JEE Main · 2022 · Shift-IImediumTHERMO-104

At 25C and 1 atm pressure, the enthalpies of combustion are as given below: | Substance | H2 | C (graphite) | C2H6(g) |…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

At 25∘C25^\circ\text{C} and 1 atm pressure, the enthalpies of combustion are as given below:

| Substance | H2\text{H}_2 | C (graphite) | C2H6(g)\text{C}_2\text{H}_6(g) | |-----------|------|------|------| | ΔcH⊖/kJ mol−1\Delta_c H^\ominus/\text{kJ mol}^{-1} | −286.0-286.0 | −394.0-394.0 | −1560.0-1560.0 |

The enthalpy of formation of ethane is:

Options
  1. a

    +54.0 kJ mol−1+54.0\,\text{kJ mol}^{-1}

  2. b

    −68.0 kJ mol−1-68.0\,\text{kJ mol}^{-1}

  3. c

    −86.0 kJ mol−1-86.0\,\text{kJ mol}^{-1}

    ✓
  4. d

    +97.0 kJ mol−1+97.0\,\text{kJ mol}^{-1}

Correct Answerc

−86.0 kJ mol−1-86.0\,\text{kJ mol}^{-1}

Detailed Solution

🧠 Formation enthalpy from combustion data uses the inverse Hess path Write the formation reaction (\ce2C+3H2−>C2H6\ce{2C + 3H2 -> C2H6}) and route it through combustion: elements combust forward, product combustion reverses.

🗺️ Calculation ΔfH∘(\ceC2H6)=2 ΔcH∘(C)+3 ΔcH∘(H2)−ΔcH∘(C2H6)\Delta_f H^\circ(\ce{C2H6}) = 2\,\Delta_c H^\circ(\text{C}) + 3\,\Delta_c H^\circ(\text{H}_2) - \Delta_c H^\circ(\text{C}_2\text{H}_6)

=2(−394.0)+3(−286.0)−(−1560.0)= 2(-394.0) + 3(-286.0) - (-1560.0)

=−788.0−858.0+1560.0=−86.0 kJ mol−1= -788.0 - 858.0 + 1560.0 = -86.0\,\text{kJ mol}^{-1}

⚠️ Trap: The formula is products minus reactants in the Hess route — the compound's combustion is subtracted (reversed), elements' combustions are added.

Answer: (c) −86.0 kJ mol−1\boxed{\text{Answer: (c) } -86.0\,\text{kJ mol}^{-1}}

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