# How to determine the sign of the enthalpy of formation?

During the formation of $\ce{Al2O3}$ from 5.4 grams of $\ce{Al}$ and enough oxygen, the temperature of 2 kg of water climbs by 20 degrees. What’s the enthalpy of formation of $\ce{Al2O3}$ (per mole) ?

What I tried:

The balanced reaction:

$$\ce{4Al + 3 O2 -> 2 Al2O3}$$

5.4 grams of $\ce{Al}$ are 0.2 moles. If 2 kg of water climbed by 20 degrees then 40 000 calories were invested during the reaction. To get one mole of $\ce{Al2O3}$ we need 2 moles of $\ce{Al}$ , therefore the enthalpy of formation (per mole) of $\ce{Al2O3}$ is 400 000 calories. However, the correct answer is −400 000.

Can someone please explain the negative sign?

$\ce{Al2O3}$, upon the formation of its lattice, releases heat. This heats up water. The increase in the temperature is in water, not $\ce{Al2O3}$. We usually refer to the water as the surroundings and $\ce{Al2O3}$ as the thermodynamic system.
So, upon the formation of lattice (among with other, mostly, endothermic reactions) a lot of heat is released and that's what you have estimated. Since $\ce{Al2O3}$ is the system, it loses heat. Thus, $\Delta H_f$ is negative.