Question

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 :

`2SO_2(g) + O_2(g) rightleftharpoons 2SO_3(g)`

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.

Answer 1

Well, we can always check our answer. To check, it should be `(-296.81 pm 0.20)` `"kJ/mol"`. You should use NIST more often.

I actually got `-"310.17 kJ/mol"` though.

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Supposedly, the `DeltaH_(rxn)^@` is `-"171.2 kJ"` for this reaction as-written:

`2"SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO"_3(g)`

You have to look up `DeltaH_f^@` for `"SO"_3(g)` first. I got `-"395.77 kJ/mol"` from NIST.

Now, `DeltaH_(rxn)^@` can be calculated from tabulated enthalpies of formation.

`DeltaH_(rxn)^@ = sum_P n_P DeltaH_(f,P)^@ - sum_R n_R DeltaH_(f,R)^@`

where `P` stands for products, `R` for reactants, `n` for the mols of stuff, and `DeltaH_f^@` is the enthalpy of forming 1 mol of the substance from its elements in their elemental states at `25^@ "C"` and `"1 bar"`.

We should know that `DeltaH_(f,O_2(g))^@ = "0 kJ/mol"`... no energy is needed to form something by doing nothing to it. We have `"O"_2(g)` 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:

`-"171.2 kJ" = -"791.54 kJ" - 2DeltaH_(f,SO_2(g))^@ "kJ"`

And so:

`color(blue)(DeltaH_(f,SO_2(g))^@ = -"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. :)