JEE Main 2026 Apr 8 Shift 2, Chemistry Q24: Hess’s Law and Bond Energy

JEE Main2026Apr 8, Shift 2Chemistry

Consider the reaction :

$2 H_{2} S (g) + 3 O_{2} (g) \to 2 H_{2} O (l) + 2 SO_{2} (g)$

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

Given : $Δ_{f} H^{\circ} (H_{2} S) = - 20.1$ kJ mol $^{- 1}$

$Δ_{f} H^{\circ} (H_{2} O) = - 286.0$ kJ mol $^{- 1}$

$Δ_{f} H^{\circ} (SO_{2}) = - 297.0$ kJ mol $^{- 1}$

Your answer

Solution

Use Hess's law: $Δ_{r} H^{\circ} = \sum ν_{p} Δ_{f} H^{\circ} (products) - \sum ν_{r} Δ_{f} H^{\circ} (reactants)$ .

O $_{2}$ (g) has $Δ_{f} H^{\circ} = 0$ (elemental standard state).

$Δ_{r} H^{\circ} = [2 (- 297) + 2 (- 286)] - [2 (- 20.1) + 3 (0)]$

$= (- 594 - 572) - (- 40.2) = - 1166 + 40.2 = - 1125.8$ kJ/mol.

$∣ Δ_{r} H^{\circ} ∣ \approx 1126$ kJ/mol.

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