Hess’s Law Example Problem

Hess’s law of constant heat summation, or Hess’s law for short is a relationship describing the enthalpy change of a reaction. The total enthalpy change of a reaction is the sum of total enthalpies for each step of the reaction and is independent of the order of the steps. Basically, calculate the total enthalpy by breaking a reaction down to simple component steps of known enthalpy values. This Hess’s Law example problem shows how to manipulate reactions and their enthalpy values to find the total change of enthalpy of a reaction.

First, there are a couple notes to keep straight before beginning.

1. If a reaction is reversed, the sign of the change in enthalpy (ΔH_f) changes.
   For example: the reaction C(s) + O₂(g) → CO₂(g) has an ΔH_f of -393.5 kJ/mol.
   The reverse reaction CO₂(g) → C(s) + O₂(g) has a ΔH_f of +393.5 kJ/mol.
2. If a reaction is multiplied by a constant, the change in enthalpy is changed by the same constant.
   Example, for the previous reaction, if three times the reactants are allowed to react, ΔH_f is changed by three times.
3. If ΔH_f is positive, the reaction is endothermic. If ΔH_f is negative, the reaction is exothermic.

Hess’s Law Example Problem

Question: Find the enthalpy change for the reaction

CS₂(l) + 3 O₂(g) → CO₂(g) + 2 SO₂(g)
when:
C(s) + O₂(g) → CO₂(g); ΔH_f = -393.5 kJ/mol
S(s) + O₂(g) → SO₂(g); ΔH_f = -296.8 kJ/mol
C(s) + 2 S(s) → CS₂(l); ΔH_f = 87.9 kJ/mol

Solution: Hess’s Law problems can take a little trial and error to get started. One of the best places to begin is with a reaction with only one mole of reactant or product in the reaction.

Our reaction needs one CO₂ in the product and the first reaction also has one CO₂ product.

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

This reaction gives us the CO₂ needed on the product side and one of the O₂ needed on the reactant side. The other two O₂ can be found in the second reaction.

S(s) + O₂(g) → SO₂(g) ΔH_f = -296.8 kJ/mol

Since only one O₂ is in the reaction, multiply the reaction by two to get the second O₂. This doubles the ΔH_f value.

2 S(s) + 2 O₂(g) → 2 SO₂(g) ΔH_f = -593.6 kJ/mol

Combining these equations gives

2 S(s) + C(s) + 3 O₂(g) → CO₂(g) + SO₂(g)

The enthalpy change is the sum of the two reactions: ΔH_f = -393.5 kJ/mol + -593.6 kJ/mol = -987.1 kJ/mol

This equation has the product side needed in the problem but contains an extra two S and one C atom on the reactant side. Fortunately, the third equation has the same atoms. If the reaction is reversed, these atoms are on the product side. When the reaction is reversed, the sign of the change in enthalpy is reversed.

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

Add these two reactions together and the extra S and C atoms cancel out. The remaining reaction is the reaction needed in the question. Since the reactions were added together, their ΔH_f values are added together.

ΔH_f = -987.1 kJ/mol + -87.9 kJ/mol
ΔH_f = -1075 kJ/mol

Answer: The change in enthalpy for the reaction

CS₂(l) + 3 O₂(g) → CO₂(g) + 2 SO₂(g)

is ΔH_f = -1075 kJ/mol.

Hess’s law problems require reassembling the component reactions until the needed reaction is achieved. While Hess’s law applies to changes in enthalpy, this law can be used for other thermodynamic state equations such as Gibbs energy and entropy.