AQA A-Level Chemistry (7405)

## Learning Outcomes

- Understand why Hess's law works and why some enthalpy changes must be found indirectly
- Define standard enthalpy changes of formation and combustion accurately
- Calculate reaction enthalpies using enthalpies of formation and enthalpies of combustion
- Apply bond enthalpy data correctly and explain why bond enthalpy answers are approximate
- Avoid common exam errors involving signs, coefficients, and state symbols
- Link calorimetry results to Hess cycles in synoptic Paper 3 questions

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand Hess's law and indirect enthalpy calculations, with a focus on the following syllabus points:

- Know that the enthalpy change of a reaction is independent of route, provided the initial and final conditions are the same
- Identify and use enthalpy cycles based on standard enthalpy changes of formation
- Identify and use enthalpy cycles based on standard enthalpy changes of combustion
- Understand the exact definitions of standard enthalpy change, standard conditions, and standard state
- Calculate reaction enthalpies using the formulas based on formation and combustion data
- Use mean bond enthalpies to estimate enthalpy changes of reactions
- Explain why bond enthalpy calculations are approximate and why state symbols matter
- Recognise Born–Haber cycles as a later application of Hess's law in synoptic chemistry

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. What does Hess's law state, and why does it allow indirect calculation of enthalpy changes?
2. What is the difference between standard enthalpy of formation and standard enthalpy of combustion?
3. Which formula uses products minus reactants, and which uses reactants minus products?
4. Why do bond enthalpy calculations often give a value different from the data-book enthalpy change?
5. What happens to the sign of ΔH if you reverse an equation in a Hess cycle?

## Introduction

Hess's law is one of the most useful ideas in energetics because many enthalpy changes cannot be measured directly in the lab. Some reactions are too slow, do not go to completion cleanly, or are not practical under normal laboratory conditions. Instead of measuring the target reaction itself, you measure other reactions and combine them.

This works because enthalpy is a state function. That means the total enthalpy change depends only on the starting substances and the final substances, not on the route between them. In exams, this topic tests both knowledge and method: students often know the definition but lose marks through wrong signs, missed coefficients, or incorrect state symbols.

> **Key Term: Hess's law**
> The enthalpy change of a reaction is the same whatever route is taken, provided the initial and final conditions are the same.

### Why Hess's law works and when to use it

The reason Hess's law works is the conservation of energy. If a reaction from A to B released different amounts of energy by different routes, energy would appear or disappear for no reason. Since that cannot happen, all valid routes between the same start and finish must give the same overall enthalpy change.

A simple way to picture this is to imagine going from one floor of a building to another. Whether you use the stairs or the lift, your change in height is the same. Hess cycles do the same job for enthalpy.

This matters when the direct route is hard to measure. For example, forming one mole of methane directly from carbon and hydrogen is not a simple calorimetry experiment under standard lab conditions. But if you know the enthalpy changes for combustion of carbon, hydrogen and methane, you can build an alternative route through the same products and calculate the missing value.

> **Key Term: enthalpy cycle**
> A diagram or set of equations that uses Hess's law to relate a target enthalpy change to other known enthalpy changes.

For Hess's law questions, always start by writing the target equation clearly, with state symbols. This is not just presentation. It controls the whole calculation. If the equation is wrong, the cycle and the numbers will also be wrong.

State symbols matter because enthalpy depends on physical state. For example, **H₂O(l)** and **H₂O(g)** do not have the same enthalpy. A formation equation ending in liquid water gives a different enthalpy value from one ending in steam.

> **Key Term: standard conditions**
> Conditions of **298 K**, pressure **100 kPa**, and solutions with concentration **1.0 mol dm⁻³**.
>
> **Key Term: standard state**
> The physical state of a substance under standard conditions; for an element, this is its most stable form under those conditions.

A common mistake is to think Hess's law is just a memory fact. In fact, it is a working method. You use it every time you:

- reverse an equation and reverse the sign of ΔH
- multiply an equation and multiply ΔH by the same number
- add equations so unwanted substances cancel
  > **Test Tip:** In any Hess calculation, check three things before doing arithmetic: direction of each equation, stoichiometric coefficients, and state symbols. Most lost marks come from one of these, not from difficult maths.

### Using standard enthalpies of formation and combustion

Two standard methods appear often in AQA questions: using standard enthalpies of formation and using standard enthalpies of combustion. Students confuse them because both use Hess cycles, but the formulas are different because the routes are built differently.

> **Key Term: standard enthalpy change of formation, ΔH°f**
> The enthalpy change when **1 mole** of a compound is formed from its constituent elements in their standard states under standard conditions.
>
> **Key Term: standard enthalpy change of combustion, ΔH°c**
> The enthalpy change when **1 mole** of a substance is completely burned in oxygen under standard conditions, with all reactants and products in their standard states.

The phrase **1 mole** is essential in both definitions. For formation, the product must be exactly one mole of the compound. For combustion, exactly one mole of the substance burns. That is why fractional coefficients are often allowed in formation equations.

