At $25 °\mathrm{C}$ and $1 \mathrm{atm}$ pressure, the enthalpy of combustion of benzene $\left(1\right)$ and acetylene $\left(\mathrm{g}\right)$ are $-3268 \mathrm{kJ}{ \mathrm{mol}}^{-1}$ and $-1300 \mathrm{kJ}{ \mathrm{mol}}^{-1}$, respectively. The change in enthalpy for the reaction $3{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)$, is

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burnt in oxygen with all the reactants and products in their standard state under standard conditions (298K and 1 bar pressure)

${\mathrm{\Delta H}}_{\mathrm{r}}={\left({\mathrm{\Delta H}}_{\mathrm{C}}\right)}_{\mathrm{reactant}}-{\left({\mathrm{\Delta H}}_{\mathrm{C}}\right)}_{\mathrm{product}}$

$=3\left[{\mathrm{\Delta Hc}}_{\mathrm{c}}\left({\mathrm{C}}_2{\mathrm{H}}_2,\mathrm{g}\right)\right]-\left[{\mathrm{\Delta H}}_{\mathrm{c}}\left({\mathrm{C}}_6{\mathrm{H}}_6,\mathrm{l}\right)\right]$

$=3\times \left(-1300\right)-\left(-3268\right)$

$=-3900+3268=-632 \mathrm{kJ}/\mathrm{mole}$