Question

How are enthalpy changes expressed in chemical equations?

Answer 1

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:

`"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 `"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 `"CO"` with `"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

`"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 `"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 `"N"_2` to moles of `"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"`