Skip to main content
EnglandChemistrySyllabus dot point

How do we measure and calculate the energy released or absorbed in a reaction?

Enthalpy change, exothermic and endothermic reactions, standard enthalpy changes, calorimetry, Hess's law and enthalpy cycles, and mean bond enthalpy calculations.

An Edexcel 9CH0 Topic 7 answer covering enthalpy change, exothermic and endothermic reactions, standard enthalpies, calorimetry, Hess's law cycles and mean bond enthalpies.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. The answer
  3. Examples in context
  4. Try this

What this topic is asking

Edexcel Topic 7 wants you to define enthalpy change and the standard enthalpy terms, measure enthalpy changes by calorimetry using q=mcΔTq = mc\Delta T, apply Hess's law through enthalpy cycles built from formation or combustion data, and estimate ΔH\Delta H from mean bond enthalpies. Almost every question is quantitative, so the marks come from clean working with correct signs and units.

The answer

Enthalpy change and standard conditions

The key standard terms you must define precisely:

  • Standard enthalpy of formation ΔHf\Delta H_f^{\ominus}: the enthalpy change when one mole of a compound forms from its elements in their standard states under standard conditions. By definition ΔHf\Delta H_f^{\ominus} of an element in its standard state is zero.
  • Standard enthalpy of combustion ΔHc\Delta H_c^{\ominus}: the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
  • Standard enthalpy of reaction ΔHr\Delta H_r^{\ominus}: the enthalpy change for the reaction as written in moles.

Reaction profiles

In an exothermic reaction the products sit lower in energy than the reactants, so ΔH\Delta H is negative; the activation energy EaE_a is the hump that must be climbed first. In an endothermic reaction the products sit higher. The difference between reactant and product energy levels is ΔH\Delta H.

Calorimetry

Measure the temperature rise of a known mass of water (or aqueous solution) and apply the heat equation.

For enthalpy of combustion, the fuel heats the water in a calorimeter. For enthalpy of neutralisation or solution, the temperature change of the reacting solution itself is used, taking the solution density as 1 g cm31\ \text{g cm}^{-3}.

Hess's law and enthalpy cycles

From formation data, the arrows in the cycle point up from the elements to both reactants and products, giving

ΔHr=ΔHf(products)ΔHf(reactants).\Delta H_r = \sum \Delta H_f^{\ominus}(\text{products}) - \sum \Delta H_f^{\ominus}(\text{reactants}).

From combustion data, the arrows point down to the common combustion products, giving
ΔHr=ΔHc(reactants)ΔHc(products).\Delta H_r = \sum \Delta H_c^{\ominus}(\text{reactants}) - \sum \Delta H_c^{\ominus}(\text{products}).

Watch the arrow directions: this is the most common place to drop a sign.

Mean bond enthalpies

A mean bond enthalpy is the average energy to break one mole of a particular bond in the gas phase, averaged over many compounds.

ΔH(bonds broken)(bonds formed).\Delta H \approx \sum(\text{bonds broken}) - \sum(\text{bonds formed}).

Because the values are averages, the result is only an estimate, and it strictly applies only to gaseous species.

Examples in context

Example 1. Self-heating cans and exothermic chemistry. Self-heating coffee cans exploit the strongly exothermic reaction of calcium oxide with water, CaO+H2OCa(OH)2\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2, with ΔH65 kJ mol1\Delta H \approx -65\ \text{kJ mol}^{-1}. Engineers use the same q=mcΔTq = mc\Delta T relationship to size the reagent quantity so the drink reaches a target temperature, exactly the calorimetry maths Edexcel tests in Topic 7.

Example 2. Fuel comparison by bomb calorimetry. Combustion enthalpies quoted for fuels (petrol, ethanol, hydrogen) are measured in a sealed bomb calorimeter, which minimises the heat loss and incomplete combustion that make a simple school spirit-burner experiment give values far smaller in magnitude than the data book. Edexcel evaluation questions reward naming heat loss, incomplete combustion and evaporation as the reasons the school value is too small.

Try this

Q1. Define the standard enthalpy of formation of a compound. [2 marks]

  • Cue. Enthalpy change when one mole of the compound forms from its elements in their standard states under standard conditions (100 kPa100\ \text{kPa}, stated temperature).

Q2. 0.0125 mol0.0125\ \text{mol} of a fuel raises the temperature of 200 g200\ \text{g} of water by 14.0 K14.0\ \text{K}. Calculate ΔHc\Delta H_c. [3 marks]

  • Cue. q=200×4.18×14.0=11704 Jq = 200 \times 4.18 \times 14.0 = 11704\ \text{J}; ΔHc=11704/0.0125=936 kJ mol1\Delta H_c = -11704 / 0.0125 = -936\ \text{kJ mol}^{-1}.

Q3. Using ΔHf\Delta H_f^{\ominus} values, write the Hess expression for ΔHr\Delta H_r and explain the sign convention. [3 marks]

  • Cue. ΔHr=ΔHf(products)ΔHf(reactants)\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}); elements have ΔHf=0\Delta H_f = 0.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20195 marksA student burned 0.0115 mol0.0115\ \text{mol} of methanol and used the heat to raise the temperature of 150 g150\ \text{g} of water by 18.5 K18.5\ \text{K}. Calculate the enthalpy of combustion of methanol, and explain why the value differs from the data-book value of 726 kJ mol1-726\ \text{kJ mol}^{-1}.
Show worked answer →

Calculate the heat gained by the water, then divide by moles of fuel.

Heat: q=mcΔT=150×4.18×18.5=11600 J=11.6 kJq = mc\Delta T = 150 \times 4.18 \times 18.5 = 11600\ \text{J} = 11.6\ \text{kJ} (1 mark).

Per mole: ΔHc=11.60.0115=1009 kJ mol1\Delta H_c = -\dfrac{11.6}{0.0115} = -1009\ \text{kJ mol}^{-1} (2 marks; the negative sign because combustion is exothermic).

Evaluation: the experimental magnitude is much smaller than 726 kJ mol1-726\ \text{kJ mol}^{-1} because of heat loss to the surroundings, incomplete combustion, and evaporation of methanol (2 marks). Note the experimental value here is actually larger in magnitude only if heat loss were negative; markers accept any valid named sources of error with a sign-consistent statement.

Edexcel 20214 marksUsing mean bond enthalpies E(CH)=+413E(\text{C}{-}\text{H}) = +413, E(O=O)=+498E(\text{O}{=}\text{O}) = +498, E(C=O)=+805E(\text{C}{=}\text{O}) = +805 and E(OH)=+464 kJ mol1E(\text{O}{-}\text{H}) = +464\ \text{kJ mol}^{-1}, calculate ΔH\Delta H for CH4+2O2CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} and explain why it is only an estimate.
Show worked answer →

Bonds broken: 4×413+2×498=1652+996=2648 kJ4 \times 413 + 2 \times 498 = 1652 + 996 = 2648\ \text{kJ} (1 mark).

Bonds formed: 2×805+4×464=1610+1856=3466 kJ2 \times 805 + 4 \times 464 = 1610 + 1856 = 3466\ \text{kJ} (1 mark).

ΔH=brokenformed=26483466=818 kJ mol1\Delta H = \text{broken} - \text{formed} = 2648 - 3466 = -818\ \text{kJ mol}^{-1} (1 mark).

It is an estimate because mean bond enthalpies are averaged over many different molecules, so the actual bonds in CH4\text{CH}_4 and H2O\text{H}_2\text{O} differ slightly; the data also assume gaseous species (1 mark).

Related dot points

Sources & how we know this