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Enthalpy changes

What you'll learn

  • How to decide whether a reaction is exothermic or endothermic from its enthalpy change.
  • How to draw and interpret enthalpy profile diagrams, including activation energy.
  • How to calculate enthalpy changes from experiments using q=mcΔTq = mc\Delta Tq=mcΔT.
  • How to use bond enthalpies and Hess’ law cycles to find enthalpy changes indirectly.

1. Enthalpy changes: the basic idea

Chemical reactions often involve heat energy being transferred between the reaction mixture and its surroundings.

Definition

Enthalpy change, ΔH

An enthalpy change, ΔH\Delta HΔH, is the heat energy change for a reaction at constant pressure. At A-Level, it is usually given in kJ mol⁻¹.

If a reaction releases heat to the surroundings, the surroundings get warmer. This is exothermic, and ΔH\Delta HΔH is negative.

If a reaction takes in heat from the surroundings, the surroundings get cooler. This is endothermic, and ΔH\Delta HΔH is positive.

Key Idea

Sign of ΔH

For an exothermic reaction, products have lower enthalpy than reactants, so ΔH<0\Delta H < 0ΔH<0. For an endothermic reaction, products have higher enthalpy than reactants, so ΔH>0\Delta H > 0ΔH>0.

2. Enthalpy profile diagrams

An enthalpy profile diagram shows how enthalpy changes as a reaction proceeds. The vertical axis is enthalpy, HHH, and the horizontal axis is reaction progress.

The diagram compares the enthalpy of the reactants with the enthalpy of the products. It also shows the “hill” that particles must get over for reaction to occur.

Enthalpy profile diagrams for exothermic and endothermic reactions showing activation energy and ΔH

Definition

Activation energy

The activation energy, EaE_aEa​, is the minimum energy required for a reaction to take place.

Even exothermic reactions need activation energy. Bonds in the reactants must start to break before new bonds can form, so the reaction has to pass through a high-energy stage.

Example

Interpreting an enthalpy profile

A reaction has reactants at 80 kJ mol⁻¹, products at 25 kJ mol⁻¹, and a peak at 140 kJ mol⁻¹.

  1. Compare products with reactants: products are lower in enthalpy, so the reaction is exothermic.

  2. Calculate the enthalpy change:

ΔH=25−80=−55 kJ mol−1 \Delta H = 25 - 80 = -55\ \text{kJ mol}^{-1} ΔH=25−80=−55 kJ mol−1
  1. Calculate the activation energy from reactants to the peak:
Ea=140−80=60 kJ mol−1 E_a = 140 - 80 = 60\ \text{kJ mol}^{-1} Ea​=140−80=60 kJ mol−1
Common Mistake

Measuring activation energy from the wrong place

Activation energy is measured from the reactants up to the peak, not from the products up to the peak.

3. Standard conditions and standard states

Enthalpy changes depend on conditions such as temperature, pressure, concentration, and physical state. To make values comparable, chemists use standard conditions.

Definition

Standard conditions and standard states

For OCR A, standard conditions can be taken as 100 kPa and a stated temperature, usually 298 K. A standard state is the physical state of a substance under those standard conditions.

For example, under standard conditions, water is H₂O(l), oxygen is O₂(g), and carbon is C(s, graphite).

Tip

Always include state symbols

For enthalpy definitions and equations, state symbols matter. H₂O(l) and H₂O(g) have different enthalpies.

4. Types of enthalpy change

Enthalpy change of reaction, ΔᵣH

Definition

Enthalpy change of reaction

The enthalpy change of reaction, ΔrH\Delta_rHΔr​H, is the enthalpy change associated with a stated chemical equation, in the molar quantities shown by that equation.

For example:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

The value of ΔrH\Delta_rHΔr​H applies to exactly that equation as written.

Enthalpy change of formation, ΔfH

Definition

Enthalpy change of formation

The enthalpy change of formation, ΔfH\Delta_fHΔf​H, is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states.

Example formation equation:

C(s, graphite) + 2H₂(g) → CH₄(g)

Notice that exactly 1 mol of CH₄ is formed.

Enthalpy change of combustion, ΔcH

Definition

Enthalpy change of combustion

The enthalpy change of combustion, ΔcH\Delta_cHΔc​H, is the enthalpy change when 1 mol of a substance is completely burned in oxygen.

Example:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Complete combustion of carbon-containing substances forms CO₂, not CO.

Enthalpy change of neutralisation, ΔneutH

Definition

Enthalpy change of neutralisation

The enthalpy change of neutralisation, ΔneutH\Delta_{\text{neut}}HΔneut​H, is the enthalpy change when 1 mol of water is formed in a neutralisation reaction.

For a strong acid and strong alkali, the ionic equation is:

H⁺(aq) + OH⁻(aq) → H₂O(l)

5. Measuring enthalpy changes directly: calorimetry

In school laboratory calorimetry, the heat energy change is often found from the temperature change of water or an aqueous solution.

