Enthalpy change, exothermic and endothermic reactions, standard enthalpy changes (formation, combustion), calorimetry and the equation q = mcDeltaT, Hess's law and enthalpy cycles, mean bond enthalpies.

What this dot point is asking

AQA wants you to define enthalpy change and the standard enthalpies of formation and combustion, classify reactions as exothermic or endothermic, calculate heat released using calorimetry, apply Hess's law with enthalpy cycles, and calculate enthalpy changes from mean bond enthalpies.

Enthalpy change and reaction types

In an exothermic reaction, heat is released to the surroundings and $\Delta H$ is negative (products are at lower energy). In an endothermic reaction, heat is absorbed and $\Delta H$ is positive.

The activation energy is the minimum energy needed for a reaction to occur; it appears as the energy barrier on a reaction profile.

Standard enthalpy definitions

• Standard enthalpy of formation ($\Delta_f H^\ominus$): enthalpy change when one mole of a compound forms from its elements in their standard states.
• Standard enthalpy of combustion ($\Delta_c H^\ominus$): enthalpy change when one mole of a substance burns completely in oxygen, under standard conditions.

Calorimetry

Heat energy is found from:

Divide $q$ (in kJ) by the moles of the limiting reactant to get $\Delta H$ in $\text{kJ mol}^{-1}$. Add a minus sign for an exothermic temperature rise.

Hess's law and bond enthalpies

There are two standard cycle types. Using formation data, the arrows point up from the elements to both reactants and products, so $\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$. Using combustion data, the arrows point down from reactants and products to the combustion products, so $\Delta_r H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products})$. The two have opposite sign conventions because the data point in opposite directions, which is the most common AQA trap. Drawing the cycle with clearly labelled arrows, then following the alternative route, avoids sign errors.

Mean bond enthalpy is the average energy to break one mole of a given covalent bond in the gaseous state, averaged over many different compounds (so the value for $\text{C-H}$ is a mean over methane, ethane, and so on). Because breaking bonds is endothermic and making bonds is exothermic, $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$. Bond-enthalpy answers are only estimates: they assume every bond of a given type is identical, which is why a calorimetry or formation-data value is more accurate for a specific compound.

Try this

Q1. State whether a reaction with a positive $\Delta H$ is exothermic or endothermic. [1 mark]

• Cue. Endothermic (heat absorbed, products at higher energy).

Q2. $50 \text{ cm}^3$ of solution rises by $8.0 \text{ K}$. Calculate $q$ in kJ. [2 marks]

• Cue. $q = 50 \times 4.18 \times 8.0 = 1672 \text{ J} = 1.67 \text{ kJ}$.