i) $\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2 \mathrm{\Delta H}=-298.2 \mathrm{kJ}$

ii) $S{\text{O}}_2+1/2{\mathrm{O}}_2\to {\mathrm{SO}}_3 \mathrm{\Delta H}=-98.7 \mathrm{kJ}$

iii) ${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4 \mathrm{\Delta H}=-130.2 \mathrm{kJ}$

iii) ${\mathrm{H}}_2+1/2{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O} \mathrm{\Delta H}=-287.3 \mathrm{kJ}$

The enthalpy of formation of ${\mathrm{H}}_2{\mathrm{SO}}_4$ at $298\mathrm{ }\mathrm{K}$ is:

Enthalpy of formation of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is represented by the reaction as follows:

${\mathrm{H}}_2+\mathrm{S}+2{\mathrm{O}}_2\to {\mathrm{H}}_2{\mathrm{SO}}_4$

From the given reactions,

$\begin{array}{l}\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2{\mathrm{\Delta H}}_1=-298.2 \mathrm{kJ}\\ {\mathrm{SO}}_2+1/2{\mathrm{O}}_2\to {\mathrm{SO}}_3{\mathrm{\Delta H}}_2=-98.7 \mathrm{kJ}\\ {\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4{\mathrm{\Delta H}}_3=-130.2 \mathrm{kJ}\\ {\mathrm{H}}_2+1/2{\mathrm{O}}_2\to {\mathrm{H}}_2{\mathrm{O\Delta H}}_4=-287.3 \mathrm{kJ}\end{array}$

On (i) $+($ ii $)+($ iii $)+($ iv) we get,

$\begin{array}{c}{\mathrm{H}}_2+\mathrm{S}+2{\mathrm{O}}_2\to {\mathrm{H}}_2{\mathrm{SO}}_4 \\ \mathrm{\Delta H}={\mathrm{\Delta H}}_1+{\mathrm{\Delta H}}_2+{\mathrm{\Delta H}}_3+{\mathrm{\Delta H}}_4\\ \Rightarrow \mathrm{\Delta H}=-298.2-98.7-130.2-287.3\\ \Rightarrow \mathrm{\Delta H}=-814.4\end{array}$

Hence, option (A) is correct.