Question

From the following bond energies:
$H - H$ bond energy $: 431.37\, kJ\, mol ^{-1}$
$C = C$ bond energy $: 606.10\, kJ\, mol^{-1}$
$C - C$ bond energy $: 336.49\, kJ \,mol ^{-1}$
$C - H$ bond energy $: 410.50\, kJ \,mol ^{-1}$
Enthalpy for the reaction,

will be

• A. $-243.6\, kJ \,mol ^{-1}$
• B. $-120.0\, kJ \,mol ^{-1}$
• C. $553.0\, kJ \,mol ^{-1}$
• D. $1523.6\, kJ \,mol ^{-1}$

Answer 1

Answer: (B)

For the given reaction, enthalpy of reaction can be calculated as $=B . E .($ reactant $)-B . E .$ (product) $=\left[B. E._{(C=C)}+B. E_{\cdot(H-H)}+4 \times B. E._{(C-H)}\right]$

$-[B. E.( C - C )+6 \times B. E.( C - H )$

$=[606.10+431.37+4 \times 410.50]-[336.49+6 \times 410.50]$

$=2679.47-2799.49=-120.02 \,kJ\, mol ^{-1}$