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JEE MAIN 2026 · Chemical Thermodynamics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 74

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

Consider the reaction

2H2S(g)+3O2(g)→2H2O(l)+2SO2(g)2\text{H}_2\text{S(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)} + 2\text{SO}_2\text{(g)}

The magnitude of enthalpy change for the reaction in kJ mol⁻¹ is ______. (Nearest integer)

Given: ΔfH⊖(H2S)=−20.1\Delta_f H^\ominus(\text{H}_2\text{S}) = -20.1 kJ mol⁻¹, ΔfH⊖(H2O)=−286.0\Delta_f H^\ominus(\text{H}_2\text{O}) = -286.0 kJ mol⁻¹, ΔfH⊖(SO2)=−297.0\Delta_f H^\ominus(\text{SO}_2) = -297.0 kJ mol⁻¹

Your answer

Solution

Solution

ΔrH=∑ΔfH(products)−∑ΔfH(reactants)\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants}), with ΔfH(O2)=0\Delta_f H(\text{O}_2) = 0:

ΔrH=[2(−286.0)+2(−297.0)]−[2(−20.1)]=−1166.0+40.2=−1125.8 kJ mol−1.\Delta_r H = [2(-286.0) + 2(-297.0)] - [2(-20.1)] = -1166.0 + 40.2 = -1125.8 \text{ kJ mol}^{-1}.

Magnitude to the nearest integer: 1126.

Answer: 1126.

Answer checked against the official answer key. The solution is our own working.

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