For example, the standard enthalpy of formation of water is:

H₂(g) + 1/2 O₂(g) → H₂O(l)

The oxygen coefficient is 1/2 because the definition requires one mole of water formed.

Another essential point: elements in their standard states have **ΔH°f = 0**. So **H₂(g)**, **O₂(g)** and **C(graphite)** each have formation enthalpy zero. Students sometimes try to insert values for these and create extra errors.

If you are using formation data, the formula is:

ΔH°reaction = ΣΔH°f(products) − ΣΔH°f(reactants)

Why this order? Because formation enthalpies measure energy change from elements to substances. In the cycle, both reactants and products connect to the elements, so to go from reactants to products you subtract the reactant side from the product side.

Worked example using formation data:

Calculate ΔH° for

2CaO(s) + SiO₂(s) → Ca₂SiO₄(s)

Given:

- ΔH°f[CaO(s)] = -635 kJ mol⁻¹
- ΔH°f[SiO₂(s)] = -910 kJ mol⁻¹
- ΔH°f[Ca₂SiO₄(s)] = -2307 kJ mol⁻¹

Step 1: Write the formula.

ΔH°reaction = ΣΔH°f(products) − ΣΔH°f(reactants)

Step 2: Insert values with coefficients.

ΔH°reaction = [1 × (-2307)] − [(2 × -635) + (1 × -910)]

Step 3: Work out the reactant total.

(2 × -635) + (-910) = -1270 - 910 = -2180

Step 4: Subtract.

ΔH°reaction = -2307 - (-2180) = -127 kJ mol⁻¹

So the reaction is **exothermic**, with:

ΔH°reaction = **-127 kJ mol⁻¹**

Now compare that with combustion data. The formula is:

ΔH°reaction = ΣΔH°c(reactants) − ΣΔH°c(products)

This order is the opposite pattern many students expect. The reason is that combustion values take every substance down to common combustion products such as CO₂ and H₂O. In the cycle, both sides burn to the same final products, so you compare how far the reactants fall with how far the products fall.

Worked example using combustion data:

Calculate ΔH° for

CO(g) + 1/2 O₂(g) → CO₂(g)

Given:

- ΔH°c[CO(g)] = -283 kJ mol⁻¹
- ΔH°c[CO₂(g)] = 0 kJ mol⁻¹
- ΔH°c[O₂(g)] = 0 kJ mol⁻¹

Step 1: Use reactants minus products.

ΔH°reaction = ΣΔH°c(reactants) − ΣΔH°c(products)

Step 2: Add reactant combustion values.

Reactants total = (-283) + (1/2 × 0) = -283

Step 3: Add product combustion values.

Products total = 0

Step 4: Subtract.

ΔH°reaction = -283 - 0 = -283 kJ mol⁻¹

This makes sense chemically: carbon monoxide combusting to carbon dioxide is exothermic.

Common student errors here:

- using products minus reactants for combustion values
- forgetting to multiply by coefficients
- forgetting that an element in its standard state has ΔH°f = 0
- writing the wrong combustion products, especially for organic compounds
- missing state symbols
  > **Test Tip:** Learn the pair as a contrast:
  > formation = products minus reactants
  > combustion = reactants minus products

### Using mean bond enthalpies

Bond enthalpy questions are another Hess-type method, but they work in a different way. Instead of using tabulated formation or combustion values, you imagine all bonds in the reactants being broken into gaseous atoms, then new bonds being formed to make the products.

> **Key Term: mean bond enthalpy**
> The average enthalpy change when one mole of a stated bond is broken in gaseous molecules.

Bond breaking is always endothermic, so those values are positive. Bond making is always exothermic, so that part of the calculation is treated as energy released.

The formula is:

ΔH = Σ(bonds broken) − Σ(bonds made)

This is not just a formula to memorise. It follows from the physical steps:

1. Energy is put in to break reactant bonds.
2. Energy is released when product bonds form.
3. The overall enthalpy change is input minus output.

Worked example:

Calculate the enthalpy change for

H₂(g) + Cl₂(g) → 2HCl(g)

Given mean bond enthalpies:

- H–H = 436 kJ mol⁻¹
- Cl–Cl = 243 kJ mol⁻¹
- H–Cl = 432 kJ mol⁻¹

Step 1: Count bonds broken.

In reactants:

- 1 × H–H
- 1 × Cl–Cl

Energy to break bonds:
= 436 + 243
= 679 kJ mol⁻¹

Step 2: Count bonds made.

In products:

- 2 × H–Cl

Energy released on bond formation:
= 2 × 432
= 864 kJ mol⁻¹

Step 3: Apply the formula.

ΔH = bonds broken − bonds made
ΔH = 679 − 864
ΔH = -185 kJ mol⁻¹

So the reaction is exothermic.

This method is useful, but it gives an estimate rather than an exact value.

