JEE Main 2026 Chemistry question
Consider the reaction 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g). The magnitude of enthalpy change for the reaction in kJ mol⁻¹ is . (Nearest integer) Given: Δ_fH°(H₂S)=−20.1 kJ mol⁻¹, Δ_fH°(H₂O)=−286.0 kJ mol⁻¹, Δ_fH°(SO₂)=−297.0 kJ mol⁻¹
Answer: 1126
Solution
ΔH=2(−286.0)+2(−297.0)−2(−20.1)−3(0)=−572−594+40.2=−1125.8 kJ, so the magnitude is ≈1126 kJ.