The key equation is:

q=mcΔT q = mc\Delta T q=mcΔT

where:

  • qqq is heat energy transferred, usually in J
  • mmm is the mass of solution in g
  • ccc is the specific heat capacity, usually 4.18 J g⁻¹ K⁻¹ for aqueous solutions
  • ΔT\Delta TΔT is the temperature change in K or °C

Because a temperature change of 1 K is the same size as a temperature change of 1 °C, you can use °C for ΔT\Delta TΔT.

Example

Calculating an enthalpy change of neutralisation

25.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 25.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.8 °C. Assume the solution has density 1.00 g cm⁻³ and c=4.18 J g−1K−1c = 4.18\ \text{J g}^{-1}\text{K}^{-1}c=4.18 J g−1K−1.

  1. Find the total mass of solution. The total volume is 50.0 cm³, so the mass is 50.0 g.

  2. Calculate the heat released to the solution:

q=mcΔT=50.0×4.18×6.8=1421.2 J q = mc\Delta T = 50.0 \times 4.18 \times 6.8 = 1421.2\ \text{J} q=mcΔT=50.0×4.18×6.8=1421.2 J
  1. Find the amount of water formed. HCl and NaOH react 1:1, so:
n=cV=1.00×0.0250=0.0250 mol n = cV = 1.00 \times 0.0250 = 0.0250\ \text{mol} n=cV=1.00×0.0250=0.0250 mol
  1. Convert heat energy to kJ mol⁻¹ and use a negative sign because the temperature increased, meaning the reaction is exothermic:
ΔneutH=−1.42120.0250=−56.8 kJ mol−1 \Delta_{\text{neut}}H = -\frac{1.4212}{0.0250} = -56.8\ \text{kJ mol}^{-1} Δneut​H=−0.02501.4212​=−56.8 kJ mol−1
Common Mistake

Forgetting the sign

If the solution temperature rises, the reaction released heat, so the reaction enthalpy change is negative.

Tip

Practical evaluation

Simple calorimetry often gives less exothermic values than data-book values because heat is lost to the surroundings and absorbed by the apparatus.

6. Bond enthalpies

A chemical reaction involves bonds being broken and new bonds being made.

Breaking bonds requires energy, so it is endothermic. Making bonds releases energy, so it is exothermic.

Definition

Average bond enthalpy

An average bond enthalpy is the energy needed to break 1 mol of a particular type of bond in gaseous molecules.

OCR does not require a formal definition beyond this idea, but you do need to know that bond enthalpies are average values. A C–H bond in methane is not exactly identical to a C–H bond in ethane, so calculated values are approximate.

The calculation is:

ΔH=∑bond enthalpies of bonds broken−∑bond enthalpies of bonds made \Delta H = \sum \text{bond enthalpies of bonds broken} - \sum \text{bond enthalpies of bonds made} ΔH=∑bond enthalpies of bonds broken−∑bond enthalpies of bonds made
Key Idea

Bonds broken minus bonds made

Breaking bonds costs energy. Making bonds releases energy. So use:
ΔH=energy in−energy out\Delta H = \text{energy in} - \text{energy out}ΔH=energy in−energy out.

Example

Calculating ΔH from bond enthalpies

Estimate ΔH\Delta HΔH for:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Use bond enthalpies: C–H = 413, O=O = 498, C=O in CO₂ = 805, O–H = 464 kJ mol⁻¹.

  1. Count the bonds broken in the reactants: four C–H bonds in CH₄ and two O=O bonds in 2O₂.
Energy broken=4(413)+2(498)=2648 kJ mol−1 \text{Energy broken} = 4(413) + 2(498) = 2648\ \text{kJ mol}^{-1} Energy broken=4(413)+2(498)=2648 kJ mol−1
  1. Count the bonds made in the products: two C=O bonds in CO₂ and four O–H bonds in 2H₂O.
Energy made=2(805)+4(464)=3466 kJ mol−1 \text{Energy made} = 2(805) + 4(464) = 3466\ \text{kJ mol}^{-1} Energy made=2(805)+4(464)=3466 kJ mol−1
  1. Subtract bonds made from bonds broken:
ΔH=2648−3466=−818 kJ mol−1 \Delta H = 2648 - 3466 = -818\ \text{kJ mol}^{-1} ΔH=2648−3466=−818 kJ mol−1
Common Mistake

Bond enthalpies need gaseous molecules

Average bond enthalpies apply to bonds in gaseous molecules, so equations using bond enthalpies often use H₂O(g), not H₂O(l).

7. Hess’ law and enthalpy cycles

Some enthalpy changes cannot be measured directly. Hess’ law lets you calculate them indirectly.

The important principle is conservation of energy: if reactants become products, the overall enthalpy change is the same no matter which route is taken.

Hess cycles using enthalpies of formation and combustion

Using enthalpies of formation

For formation data:

ΔrH=∑ΔfH(products)−∑ΔfH(reactants) \Delta_rH = \sum \Delta_fH(\text{products}) - \sum \Delta_fH(\text{reactants}) Δr​H=∑Δf​H(products)−∑Δf​H(reactants)

Remember: elements in their standard states have ΔfH=0\Delta_fH = 0Δf​H=0.

