JEE Main 2026 — Chemistry Physical Chemistry

The standard enthalpy of reaction is given by:

$$\Delta_r H^{\ominus} = \sum \Delta_f H^{\ominus}(\text{products}) - \sum \Delta_f H^{\ominus}(\text{reactants})$$Δr​H⊖=∑Δf​H⊖(products)−∑Δf​H⊖(reactants)

For the given reaction:

$$2\text{H}_2\text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) + 2\text{SO}_2(g)$$2H2​S(g)+3O2​(g)→2H2​O(l)+2SO2​(g)

$$\Delta_r H^{\ominus} = [2 \Delta_f H^{\ominus}(\text{H}_2\text{O}, l) + 2 \Delta_f H^{\ominus}(\text{SO}_2, g)] - [2 \Delta_f H^{\ominus}(\text{H}_2\text{S}, g) + 3 \Delta_f H^{\ominus}(\text{O}_2, g)]$$Δr​H⊖=[2Δf​H⊖(H2​O,l)+2Δf​H⊖(SO2​,g)]−[2Δf​H⊖(H2​S,g)+3Δf​H⊖(O2​,g)]

Substituting the given values:

$$\Delta_r H^{\ominus} = [2(-286.0) + 2(-297.0)] - [2(-20.1) + 3(0)]$$Δr​H⊖=[2(−286.0)+2(−297.0)]−[2(−20.1)+3(0)]

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

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

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

Rounding to the nearest integer, we get $1126$1126.

Answer: $1126$1126