If at 298 ~K the bond energies of C-H, C-C, C=C and H-H bonds are respectively 414,347,615 and 435 ~kJ mol^-1 , the value of enthalpy change for the reaction H_2 C=CH_2(~g)+H_2(~g) arrow H_3 C-CH_3(~g) at 298 ~K will be

Correct answer: (C) -125 ~kJ. CH_2=CH_2(~g)+H_2(~g) arrow CH_3-CH_3 Δ H=1(C=C)+4(C-H)+1(H-H)-1(C-C)-6(C-H)=1(C=C)+1(H-H)-1(C-C)-2(C-H) =615+435-347-2 × 414=1050-1175=-125 ~kJ

Q7 / 225 · JEE Main 2003 · Chemistry

If at $298 \mathrm{~K}$ the bond energies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}$ and $\mathrm{H}-\mathrm{H}$ bonds are respectively $414,347,615$ and $435 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the value of enthalpy change for the reaction $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_3(\mathrm{~g})$ at $298 \mathrm{~K}$ will be

A
$-250 \mathrm{~kJ}$
B
$+125 \mathrm{~kJ}$
Correct answer
C
$-125 \mathrm{~kJ}$
D
$+250 \mathrm{~kJ}$

Solution

$\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3$ $\Delta H=1(\mathrm{C}=\mathrm{C})+4(\mathrm{C}-\mathrm{H})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-6(\mathrm{C}-\mathrm{H})=1(\mathrm{C}=\mathrm{C})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-2(\mathrm{C}-\mathrm{H})$ $=615+435-347-2 \times 414=1050-1175=-125 \mathrm{~kJ}$

Hence, the correct answer is option: 'C'.

Solution

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