# 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) &-> 8SO2(g)} &\Delta H_\mathrm{f} &= \pu{-70.9 kcal mol^{-1}}\tag{1} \\ \ce{2SO2(g) + O2(g) &->[V2O5] 2SO3(g)} &\Delta H_\mathrm{f} &= \pu{-94.5 kcal mol^{-1}}\tag{2} \\ \ce{SO3(g) + H2O(g) &-> H2SO4(g)} &\Delta H_\mathrm{f} &= \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:

1. Summing all the enthalpies, I arrived at to $\Delta H_\mathrm{f} = \pu{-427.6 kcal mol^{-1}}$.
2. Using the data from equation $(3)$, I conclude that the answer is $\Delta H_\mathrm{f} = \pu{-193.9 kcal}$.
• The second point in what you have tried that is not $d$H$_f but only dH – DSinghvi May 11 '15 at 15:51 • @DSinghvi actually the question stated like this (given that the standard formation enthalpy of SO2, SO3, H2O, and H2SO4 consecutively are -70,9 kkal/mol, -94,5 kkal/mol, -68,3 kkal/mol, and -193,9 kkal/mol) I moved the enthalpy to the reaction. – Prajogo Atmaja May 11 '15 at 15:55 • @DSinghvi By the way, from your response, that means I have to conclude every steps In order for me to have the ΔHf of H2SO4? – Prajogo Atmaja May 11 '15 at 16:03 • what do you understand by$/Delta H_f/$– DSinghvi May 12 '15 at 7:08 ## 1 Answer Since enthalpy is a state function, all that is needed is the$\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:

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

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

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

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

Combining these equations yields:

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

(according to the equations and numerical values provided)