At 25$^\circ$C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are $-$ 3268 kJ mol^$-$1 and $-$1300 kJ mol^$-$1, respectively. The change in enthalpy for the reaction 3 C₂H₂(g) $\to$ C₆H₆ (I), is :

I. $\mathrm{C}_{6} \mathrm{H}_{6}(\ell)+\frac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ <br/><br/> $\Delta \mathrm{H}_{1}=-3268 \mathrm{~kJ} / \mathrm{mol}$ <br/><br/> II. $\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ <br/><br/> $\Delta \mathrm{H}_{2}=-1300 \mathrm{~kJ} / \mathrm{mol}$ <br/><br/> III. $3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\ell) \quad \Delta \mathrm{H}_{3}$ <br/><br/> Applying Hess's law of constant heat summation <br/><br/> $$ \begin{aligned} \Delta \mathrm{H}_{3} &=3 \times \Delta \mathrm{H}_{2}-\Delta \mathrm{H}_{1} \\\\ &=3 \times(-1300)-(-3268) \\\\ &=-632 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$