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← JEE Main 2026 · 8 Apr · Shift 2
ChemistryInteger+4 / -1Question 74 of 75

JEE Main 2026 · 8 Apr · Shift 2, Chemistry question 74

Consider the reaction

2 HX2S(g)+3 OX2(g)→2 HX2O(l)+2 SOX2(g)\ce{2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g)}

The magnitude of enthalpy change for the reaction in kJ mol−1^{-1} is ________. (Nearest integer)

Given : ΔfH⊖(HX2S)=−20.1\Delta_fH^\ominus(\ce{H2S})=-20.1 kJ mol−1^{-1}, ΔfH⊖(HX2O)=−286.0\Delta_fH^\ominus(\ce{H2O})=-286.0 kJ mol−1^{-1}, ΔfH⊖(SOX2)=−297.0\Delta_fH^\ominus(\ce{SO2})=-297.0 kJ mol−1^{-1}

Answer and worked solution

Official answer: 1126

Hess's law.
ΔH=2(−286.0)+2(−297.0)−2(−20.1)=−1125.8 kJ.\Delta H=2(-286.0)+2(-297.0)-2(-20.1)=-1125.8\text{ kJ}.

Magnitude: about 1126.

Answer: 1126