OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीThe stepwise bond dissociation enthalpies for methane are: \( \text{CH}_4(g) \rightarrow \text{CH}_3(g) + \text{H}(g) \); \( \Delta_{\text{bond}} H_1^\ominus = +427\text{ kJ mol}^{-1} \) \( \text{CH}_3(g) \rightarrow \text{CH}_2(g) + \text{H}(g) \); \( \Delta_{\text{bond}} H_2^\ominus = +439\text{ kJ mol}^{-1} \) \( \text{CH}_2(g) \rightarrow \text{CH}(g) + \text{H}(g) \); \( \Delta_{\text{bond}} H_3^\ominus = +452\text{ kJ mol}^{-1} \) \( \text{CH}(g) \rightarrow \text{C}(g) + \text{H}(g) \); \( \Delta_{\text{bond}} H_4^\ominus = +347\text{ kJ mol}^{-1} \) What is the mean \( \text{C}-\text{H} \) bond enthalpy in methane?
Correct answer · Option 2
\( 416.25\text{ kJ mol}^{-1} \)
Explanation
Mean bond enthalpy = \( \frac{1}{4} \sum \Delta_{\text{bond}} H^\ominus = \frac{427 + 439 + 452 + 347}{4} = \frac{1665}{4} = 416.25\text{ kJ mol}^{-1} \).
