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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2026 · 8 April · Evening Shift · Chemistry · Thermodynamics (C)

Consider the reaction 2H_2S(g) + 3O_2(g) arrow 2H_2O(l) + 2SO_2(g)...

Consider the reaction
2H2S(g)+3O2(g)→2H2O(l)+2SO2(g)2\text{H}_2\text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) + 2\text{SO}_2(g)
The magnitude of enthalpy change for the reaction in kJ mol−1^{-1} is ________. (Nearest integer)
Given: ΔfH⊖(H2S)=−20.1\Delta_fH^{\ominus}(\text{H}_2\text{S}) = -20.1 kJ mol−1^{-1}
ΔfH⊖(H2O)=−286.0\Delta_fH^{\ominus}(\text{H}_2\text{O}) = -286.0 kJ mol−1^{-1}
ΔfH⊖(SO2)=−297.0\Delta_fH^{\ominus}(\text{SO}_2) = -297.0 kJ mol−1^{-1}

Correct answer

0

Step-by-step solution

The standard enthalpy of reaction is given by:

ΔrH⊖=∑ΔfH⊖(products)−∑ΔfH⊖(reactants)\Delta_r H^{\ominus} = \sum \Delta_f H^{\ominus}(\text{products}) - \sum \Delta_f H^{\ominus}(\text{reactants})

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

ΔrH⊖=[2ΔfH⊖(H2O,l)+2ΔfH⊖(SO2,g)]−[2ΔfH⊖(H2S,g)+3ΔfH⊖(O2,g)]\Delta_r H^{\ominus} = [2 \Delta_f H^{\ominus}(\text{H}_2\text{O}, l) + 2 \Delta_f H^{\ominus}(\text{SO}_2, g)] - [2 \Delta_f H^{\ominus}(\text{H}_2\text{S}, g) + 3 \Delta_f H^{\ominus}(\text{O}_2, g)]

Substituting the given values:
ΔrH⊖=[2(−286.0)+2(−297.0)]−[2(−20.1)+3(0)]\Delta_r H^{\ominus} = [2(-286.0) + 2(-297.0)] - [2(-20.1) + 3(0)]

ΔrH⊖=[−572.0−594.0]−[−40.2]\Delta_r H^{\ominus} = [-572.0 - 594.0] - [-40.2]

ΔrH⊖=−1166.0+40.2=−1125.8 kJ mol−1\Delta_r H^{\ominus} = -1166.0 + 40.2 = -1125.8 \text{ kJ mol}^{-1}

The magnitude of the enthalpy change is 1125.8 kJ mol−11125.8 \text{ kJ mol}^{-1}.

Rounding to the nearest integer, we get 11261126.

Answer: 11261126
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