# Hess’s law type 1

Some reactions’ enthalpy changes are difficult to measure, usually because they are very slow (e.g., iron rusting). Instead of measuring it directly, we can use Hess’s law.

The net changes in the properties of a system are independent of the way the system changes from the initial state to the final state. We can apply this like so: if a set of component reactions add up to equal a target reaction, then the components’ enthalpy changes must add up to equal the target’s enthalpy change:

$\displaystyle \Delta{} H_{\text{target}} = \sum \Delta{} H_{\text{known}}$.

## Example

What is the enthalpy change for the formation of two moles of nitrogen monoxide given the following reference equations?

½ N_{2(g)} + O_{2(g)} → NO_{2(g)},
$\displaystyle \Delta{} H_{1} = 34 \, \text{kJ}$

½ O_{2(g)} + NO_{(g)} → NO_{2(g)},
$\displaystyle \Delta{} H_{2} = - 56 \, \text{kJ}$

First, we should identify our target equation: N_{2(g)} + O_{2(g)} → 2 NO_{(g)}. We need one mole of N_{2} on the left side, so we need to multiply the first reference equation by 2. We need two moles of NO on the right side, so we must multiply the second equation by −2 (multiplying by a negative flips the equation). Remember to multiply the $\displaystyle \Delta{} H$ values as well.

N_{2(g)} + ~~2~~ O_{2(g)} → ~~2
NO~~,
$\displaystyle \Delta{} H_{1} = 68 \, \text{kJ}$_{2(g)}

~~2 NO~~ → _{2(g)}~~O~~ + 2
NO_{2(g)}_{(g)},
$\displaystyle \Delta{} H_{2} = 112 \, \text{kJ}$

When we add these, nitrogen dioxide and one mole of oxygen cancel out, leaving us with our target equation with the correct enthalpy change:

N_{2(g)} + O_{2(g)} → 2 NO_{(g)},
$\displaystyle \Delta{} H = 180 \, \text{kJ}$.