Enthalpy changes and standard conditions, exothermic and endothermic reactions, enthalpy of combustion, formation and neutralisation, calorimetry, Hess's law and enthalpy cycles, and mean bond enthalpy calculations.

What this dot point is asking

CCEA wants you to define enthalpy change and standard conditions, distinguish exothermic and endothermic reactions, define the standard enthalpies of combustion, formation and neutralisation, carry out calorimetry calculations, apply Hess's law using enthalpy cycles, and estimate enthalpy changes from mean bond enthalpies.

Enthalpy changes and standard conditions

Exothermic reactions release energy, so $\Delta H$ is negative; endothermic reactions absorb energy, so $\Delta H$ is positive.

Calorimetry

The heat change is calculated from:

where $q$ is heat in J, $m$ is the mass of solution in g, $c$ is the specific heat capacity ($4.18\ \text{J g}^{-1}\,\text{K}^{-1}$ for water) and $\Delta T$ is the temperature change. Dividing $q$ by the moles reacting and converting to kJ gives the molar enthalpy change.

Hess's law

Using enthalpy cycles with formation or combustion data:

The reason this works is that the formation route and the direct route both start from the same elements and end at the same products, so the energy change must be identical whichever path is taken. With combustion data the cycle is built the other way round, because both reactants and products burn to the same final oxides; the working rule there is reactants minus products. Drawing the cycle with arrows pointing in the right directions is the surest way to avoid a sign slip: formation arrows point up from the elements, combustion arrows point down to the oxides.

Mean bond enthalpies

Bond breaking is endothermic and bond making is exothermic, so:

Values from bond enthalpies are only estimates because mean bond enthalpies are averaged over many compounds. The actual $\text{C} - \text{H}$ bond strength in methane differs slightly from the one in ethanol, but the tabulated mean bond enthalpy uses a single averaged figure. Bond enthalpy calculations also assume every substance is gaseous, so they cannot account for the energy released when a product condenses to a liquid. For these reasons a bond enthalpy estimate is acceptable as an approximation but a Hess's law calculation from formation data is more accurate.

Examples in context

Example 1. Comparing fuels for a camping stove. Manufacturers quote the enthalpy of combustion per gram (the specific energy) when comparing fuels. Methylated spirit (largely ethanol, $-1367\ \text{kJ mol}^{-1}$, $M = 46$) releases about $29.7\ \text{kJ g}^{-1}$, while butane ($-2877\ \text{kJ mol}^{-1}$, $M = 58$) releases about $49.6\ \text{kJ g}^{-1}$. A CCEA candidate uses the calorimetry method to measure these values in the laboratory, then explains the gap to data-book figures in terms of heat loss and incomplete combustion, which is exactly the experimental skill the specification assesses.

Example 2. Hess's law where direct measurement is impossible. The enthalpy of formation of carbon monoxide cannot be measured directly, because burning carbon in a limited supply of oxygen always gives a mixture of $\text{CO}$ and $\text{CO}_2$. Instead it is found indirectly: the enthalpies of combustion of carbon to $\text{CO}_2$ and of carbon monoxide to $\text{CO}_2$ are both measurable, and Hess's law combines them ($\Delta H_f(\text{CO}) = \Delta H_c(\text{C}) - \Delta H_c(\text{CO})$). This is the classic illustration of why Hess's law matters: it reaches values that no single experiment can.

Try this

Q1. State whether an endothermic reaction has a positive or negative $\Delta H$. [1 mark]

• Cue. Positive, because energy is absorbed from the surroundings.

Q2. Burning a fuel raises the temperature of $100\ \text{g}$ of water by $20\ \text{K}$. Calculate the heat released. [2 marks]

• Cue. $q = 100 \times 4.18 \times 20 = 8360\ \text{J} = 8.36\ \text{kJ}$.