IB Chemistry HL

Revision Notes

15.2.2 Calculating Entropy Change

Calculating Standard Entropy Change

  • The standard molar enthalpy values, S, relate to standard conditions of temperature and pressure
  • The entropy change, ΔS, can be calculated from thermodynamic data using the following equation:

ΔS298(reaction) = ΣS298(products) – ΣS298(reactants)

    • This equation is provided in the data booklet
  • The units of ΔSsystem are in J K-1 mol1
  • 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
    • S298(H2O (l)) = 70.0 J K-1 mol1
    • S298(H2O (g)) = 188.8 J K-1 mol1
  • When calculating ΔS, 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 S(NO (g))
    • N2O4 (g) → 2NO2 (g)
    • ΔS298(reaction) = ΣS298(products) – ΣS298(reactants)
    • ΔS= [(2 x S298(NO2)] – S298(N2O4)

Worked Example

What is the entropy change when calcium carbonate decomposes?

CaCO3 (s) → CaO (s) + CO2 (g)

  • S298(CaCO3 (s)) = 92.9 J K-1 mol1
  • S298(CaO (s)) = 39.7 J K-1 mol1
  • S298(CO2 (g)) = 213.6 J K-1 mol1

Answer:

Step 1: Write out equation to calculate ΔS298(reaction)

      • ΔS298(reaction) = ΣS298(products) – ΣS298(reactants)

Step 2: Substitute in formulas and then values for S

      • ΔS298(reaction) = [S298(CaO) + S298(CO2)] – S298(CaCO3)
      • ΔS(reaction) = (39.7 + 213.6) – 92.9
      • ΔS(reaction) = +160.4 J K-1 mol1

Worked Example

What is the entropy change when ammonia is formed from nitrogen and hydrogen?

N2 (g) + 3H2 (g) 2NH3 (g)

  • S298(N2 (g)) = 191.6 J K-1 mol1
  • S298(H2 (g)) = 131 J K-1 mol1
  • S298(NH3) = 192.3 J K-1 mol1

Answer:

Step 1: Write out equation to calculate ΔS298(reaction)

      • ΔS298(reaction) = ΣS298(products) – ΣS298(reactants)

Step 2: Substitute in formulas and then values for Staking into account the coefficients

      • ΔS298(reaction) = [2 x S298(NH3)] – [S298(N2)+ (3 x S298(H2 ))]
      • ΔS298(reaction) = [2 x 192.3] – [191.6 + (3 x 131)]
      • ΔS298(reaction) = 384.6 – 584.6
      • ΔS298(reaction) = -200 J K-1 mol1
Close

Join Save My Exams

Download all our Revision Notes as PDFs

Try a Free Sample of our revision notes as a printable PDF.

Join Now
Already a member?
Go to Top