What is the enthalpy of formation of 1 mol of sulfuric acid? In the contact process, the commercial process of producing $\ce{H2SO4}$, several reactions occur before the final product is produced:
  
  \begin{align}
\ce{S8(s) + 8O2(g) &amp;-&gt; 8SO2(g)} &amp;\Delta H_\mathrm{f} &amp;= \pu{-70.9 kcal mol^{-1}}\tag{1} \\
\ce{2SO2(g) + O2(g) &amp;-&gt;[V2O5] 2SO3(g)} &amp;\Delta H_\mathrm{f} &amp;= \pu{-94.5 kcal mol^{-1}}\tag{2} \\
\ce{SO3(g) + H2O(g) &amp;-&gt; H2SO4(g)} &amp;\Delta H_\mathrm{f} &amp;= \pu{-193.9 kcal mol^{-1}}\tag{3}
\end{align}
  
  Given that $\Delta H_\mathrm{f}$ for $\ce{H2O(g)}$ is $\pu{-68.3 kcal mol^{-1}}$, what is the enthalpy of formation of $\pu{1 mol}$ of $\ce{H2SO4}$?


What I've tried to do:


Summing all the enthalpies, I arrived at to $\Delta H_\mathrm{f} = \pu{-427.6 kcal mol^{-1}}$.
Using the data from equation $(3)$, I conclude that the answer is $\Delta H_\mathrm{f} = \pu{-193.9 kcal}$.
Since enthalpy is a state function, all that is needed is the $\Delta H_{\mathrm{f}}$ of each reactants that form $\ce{H2SO4}$. Multiply the equations through by the correct fractions (including the enthalpies) to result in the correct coefficients and enthalpy for the desired reaction:

$$\ce{SO3{(g)} +H2O{(g)} -&gt; H2SO4{(g)}}~~~~~\Delta H_{\mathrm{f}}=-193.9~\mathrm{kcal~mol^{-1}}$$

$$\ce{H2{(g)} + \frac{1}{2}O2{(g)}-&gt;H2O{(g)}}~~~~~\Delta H_{\mathrm{f}}=-68.3~\mathrm{kcal~mol^{-1}}$$

$$\ce{SO2{(g)} + \frac{1}{2}O2{(g)} \xrightarrow{\ce{V2O5}} SO3{(g)}}~~~~~\Delta H_{\mathrm{f}}=-47.25~\mathrm{kcal~mol^{-1}}$$

$$\ce{\frac{1}{8}S8{(s)} +O2{(g)} -&gt; SO2{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-8.8625~\mathrm{kcal~mol^{-1}}$$

Combining these equations yields:

$$\ce{\frac{1}{8}S8{(s)} +2O2{(g)} +H2{(g)} -&gt; H2SO4{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-318~\mathrm{kcal~mol^{-1}}$$

(according to the equations and numerical values provided)