At $25 ° C$ and 1 atm pressure, the enthalpy of combustion of benzene (1) and acetylene (g) are $- 3268 kJ {mol}^{- 1}$ and $- 1300 kJ {mol}^{- 1}$ , respectively. The change in enthalpy for the reaction $3 C_{2} H_{2} (g) \to C_{6} H_{6} (l)$ , is

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$- 632 kJ {mol}^{- 1}$

$- 732 kJ {mol}^{- 1}$

$+ 632 kJ {mol}^{- 1}$

$+ 324 kJ {mol}^{- 1}$

answer is C.

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Detailed Solution

$ΔH = \sum {ΔH}_{C} (Reactant) - \sum {ΔH}_{C} (Product)$

$\begin{aligned} = [3 \times (- 1300) - (- 3268)] \\ = - 632 {kJmol}^{- 1} \end{aligned}$

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