# 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$_f but only dH Commented May 11, 2015 at 15:51 • @DSinghvi actually the question stated like this (given that the standard formation enthalpy of SO2, SO3, H2O, and H2SO4 consecutively are -70,9 kkal/mol, -94,5 kkal/mol, -68,3 kkal/mol, and -193,9 kkal/mol) I moved the enthalpy to the reaction. Commented May 11, 2015 at 15:55 • @DSinghvi By the way, from your response, that means I have to conclude every steps In order for me to have the ΔHf of H2SO4? Commented May 11, 2015 at 16:03 • what do you understand by$/Delta H_f/$Commented May 12, 2015 at 7:08 • The enthalpy of the second and third reaction should not be labeled formation enthalpy because the reactants at not elements (and in the second reaction, the stoichiometric coefficient of the product is different from one). – Karsten Commented Feb 20, 2020 at 16:41 ## 1 Answer Since enthalpy is a state function, all that is needed is the$\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)