is a gas and can therefore be separated from solid impurities: $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ Given that the standard free energies of formation of \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) are \(-137.3 \mathrm{~kJ} / \mathrm{mol}\) and \(-587.4 \mathrm{~kJ} / \mathrm{mol},\) respectively, calculate the equilibrium constant of the reaction at \(80^{\circ} \mathrm{C}\). Assume that \(\Delta G_{\mathrm{f}}^{\circ}\) is temperature independent.

Short Answer

Expert verified
The equilibrium constant for the reaction is approximately \(6.1 * 10^7\).

Step by step solution

01

Convert temperature to Kelvin

Firstly, convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is \(K = \degree C + 273.15\). So, \(80 \degree C = 80 + 273.15 = 353.15 K\).
02

Calculate the standard free-energy change

We'll use the formula for the standard free-energy change for the reaction, \(\Delta G^{\circ} = \Delta G_{f}^{\circ} (\text{products}) - \Delta G_{f}^{\circ} (\text{reactants})\). Given that \(\Delta G_{f}^{\circ} (\mathrm{CO}) = -137.3 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta G_{f}^{\circ} (\mathrm{Ni}(\mathrm{CO})_{4}) = -587.4 \mathrm{~kJ} / \mathrm{mol}\), we get \(\Delta G^{\circ} = -587.4 \mathrm{~kJ} / \mathrm{mol} - [1*(-237.1 \mathrm{~kJ} / \mathrm{mol}) + 4*(-137.3 \mathrm{~kJ} / \mathrm{mol})] = -75.1 \, \mathrm{kJ} / \mathrm{mol}\).
03

Calculate the equilibrium constant

Use the formula that relates standard free-energy change with the equilibrium constant: \(\Delta G = -RT \ln K\). Rearrange to solve for K: \(K = e^{(-\Delta G/RT)}\). Substitute the values of \(\Delta G, R, T\) into the equation: \(K = e^{[-(-75.1 \mathrm{~kJ} / \mathrm{mol}) / (8.314 \mathrm{~J} / \mathrm{K} \mathrm{mol} * 353.15 \mathrm{~K})]}\). Convert \(\Delta G\) to J/mol by multiplying with 1000: \( K = e^{[(75.1*10^3) / (8.314 * 353.15)]} \approx 6.1 * 10^7\).

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Most popular questions from this chapter

The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial pressures are \(P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm}\) \(P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm},\) and \(P_{\mathrm{CO}}=1.20 \mathrm{~atm}\)

For a reaction with a negative \(\Delta G^{\circ}\) value, which of the following statements is false? (a) The equilibrium constant \(K\) is greater than one. (b) The reaction is spontaneous when all the reactants and products are in their standard states. (c) The reaction is always exothermic.

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) dissolves spontaneously and endothermically in water. What can you deduce about the sign of \(\Delta S\) for the solution process?

The molar heat of vaporization of ethanol is 39.3 kJ/mol and the boiling point of ethanol is \(78.3^{\circ} \mathrm{C}\). Calculate \(\Delta S\) for the vaporization of 0.50 mol ethanol.

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ is \(5.62 \times 10^{35}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C}\)
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