From the following bond energies : $\mathrm{H}-\mathrm{H}$ bond energy : $431.37 \mathrm{~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, will be

1. $1523.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2. $-243.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3. $-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4. $553.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

Key Idea Enthalpy of reaction $=\sum \mathrm{BE}_{\text {Reactants }}-\sum \mathrm{BE}_{\text {Products }}$ For reaction, $\begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\left[4 \times \mathrm{BE}_{(\mathrm{C}-\mathrm{H})}+1 \times \mathrm{BE}_{(\mathrm{C}=\mathrm{C})}+1 \times \mathrm{BE}_{(\mathrm{H}-\mathrm{H})}\right] \\ & -\left[6 \times \mathrm{BE}_{(\mathrm{C}-\mathrm{H})}+1 \times \mathrm{BE}_{(\mathrm{C}-\mathrm{C})}\right] \\ & =(4 \times 410.50+1 \times 606.10+1 \times 431.37) \\ & \quad-[(6 \times 410.50)+(1 \times 336.49)] \\ & =-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

Asked in: NEET 2009 (Screening)

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