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  • From the following bond energies: H - H bond energy: 431.37 kJ mol-1 <p style="margin-left:0in; margi

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:

Option: 1

-243.6 kJ mol-1


Option: 2

-120.0 kJ mol- 1


Option: 3

553.0 kJ mol-1


Option: 4

1523.6 kJ mol- 1


Answers (1)

best_answer

We know\\ \mathrm{\Delta H = \text{Bond dissociation energy of reactant - Bond dissociation energy of product}} \\\mathrm{ \Delta H = \text { B.E. (reactant) }-\text { B.E. (product) } }

                                               

\\ \Delta \mathrm{H}=\left[4 \times \mathrm{B.E}_{(\mathrm{C}-\mathrm{H})}+1 \times \mathrm{B.E}_{(\mathrm{C}=\mathrm{C})}+1 \times \mathrm{B.E}_{(\mathrm{H}-\mathrm{H})}-6 \times \mathrm{B.E}_{(\mathrm{C}-\mathrm{H})}+1 \times \mathrm{B.E}_{(\mathrm{C}-\mathrm{C})}]\right.\\ \\ \begin{array}{l} \ \ \ \ \ {=[4 \times 410.50+1 \times 606.10+1 \times 431.37]-[6 \times 410.50+1 \times 336.49]} \\ \\ \ \ \ \ \ {=-120.0 \mathrm{kJ} \mathrm{mol}^{-}} \end{array}

Posted by

himanshu.meshram

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