# Why is delta-H negative here?

Very basic question here, but I'm confused why Delta-H seems to be the reverse of what I would expect for bond enthalpy. For example, here's a problem in my textbook where the goal is to find delta-H:

$$\ce{H2(g) + Br2(l) -> 2HBr(g)}$$

Using the table I have, I see that an $$\ce{H-H}$$ bond has an average enthalpy of 436, $$\ce{Br-Br}$$ has 193, and $$\ce{H-Br}$$ has 366. 436+193 = 629, and 2*366 = 732, so the reactants have a total of 629, compared to the products, which have 732.

My intuition (and Google, as far as I can tell) would tell me that, because the H value increased when the reaction occured, the answer would be positive 103. But my textbook says the answer is -103. This makes no sense to me - what am I missing here? Why do all the Delta-H (AKA Change in H) values seem to be negative what you'd expect?

• You can also handle the issue this way. In an exothermic reaction the enthalpy is negative which means the bonds formed (negative) outweigh the bonds formed (positive). Just add the two values. Apr 7, 2021 at 18:10

The bond enthalpy is conventionally the enthalpy change of bond breaking, not of bond forming, in the contrary to formation enthalpy of compounds. This way they are all positive.

The consequence is, the order of products and reactants in computation the enthalpy difference has to be switched.

So if the summary bond enthalpy of products is higher than of reactants, the difference of enthalpy as the reaction enthalpy is negative, not positive, and the reaction is exothermic.

Formation enthalpies:

$$\Delta H_\mathrm{r}=\Sigma \left( \Delta H_\mathrm{f, products} \right) - \Sigma \left( \Delta H_\mathrm{f, reactants} \right)$$

Bond enthalpies:

$$\Delta H_\mathrm{r}= \Sigma \left( \Delta H_\mathrm{b, reactants} \right) - \Sigma \left( \Delta H_\mathrm{b, products} \right)$$