Why? Because the values are **mean** bond enthalpies. A C–H bond in one molecule is not exactly identical in enthalpy to a C–H bond in another molecule. The table gives an average from different compounds. Also, bond enthalpy data apply to **gaseous** species. If your actual reaction includes liquids, such as water(l), then using bond enthalpies means you are really calculating a gas-phase approximation unless extra phase-change data are included.

That is why bond enthalpy answers often differ from experimental or data-book values.

Common errors in bond enthalpy questions:

- counting the number of bonds wrongly
- using the unbalanced equation
- using H₂O(l) instead of H₂O(g) in reasoning
- doing bonds made minus bonds broken
- forgetting to multiply by the number of molecules or coefficient

A good habit is to draw displayed or structural formulas before counting bonds. This slows you down slightly, but it prevents more expensive mistakes.

### Calorimetry links, synoptic extension, and exam technique

Paper 3 often links Hess's law to practical energetics. You may be given calorimetry data, asked to calculate an experimental enthalpy change, and then expected to use it in a Hess cycle.

> **Key Term: calorimetry**
> Measuring energy changes in reactions by recording temperature changes and using the equation q = mcΔT.

The basic equations are:

q = mcΔT

and

ΔH = -q/n

where:

- q is heat energy in J
- m is mass in g
- c is specific heat capacity, often 4.18 J g⁻¹ K⁻¹ for aqueous solutions
- ΔT is temperature change
- n is moles reacted

Worked example:

50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.8 K.

Step 1: Find total mass of solution.

Assume density is 1.00 g cm⁻³, so mass = 100.0 g

Step 2: Calculate q.

q = mcΔT
q = 100.0 × 4.18 × 6.8
q = 2842.4 J
= 2.8424 kJ

Step 3: Find moles reacting.

Moles of HCl = 0.0500 mol
Moles of NaOH = 0.0500 mol

They react 1:1, so moles of water formed = 0.0500 mol

Step 4: Calculate ΔH.

ΔH = -q/n
ΔH = -2.8424 / 0.0500
ΔH = -56.8 kJ mol⁻¹

That is close to the expected value for strong acid-strong alkali neutralisation.

In evaluation questions, experimental values are often **less exothermic** than accepted values. That does not mean the chemistry changed. It means the method did not capture all the heat released.

Typical reasons:

- heat lost to surroundings
- heat absorbed by apparatus
- incomplete combustion in flame experiments
- fuel evaporation before complete burning
- assuming solution has the same specific heat capacity as water
- assuming density is exactly 1.00 g cm⁻³

A bomb calorimeter gives more accurate combustion values because the system is better insulated and combustion is more complete in excess oxygen.

Synoptically, Born–Haber cycles are also applications of Hess's law. The same route-independence idea applies, but the enthalpy terms are different, such as atomisation, ionisation energy, electron affinity and lattice enthalpy. The key point is to keep the terminology separate. Do not confuse a standard combustion cycle with a Born–Haber cycle just because both are Hess cycles.

> **Key Term: Born–Haber cycle**
> An enthalpy cycle that applies Hess's law to ionic compounds, linking formation enthalpy with atomisation, ionisation, electron affinity, and lattice enthalpy terms.

Final exam checklist for method:

1. Write the target equation first.
2. Balance it fully.
3. Include correct state symbols.
4. Decide which data type you are using.
5. Choose the correct formula or cycle logic.
6. Apply coefficients carefully.
7. Check signs after reversing or subtracting.
8. Give units as **kJ mol⁻¹**.

## Key Point Checklist

_This article has covered the following key knowledge points:_

- Hess's law states that the enthalpy change of a reaction is independent of route, provided the initial and final conditions are the same
- Hess's law is based on conservation of energy and the fact that enthalpy is a state function
- Standard conditions are **298 K**, **100 kPa**, and solutions of **1.0 mol dm⁻³**
- Standard enthalpy of formation is for forming **1 mole** of a compound from its elements in their standard states
- Standard enthalpy of combustion is for complete burning of **1 mole** of a substance in oxygen under standard conditions
- Elements in their standard states have **ΔH°f = 0**
- Using formation data: **ΔH°reaction = ΣΔH°f(products) − ΣΔH°f(reactants)**
- Using combustion data: **ΔH°reaction = ΣΔH°c(reactants) − ΣΔH°c(products)**
- Using bond enthalpies: **ΔH = Σ(bonds broken) − Σ(bonds made)**
- Bond enthalpy values give estimates because they are mean values for gaseous molecules
- Reversing an equation changes the sign of ΔH, and multiplying an equation multiplies ΔH by the same factor
- In calorimetry, use **q = mcΔT** and **ΔH = -q/n**, with practical values often less exothermic because of heat loss and other method limits

## Key Terms and Concepts

- Hess's law
- enthalpy cycle
- standard conditions
- standard state
- standard enthalpy change of formation, ΔH°f
- standard enthalpy change of combustion, ΔH°c
- mean bond enthalpy
- calorimetry
- Born–Haber cycle


Energetics - Hess's law