Question:
What is the bond enthalpy (in kJ $mol^{-1}$) of C-H in ethane? ($\Delta_f H^{\ominus}(C_2H_6(g)) = -85$ kJ $mol^{-1}$; $H_2(g) \rightarrow 2H(g); \Delta_a H^{\ominus} = 435$ kJ $mol^{-1}$; $C(s) \rightarrow C(g); \Delta_a H^{\ominus} = 715$ kJ $mol^{-1}$; $\Delta_a H^{\ominus}(C-C) = 347$ kJ $mol^{-1}$)
What is the bond enthalpy (in kJ $mol^{-1}$) of C-H in ethane? ($\Delta_f H^{\ominus}(C_2H_6(g)) = -85$ kJ $mol^{-1}$; $H_2(g) \rightarrow 2H(g); \Delta_a H^{\ominus} = 435$ kJ $mol^{-1}$; $C(s) \rightarrow C(g); \Delta_a H^{\ominus} = 715$ kJ $mol^{-1}$; $\Delta_a H^{\ominus}(C-C) = 347$ kJ $mol^{-1}$)
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Heat of formation = Energy to break bonds of elements - Energy to break bonds of the molecule.
Updated On: Oct 15, 2026
- 412.2
- 402.8
- 390.7
- 380.6
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Bond enthalpy calculation from atomization energies.
Step 2: Meaning
$\Delta_f H = \Sigma \Delta_a H(reactants) - \Sigma \text{Bond Enthalpies}(products)$.
Step 3: Analysis
$\Delta_f H(C_2H_6) = [2\Delta_a H(C) + 3\Delta_a H(H_2)] - [\Delta_a H(C-C) + 6\Delta_a H(C-H)]$. $-85 = [2(715) + 3(435)] - [347 + 6E_{C-H}]$. $-85 = [1430 + 1305] - [347 + 6E_{C-H}]$. $-85 = 2735 - 347 - 6E_{C-H}$. $6E_{C-H} = 2388 + 85 = 2473$. $E_{C-H} = 2473 / 6 \approx 412.2$ kJ/mol.
Step 4: Conclusion
The bond enthalpy is 412.2 kJ $mol^{-1}$.
Final Answer: (A)
Bond enthalpy calculation from atomization energies.
Step 2: Meaning
$\Delta_f H = \Sigma \Delta_a H(reactants) - \Sigma \text{Bond Enthalpies}(products)$.
Step 3: Analysis
$\Delta_f H(C_2H_6) = [2\Delta_a H(C) + 3\Delta_a H(H_2)] - [\Delta_a H(C-C) + 6\Delta_a H(C-H)]$. $-85 = [2(715) + 3(435)] - [347 + 6E_{C-H}]$. $-85 = [1430 + 1305] - [347 + 6E_{C-H}]$. $-85 = 2735 - 347 - 6E_{C-H}$. $6E_{C-H} = 2388 + 85 = 2473$. $E_{C-H} = 2473 / 6 \approx 412.2$ kJ/mol.
Step 4: Conclusion
The bond enthalpy is 412.2 kJ $mol^{-1}$.
Final Answer: (A)
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