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Hess's Law Example Problem

Calculating Enthalpy Changes using Hess's Law


Enthalpy is a measure of heat of a reaction.

Enthalpy is a measure of heat of a reaction.

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This example problem demonstrates strategies and how to use Hess's Law to find the enthalpy change of a reaction using enthalpy data from similar reactions.


What is the value for ΔH for the following reaction?

CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g)

C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol
C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol


Hess's law says the total enthalpy change does not rely on the path taken from beginning to end. Enthalpy can be calculated in one grand step or multiple smaller steps.

To solve this type of problem, we need to organize the given chemical reactions where the total effect yields the reaction needed. There are a few rules that must be followed when manipulating a reaction.
  1. The reaction can be reversed. This will change the sign of ΔHf.
  2. The reaction can be multiplied by a constant. The value of ΔHf must be multiplied by the same constant.
  3. Any combination of the first two rules may be used.
Finding a correct path is different for each Hess's law problem and may require some trial and error. A good place to start is to find one of the reactants or products where there is only one mole in the reaction.

We need one CO2 and the first reaction has one CO2 on the product side.

C(s) + O2(g) → CO2(g), ΔHf = -393.5 kJ/mol

This gives us the CO2 we need on the product side and one of the O2 moles we need on the reactant side.

To get two more O2 moles, use the second equation and multiply it by two. Remember to multiply the ΔHf by two as well.

2 S(s) + 2 O2(g) → 2 SO2(g), ΔHf = 2(-326.8 kJ/mol)

Now we have two extra S and one extra C molecule on the reactant side we don't need. The third reaction also has two S and one C on the reactant side. Reverse this reaction to bring the molecules to the product side. Remember to change the sign on ΔHf.

CS2(l) → C(s) + 2 S(s), ΔHf = -87.9 kJ/mol

When all three reactions are added, the extra two sulfur and one extra carbon atoms are cancelled out, leaving the target reaction. All that remains is adding up the values of ΔHf

ΔH = -393.5 kJ/mol + 2(-296.8 kJ/mol) + (-87.9 kJ/mol)
ΔH = -393.5 kJ/mol - 593.6 kJ/mol - 87.9 kJ/mol
ΔH = -1075.0 kJ/mol


The change in enthalpy for the reaction is -1075.0 kJ/mol.

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