Thermochemical equations are just like other balanced equations except they also specify the heat flow for the reaction. The heat flow is listed to the right of the equation using the symbol ΔH. The most common units are kilojoules, kJ. Here are two thermochemical equations:

H_{2} (g) + ½ O_{2} (g) → H_{2}O (l); ΔH = -285.8 kJ

HgO (s) → Hg (l) + ½ O_{2} (g); ΔH = +90.7 kJ

### Writing Thermochemical Equations

When you write thermochemical equations, be sure to keep the following points in mind:

- Coefficients refer to the number of moles. Thus, for the first equation, -282.8 kJ is the ΔH when 1 mol of H
_{2}O (l) is formed from 1 mol H_{2}(g) and ½ mol O_{2}. - Enthalpy changes for a phase change, so the enthalpy of a substance depends on whether is it is a solid, liquid, or gas. Be sure to specify the phase of the reactants and products using (s), (l), or (g) and be sure to look up the correct ΔH from the heat of formation tables. The symbol (aq) is used for species in water (aqueous) solution.
- The enthalpy of a substance depends upon temperature. Ideally, you should specify the temperature at which a reaction is carried out. When you look at a table of heats of formation, notice that the temperature of the ΔH is given. For homework problems, and unless otherwise specified, the temperature is assumed to be 25°C. In the real world, the temperature may different and thermochemical calculations can be more difficult.

### Properties of Thermochemical Equations

Certain laws or rules apply when using thermochemical equations:

**ΔH is directly proportional to the quantity of a substance that reacts or is produced by a reaction.**Enthalpy is directly proportional to mass. Therefore, if you double the coefficients in an equation, then the value of ΔH is multiplied by two. For example:- H
_{2}(g) + ½ O_{2}(g) → H_{2}O (l); ΔH = -285.8 kJ - 2 H
_{2}(g) + O_{2}(g) → 2 H_{2}O (l); ΔH = -571.6 kJ

- H
**ΔH for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction.**For example:- HgO (s) → Hg (l) + ½ O
_{2}(g); ΔH = +90.7 kJ - Hg (l) + ½ O
_{2}(l) → HgO (s); ΔH = -90.7 kJ - This law is commonly applied to phase changes, although it is true when you reverse any thermochemical reaction.

- HgO (s) → Hg (l) + ½ O
**ΔH is independent of the number of steps involved.**This rule is called**Hess's Law**. It states that ΔH for a reaction is the same whether it occurs in one step or in a series of steps. Another way to look at it is to remember that ΔH is a state property, so it must be independent of the path of a reaction.- If Reaction (1) + Reaction (2) = Reaction (3), then ΔH
_{3}= ΔH_{1}+ ΔH_{2}

- If Reaction (1) + Reaction (2) = Reaction (3), then ΔH