Calculating Standard Entropy Changes (HL) (DP IB Chemistry)
Revision Note
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)
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
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
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