The enthalpies of combustion of $S (s), {SO}_{2} (g)$ and $H_{2} (g)$ are $- 298.2, - 98.7$ and $- 287.3 kJ {mol}^{- 1}$ . If the enthalpy of reaction ${SO}_{3} (g) + H_{2} O (l) ⟶ H_{2} {SO}_{4} (l)$ is $- 130.2 kJ {mol}^{- 1}$ the enthalpy of formation of $H_{2} {SO}_{4} (1)$ would be

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

$- 554.2 kJ {mol}^{- 1}$

$- 814.4 kJ {mol}^{- I}$

$- 435.5 {kJmol}^{- 1}$

answer is A.

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Detailed Solution

Given data are

$\begin{array}{ll} S (s) + O_{2} (g) ⟶ {SO}_{2} (g) Δ H = - 298.2 kJ {mol}^{- 1} \\ {SO}_{2} + \frac{1}{2} O_{2} (g) ⟶ {SO}_{3} (g) Δ H = - 98.7 kJ {mol}^{- 1} \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l) Δ H = - 287.3 kJ {mol}^{- 1} \\ {SO}_{3} (g) + H_{2} O (l) ⟶ H_{2} {SO}_{4} (l) Δ H = - 130.2 kJ {mol}^{- 1} \end{array}$

The formation of $H_{2} {SO}_{4} (l)$ implies the reaction $H_{2} (g) + S (s) + 2 O_{2} (g) ⟶ H_{2} {SO}_{4} (l)$

This reaction is obtained by adding the above four reactions. Hence, the enthalpy of formation of $H_{2} {SO}_{4} (1)$ is $Δ H = - (298.2 + 98.7 + 287.3 + 130.2) kJ {mol}^{- 1} = - 814.4 kJ {mol}^{- 1}$