Hess's Law and Enthalpy Calculations — AQA A-level Chemistry

SPEC 3.1.4.3 • ENERGETICS

Hess's Law

Calculating enthalpy changes using enthalpy cycles with formation and combustion data

Why Hess's Law?

Many enthalpy changes cannot be measured directly in the laboratory. For example, we cannot easily measure the enthalpy of formation of methane by reacting carbon and hydrogen gas — the reaction is too slow and incomplete under standard conditions.

Hess's Law provides a powerful indirect method: if we know the enthalpy changes for related reactions, we can calculate the enthalpy change we cannot measure.

The Core Idea

Energy is conserved. If a reaction can happen by two different routes, the total energy change must be the same for both routes — because energy cannot be created or destroyed.

This lesson will teach you how to construct enthalpy cycles and use them to calculate unknown enthalpy changes — a skill that appears in almost every A-Level Chemistry exam.

Hess's Law — The Principle

Hess's Law

The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

Hess's Law is a direct consequence of the First Law of Thermodynamics (conservation of energy). If you start with the same reactants and end with the same products, the total energy change must be identical — regardless of whether the reaction happens in one step or multiple steps.

The Equality

ΔH_r = ΔH₂ − ΔH₁

ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

Two Types of Enthalpy Cycles

In A-Level Chemistry, you will use two main types of enthalpy cycles, depending on what data you are given:

Cycle TypeData GivenFormula

Formation CycleEnthalpies of formation (ΔH_f°)ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

Combustion CycleEnthalpies of combustion (ΔH_c°)ΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products)

Critical Difference

Notice the sign reversal! For formation: products − reactants. For combustion: reactants − products. This is the most common source of errors in Hess's Law calculations.

Examiner Tip

Take into account the number of moles of each reactant and product. For example, if there are 2 moles of NaHCO₃(s) in the equation, the ΔH_f value must be multiplied by 2.

The reason for this difference relates to the direction of arrows in the enthalpy cycle — we will see this clearly in the following slides.

Enthalpy Cycles Using Formation Data

Knowing the enthalpy change of formation (ΔH_f) allows us to determine the overall enthalpy change of a reaction (ΔH_r). When given enthalpies of formation, we construct a cycle where the elements in their standard states form the base of the triangle.

The enthalpy change from elements to products (direct route) is equal to the enthalpy change of elements forming reactants and then products (indirect route). Since ΔH₂ = ΔH₁ + ΔH_r, therefore:

Formation Cycle Formula

Δ H_{r} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)

Why This Works

Route 1 (direct): Reactants → Products gives ΔH_r

Route 2 (via elements): Reactants → Elements → Products

Going "backwards" along ΔH_f(reactants) means we subtract it!

Formation Cycle — Step by Step

Calculate ΔH_r° for: SO₂(g) + 2H₂S(g) → 3S(s) + 2H₂O(l)

SubstanceΔH_f° / kJ mol⁻¹

SO₂(g)−297

H₂S(g)−20

H₂O(l)−286

Step 1: Draw the enthalpy cycle

Step 2: Apply Hess's Law — ΔH_r° = ΣΔH_f°(products) − ΣΔH_f°(reactants)

In a cycle, go from reactants to products. The sign on ΔH_f°(reactants) is reversed because the arrow points in the opposite direction (we're going UP from elements to reactants, but our route goes DOWN from reactants to elements).

Step 3: Substitute and evaluate

ΣΔH_f°(products) = 3(0) + 2(−286) = −572 kJ(Note: ΔH_f° for S(s) = 0 as it is an element in its standard state)

ΣΔH_f°(reactants) = (−297) + 2(−20) = −337 kJ

ΔH_r° = (−572) − (−337) = −572 + 337 = −235 kJ mol⁻¹

01Worked Example — Formation Cycle

Calculate the standard enthalpy change for the combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Given:

SubstanceΔH_f° / kJ mol⁻¹

CH₄(g)−74.8

O₂(g)0

CO₂(g)−393.5

H₂O(l)−285.8

1

Apply the formation cycle formula:

Δ H_{r} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)

2

Products: 1 mol CO₂ + 2 mol H₂O

ΣΔH_f(products) = (1 × −393.5) + (2 × −285.8) = −393.5 + (−571.6) = −965.1 kJ

3

Reactants: 1 mol CH₄ + 2 mol O₂

ΣΔH_f(reactants) = (1 × −74.8) + (2 × 0) = −74.8 kJ

4

ΔH_r = (−965.1) − (−74.8) = −965.1 + 74.8 = −890.3 kJ mol⁻¹

Answer: ΔH_c°[CH₄] = −890 kJ mol⁻¹

Enthalpy Cycles Using Combustion Data

It can be difficult to find the enthalpy change of formation of compounds experimentally. However, many enthalpies of combustion can be measured experimentally, so these can be used to find the enthalpy of formation. We construct a cycle where the combustion products (CO₂ and H₂O) form the base.

