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

(C) $-632 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(A) $+324 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(B) $+632 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(D) $-732 \mathrm{~kJ} \mathrm{~mol}^{-1}$

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Chemical Thermodynamics

2 5^{\circ} C

and 1 atm pressure, the enthalpy of combustion of benzene (l) 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

Answer: (C)

Using Hess's law,

Δ H = \sum Δ H_{combustion} (reactants) - \sum Δ H_{combustion} (products)

. For the reaction

3 C_{2} H_{2} (g) \to C_{6} H_{6} (l)

Δ H = 3 \times (- 1300) - (- 3268) = - 3900 + 3268 = - 632 kJ mol^{- 1}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option