The enthalpy of formation of ethane (C_2 H_6 ) from ethylene by addition of hydrogen where the bond-energies of C-H, C-C, C=C, H-H are 414 ~kJ, 347 ~kJ, 615 ~kJ and 435 ~kJ respectively is kJ

Correct answer: 125. C_2 H_4(~g)+H_2(~g) arrow C_2 H_6(~g) Δ H=BE(C=C)+4 BE(C-H)+BE(H-H) -BE(C-C)-6 BE(C-H) Δ H=BE(C=C)+BE(H-H)-BE(C-C) -2 BE(C-H) =615+435-347-2 × 414 =-125 ~kJ

Q30 / 90 · JEE Main 2024 (04 Apr Shift 1) · Chemistry

The enthalpy of formation of ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$ from ethylene by addition of hydrogen where the bond-energies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}, \mathrm{H}-\mathrm{H}$ are $414 \mathrm{~kJ}, 347 \mathrm{~kJ}, 615 \mathrm{~kJ}$ and $435 \mathrm{~kJ}$ respectively is $\qquad$ $\mathrm{kJ}$

Correct answer: 125

Solution

$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \\ & \Delta \mathrm{H}=\mathrm{BE}(\mathrm{C}=\mathrm{C})+4 \mathrm{BE}(\mathrm{C}-\mathrm{H})+\mathrm{BE}(\mathrm{H}-\mathrm{H}) \\ & -\mathrm{BE}(\mathrm{C}-\mathrm{C})-6 \mathrm{BE}(\mathrm{C}-\mathrm{H}) \\ & \Delta \mathrm{H}=\mathrm{BE}(\mathrm{C}=\mathrm{C})+\mathrm{BE}(\mathrm{H}-\mathrm{H})-\mathrm{BE}(\mathrm{C}-\mathrm{C}) \\ & -2 \mathrm{BE}(\mathrm{C}-\mathrm{H}) \\ & =615+435-347-2 \times 414 \\ & =-125 \mathrm{~kJ}\end{aligned}$

Hence, the correct answer is 125.

Solution

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