Standard Enthalpy Change (Cambridge International A-Level Chemistry)

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

  • ΔH^θ = 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

• All standard enthalpies are performed under standard conditions

• Standard enthalpy change of reaction, ΔH^θ_r

  • The enthalpy change when the reactants in the stoichiometric equation react to form the products, with all species in their standard states
  • Can be both exothermic and endothermic

• Standard enthalpy change of formation, ΔH^θ_f

  • The enthalpy change when one mole of a compound is formed from its elements in their standard states
  • Can be both exothermic and endothermic

• Standard enthalpy change of combustion, ΔH^θ_c

  • The enthalpy change when **one mole **of a substance, in its standard state, is burnt in excess oxygen
  • It is an** exothermic** change

• Standard enthalpy change of neutralisation, ΔH^θ_neut

  • The enthalpy change when one mole of water is formed by reacting an acid and an alkali
  • It is an exothermic change

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) ΔH^θ_r* *= –286 kJ mol^-1

Calculate ΔH_r* *for the reaction below:

2H₂ (g) + O₂ (g) → 2H₂O (l)

Answer

• The ΔH^θ_r* *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:

  • ΔH_r* * = 2 mol x (–286 kJ mol^-1)
  • ΔH_r* * = –572 kJ mol^-1

Worked Example Calculating the enthalpy change of formation

Calculate ΔH^θ_f* *of the reaction below:

4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s) ΔH^θ_f* [Fe₂O₃ (s)] *= –824.2 kJ mol^-1

Answer

• The ΔH^θ_f* *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:

  • ΔH^θ_f* * = 2 mol x (–824.2kJ mol^-1)
  • ΔH^θ_f* * = –1648.4 kJ mol^-1

Worked Example Calculating enthalpy changes

Identify each of the following as ΔH^θ_r, ΔH^θ_f, ΔH^θ_c* or ΔH*^θ_neut:

1. MgCO₃ (s) → MgO (s) + CO₂ (g)
2. C (graphite) + O₂ (g) → CO₂ (g)
3. HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

Answers:

Answer 1

• It cannot be ΔH^θ_f as there is more than one compound being formed
• It cannot be ΔH^θ_c as there is no reaction with oxygen
• It cannot be ΔH^θ_neut as there is no acid or alkali involved* *
• This is just a chemical reaction, therefore, it is ΔH^θ_r

Answer 2

• ΔH^θ_f as one mole of CO₂ is formed from its elements in their standard states AND ΔH^θ_c as one mole of carbon is burnt completely in oxygen

Answer 3

• ΔH_neut^Ꝋ as one mole of water is formed from the reaction of an acid and alkali

Exam Hint The ΔH^θ_f of an element in its standard state is zero. For example, ΔH^θ_f of O₂ (g) is 0 kJ mol^-1