AP®ChemistryCollege BoardStudy GuidesUnit 6: ThermochemistryEnthalpy ChangesHess's Law

Hess's Law (College Board AP® Chemistry): Study Guide

Philippa Platt

Written by: Philippa Platt

Reviewed by: Stewart Hird

Updated on

Hess's law

  • Hess’s law (Germain Hess, 1840) states that the enthalpy change for a reaction is the same whatever route is taken

  • Hess’s law follows from the first law of thermodynamics because energy is conserved

    • The enthalpy change for an overall process equals the sum of the enthalpy changes of its steps

  • This means that in an isolated system, the total amount of energy is constant

  • Hess’s law can be used to calculate the standard enthalpy change of a reaction from known standard enthalpy changes

  • Hess’s Law tells us that:

If the coefficients of a chemical equation are all multiplied by a constant, the ΔH°r is multiplied by that same constant

If two or more equations are added together to obtain an overall reaction, the heats of these equations are also added to give the heat of the overall reaction

  • This means that whether the reaction takes place in one or two steps, the total enthalpy change of the reaction will still be the same

Reversing a reaction

  • The combustion of 1 mole of propane, with water formed as a gas, has an enthalpy change of −2044 kJ

    • The reverse reaction cannot be carried out in one step, but its enthalpy change must be +2044 kJ

    • ΔH°r forward reaction = – ΔH°r reverse reaction 

Combustion of propane

Energy diagram for propane: CH3CH2CH3 + 5O2 on the higher level and 3CO2 + 4H2O on the lower level; a downward arrow labeled heat of combustion, delta H = −2044 kJ, and an upward arrow labeled heat of synthesis, delta H = +2044 kJ

The heat of reaction has the same magnitude for a reaction in both the forward or reverse reaction. In one direction the enthalpy change will be positive and the other direction the enthalpy change will be negative

  • If 2 moles of propane are burned in oxygen the enthalpy change will also double

2CH3CH2CH3 (g) + 10O2 (g) → 6CO2 (g) + 8H2O (g) ΔH°rxn = −4088 kJ/molrxn

  • If 0.5 moles of propane are burned then the enthalpy change will be half

½CH3CH2CH3 (g) + 2½O2 (g) → 1½CO2 (g) + 2H2O (g) ΔH°rxn = −1022 kJ/molrxn

Worked Example

Consider the following reactions:

  • N2 (g) + O2 (g) → 2NO (g)   ΔH = +180 kJ/molrxn

  • 2NO2 (g) → 2NO (g) + O2 (g)   ΔH = +112 kJ/molrxn

What is the ΔH value, in kJ/molrxn, for the following reaction?

N2 (g) + 2O2 (g) → 2NO2 (g)

Answer:

  • Step 1: Identify which given equation contains the product you want:

    • This equation contains the desired product on the left side:

2NO2 (g) → 2NO (g) + O2 (g)   ΔH = +112 kJ/molrxn

  • Step 2: Adjust the equation if necessary, to give the same product:

    • If you reverse it, reverse the ΔH value

2NO (g) + O2 (g) → 2NO2 (g)   ΔH = −112 kJ/molrxn

  • Step 3:  Adjust the equation if necessary, to give the same number of moles of product:

    • The equation contains the same number of moles as in the question, so no need to adjust the moles

  • Step 4: Identify which given equation contains your reactant:

N2 (g) + O2 (g) → 2NO (g)   ΔH = +180 kJ/molrxn

  • Step 5: Adjust the equation if necessary, to give the same reactant. If you reverse it, reverse the ΔH value:

    • There is no need to reverse it as the reactant is already on the left side

  • Step 6: Adjust the equation if necessary, to give the same number of moles of reactant

  • Step 7: Add the two equations together

N2 (g) + O2 (g) → 2NO (g)   ΔH = +180 kJ/molrxn

2NO (g) + O2 (g) → 2NO2 (g)   ΔH = −112 kJ/molrxn

  • Step 8: Cancel the common items

N2 (g) + O2 (g) + 2NO (g) + O2 (g) → 2NO (g) + 2NO2 (g)

  • Step 9: Add the two ΔH values together to get the one you want

N2 (g) + 2O2 (g) → 2NO2 (g)   ΔH = +180 − 112 = +68 kJ/molrxn

Worked Example

Consider the following reactions

  • C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l)   ΔH = −1411 kJ/molrxn

  • C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (l)   ΔH = −1560 kJ/molrxn

  • H2 (g) + 0.5O2 (g) → H2O (l)   ΔH = −286 kJ/molrxn

Calculate the enthalpy change, ΔH, for the conversion of one mole of ethene and one mole of hydrogen to one mole of ethane.

C2H4 (g) + H2 (g) → C2H6 (g)

Answer:

  • Step 1: Reverse the second equation to form ethane as a product

C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l)   ΔH = −1411 kJ/molrxn

2CO2 (g) + 3H2O (l) → C2H6 (g) + 3.5O2 (g)   ΔH = +1560 kJ/molrxn

H2 (g) + 0.5O2 (g) → H2O (l)   ΔH = −286 kJ/molrxn

  • Step 2: Cancel the species that appear on both sides, leaving the target equation, C2H4 (g) + H2 (g) → C2H6 (g)

C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l)  

2CO2 (g) + 3H2O (l) → C2H6 (g) + 3.5O2 (g)              

H2 (g) + 0.5O2 (g) → H2O (l)                   

  • Step 3: The enthalpy change, ΔH, is the sum of the enthalpy changes

ΔH = (−1411) + 1560 + (−286) = −137 kJ/molrxn

Examiner Tips and Tricks

Before adding ΔH values, check that your manipulated equations add up to exactly the target equation

ΔH as written is already per mole of reaction, as the equation is balanced. So, check the coefficients before scaling anything

You may need to reverse equations to get the reactants and products on the correct side or multiply the equations in order for them to be balanced

If you reverse an equation, you must also reverse the sign and use this in the calculation

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Stewart Hird

Reviewer: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.