Calculate the standard free energy of the reaction at for the combustion of methane using the given — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate the standard free energy of the reaction at $27^\circ\text{C}$ for the combustion of methane using the given data: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$<br>$\Delta_fH^\circ(\text{kJ/mol}): \text{CH}_4(-74.5), \text{O}_2(0), \text{CO}_2(-393.5), \text{H}_2\text{O}(-286.0)$<br>$S^\circ(\text{J/K-mol}): \text{CH}_4(186), \text{O}_2(205), \text{CO}_2(212), \text{H}_2\text{O}(70)$
💡 Solution & Explanation
$\Delta H = [-393.5 + 2(-286)] - [-74.5] = -891.0\text{ kJ/mol}$. $\Delta S = [212 + 2(70)] - [186 + 2(205)] = 352 - 596 = -244\text{ J/K-mol}$. $\Delta G = \Delta H - T\Delta S = -891.0 - 300(-0.244) = -891.0 + 73.2 = -817.8\text{ kJ/mol}$. The closest option strictly provided is -819 kJ/mol.