# 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 (ΔHf) changes.
For example: the reaction C(s) + O2(g) → CO2(g) has an ΔHf of -393.5 kJ/mol.
The reverse reaction CO2(g) → C(s) + O2(g) has a ΔHf 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, ΔHis changed by three times.
3. If ΔHf is positive, the reaction is endothermic. If ΔHf 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); 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

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) ΔHf = -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) ΔHf = -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 ΔHf value.

2 S(s) + 2 O2(g) → 2 SO2(g) ΔHf = -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: ΔHf = -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); ΔHf = -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 ΔHf values are added together.

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

Answer: The change in enthalpy for the reaction

CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)

is ΔHf = -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.

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