Calculate enthalpy change for following reaction. $\mathrm{H}_2…

Calculate enthalpy change for following reaction. $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_{3(\mathrm{~g})}$ [The bond energy of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}$ and $\mathrm{H}-\mathrm{H}$ is $414,347,615$ and 435 kJ respectively]
  1. 125 kJ
  2. -125 kJ
  3. 250 kJ
  4. $\quad-250 \mathrm{~kJ}$

Solution

For the reaction, $\begin{aligned} & \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_{3(\mathrm{~g})} \\ & \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=\sum \Delta \mathrm{H}_{\text {reactant }}^{\circ}-\sum \Delta \mathrm{H}_{\text {product }}^{\circ} \\ & =\left[4 \Delta \mathrm{H}_{(\mathrm{C}-\mathrm{H})}^{\circ}+\Delta \mathrm{H}_{(\mathrm{C=C})}^{\circ}+\Delta \mathrm{H}_{(\mathrm{H}-\mathrm{H})}^{\circ}\right] \\ & -\left[6 \Delta \mathrm{H}_{(\mathrm{C}-\mathrm{H})}^{\circ}+\Delta \mathrm{H}_{(\mathrm{C}-\mathrm{C})}^{\circ}\right] \\ & =[(4 \times 414 \mathrm{~kJ})+615 \mathrm{~kJ}+435 \mathrm{~kJ}]-[(6 \times 414 \mathrm{~kJ}) \\ & +347 \mathrm{~kJ}] \\ & =(1656 \mathrm{~kJ}+615 \mathrm{~kJ}+435 \mathrm{~kJ})-(2484 \mathrm{~kJ}+347 \mathrm{~kJ}) \\ & =2706 \mathrm{~kJ}-2831 \mathrm{~kJ}=-125 \mathrm{~kJ} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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