Verified Solution • 2025–26 Session moderate

QJEEM22Chcecd: JEE Main 2022 Chemistry - Class 11 JEEM22Chcecd Solution: At 25 C and 1 atm pressure, the enthalpy of combustion of…

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 :

Multiple Choice Options

  • A +324 kJ mol −- 1
  • B +632 kJ mol −- 1
  • C −- 632 kJ mol −- 1
  • D −- 732 kJ mol −- 1

Step-by-Step Verified Solution

I. C6H6(ℓ)+152O2( g)→6CO2( g)+3H2O(g)\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})

ΔH1=−3268 kJ/mol\Delta \mathrm{H}_{1}=-3268 \mathrm{~kJ} / \mathrm{mol}

II. C2H2( g)+52O2( g)→2CO2( g)+H2O(g)\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})

ΔH2=−1300 kJ/mol\Delta \mathrm{H}_{2}=-1300 \mathrm{~kJ} / \mathrm{mol}

III. 3C2H2( g)→C6H6(ℓ)ΔH33 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\ell) \quad \Delta \mathrm{H}_{3}

Applying Hess's law of constant heat summation

ΔH3amp;=3×ΔH2−ΔH1amp;=3×(−1300)−(−3268)amp;=−632 kJ/mol\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}

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