Defining Enthalpy Changes
- To fairly compare the changes in enthalpy between reactions, all reactions should be carried out under standard conditions
- These standard conditions are:
- A pressure of 101 kPa
- A temperature of 298 K (25 oC)
- Each substance involved in the reaction is in its normal physical state (solid, gas or liquid)
- To show that a reaction has been carried out under standard conditions, the symbol θ is used
- ΔHθ = the standard enthalpy change
- These are a number of key definitions for common language relating to enthalpy change that all chemists need to know
Enthalpy definitions table
| Standard Enthalpy Change of... | Definition | Symbol | Exothermic / Endothermic |
| Reaction | The enthalpy change when the reactants in the stoichiometric equation react to form the products under standard conditions | ΔHθr | Both |
| Formation | The enthalpy change when one mole of a compound is formed from its elements under standard conditions | ΔHθf | Both |
| Combustion | The enthalpy change when one mole of a substance is burnt in excess oxygen under standard conditions | ΔHθc | Exothermic |
| Neutralisation | The enthalpy change when one mole of water is formed by reacting an acid and an alkali under standard conditions | ΔHθneut | Exothermic |
Worked example
Calculating the enthalpy change of reaction of water
One mole of water is formed from hydrogen and oxygen, releasing 286 kJ of energy.
H2 (g) + ½O2 (g) → H2O (l) ΔHθr = –286 kJ mol-1
Calculate ΔHr for the reaction below:
2H2 (g) + O2 (g) → 2H2O (l)
Answer
- The ΔHθr value of –286 kJ mol-1 is for one mole of water being formed
- Two moles of water molecules are formed for the equation in question
- So, the energy released is simply:
- ΔHr = 2 mol x (–286 kJ mol-1)
- ΔHr = –572 kJ mol-1
Worked example
Calculating the enthalpy change of formation
Calculate ΔHθf of the reaction below:
4Fe (s) + 3O2 (g) → 2Fe2O3 (s) ΔHθf [Fe2O3 (s)] = –824.2 kJ mol-1
Answer
- The ΔHθf value of –824.2 kJ mol-1 is for one mole of Fe2O3 (s)being formed
- Two moles of Fe2O3 (s) are formed for the equation in question
- So, the energy released is simply:
- ΔHθf = 2 mol x (–824.2kJ mol-1)
- ΔHθf = –1648.4 kJ mol-1
Worked example
Calculating enthalpy changes
Identify each of the following as ΔHθr, ΔHθf, ΔHθc or ΔHθneut:
- MgCO3 (s) → MgO (s) + CO2 (g)
- C (graphite) + O2 (g) → CO2 (g)
- HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
Answers:
Answer 1
- It cannot be ΔHθf as there is more than one compound being formed
- It cannot be ΔHθc as there is no reaction with oxygen
- It cannot be ΔHθneut as there is no acid or alkali involved
- This is just a chemical reaction, therefore, it is ΔHθr
Answer 2
- ΔHθf as one mole of CO2 is formed from its elements in their standard states
AND
ΔHθc as one mole of carbon is burnt completely in oxygen
Answer 3
- ΔHneutꝊ as one mole of water is formed from the reaction of an acid and alkali
Exam Tip
The ΔHθf of an element in its standard state is zero. For example, ΔHθf of O2 (g) is 0 kJ mol-1
