If $S + O_{2} ⟶ {SO}_{2}, Δ H = - 298.2 kJ {mole}^{- 1}$
$\begin{array}{l} {SO}_{2} + \frac{1}{2} O_{2} ⟶ {SO}_{3}, Δ H = - 98.7 kJ {mole}^{- 1} \\ {SO}_{3} + H_{2} O ⟶ H_{2} {SO}_{4}, Δ H = - 130.2 kJ {mole}^{- 1} \\ H_{2} + \frac{1}{\circ} O_{2} ⟶ H_{2} O, Δ H = - 287.3 kJ {mole}^{- 1} \end{array}$
The enthalpy of formation of $H_{2} {SO}_{4}$ at 298 K will be:

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$- 814.4 kJ {mole}^{- 1}$

$+ 814.4 kJ {mole}^{- 1}$

$- 650.3 kJ {mole}^{- 1}$

$- 433.7 kJ {mole}^{- 1}$

answer is A.

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Detailed Solution

The formation of $H_{2} {SO}_{4}$

$H_{2} + S + 2 O_{2} \to H_{7} {SO}_{4}$

enthalpy of formation of $H_{2} {SO}_{4}$ is equal to the sum of enthalpy of all other four equations,

$ΔH = - (298.2 + 98.7 + 130.2 + 287.3) = - 814.4 kJ / mole$

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