Example

Using formation enthalpies

Calculate ΔrH\Delta_rHΔr​H for:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Use ΔfH\Delta_fHΔf​H: CH₄(g) = −75, CO₂(g) = −394, H₂O(l) = −286 kJ mol⁻¹. Oxygen is an element in its standard state.

  1. Add the formation enthalpies of the products, including coefficients:
∑ΔfH(products)=−394+2(−286)=−966 kJ mol−1 \sum \Delta_fH(\text{products}) = -394 + 2(-286) = -966\ \text{kJ mol}^{-1} ∑Δf​H(products)=−394+2(−286)=−966 kJ mol−1
  1. Add the formation enthalpies of the reactants:
∑ΔfH(reactants)=−75+2(0)=−75 kJ mol−1 \sum \Delta_fH(\text{reactants}) = -75 + 2(0) = -75\ \text{kJ mol}^{-1} ∑Δf​H(reactants)=−75+2(0)=−75 kJ mol−1
  1. Apply products minus reactants:
ΔrH=−966−(−75)=−891 kJ mol−1 \Delta_rH = -966 - (-75) = -891\ \text{kJ mol}^{-1} Δr​H=−966−(−75)=−891 kJ mol−1

Using enthalpies of combustion

For combustion data, both reactants and products are taken down to the same combustion products, usually CO₂ and H₂O.

For combustion cycles:

ΔrH=∑ΔcH(reactants)−∑ΔcH(products) \Delta_rH = \sum \Delta_cH(\text{reactants}) - \sum \Delta_cH(\text{products}) Δr​H=∑Δc​H(reactants)−∑Δc​H(products)
Example

Using combustion enthalpies

Calculate ΔrH\Delta_rHΔr​H for:

C₂H₄(g) + H₂(g) → C₂H₆(g)

Use ΔcH\Delta_cHΔc​H: C₂H₄(g) = −1411, H₂(g) = −286, C₂H₆(g) = −1560 kJ mol⁻¹.

  1. Add the combustion enthalpies of the reactants:
∑ΔcH(reactants)=−1411+(−286)=−1697 kJ mol−1 \sum \Delta_cH(\text{reactants}) = -1411 + (-286) = -1697\ \text{kJ mol}^{-1} ∑Δc​H(reactants)=−1411+(−286)=−1697 kJ mol−1
  1. Add the combustion enthalpies of the products:
∑ΔcH(products)=−1560 kJ mol−1 \sum \Delta_cH(\text{products}) = -1560\ \text{kJ mol}^{-1} ∑Δc​H(products)=−1560 kJ mol−1
  1. Apply reactants minus products:
ΔrH=−1697−(−1560)=−137 kJ mol−1 \Delta_rH = -1697 - (-1560) = -137\ \text{kJ mol}^{-1} Δr​H=−1697−(−1560)=−137 kJ mol−1
Common Mistake

Mixing up the Hess formulae

For formation cycles, use products minus reactants. For combustion cycles, use reactants minus products. This happens because the arrows point in different directions.

8. Direct and indirect determination: choosing the method

You can determine enthalpy changes directly by experiment, or indirectly using data.

Direct methods include:

  • measuring temperature changes in solution reactions, such as neutralisation
  • burning fuels and measuring heat transferred to water
  • using q=mcΔTq = mc\Delta Tq=mcΔT and then dividing by the amount in mol

Indirect methods include:

  • using average bond enthalpies
  • using enthalpies of formation
  • using enthalpies of combustion
  • constructing unfamiliar Hess cycles from given arrows and data

In practical work, you should be able to describe the procedure, process the data, and evaluate limitations such as heat loss, incomplete combustion, evaporation, and uncertainty in temperature readings.

Exam technique

In the exam

  1. Check exactly what 1 mol refers to: fuel burned, compound formed, or water produced.

  2. Keep qqq in J for q=mcΔTq = mc\Delta Tq=mcΔT, then convert to kJ before calculating kJ mol⁻¹.

  3. For Hess cycles, follow the arrows and coefficients carefully; multiply enthalpy values by the number of moles in the balanced equation.

Self review

Check yourself

  • Why is ΔH\Delta HΔH negative for an exothermic reaction?
  • How would you calculate ΔneutH\Delta_{\text{neut}}HΔneut​H from masses, temperature change, and concentration data?
  • When using combustion enthalpies in a Hess cycle, why is the formula reactants minus products?
Recap questions

1 of 5

On an enthalpy profile, the reactants are at 70 kJ mol⁻¹ and the products are at 40 kJ mol⁻¹. Which statement is correct?

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In an exothermic reaction, products have [     ] enthalpy than reactants, so ΔH\Delta HΔH is [     ].

Enthalpy changes Revision Guide

  1. A Level
  2. /Chemistry
  3. /Enthalpy changes