At 298 , K,the bond energies of C-H,C-C,C=C,a | vedclass

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(B) The enthalpy change of the reaction is calculated using the formula: $\Delta H = \sum (B.E.)_{	ext{reactants}} - \sum (B.E.)_{	ext{products}}$.F

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$-125 \, kJ$

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(B) The enthalpy change of the reaction is calculated using the formula: $\Delta H = \sum (B.E.)_{	ext{reactants}} - \sum (B.E.)_{	ext{products}}$.For the reaction $H_2C=CH_{2(g)} + H_{2(g)} ightarrow H_3C-CH_{3(g)}$:Bonds broken: $1 	imes (C=C)$ and $1 	imes (H-H)$.Bonds formed: $1 	imes (C-C)$ and $2 	imes (C-H)$.Note: The reactants have $4 	imes (C-H)$ bonds and the product has $6 	imes (C-H)$ bonds. The net change is the formation of $2 	imes (C-H)$ bonds.$\Delta H = [1(B.E.)_{C=C} + 4(B.E.)_{C-H} + 1(B.E.)_{H-H}] - [1(B.E.)_{C-C} + 6(B.E.)_{C-H}]$$\Delta H = (615 + 4 	imes 414 + 435) - (347 + 6 	imes 414)$$\Delta H = (615 + 1656 + 435) - (347 + 2484)$$\Delta H = 2706 - 2831 = -125 \, kJ \, mol^{-1}$.

At $298 \, K$,the bond energies of $C-H$,$C-C$,$C=C$,and $H-H$ bonds are $414$,$347$,$615$,and $435 \, kJ \, mol^{-1}$ respectively. What is the enthalpy change for the reaction $H_2C=CH_{2(g)} + H_{2(g)} \rightarrow H_3C-CH_{3(g)}$ at $298 \, K$?

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