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(A) For the reaction:$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$$\Delta_{r}H^{\circ} = \sum \Delta H_{	ext{bonds broken}} - \sum \Delt

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$-125$

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(A) For the reaction:$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$$\Delta_{r}H^{\circ} = \sum \Delta H_{	ext{bonds broken}} - \sum \Delta H_{	ext{bonds formed}}$$\Delta_{r}H^{\circ} = [4 	imes \Delta H_{(C-H)} + 1 	imes \Delta H_{(C=C)} + 1 	imes \Delta H_{(H-H)}] - [6 	imes \Delta H_{(C-H)} + 1 	imes \Delta H_{(C-C)}]$$\Delta_{r}H^{\circ} = [4 	imes 414 + 615 + 435] - [6 	imes 414 + 347]$$\Delta_{r}H^{\circ} = [1656 + 615 + 435] - [2484 + 347]$$\Delta_{r}H^{\circ} = 2706 - 2831 = -125 \ kJ$

Calculate the enthalpy change for the following reaction:
$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$
[The bond energies of $C-H, C-C, C=C$ and $H-H$ are $414, 347, 615$ and $435 \ kJ/mol$ respectively.] (in $kJ$)

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Calculate the enthalpy change for the following reaction:$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$[The bond energies of $C-H, C-C, C=C$ and $H-H$ are $414, 347, 615$ and $435 \ kJ/mol$ respectively.] (in $kJ$)