# How are enthalpy changes expressed in chemical equations?

Dec 28, 2013

The enthalpy change for a given reaction depends on the stoichiometry of the reaction.

#### Explanation:

For example, 2 mol of carbon monoxide react with 2 mol of nitrogen monoxide according to the balanced equation:

$\text{2CO(g)" + "2NO(g)" → "2CO"_2(g) + "N"_2"(g); ΔH = -746 kJ}$

The process is exothermic, releasing 746 kJ of energy. This amounts to 373 kJ of energy per mole of $\text{CO}$.

Consider now the reaction of 0.250 mol carbon monoxide.

We can calculate the amount of energy released by using the enthalpy of reaction (ΔH = "-373 kJ/mol CO").

Since enthalpy $H$ is an extensive variable, it depends on the amount of substance present.

Therefore, a change in enthalpy ΔH also depends on the amount of substance that reacts.

The reaction of 0.250 mol of $\text{CO}$ with $\text{NO}$, for example, results in the release of

0.250 color(red)(cancel(color(black)("mol CO"))) × "373 kJ"/(1 color(red)(cancel(color(black)("mol CO")))) = "93.3 kJ"

Now consider the reaction

$\text{N"_2 + "O"_2 → "2NO"; ΔH = "+180.6 kJ}$

a) What is the enthalpy change for the formation of one mole of nitrogen monoxide?

Here we use the conversion factor $\text{180.6 kJ"/"2 mol NO}$.

ΔH = 1 color(red)(cancel(color(black)("mol NO"))) × "180.6 kJ"/(2 color(red)(cancel(color(black)("mol NO")))) = "95.30 kJ"

b) What is the enthalpy change for the reaction of 100.0 g of nitrogen with excess oxygen?

Here, we need to convert grams of ${\text{N}}_{2}$ to moles of ${\text{N}}_{2}$ and use the conversion factor "180.6 kJ"/("1 mol N"_2).

100.0 color(red)(cancel(color(black)("g N"_2))) × (1 color(red)(cancel(color(black)("mol N"_2))))/(28.01 color(red)(cancel(color(black)("g N"_2)))) × "180.6 kJ"/(1 color(red)(cancel(color(black)("mol N"_2)))) "= 1164 kJ"