Consider the reaction : $2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{SO}_{2}(\mathrm{g})$ The magnitude of enthalpy change for the reaction in $\mathrm{kJ}\,\mathrm{mol}^{-1}$ is ____. (Nearest integer) Given : $\Delta_{f} \mathrm{H}^{\ominus}\left(\mathrm{H}_{2} \mathrm{S}\right)=-20.1\,\mathrm{kJ}\,\mathrm{mol}^{-1}$ $\Delta_{f} \mathrm{H}^{\Theta}\left(\mathrm{H}_{2} \mathrm{O}\right)=-286.0\,\mathrm{kJ}\,\mathrm{mol}^{-1}$ $\Delta_{f} \mathrm{H}^{\Theta}\left(\mathrm{SO}_{2}\right)=-297.0\,\mathrm{kJ}\,\mathrm{mol}^{-1}$

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Chemical Thermodynamics

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^{⊖} (H_{2} S) = - 20.1 kJ mol^{- 1}

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

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

Answer: 1126

The reaction enthalpy is the sum of the standard enthalpies of formation of the products minus those of the reactants, with

Δ_{f} H^{⊖} (O_{2}) = 0

. Thus

Δ_{r} H^{⊖} = [2 (- 297.0) + 2 (- 286.0)] - [2 (- 20.1)] = (- 1166.0) - (- 40.2) = - 1125.8 kJ mo l^{- 1}

. Its magnitude is

1125.8

, which to the nearest integer is

1126 kJ mo l^{- 1}

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Integer answer