Draw downward pointing arrows from each substance to its combustion products. Write values on the arrows and multiply by the number of moles. In a cycle, go from reactants to products, changing the sign if the arrow points in the opposite direction.

Combustion Cycle Formula

Δ H_{r} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)

Why Reactants − Products?

Route 1 (direct): Reactants → Products gives ΔH_r

Route 2 (via combustion products): Both reactants and products burn to give CO₂ + H₂O

Following arrows: ΔH_c(reactants) then BACKWARDS along ΔH_c(products)

Combustion Cycle — Step by Step

Calculate ΔH_f° for ethanol: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

SubstanceΔH_c° / kJ mol⁻¹

C₂H₅OH(l)−1367

C(s)−394

H₂(g)−286

Step 1: Draw the enthalpy cycle

Step 2: Apply Hess's Law — ΔH_f° = ΣΔH_c°(reactants) − ΣΔH_c°(products)

The sign on ΔH_c°(products) is reversed because the arrow pointing to combustion products goes in the opposite direction to our route (we need to go UP from combustion products to products).

Step 3: Substitute and evaluate

ΣΔH_c°(reactants) = 2(−394) + 3(−286) = −788 + (−858) = −1646 kJ

ΣΔH_c°(products) = −1367 kJ

ΔH_f° = (−1646) − (−1367) = −1646 + 1367 = −279 kJ mol⁻¹

02Worked Example — Combustion Cycle

Calculate the standard enthalpy of formation of propane (C₃H₈).

Formation equation: 3C(s) + 4H₂(g) → C₃H₈(g)

Given:

SubstanceΔH_c° / kJ mol⁻¹

C(s, graphite)−394

H₂(g)−286

C₃H₈(g)−2220

1

This is a combustion cycle — we're finding ΔH_f from ΔH_c data.

2

Apply:

Δ H_{f} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)

3

Reactants: 3 mol C + 4 mol H₂

ΣΔH_c(reactants) = (3 × −394) + (4 × −286) = −1182 + (−1144) = −2326 kJ

4

Products: 1 mol C₃H₈

ΣΔH_c(products) = 1 × −2220 = −2220 kJ

5

ΔH_f = (−2326) − (−2220) = −2326 + 2220 = −106 kJ mol⁻¹

Answer: ΔH_f°[C₃H₈] = −106 kJ mol⁻¹

Key Rules for Hess's Law Calculations

Rule 1: Elements Have Zero ΔH_f

The standard enthalpy of formation of any element in its standard state is zero by definition.

Examples: ΔH_f°[O₂(g)] = 0, ΔH_f°[C(graphite)] = 0, ΔH_f°[Fe(s)] = 0

Rule 2: Multiply by Stoichiometric Coefficients

Always multiply ΔH values by the number of moles shown in the balanced equation.

If the equation shows 2H₂O, use 2 × ΔH_f[H₂O], not just ΔH_f[H₂O].

Rule 3: Watch the Signs

Formation: Products − Reactants (P − R)

Combustion: Reactants − Products (R − P)

Most Common Exam Errors

Forgetting to multiply by coefficients (e.g., using −286 instead of 2 × −286 for 2H₂O)

Using the wrong formula for combustion cycles

Including ΔH_f values for elements (they should be zero!)

Sign errors when subtracting negative numbers

03Worked Example — With Coefficients

Calculate ΔH for the thermal decomposition of sodium hydrogencarbonate:

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(l) + CO₂(g)

Given enthalpies of formation:

SubstanceΔH_f° / kJ mol⁻¹

NaHCO₃(s)−951

Na₂CO₃(s)−1131

H₂O(l)−286

CO₂(g)−394

1

This is a formation cycle — we have ΔH_f data.

2

Products: 1 mol Na₂CO₃ + 1 mol H₂O + 1 mol CO₂

ΣΔH_f(products) = (−1131) + (−286) + (−394) = −1811 kJ

3

Reactants: 2 mol NaHCO₃ (note the coefficient!)

ΣΔH_f(reactants) = 2 × (−951) = −1902 kJ

4

ΔH_r = (−1811) − (−1902) = −1811 + 1902 = +91 kJ mol⁻¹

Answer: ΔH = +91 kJ mol⁻¹ (endothermic)

This makes sense — thermal decomposition requires heating!

