# Can absolute enthalpy values be used to determine reaction enthalpy?

Our teacher added this question to an exam, and there was a huge disagreement with my classmates and teacher about the correct answer. One side favoured answer 2, the other side favoured answer 4 being the false statement.

I personally selected answer 4, as it did contradict with the other 3: option 1 has $\Delta H < 0$, option 2 has $\Delta H = H_\mathrm{f} -H_\mathrm{i} = \pu{-500kJ}$ and option 3 has $\Delta H = \pu{-500 kJ}$, so option 4 contradicts the above. However I would like to hear your thoughts.

Which of the following options that refer to the enthalpy of a specific chemical reaction is not correct?

1. $H_\mathrm{i} > H_\mathrm{f}$
2. The enthalpy of the system has altered from the initial value of $\pu{1600 kJ}$ ($H_\mathrm{i}$) to the final value of $\pu{1100 kJ}$ ($H_\mathrm{f}$)
3. $\Delta H = \pu{-500 kJ}$
4. $\Delta H = \pu{500 kJ}$
• You can't measure an absolute value of enthalpy. – wolphram Oct 16 '17 at 20:38
• Notice also that 1, 3, and 4 all refer to delta values, explicitly or implicitly. – Zhe Oct 16 '17 at 20:43
• @wolphram That's a legend from dark times before relativity, another outdated theory needing debunking. chemistry.stackexchange.com/a/20103/9961 – Mithoron Oct 16 '17 at 22:57

Assuming that 3 are true, the only self-consistent set are 1,2 and 3.

Importantly, statements 3 and 4 contradict one another, so one of them must be the incorrect statement.

Consider a reaction with a change in enthalpy ($\Delta H$) of $\pu{-500 kJ/mol}$, and statement 3 is true:

\begin{align} \Delta H &= H_\mathrm{f} - H_\mathrm{i}\\ \implies H_\mathrm{i} &= H_\mathrm{f} - \Delta H\\ H_\mathrm{i} &= H_\mathrm{f} + \pu{500 kJ/mol}\\ \implies H_\mathrm{i} &> H_\mathrm{f} \end{align} We've shown that statement 1 is true also.

As wolphram says in the comments, you can't measure an absolute value of enthalpy, but you can measure it relative to an chosen zero enthalpy. If you chose zero enthalpy such that $H_\mathrm{i} = \pu{1600 kJ/mol}$, then by the first equation above we can say that: \begin{align} H_\mathrm{f} &= \Delta H + H_\mathrm{i}\\ \implies H_\mathrm{f} &= \pu{-500 kJ/mol} + \pu{1600 kJ/mol}\\ \implies H_\mathrm{f} &= \pu{1100 kJ/mol} \end{align} We've shown that statement 2 is true as well.

Statement 4 contradicts statement 3 (as well as the other two), so statement 4 must be incorrect.

• We still refer to the referenced values as deltas. All reported values are already referenced this way and those values are indeed reported as $\Delta H$ and not $H$. I agree that your answer and rationale are correct, but I have serious issues with the wording of the question then... – Zhe Oct 16 '17 at 21:54
• @Zhe The system itself can't have a $\Delta H$ without reference to something, but it can have a $\Delta H_{form}$ (of formation) which, as you state, would be a vastly more physically sensible thing to talk about that $H_f =1100$. I definitely agree that statement 2 is worded very badly and the general phrasing of the question ought to be serious rethought. – user213305 Oct 16 '17 at 21:58