Topic 7 · Chemical Energetics II · spread

Chemical Energetics II

7

Spontaneity and Entropy

Spontaneous reaction: a reaction that is energetically feasible, i.e. ΔG⦵ < 0. Once started, it proceeds on its own without a continuous input of energy from the surroundings.

NoteSpontaneous says nothing about rate: a spontaneous reaction can be very slow.

7.1Entropy

DefinitionEntropy, S: a measure of the disorder in a system. The more disordered a system, the greater the number of ways for particles and energy to be dispersed, therefore the larger is the entropy.
ΔS = Sfinal − Sinitial units: J mol⁻¹ K⁻¹
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Factors Affecting Entropy

Temperature ↑

ΔS > 0: particles have more average kinetic energy and there are more ways to distribute the larger number of quanta of energy.

Change in phase

S: solid < liquid ≪ gas. Melting: particles no longer in fixed positions → more ways to arrange particles and distribute energy. Boiling: particles move randomly, large ↑ in volume.

More gas particles

ΔS > 0: disorder increases as the number of moles of gas molecules increases, e.g. 2NaN₃(s) → 2Na(s) + 3N₂(g). No change in moles of gas → ΔS ≈ 0.

Predict ΔSSignReason
2N₂O(g) + O₂(g) → 4NO(g)+moles of gas ↑ (3 → 4) → more ways to distribute particles and energy
2H₂O₂(l) → 2H₂O(l) + O₂(g)+a gas forms from liquids
3O₂(g) → 2O₃(g)−moles of gas ↓ (3 → 2)
2SO₂(g) + O₂(g) → 2SO₃(g)−moles of gas ↓ (3 → 2) → fewer ways to distribute
NoteDissolving an ionic solid: two opposing effects: ions become free to move (S ↑), but water molecules become more ordered around the ions (S ↓). Usually the first wins, e.g. NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq), ΔS > 0.
34Topic 07 · Chemical Energetics II
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Gibbs Free Energy

ΔG⦵ = ΔH⦵ − TΔS⦵T in kelvin, convert ΔS to kJ mol⁻¹ K⁻¹ before use
DefinitionThe standard Gibbs free energy, ΔG⦵, is used to predict the feasibility of a reaction.
ΔG < 0spontaneous, K > 1, products favoured
ΔG = 0at equilibrium (e.g. at m.p. or b.p.)
ΔG > 0non-spontaneous, spontaneous in reverse
ΔGr = ΣΔGf(products) − ΣΔGf(reactants)

Worked example: RustingIron rusts in air:
2Fe(s) + 32O₂(g) → Fe₂O₃(s)
ΔH = −825 kJ mol⁻¹ and ΔS = −272 J mol⁻¹ K⁻¹. Is rusting spontaneous at 298 K?

  1. Convert ΔS to kJ: −272 J mol⁻¹ K⁻¹ = −0.272 kJ mol⁻¹ K⁻¹
  2. ΔG = ΔH − TΔS = −825 − 298 × (−0.272)
  3. ΔG = −744 kJ mol⁻¹
  4. ΔG < 0 → spontaneous at 298 K

Worked example: Melting iceIce melts at 273 K. The enthalpy change of fusion is +6.0 kJ mol⁻¹. Find ΔS for melting.

  1. At the m.p., ice and water are at equilibrium → ΔG = 0
  2. 0 = ΔH − TΔS → ΔS = ΔH ÷ T
  3. ΔS = 6000 ÷ 273 = +22.0 J mol⁻¹ K⁻¹
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10

Effect of Temperature on Spontaneity

ΔHΔSSpontaneity
−+Since ΔS > 0, −TΔS < 0 → ΔG negative at all temperatures: always spontaneous
+−Since ΔS < 0, −TΔS > 0 → ΔG positive at all temperatures: never spontaneous
−−Since ΔS < 0, −TΔS > 0. As T ↑, −TΔS becomes more positive → ΔG positive eventually: spontaneous only at low T
++Since ΔS > 0, −TΔS < 0. As T ↑, −TΔS becomes more negative → ΔG negative eventually: spontaneous at high T
TΔGΔG > 0ΔG < 0: spontaneous0ΔH −, ΔS +ΔH +, ΔS −ΔH −, ΔS −ΔH +, ΔS +
CaseExamples
ΔH −, ΔS +combustion of octane, decomposition of nitroglycerine
ΔH +, ΔS −photosynthesis: must be driven by a continuous energy input (sunlight)
ΔH −, ΔS −precipitation of AgCl, condensation, freezing (enthalpy driven)
ΔH +, ΔS +melting, boiling, decomposition of Group 2 carbonates (entropy driven)

Worked example: Minimum temperatureIn a blast furnace:
C(s) + CO₂(g) → 2CO(g)
ΔH = +172.5 kJ mol⁻¹ and ΔS = +176.2 J mol⁻¹ K⁻¹. Find the minimum temperature for the reaction to be spontaneous.

  1. At 298 K: ΔG = 172.5 − 298 × 0.1762 = +120 kJ mol⁻¹ > 0 → not spontaneous
  2. Spontaneous when ΔG < 0: ΔH − TΔS < 0 → T > ΔH ÷ ΔS
  3. T > 172.5 ÷ 0.1762 = 979 K

Worked exampleBaking powder releases CO₂ on heating:
2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)
ΔH = +130 kJ mol⁻¹ and ΔS = +335 J mol⁻¹ K⁻¹. Above what temperature is this spontaneous?

  1. At 298 K: ΔG = 130 − 298 × 0.335 = +30.2 kJ mol⁻¹ > 0 → not spontaneous
  2. Spontaneous when ΔG < 0 → T > ΔH ÷ ΔS
  3. T > 130 ÷ 0.335 = 388 K

Likewise MgCO₃ decomposes above 576 K but BaCO₃ only above 1566 K: ΔH(MgCO₃) is less positive, ΔS similar.

36Topic 07 · Chemical Energetics II
11

Limitations of ΔG⦵

11.1Standard states

ΔG⦵ is worked out for standard conditions only (1 bar, 1 mol dm⁻³). Real conditions are often different, so ΔG⦵ can give the wrong prediction.

e.g. H₂O(l) → H₂O(g) at 298 K: ΔG⦵ = 44.0 − 298 × 0.118 = +8.84 kJ mol⁻¹, so water should not boil. Yet at a low enough pressure, water does boil at 298 K.

11.2Temperature

Using ΔG = ΔH − TΔS at a new temperature assumes ΔH and ΔS stay constant as T changes. This is not true if a substance changes phase in between, as ΔH and ΔS then change a lot.

e.g. H₂ + I₂ → 2HI: ΔG = +3.5 kJ mol⁻¹ at 298 K (I₂ solid) but −20.0 kJ mol⁻¹ at 500 K (I₂ gas), so 298 K data cannot predict ΔG at 500 K.

11.3Kinetics

ΔG only tells us whether a reaction is energetically feasible, not how fast it goes.

C(diamond) → C(graphite) has ΔG < 0, so it is spontaneous. But its Ea is very high, so almost no particles have enough energy to react at room temperature, and the change is far too slow to observe.

→ Diamond is energetically unstable but kinetically stable relative to graphite.

energyreaction coordinateC(diamond)C(graphite)EₐΔG < 0
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