Step-by-Step Method

Follow this systematic approach for any Hess's Law calculation:

5-Step Method

Step 1: Write the balanced equation for the target reaction at the top

Step 2: Identify what data type you have (ΔH_f or ΔH_c) and draw the cycle with elements (formation) or combustion products (combustion) at the bottom

Step 3: Draw arrows in the correct directions — for formation, arrows point UP from elements; for combustion, arrows point DOWN to combustion products

Step 4: Write values on arrows and multiply by the number of moles from the balanced equation

Step 5: Go from reactants to products around the cycle — change the sign if the arrow points in the opposite direction to your route

The Arrow Rule

When following a route around the cycle: if you travel with an arrow, add its value. If you travel against an arrow, subtract its value (change the sign). This is why formation uses P − R and combustion uses R − P.

04Worked Example — Finding a Missing Value

The enthalpy of combustion of ethene (C₂H₄) is −1411 kJ mol⁻¹. Calculate the enthalpy of formation of ethene.

Given:

SubstanceΔH_f° / kJ mol⁻¹

CO₂(g)−394

H₂O(l)−286

1

Write the combustion equation: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l)

2

Using the formation cycle formula for combustion:

ΔH_c = ΣΔH_f(products) − ΣΔH_f(reactants)

3

Products: ΣΔH_f = 2(−394) + 2(−286) = −788 − 572 = −1360 kJ

4

Reactants: ΣΔH_f = ΔH_f[C₂H₄] + 3(0) = ΔH_f[C₂H₄]

5

Substitute: −1411 = (−1360) − ΔH_f[C₂H₄]

6

Rearrange: ΔH_f[C₂H₄] = −1360 − (−1411) = −1360 + 1411 = +51 kJ mol⁻¹

Answer: ΔH_f°[C₂H₄] = +51 kJ mol⁻¹

Summary of Key Formulas

Formation Cycle

Δ H_{r} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)

Remember: P − R (Products minus Reactants)

Combustion Cycle

Δ H_{r} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)

Remember: R − P (Reactants minus Products)

Quick Memory Aid

"Formation: Products First" — in the formation formula, products come first (P − R)

"Combustion: Reactants First" — in the combustion formula, reactants come first (R − P)

Practice

Question 1: Calculate the enthalpy change for the reaction:

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Given ΔH_f° values: C₂H₅OH(l) = −277 kJ mol⁻¹, CO₂(g) = −394 kJ mol⁻¹, H₂O(l) = −286 kJ mol⁻¹

1

This is formation data, so use: ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

2

Products: 2(−394) + 3(−286) = −788 + (−858) = −1646 kJ

3

Reactants: (−277) + 3(0) = −277 kJ

4

ΔH_r = (−1646) − (−277) = −1646 + 277 = −1369 kJ mol⁻¹

Practice

Question 2: Calculate the enthalpy of formation of butane (C₄H₁₀).

Given ΔH_c° values:

C(graphite) = −394 kJ mol⁻¹

H₂(g) = −286 kJ mol⁻¹

C₄H₁₀(g) = −2877 kJ mol⁻¹

1

Formation equation: 4C(s) + 5H₂(g) → C₄H₁₀(g)

2

This is combustion data, so use: ΔH_f = ΣΔH_c(reactants) − ΣΔH_c(products)

3

Reactants: 4(−394) + 5(−286) = −1576 + (−1430) = −3006 kJ

4

Products: 1(−2877) = −2877 kJ

5

ΔH_f = (−3006) − (−2877) = −3006 + 2877 = −129 kJ mol⁻¹

Practice

Question 3: The enthalpy of formation of ammonia (NH₃) is −46 kJ mol⁻¹.

Calculate the enthalpy change for: 2NH₃(g) → N₂(g) + 3H₂(g)

Hint: This is the reverse of formation!

1

Use the formation cycle: ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

2

Products: N₂ and H₂ are elements, so ΣΔH_f(products) = 0 + 0 = 0 kJ

3

Reactants: 2 mol NH₃, so ΣΔH_f(reactants) = 2 × (−46) = −92 kJ

4

ΔH_r = (0) − (−92) = +92 kJ mol⁻¹

The decomposition of ammonia is endothermic — the reverse of formation!

Summary

Key Concepts

ConceptKey Point

Hess's LawTotal enthalpy change is independent of route taken

Formation CycleΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

Combustion CycleΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products)

ElementsΔH_f° of any element in standard state = 0

CoefficientsAlways multiply ΔH values by stoichiometric coefficients

Exam Checklist

Identify data type: formation or combustion?

Choose correct formula: P − R for formation, R − P for combustion

Multiply by coefficients from balanced equation

Remember: ΔH_f° of elements = 0

Watch signs when subtracting negative numbers