How are enthalpy changes expressed in chemical equations?

1 Answer
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:

"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"