So I am talking about complicated compounds and not simple reactions where you subtract heats of formation. I was trying to find the heat of reaction for an aryl substituted methyl aryl sulfide being oxidised to sulfone group. Now, I calculated BDE for S=O bond and subtracted O-O energy to find heat of reaction for oxidation to sulfoxide and then repeated the step for going to sulfone. To calculate that heat experimentally, I insulated my reaction vessel and bought it to reflux temperature of the solvent. I then added hydrogen peroxide and collected the evaporated solvent and measured its volume. I used its heat of vaporization as a proxy to calculate the heat of reaction and it came out to only 20% of the Heat of formation calculated from Hess's law

I would love to know where I went wrong and if my knowledge of Hess's law is wrong.

  • 1
    $\begingroup$ Try to summarize your procedure mathematically. Verbose description is hard to follow. $\endgroup$
    – Poutnik
    Nov 18, 2022 at 17:11
  • $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$
    – Poutnik
    Nov 18, 2022 at 17:12
  • $\begingroup$ I assume BDE = Bond dissociation Energy. $\endgroup$
    – Poutnik
    Nov 18, 2022 at 17:37
  • $\begingroup$ It might help if you explain how you did this in more detail: " I insulated my reaction vessel and bought it to reflux temperature of the solvent". How did you heat the vessel to the reflux point exactly? And then you added a reagent that caused a rapid reaction (adiabatic conditions) releasing heat into the vessel and causing boiling. You trapped the boiled solvent and from its volume determined the total heat released. Looks like a lot of possible heat losses may have occurred along the way? $\endgroup$
    – Buck Thorn
    Nov 19, 2022 at 9:05

1 Answer 1


The Hess' law is the particular application of the fundamental law of energy conservation, applied to the total enthalpy.

It says the enthalpy change depends only on initial and final state, but not on the path/way how the final state was reached.

Particularly if we have a complicated process $$\ce{A -> D},$$

which can be in reality or formally divided to steps in 2 alternative ways, like:

$$\ce{A -> B -> C -> D},$$

$$\ce{A -> E -> D},$$


$$\Delta_{\ce{A->D}} H = \Delta_{\ce{A->B}} H + \Delta_{\ce{B->C}} H + \Delta_{\ce{C->D}} H\\ = \Delta_{\ce{A->E}} H + \Delta_{\ce{E->D}} H$$

If it was not so, you could generate energy from nothing, by cyclical switching between 2 states along different ways, 1 way releasing more energy than the other would consume.

E.g.: $$\ce{A -> B -> C -> D -> E -> A}$$

If we consider your specific case

$$\ce{R_1-S-R_2(l?) + 2 H2O2(aq) -> R_1-(S=O)-R_2(l?) + 2 H2O(l)}$$

divided to formal reaction steps as:

\begin{align} \ce{2 H2O2(aq) &-> 4 ^{.}OH(aq)}\\ \ce{2 ^{.}OH(aq) &-> 2 ^{.}O^{.}(aq) + 2 ^{.}H(aq)}\\ \ce{2 ^{.}OH(aq) + 2 ^{.}H(aq) &-> 2 H2O(l)}\\ \ce{R_1-S-R_2(l?) + ^{.}O^{.}(aq) &-> R_1-(S=O)-R_2(l?)}\\ \ce{R_1-(S=O)-R_2(l?) + ^{.}O^{.}(aq) &-> R_1-(O=S=O)-R_2(l?)} \end{align}

then the reaction enthalpy would be — by the Hess law — the sum of reaction enthalpies of particular steps. It does not matter if the steps are formal or real, as the path does not matter.

$$\Delta_\mathrm{r} H = \sum_i{\Delta_{\mathrm{r},i} H}$$

  • $\begingroup$ I think you make sense and lemme just calculate the deltaH with your method, confirm on monday and I'll mark my question solved. $\endgroup$ Nov 19, 2022 at 17:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.