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. $553.0\, kJ\, mol^{-1}$
• B. $1523.6\, kJ\, mol^{-1}$
• C. $-2 43.6\, kJ\, mol^{-1}$
• D. $-120.0\, kJ\, mol^{-1}$

Answer 1

Answer: (D)

$\Delta H =$ dissociation energy of reactant $-$ Bond dissociation of energy of product.

$\Delta H =(606.10+4 \times 410.5+431.37)-(6 \times 410.50+336.49)$

$=-120.0\, kJ\, / mol$