Question
Calculate $\Delta_\text{r}\text{H}^\circ$ for the reaction,
$\text{H}-\text{C}=\text{C}-\text{H}+30=0\overrightarrow{\ \ \ \ \ \ \ }\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{H}\\20=\text{C}=0+2\text{H}-0-\text{H}$
The average bond enthalpies of various bonds are as follows.
Bond
Bond enthalpy ($\mathrm{kJ} \mathrm{~mol}^{-1}$)
C-H
414
O=O
499
C=O
724
O-H
460
C=C
619

Answer

In this reaction, four $\mathrm{C}-\mathrm{H}$ bonds, one $\mathrm{C}=\mathrm{C}$ bond and three $\mathrm{O}=\mathrm{O}$ bonds are broken and four $\mathrm{C}=\mathrm{O}$ bonds and four $\mathrm{O}-\mathrm{H}$ bonds are formed. Thus, $\Delta \mathrm{H}^{\circ}=[4 \times$ (bond enthalpy of $\mathrm{C}-\mathrm{H})+$ (bond enthalpy) of $\mathrm{C}=\mathrm{C}$ ) $+3[$ bond enthalpy of $\mathrm{O}=\mathrm{O}]$ $-4 \times[$ bond enthalpy of $\mathrm{C}=\mathrm{O}]+4$ (bond enthalpy of $\mathrm{O}-\mathrm{H}$ ) $]$
$\Delta \mathrm{H}=(4 \times 414+619+3 \times 499)-(4 \times 724+4 \times 460)$
$=-964 \mathrm{~kJ} \mathrm{~mol}^{-1}$
 

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