AP EAMCET · Chemistry · Thermodynamics (C)
The standard enthalpy of atomization of ethane according to the equation \(\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \rightarrow 2 \mathrm{C}(\mathrm{g})+6 \mathrm{H}(\mathrm{g})\) is \(622 \mathrm{~kJ} \mathrm{~mol}^{-1}\). If standard mean \(\mathrm{C}-\mathrm{H}\) bond dissociation enthalpy is \(90 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), the standard mean dissociation enthalpy of \(\mathrm{C}-\mathrm{C}\) bond (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) is
- A 540
- B 90
- C 85
- D 82
Answer & Solution
Correct Answer
(D) 82
Step-by-step Solution
Detailed explanation
\begin{aligned} & \Delta_{\mathrm{a}} \mathrm{H}^{\circ}=\Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\circ}+\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ} \\ & 622=[6 \times 90]+\left[\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}\right]=540+\Delta_{\mathrm{C}-\mathrm{C}}…
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