Standard Enthalpy Change (Cambridge (CIE) AS Chemistry): Revision Note

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 

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 

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

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