IB DP Chemistry: HL復(fù)習(xí)筆記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:
ΔS?298(reaction) = ΣS?298(products) - ΣS?298(reactants)
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- This equation is provided in the data booklet
- The units of ΔSsystem??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?S??will be different for the liquid and gaseous phases
- S?298(H2O (l)) = 70.0 J K-1?mol–1
- S?298(H2O (g)) = 188.8 J K-1?mol–1
- When calculating ΔS?, the coefficients used to balance the equation must be applied when calculating the overall entropy change
- For example, when calculating the ΔS??for the reaction below we need to double the value for?S??(NO (g))
- N2O4?(g) →?2NO2?(g)
- ΔS?298(reaction) = ΣS?298(products) - ΣS?298(reactants)
- ΔS??= [(2?x?S?298(NO2)] -?S?298(N2O4)
Worked Example
What is the entropy change when calcium carbonate decomposes?
CaCO3?(s) → CaO (s) + CO2?(g)
- S?298(CaCO3?(s)) = 92.9 J K-1?mol–1
- S?298(CaO (s)) = 39.7 J K-1?mol–1
- S?298(CO2?(g)) = 213.6 J K-1?mol–1
Answer:
Step 1:?Write out equation to calculate ΔS?298(reaction)
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- ΔS?298(reaction) = ΣS?298(products) - ΣS?298(reactants)
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Step 2:?Substitute in formulas and then values for?S?
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- ΔS?298(reaction) = [S?298(CaO) +?S?298(CO2)] -?S?298(CaCO3)
- ΔS?(reaction) = (39.7 + 213.6) - 92.9
- ΔS?(reaction) =?+160.4 J K-1?mol–1
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Worked Example
What is the entropy change when ammonia is formed?from?nitrogen and hydrogen?
N2?(g) + 3H2?(g)???2NH3?(g)
- S?298(N2?(g)) = 191.6 J K-1?mol–1
- S?298(H2?(g)) = 131 J K-1?mol–1
- S?298(NH3) = 192.3 J K-1?mol–1
Answer:
Step 1:?Write out equation to calculate ΔS?298(reaction)
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- ΔS?298(reaction) = ΣS?298(products) - ΣS?298(reactants)
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Step 2:?Substitute in formulas and then values for?S??taking into account the coefficients
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- ΔS?298(reaction) = [2 x?S?298(NH3)]?- [S?298(N2)+ (3 x?S?298(H2?))]
- ΔS?298(reaction) = [2 x 192.3] - [191.6 + (3 x 131)]
- ΔS?298(reaction) = 384.6 - 584.6
- ΔS?298(reaction) =?-200 J K-1?mol–1
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