The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{~K}\) for difluoroacetylene \(\left(\mathrm{C}_{2} \mathrm{~F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{~F}_{6}\right)\) are $$ \begin{array}{|lcc|} & \left.\Delta G_{\mathrm{f}}^{\circ}(\mathrm{k}] / \mathrm{mol}\right) & \Delta H_{\mathrm{f}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{C}_{2} \mathrm{~F}_{2}(g) & 191.2 & 241.3 \\ \mathrm{C}_{6} \mathrm{~F}_{6}(g) & 78.2 & 132.8 \\ \hline \end{array} $$ For the following reaction: $$ \mathrm{C}_{6} \mathrm{~F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{~F}_{2}(g) $$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\). b. calculate \(K\) at \(298 \mathrm{~K}\). c. estimate \(K\) at \(3000 . \mathrm{K}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Step by step solution

01

1. Write down the given data

We are given the standard free energies of formation \(\left(\Delta G_{\mathrm{f}}^{\circ}\right)\) and the standard enthalpies of formation \(\left(\Delta H_{\mathrm{f}}^{\circ}\right)\) for both substances: \(\mathrm{C}_{2}\mathrm{F}_{2}(g): \Delta G_{\mathrm{f}}^{\circ}=191.2 \mathrm{~kJ/mol}, \Delta H_{\mathrm{f}}^{\circ}=241.3 \mathrm{~kJ/mol}\) \(\mathrm{C}_{6}\mathrm{F}_{6}(g): \Delta G_{\mathrm{f}}^{\circ}=78.2 \mathrm{~kJ/mol}, \Delta H_{\mathrm{f}}^{\circ}=132.8 \mathrm{~kJ/mol}\) The reaction is: \(\mathrm{C}_{6}\mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2}\mathrm{F}_{2}(g)\)

02

2. Calculate ΔG° and ΔH° for the reaction

We can determine the standard Gibbs free energy change of the reaction (\(\Delta G^{\circ}\)) and the standard enthalpy change of the reaction (\(\Delta H^{\circ}\)) by taking the differences of the respective values for the products and reactants. Since there are three moles of \(\mathrm{C}_{2}\mathrm{F}_{2}\) in the product, we need to multiply its values by 3 before subtracting. \(\Delta G^{\circ}=3\Delta G_{\mathrm{f}}^{\circ}(\mathrm{C}_{2}\mathrm{F}_{2})-\Delta G_{\mathrm{f}}^{\circ}(\mathrm{C}_{6}\mathrm{F}_{6})\) \(\Delta G^{\circ}=3(191.2)-78.2=495.4 \mathrm{~kJ/mol}\) \(\Delta H^{\circ}=3\Delta H_{\mathrm{f}}^{\circ}(\mathrm{C}_{2}\mathrm{F}_{2})-\Delta H_{\mathrm{f}}^{\circ}(\mathrm{C}_{6}\mathrm{F}_{6})\) \(\Delta H^{\circ}=3(241.3)-132.8=591.1 \mathrm{~kJ/mol}\)

03

3. Calculate ΔS°

We can find the standard entropy change of the reaction (\(\Delta S^{\circ}\)) by using the relationship between Gibbs free energy, enthalpy, and entropy: \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) Rearrange the equation to solve for the standard entropy change: \(\Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}\) Substitute the values at \(T=298 \mathrm{K}\): \(\Delta S^{\circ}=\frac{591.1-495.4}{298}=0.3222 \mathrm{~kJ/(mol\cdot K)}\) Ans: The standard entropy change at \(298 \mathrm{K}\) is \(\Delta S^{\circ}=0.3222 \mathrm{~kJ/(mol\cdot K)}\). #b. Calculate the Equilibrium Constant K at 298 K#

04

4. Use the Gibbs Free Energy Change to Find K

We know that: \(\Delta G^{\circ}=-RT \ln{K}\) Rearrange the equation to solve for the equilibrium constant (\(K\)): \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) Substitute the values at \(T=298 \mathrm{K}\) and \(R=8.314 \times 10^{-3} \mathrm{~kJ/(mol\cdot K)}\): \(K = e^{-\frac{495.4}{(8.314 \times 10^{-3})\cdot 298}}=1.40 \times 10^{-25}\) Ans: The equilibrium constant at \(298 \mathrm{K}\) is \(K=1.40 \times 10^{-25}\). #c. Estimate K at 3000 K#

05

5. Use the Van 't Hoff Equation

The Van 't Hoff equation relates the equilibrium constant at two different temperatures: \(\frac{d\ln{K}}{dT}=\frac{\Delta H^{\circ}}{RT^{2}}\) Integrate with respect to temperature, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are constant: \(\ln{K_2}-\ln{K_1}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) The problem asks to estimate \(K\) at \(T_2 = 3000 \mathrm{K}\). We know \(K_1\) and \(T_1 = 298 \mathrm{K}\). Substitute values and solve for \(K_2\): \(\ln{K_2}=\ln{1.40 \times 10^{-25}}+\frac{591.1}{8.314\times10^{-3}}\left(\frac{1}{3000}-\frac{1}{298}\right)\) \(\ln{K_2}=-57.689\) \(K_2=e^{-57.689}=6.98\times10^{-26}\) Ans: The estimated equilibrium constant at \(3000 \mathrm{K}\) is \(K=6.98 \times 10^{-26}\)

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