7. What is the Free Gibb's energy, kJ/mol, at 298 K for 2C0(g) + O2(g) → 2CO2(g) if the Kc is2.24 x 1022 at the same temperature?

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Chemistry

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Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter10: Liquids And Solids

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7. What is the Free Gibb's energy, kJ/mol, at 298 K for 2C0(g) + O2(g) → 2CO2(g) if the Kc is
2.24 x 1022 at the same temperature?

Expert Solution

Step 1

Note:

Hey there! Thank you for posting the question. Since the reaction is not given, we are taking the below values of standard Free gibb’s energy (ΔG^o) for solving the question. We have tried our best to solve this problem.

$\begin{array}{l} {ΔG}_{f}^{o} for CO (g) = - 137.15 kJ / mol \\ {ΔG}_{f}^{o} for O_{2} (g) = 0 kJ / mol \\ {ΔG}_{f}^{o} for {CO}_{2} (g) = - 394.36 kJ / mol \end{array}$

Step 2

The standard free energy of Oxygen is 0.

$2 CO (g) + O_{2} (g) \overset{}{\to} 2 {CO}_{2} (g) {ΔG}^{o} = \sum {n_{p} {ΔG}_{f}^{o}}_{(products)} - \sum {n_{r} {ΔG}_{f}^{o}}_{(reactants)} = [2 mol \times {ΔG}_{f}^{o}_{({CO}_{2})}] - [(2 mol \times {ΔG}_{f}^{o}_{(CO)}) + (1 mol \times {ΔG}_{f}^{o}_{(O_{2})})] = [2 mol \times (- 394.36 kJ / mol)] - [(2 mol \times (- 137.15 kJ / mol)) + (1 mol \times (0 kJ / mol))] = - 514.4 kJ$

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What is the Free Gibb's energy, kJ/mol, at 298 K for 2C0(g) + O2(g) → 2CO2(g) if the Kc is 2.24 x 1022 at the same temperature?