The enthalpy of formation steadily changes from to as we go from , to . The value of however shows o — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy of formation steadily changes from $-17.89 \text{ Kcal/mol}$ to $-49.82 \text{ Kcal/mol}$ as we go from $\text{CH}_4$, $\text{C}_2\text{H}_6$ to $\text{C}_8\text{H}_{18}$. The value of $\Delta G$ however shows opposite trend, from $-12.12 \text{ Kcal/mol}$ for $\text{CH}_4$ to $4.14 \text{ Kcal/mol}$ for $\text{C}_8\text{H}_{18}$. Why?
Answer: C
💡 Solution & Explanation
The formation reaction is $n \text{C(s)} + (n+1) \text{H}_2\text{(g)} \to \text{C}_n\text{H}_{2n+2}\text{(g)}$. The change in gaseous moles is $\Delta n_g = -n$. As alkanes get larger, this results in an increasingly larger drop in entropy ($\Delta S < 0$). In $\Delta G = \Delta H - T\Delta S$, this heavily negative $-T\Delta S$ term dominates, causing $\Delta G$ to trend positive.
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