JEE Main · 2022 · Shift-IImediumTHERMO-026

At STP, Enthalpy of combustion of benzene (l) and acetylene (g) are -3268 and -1300\,kJ mol-1 respectively. The H for…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

At STP, Enthalpy of combustion of benzene (l) and acetylene (g) are −3268-3268 and −1300 kJ mol−1-1300\,\text{kJ mol}^{-1} respectively. The ΔH\Delta H for 3C2H2(g)→C6H6(l)3\text{C}_2\text{H}_2(g) \rightarrow \text{C}_6\text{H}_6(l) is:

Options
  1. a

    +324 kJ mol−1+324\,\text{kJ mol}^{-1}

  2. b

    +632 kJ mol−1+632\,\text{kJ mol}^{-1}

  3. c

    −632 kJ mol−1-632\,\text{kJ mol}^{-1}

    ✓
  4. d

    −732 kJ mol−1-732\,\text{kJ mol}^{-1}

Correct Answerc

−632 kJ mol−1-632\,\text{kJ mol}^{-1}

Detailed Solution

🧠 Hess's Law: reverse and combine combustion equations to get the target The target reaction 3\ceC2H2(g)−>C6H6(l)3\ce{C2H2(g) -> C6H6(l)} can be built from the given combustion enthalpies.

🗺️ Calculation Write combustion equations: \ceC2H2(g)+5/2O2(g)−>2CO2(g)+H2O(l),  ΔH=−1300 kJ mol−1\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(l)},\; \Delta H = -1300\text{ kJ mol}^{-1} \ceC6H6(l)+15/2O2(g)−>6CO2(g)+3H2O(l),  ΔH=−3268 kJ mol−1\ce{C6H6(l) + 15/2 O2(g) -> 6CO2(g) + 3H2O(l)},\; \Delta H = -3268\text{ kJ mol}^{-1}

Multiply the first by 3, then subtract the second: 3×(−1300)−(−3268)=−3900+3268=−632 kJ mol−13 \times (-1300) - (-3268) = -3900 + 3268 = -632\text{ kJ mol}^{-1}

⚡ Speed check: Three acetylene molecules have 3 × (−1300) = −3900 kJ of combustion energy. Benzene releases −3268 kJ. The difference −632 kJ is the energy released on trimerisation.

⚠️ Trap: Students forget to reverse the benzene combustion (subtract, not add it again).

Answer: (c)\boxed{\text{Answer: (c)}}

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