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)}}

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Thermodynamics) inside The Crucible, our adaptive practice platform.

Practice in The Crucible →Browse Thermodynamics chapter →

More JEE Main Thermodynamics PYQs

2025hard

Total enthalpy change for freezing of 1 mol of water at 10°C to ice at -10°C is _____. Given: fusH = x\ kJ/mol Cp[H2O(l)] = y\ J mol-1K-1 Cp[H2O(s)] = z\ J mol-1K-1

2025easy

One mole of an ideal gas expands isothermally and reversibly from 10 dm3 to 20 dm3 at 300 K. U, q and work done in the process respectively are: Given: R = 8.3 JK-1 and mol-1 In 10 = 2.3, 2 = 0.30, 3…

2024medium

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Enthalpy of neutralisation of strong monobasic acid with strong monoacidic…

2023medium

Given: (A) 2CO(g) + O2(g) 2CO2(g), H1 = -x\,kJ mol-1 (B) C(graphite) + O2(g) CO2(g), H2 = -y\,kJ mol-1 The H for the reaction C(graphite) + 12O2(g) CO(g) is:

2023medium

The enthalpy change for the adsorption process and micelle formation respectively are:

View all 58 Thermodynamics PYQs →
At STP, Enthalpy of combustion of benzene (l) and acetylene (g) are -3268 and -1300\,kJ mol-1… (JEE Main 2022) | Canvas Classes