# Enthalpy of formation and reaction equations

I'm disagreeing with my chemistry teacher over this enthalpy of formation problem, so can someone tell me if I am right or why I am wrong?

The enthalpy of formation of gaseous sulfur trioxide is $-396~\mathrm{kJ/mol}$. What is the enthalpy of the reaction represented by the following balanced equation?

$$\ce {2 SO_3 (g) -> 2 S (s) + 3 O2 (g) }$$

(A) $-396~\mathrm{kJ}$   (B) $396~\mathrm{kJ}$   (C) $792~\mathrm{kJ}$   (D) $-792~\mathrm{kJ}$   (E) $198~\mathrm{kJ}$

I got (C) by flipping the sign of the enthalpy of formation given (because the equation given proceeds in the opposite direction from the formation of sulfur trioxide) and multiplying it by two (because there are 2 moles of sulfur trioxide).
My teacher and most of my other classmates said that it was (B) instead. They said that I shouldn't multiply by 2 because the equation given is at it's lowest realistic coefficient values.
I don't understand this for a few reasons:

1. the enthalpy of formation is defined as the enthalpy change per mole of substance
2. in previous problems, we would change the chemical equation to make the product only have one mole

For example, here's one problem:

For which of the following reactions is the heat of reaction equal to the heat of formation of the product? \begin{align} \text{(A)} &&\ce{N_2 (g) + 3 H_2 (g) &-> 2 NH_3 (g)}\\ \text{(B)} &&\ce{1/2 N_2 (g) + O_2 (g) &-> NO_2 (g) }\\ \end{align}

We narrowed it already down to (A) and (B).

My teacher said the answer was (B), because while both equations had the elements in their elemental form, the first reaction produces 2 moles of product while the second produces 1 mole of product. Thus the fraction coefficients were necessary.

• I agree with you, but it has been some time since I needed to have the correct answer for this type of question.
– LDC3
Oct 22, 2014 at 1:32
• Wait wait hold up a second. The standard enthalpy of formation is defined as the enthalpy of formation of one mole of whatever. You simply said enthalpy of formation. That could refer to 1 mole, 2 moles, 1000 moles, or 1/10000 moles. I ctrl+F'ed your post and I don't see the word "standard" anywhere, suggesting that you aren't actually referring to the std. enthalpy of formation. Oct 22, 2014 at 2:09
• I assumed it was standard because it came from the chart "Thermodynamic Quantities for Selected Substances at 298.15 K and 1 atm" in the appendix of our book. But even if I can't assume that, the enthalpy value given is in the format of -396 kJ for every 1 mole of sulfur trioxide.
– 2mia
Oct 22, 2014 at 2:37
• If your teacher asks a similar question in the test you should ask whether the given value is the enthalpy of formation or the standard enthalpy of formation. If it is the former, then the question is ambiguous because without giving you the reaction equation for the formation process you can only guess the stoichiometric coefficients. If it is the latter then your reasoning is exactly right. Oct 22, 2014 at 2:48
• @2mia Yes, you may have been given standard thermodynamic data, but that doesn't mean you can't calculate non-standard thermodynamic information. Oct 22, 2014 at 13:09

As far as I can tell, your reasoning is sound, though the question is actually not written spectacularly well. By "enthalpy of the reaction" it seems to be implicitly assuming that the stoichiometric numbers are referring to moles (making all the answers have units of kJ instead of kJ/mol). The question doesn't specify standard enthalpy change of reaction, but even if it did, the most common convention for balancing equations is to use the smallest whole numbers for stoichiometric coefficients, which is what is written there and would give the answer you got. You can calculate an enthalpy change of reaction for any reaction conditions you like and the value you get will change depending on the quantities of the substances involved, but if the question is implicitly assuming moles for coefficients, you still get the same answer.

Incidentally, there's no reason you couldn't use some other definition of "standard" in which you set the coefficients such that the smallest ones are equal to 1, though I can't remember seeing anyone do it this way. IUPAC convention is that whenever a quantity referring to some standard state is used, the conditions of the standard should be specified (though chemists like to ignore IUPAC).

• To clarify, I'm not saying all the coefficients should be divided by two for the equation she gave. I'm saying that the enthalpy change she gave applies to an equation where the whole numbers aren't coefficents, specifically the equation $\ce{ S(s) + 3/2O2(g) -> SO3(g)}$. I infer that that enthalpy value applies to that equation because it is given as -396kJ/mol $\ce{SO3}$
– 2mia
Oct 24, 2014 at 1:10
• If the enthalpy of formation of $\ce{SO3}$ is given in kJ/mol, i.e. as the standard enthalpy of formation, then the only way for the answer to be (B) is to divide all the coefficients. Hess' law says that the difference of the enthalpies of formation multiplied by the coefficient on each side of the equation is equal to the reaction enthalpy change. Since the right side is 0, the only way to get (B) is to halve all the coefficients. Oct 24, 2014 at 2:41

I found an enthalpy problem set PDF file: $$\ce {2 S (s) + 3 O2 (g) -> 2 SO3 (g)}\qquad \Delta H = -794.45 \, \mathrm{kJ}$$

So the best answer is C.

After reading all the answers and comments, I have noticed that the question asks for the enthalpy of the reaction. Since the reverse reaction has a value of $\pu{-396 kJ/mol}$, and the current reaction has 2 moles of $\ce{SO3}$, then the answer is $\pu{792 kJ}$ (the energy for the total reaction). The answer depends on the number of moles in the reaction

While ambiguously worded, if by the "enthalpy of reaction" the teacher is implying the standard enthalpy of reaction, this being the energy to form one mole of product, then by the most common usage of the words used in the problem, "B" is a better solution than "C", as it utilizes the more precise definition of energy per mole of formation. However as the word "standard" was never used you certainly have a valid argument as to why "C" should also be considered correct.

• Energy per mole of formation of what? There are two products with different coefficients. You cannot calculate a single value that would work for both. Without being specifically instructed to calculate the enthalpy of reaction for one mole of one of the two products, it seems illogical to arbitrarily divide all the coefficients by 2 rather than use the balanced, whole number equation as written. Oct 22, 2014 at 23:14