Entropy Calculations (College Board AP® Chemistry): Study Guide
Entropy calculations
Standard entropies and entropy changes are usually given in J K-1 mol-1, while enthalpy changes are given in kJ mol-1
So, convert one of them before combining them in ΔG° = ΔH° − TΔS°
The unit of standard entropy is J K-1 mol-1 (joules per kelvin per mole)
The standard entropy change (ΔS°) for a reaction can be determined by utilizing the standard entropies (S°) of the reactants and products
The equation to calculate the standard entropy change of a system is:
ΔS°reaction = ΣS°(products) − ΣS°(reactants) (where Σ signifies the sum of)
Consider the standard entropy change for the formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2)
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
ΔS°reaction = 2S°(NH3) − [S°(N2) + 3S°(H2)]
Unlike enthalpy of formation for elements, entropy for elements is not zero
Worked Example
Calculate the entropy change of the system for the following reaction:
2Mg (s) + O2 (g) → 2MgO (s)
S°[Mg (s)] = 32.60 J K-1 mol-1
S°[O2 (g)] = 205.0 J K-1 mol-1
S°[MgO (s)] = 38.20 J K-1 mol-1
Answer:
State the general entropy equation:
ΔS°reaction = ΣS°(products) − ΣS°(reactants)
Make it reaction specific:
ΔS°reaction = 2S°(MgO) − [2S°(Mg) + S°(O2)]
Substitute the relevant values:
ΔS°reaction = (2 × 38.20) − (2 × 32.60 + 205.0)
Solve:
ΔS°reaction = −193.8 J K-1 mol-1
A negative value indicates that there has been a decrease in entropy
This matches the particle-level prediction: 1 mol of gas is consumed and only a solid forms, so matter becomes less dispersed
Examiner Tips and Tricks
Multiply each S° by its coefficient in the balanced equation, and keep the order products − reactants as reversing it reverses the sign of ΔS°.
Then check the sign against the particles. In the worked example above, 1 mol of gas is consumed and none is formed, so ΔS° must be negative.
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