for the reaction is –20.24 kcal/mol. The bond energies (in kcal/mol) of , and are 63, 85.6 and 102.6 — Thermodynamics and Thermochemistry Chemistry Question
Question
$\Delta H$ for the reaction $2\text{C(s)} + 3\text{H}_2\text{(g)} \to \text{C}_2\text{H}_6\text{(g)}$ is –20.24 kcal/mol. The bond energies (in kcal/mol) of $\text{C}-\text{C}$, $\text{C}-\text{H}$ and $\text{H}-\text{H}$ are 63, 85.6 and 102.6, respectively. The enthalpy of sublimation of C(s) is
Answer: A
💡 Solution & Explanation
$\Delta H = 2\Delta H_{\text{sub}}(\text{C}) + 3\text{BE(H-H)} - [1\text{BE(C-C)} + 6\text{BE(C-H)}]$. $-20.24 = 2\Delta H_{\text{sub}} + 3(102.6) - [63 + 6(85.6)] \implies -20.24 = 2\Delta H_{\text{sub}} + 307.8 - 576.6 \implies 2\Delta H_{\text{sub}} = 268.8 - 20.24 = 248.56 \implies \Delta H_{\text{sub}} = 124.28\text{ kcal/mol}$.
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