Chemistry 30 Thermodynamics: Hess's Law

2.5 Hess's Law - The Equation

There is another way to calculate enthalpy changes that occur during chemical reactions that is based on the principal of Hess's Law.

ΔH for a reaction may be calculated using the published ΔH_fvalues and the equation:

ΔH = ∑ΔH_products – ∑ΔH_reactants

You may not be familiar with the ∑ symbol.
It stands for "summation" or "the sum of".
To find ΔH^o for the reaction, add together all of the heats of formation, ΔH^o_f, for all of the products and subtract from that the sum of the heats of formation of all of the reactants.

Using this formula is easy. With your Table of Thermochemical Data table handy, locate ΔH_f values for all reactants and products in the reaction. The physical state is important, so check that carefully. Also pay close attention to the balancing coefficients in the equation, as you must multiply the ΔH_f values by these coefficients. Finally, be very careful with + and - values.

I suggest that you begin by writing all the ΔH_f values directly below all participants in the equation.

Example:

Using a Table of Thermochemical Data, calculate ΔH for the combustion of benzene, C₆H₆, as shown by the following reaction:

C₆H₆ (l) + 15/2 O₂ (g) → 6 CO₂ (g) + 3 H₂O (l)

Solution:

Remember that ΔH for any pure element = 0. (some exceptions)
Look up Δ H values for C₆H₆ (l) , CO₂(g), and H₂O(l).
Remember, these ΔH values are given for 1 mole. In our final reaction;
There are 6 mole of CO₂, so multiply ΔH by 6.
There are 3 mole of H₂O, so multiply ΔH by 3.

C₆H₆ (l)	ΔH = +49.0
CO₂ (g)	ΔH = -393.5
H₂O (l)	ΔH = -285.8

Using the formula ΔH = ∑ΔH_products - ∑ΔH_reactants

C₆H₆ (l)

+

15/2 O₂ (g)

→

6 CO₂ (g)

+

3 H₂O (l)

49.0

+

15/2 × (0)

6 × (-393.5)

+

3 × (-285.8)

49.0

-3218.4

Reactants

Products

ΔH = ∑ΔH_products – ∑ΔH_reactants

ΔH =-3218.4 – (+49.0)

ΔH = -3267.4 kJ answer

Common Sources of Error

While the formula is easy to use, it is also very easy to make some simple mistakes. Students often make mistakes in the following areas, so be extra careful in these areas:

Forgetting to multiply ΔH values by the appropriate coefficient.

Using the wrong value of ΔH for water:
ΔH_f° for H₂O(l) = -285.8 kJ/mol;
ΔH_f° for H₂O(g) = -241.8 kJ/mol

Solving for ΔH as "Reactants - Products" instead of "Products – Reactants".

Accidentally changing the sign for ΔH.

Lab Activity

Heats of Reaction - Hess's Law

Chemistry 30

Thermodynamics

Teacher Resources

Other Links

Key Concepts

2.5 Hess's Law - The Equation

C₆H₆ (l)	+	15/2 O₂ (g)	→	6 CO₂ (g)	+	3 H₂O (l)
49.0	+	15/2 × (0)		6 × (-393.5)	+	3 × (-285.8)

49.0				-3218.4
Reactants				Products
		ΔH = ∑ΔH_products – ∑ΔH_reactants
		ΔH =-3218.4 – (+49.0)
		ΔH = -3267.4 kJ answer