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

Correct answer: D. -120.0 ~kJ ~mol^ -1 Δ H=dissociation energy of reactant - Bond dissociation of energy of product. aligned Δ H=& (606.10+4 ×410.5+431.37) & -(6 ×410.50+336.49) & =-120.0 kJ/ mol aligned

NEET 2009

Exam: NEET
Year: 2009
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2009 · Chemistry · Thermodynamics (C)

From the following bond energies : H-H bond energy : 431.37 ~kJ...

From the following bond energies :

\mathrm{H}-\mathrm{H}

bond energy :

431.37 \mathrm{~kJ} \mathrm{~mol}^{-1}

\mathrm{C}=\mathrm{C}

bond energy :

606.10 \mathrm{~kJ} \mathrm{~mol}^{-1}

\mathrm{C}-\mathrm{C}

bond energy :

336.49 \mathrm{~kJ} \mathrm{~mol}^{-1}

\mathrm{C}-\mathrm{H}

bond energy :

410.50 \mathrm{~kJ} \mathrm{~mol}^{-1}

Enthalpy for the reaction, 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

will be :

Options

A $553.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $1523.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-243.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Correct answer

D. $-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Step-by-step solution

\Delta \mathrm{H}=dissociation energy of

reactant - Bond dissociation of energy of product.

\begin{aligned}\Delta \mathrm{H}=& (606.10+4 \times410.5+431.37) \\ & -(6 \times410.50+336.49) \\ & =-120.0 \mathrm{ kJ}/ \mathrm{mol} \end{aligned}

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