# What is the standard enthalpy of formation of SO2(g)?

## A scientist measures the standard enthalpy change for the following reaction to be -171.2 kJ : $2 S {O}_{2} \left(g\right) + {O}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 S {O}_{3} \left(g\right)$ Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of SO2(g) is ___ kJ/mol.

Jan 17, 2018

Well, we can always check our answer. To check, it should be $\left(- 296.81 \pm 0.20\right)$ $\text{kJ/mol}$. You should use NIST more often.

I actually got $- \text{310.17 kJ/mol}$ though.

Supposedly, the $\Delta {H}_{r x n}^{\circ}$ is $- \text{171.2 kJ}$ for this reaction as-written:

$2 {\text{SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO}}_{3} \left(g\right)$

You have to look up $\Delta {H}_{f}^{\circ}$ for ${\text{SO}}_{3} \left(g\right)$ first. I got $- \text{395.77 kJ/mol}$ from NIST.

Now, $\Delta {H}_{r x n}^{\circ}$ can be calculated from tabulated enthalpies of formation.

$\Delta {H}_{r x n}^{\circ} = {\sum}_{P} {n}_{P} \Delta {H}_{f , P}^{\circ} - {\sum}_{R} {n}_{R} \Delta {H}_{f , R}^{\circ}$

where $P$ stands for products, $R$ for reactants, $n$ for the mols of stuff, and $\Delta {H}_{f}^{\circ}$ is the enthalpy of forming 1 mol of the substance from its elements in their elemental states at ${25}^{\circ} \text{C}$ and $\text{1 bar}$.

We should know that $\Delta {H}_{f , {O}_{2} \left(g\right)}^{\circ} = \text{0 kJ/mol}$... no energy is needed to form something by doing nothing to it. We have ${\text{O}}_{2} \left(g\right)$ in the atmosphere naturally.

-"171.2 kJ" = ["2 mol" xx -"395.77 kJ/mol"] - ["2 mol" xx DeltaH_(f,SO_2(g))^@ + "1 mol" xx "0 kJ/mol"]

Resolving the units, we obtain:

$- \text{171.2 kJ" = -"791.54 kJ" - 2DeltaH_(f,SO_2(g))^@ "kJ}$

And so:

$\textcolor{b l u e}{\Delta {H}_{f , S {O}_{2} \left(g\right)}^{\circ} = - \text{310.17 kJ/mol}}$

Apparently, this is rather off from what NIST has (4.5% error). I wouldn't trust that scientist's measuring strategies. :)