Given the standard molar entropies () at for = , = , = , and = . Calculate the absolute magnitude of — Thermodynamics and Thermochemistry Chemistry Question
Question
Given the standard molar entropies ($S^\circ$) at $298 \text{ K}$ for $CH_4(g)$ = $186.2$, $O_2(g)$ = $205.0$, $CO_2(g)$ = $213.6$, and $H_2O(l)$ = $69.9 \text{ J K}^{-1}\text{mol}^{-1}$. Calculate the absolute magnitude of the standard entropy change ($\Delta S^\circ$) in $\text{J K}^{-1}\text{mol}^{-1}$ for the combustion equation: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
💡 Solution & Explanation
The standard entropy change of the macroscopic reaction is $\Delta S^\circ = \sum S^\circ(products) - \sum S^\circ(reactants) = [S^\circ(CO_2) + 2S^\circ(H_2O)] - [S^\circ(CH_4) + 2S^\circ(O_2)]$. Substituting experimental values: $\Delta S^\circ = [213.6 + 2(69.9)] - [186.2 + 2(205.0)] = [213.6 + 139.8] - [186.2 + 410.0] = 353.4 - 596.2 = -242.8 \text{ J K}^{-1}\text{mol}^{-1}$. The absolute magnitude evaluates to $242.8$.