JEE Main 2022 · 24 June · evening · Single correct · +4 / −1

At 25°C and 1 atm pressure, the enthalpies of combustion are as given below :
The enthalpy of formation of ethane is

SubstanceH₂C (graphite)C₂H₆(g)
(Δ _cH^Θ)/(kJ mol⁻¹)- 286.0- 394.0- 1560.0
  1. (a) +54.0 kJ mol⁻¹
  2. (b) -68.0 kJ mol⁻¹
  3. (c) -86.0 kJ mol⁻¹
  4. (d) +97.0 kJ mol⁻¹
Answer and solution

Answer (c) -86.0 kJ mol⁻¹

Solution

Chemistry ma’am · 4 steps

Formation from combustion data: reactants' combustion enthalpies minus the product's.

  1. Target reaction. Formation of ethane from its elements in their standard states is 2C(graphite) + 3H₂(g) → C₂H₆(g). Use 2 mol of C and 3 mol of H₂ for each mole of ethane.
  2. Hess's law. Burn the reactants, then run the combustion of ethane backwards. The CO₂ and H₂O cancel, which leaves the formation reaction.

    Δ_f H⊖ = 2 Δ_c H⊖(C) + 3 Δ_c H⊖(H₂) - Δ_c H⊖(C₂H₆)

  3. Put in numbers. 2(-394.0) + 3(-286.0) - (-1560.0) = -788.0 - 858.0 + 1560.0 = -86.0 kJ mol⁻¹.
  4. So. The enthalpy of formation of ethane is -86.0 kJ mol⁻¹, option (c).

The idea to keep. With combustion enthalpies, it is reactants minus products, the reverse of the formation-enthalpy rule. Multiply each value by its coefficient first.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Chemical Thermodynamics

Asked about 160 times in JEE Main since 2020. This question is from 2022.