If $\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2;\mathrm{\Delta H}=398.2\mathrm{KJ}$ 

${\mathrm{SO}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{SO}}_3;\mathrm{\Delta H}=98.7\mathrm{KJ}$ 

${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O} \to {\mathrm{H}}_2{\mathrm{SO}}_4;\mathrm{\Delta H}=130.2\mathrm{KJ}$ 

${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O};\mathrm{\Delta H}=227.3\mathrm{KJ}$ 

The enthalpy of formation of sulphuric acid at 298 K will be

$\begin{array}{r}\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2\mathrm{\Delta H}=-398.2\mathrm{kJ}\dots ..\left(1\right)\\ {\mathrm{SO}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{SO}}_3\mathrm{\Delta H}=-98.7\mathrm{kJ}\dots ..2\\ {\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{\Delta H}=-130.2\mathrm{kJ}.....3\\ {\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}\mathrm{\Delta H}=-227.3\mathrm{kJ}\dots .4\end{array}$

 The required reaction is:

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

This reaction can be obtained by adding (i), (ii), (iii), and (iv).

The enthalpy of the reaction will be:

$\mathrm{\Delta H}=-398.2-98.7-130.2-227.3=-854.4\text{ KJ}$ 

Therefore, the correct option is A.