Calculating Standard Entropy Changes
- The standard molar enthalpy values, Sꝋ, relate to standard conditions of temperature and pressure
- The entropy change, ΔSꝋ, can be calculated from thermodynamic data using the following equation:
ΔSꝋ298(reaction) = ΣSꝋ298(products) - ΣSꝋ298(reactants)
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- This equation is provided in the data booklet
- The units of ΔSsystemꝋ are in J K-1 mol–1
- Entropy will change depending on the state of the matter
- Taking water as an example the values for Sꝋ will be different for the liquid and gaseous phases
- Sꝋ298(H2O (l)) = 70.0 J K-1 mol–1
- Sꝋ298(H2O (g)) = 188.8 J K-1 mol–1
- Taking water as an example the values for Sꝋ will be different for the liquid and gaseous phases
- When calculating ΔSꝋ, the coefficients used to balance the equation must be applied when calculating the overall entropy change
- For example, when calculating the ΔSꝋ for the reaction below we need to double the value for Sꝋ (NO (g))
- N2O4 (g) → 2NO2 (g)
- ΔSꝋ298(reaction) = ΣSꝋ298(products) - ΣSꝋ298(reactants)
- ΔSꝋ = [(2 x Sꝋ298(NO2)] - Sꝋ298(N2O4)
Worked example
What is the entropy change when calcium carbonate decomposes?
CaCO3 (s) → CaO (s) + CO2 (g)
- Sꝋ298(CaCO3 (s)) = 92.9 J K-1 mol–1
- Sꝋ298(CaO (s)) = 39.7 J K-1 mol–1
- Sꝋ298(CO2 (g)) = 213.6 J K-1 mol–1
Answer:
Step 1: Write out the equation to calculate ΔSꝋ298(reaction)
- ΔSꝋ298(reaction) = ΣSꝋ298(products) - ΣSꝋ298(reactants)
Step 2: Substitute in formulas and then values for Sꝋ
- ΔSꝋ298(reaction) = [Sꝋ298(CaO) + Sꝋ298(CO2)] - Sꝋ298(CaCO3)
- ΔSꝋ(reaction) = (39.7 + 213.6) - 92.9
- ΔSꝋ(reaction) = +160.4 J K-1 mol–1
Worked example
What is the entropy change when ammonia is formed from nitrogen and hydrogen?
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
- Sꝋ298(N2 (g)) = 191.6 J K-1 mol–1
- Sꝋ298(H2 (g)) = 131 J K-1 mol–1
- Sꝋ298(NH3) = 192.3 J K-1 mol–1
Answer:
Step 1: Write out the equation to calculate ΔSꝋ298(reaction)
- ΔSꝋ298(reaction) = ΣSꝋ298(products) - ΣSꝋ298(reactants)
Step 2: Substitute in formulas and then values for Sꝋ taking into account the coefficients
- ΔSꝋ298(reaction) = [2 x Sꝋ298(NH3)] - [Sꝋ298(N2)+ (3 x Sꝋ298(H2 ))]
- ΔSꝋ298(reaction) = [2 x 192.3] - [191.6 + (3 x 131)]
- ΔSꝋ298(reaction) = 384.6 - 584.6
- ΔSꝋ298(reaction) = -200 J K-1 mol–1
