If ΔfH∘C2H6=−84kJmol−1,Δsub H∘C, graphite )=720kJmol−1,ΔatH2=435kJmol−1 and the bond enthalpy εC−H=414kJmol−1 then the bond enthalpy εC−C will be about

If $Δ_{f} H^{\circ} (C_{2} H_{6}) = - 84 kJ {mol}^{- 1}, Δ_{sub} H^{\circ} (C$ $, graphite) = 720 kJ {mol}^{- 1}, Δ_{at} (H_{2}) = 435 kJ {mol}^{- 1}$ and the bond enthalpy $ε_{C - H} = 414 kJ {mol}^{- 1}$ then the bond enthalpy $ε_{C - C}$ will be about

A
$295 k J m o l^{- 1}$
B
$315 k J m o l^{- 1}$
C
$345 k J m o l - 1$
D
$375 k J m o l^{- 1}$

Solution:

According to Hess's law $Δ_{f} H^{\circ} = 2 Δ_{sub} H^{\circ} (C) + 3 Δ_{at} H^{\circ} (H_{2}) - ε_{C - C} - 6 ε_{C - H}$ $\begin{aligned} ε_{C - C} = 2 Δ_{sub} H^{\circ} (C) + 3 Δ_{at} H^{\circ} (H_{2}) - 6 ε_{C - H} - Δ_{f} H^{\circ} = [2 \times 720 + 3 \times 435 - 6 \times 414 + 84] kJ {mol}^{- 1} \\ = 345 kJ {mol}^{- 1} \end{aligned}$

Solution:

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