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Question: If at 298K the bond energies of \(C - H,C - C,C = C\) and \(H - H\) bonds are respectively 414, 347,...

If at 298K the bond energies of C−H,C−C,C=CC - H,C - C,C = C and H−HH - H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction H2C=CH2(g)+H2(g)→H3C−CH3(g)H_{2}C = CH_{2}(g) + H_{2}(g) \rightarrow H_{3}C - CH_{3}(g) at 298 K will be

A
  • 250 kJ
B

–250 kJ

C

–250 kJ

D

– 125 kJ

Answer

– 125 kJ

Explanation

Solution

CH2=CH2(g)+H2(g)→H3C−CH3(g)CH_{2} = CH_{2}(g) + H_{2}(g)\overset{\quad\quad}{\rightarrow}H_{3}C - CH_{3}(g)

4EC−H\mathbf{4}\mathbf{E}_{\mathbf{C - H}}⇒ 414×4=1656\mathbf{414 \times 4 = 1656} 6EC−H\mathbf{6}\mathbf{E}_{\mathbf{C - H}} ⇒ 414×6=2484\mathbf{414}\mathbf{\times}\mathbf{6 = 2484}

1EC=C\mathbf{1}\mathbf{E}_{\mathbf{C = C}} ⇒ 615×1=615\mathbf{615 \times 1 = 615} 1EC−C\mathbf{1}\mathbf{E}_{\mathbf{C - C}}⇒ 347×1=347\mathbf{347}\mathbf{\times}\mathbf{1 = 347}

}}{\mathbf{6}\mathbf{\Delta}\mathbf{H}_{\mathbf{C}\mathbf{-}\mathbf{H}}\mathbf{+ 1}\mathbf{\Delta}\mathbf{H}_{\mathbf{C}\mathbf{-}\mathbf{C}}\mathbf{= 2831}}$$ $$\mathbf{\Delta}\mathbf{H = 2706}\mathbf{-}\mathbf{2831 =}\mathbf{-}\mathbf{125kJ}$$