A-Level — chemistryHard5 min read

Entropy and Free Energy

Understand entropy, the second law of thermodynamics, and Gibbs free energy calculations for A-Level Chemistry.

Tutor AI Team
Tutor AI Team

# Entropy and Free Energy

Why do some endothermic reactions happen spontaneously? Enthalpy alone cannot explain this. The answer lies in entropy — a measure of disorder — and Gibbs free energy, which combines enthalpy and entropy to determine whether a reaction is feasible.

1. Entropy ($S$)

Entropy is a measure of the disorder or randomness of a system. The higher the entropy, the more disordered the system.

Units

Entropy: J K−1mol−1\text{J K}^{-1} \text{mol}^{-1} (note: joules, not kilojoules!)

Entropy and States of Matter

Ssolid<Sliquid≪SgasS_{\text{solid}} < S_{\text{liquid}} \ll S_{\text{gas}}

Gases have much higher entropy because particles move freely in all directions.

Factors That Increase Entropy

  1. Change of state: solid → liquid → gas (increasing disorder)
  2. Increase in number of moles of gas in products
  3. Dissolving a solid in a solvent (particles spread out)
  4. Increasing temperature (particles move faster, more disorder)
  5. Mixing substances

2. Entropy Change of Reaction

ΔSsystem=∑S⊖(products)−∑S⊖(reactants)\Delta S_{\text{system}} = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})

Example

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

Given: S⊖S^\ominus(CaCO₃) = 93, S⊖S^\ominus(CaO) = 40, S⊖S^\ominus(CO₂) = 214 J K⁻¹ mol⁻¹

ΔS=(40+214)−93=+161 J K−1mol−1\Delta S = (40 + 214) - 93 = +161 \text{ J K}^{-1}\text{mol}^{-1}

Positive ΔS\Delta S — entropy increases (a gas is produced from a solid).

3. The Second Law of Thermodynamics

For a spontaneous process, the total entropy of the universe must increase: ΔStotal=ΔSsystem+ΔSsurroundings>0\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0

The entropy change of the surroundings depends on the enthalpy change: ΔSsurroundings=−ΔHT\Delta S_{\text{surroundings}} = \frac{-\Delta H}{T}

(Exothermic reactions increase surroundings' entropy; endothermic reactions decrease it.)

4. Gibbs Free Energy

Combining system and surroundings entropy gives us Gibbs free energy:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Feasibility

  • ΔG<0\Delta G < 0: reaction is feasible (spontaneous)
  • ΔG=0\Delta G = 0: system is at equilibrium
  • ΔG>0\Delta G > 0: reaction is not feasible (non-spontaneous)

Relationship with KK

ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT\ln K

  • When ΔG<0\Delta G < 0: K>1K > 1 (products favoured)
  • When ΔG>0\Delta G > 0: K<1K < 1 (reactants favoured)
  • When ΔG=0\Delta G = 0: K=1K = 1 (at equilibrium)

5. Effect of Temperature on Feasibility

ΔH\Delta H ΔS\Delta S ΔG\Delta G Feasibility
− + Always − Always feasible at all temperatures
− − − at low T, + at high T Feasible at low temperatures
+ + + at low T, − at high T Feasible at high temperatures
+ − Always + Never feasible

Finding the Temperature Where ΔG=0\Delta G = 0

0=ΔH−TΔS0 = \Delta H - T\Delta S T=ΔHΔST = \frac{\Delta H}{\Delta S}

This is the temperature at which the reaction just becomes feasible (or ceases to be feasible).

Worked Example: Calculating $\Delta G$

Problem

Question: For a reaction: ΔH=−120\Delta H = -120 kJ/mol, ΔS=+50\Delta S = +50 J/K/mol at 298 K.

ΔG=−120−298×0.050=−120−14.9=−134.9 kJ mol−1\Delta G = -120 - 298 \times 0.050 = -120 - 14.9 = -134.9 \text{ kJ mol}^{-1}

ΔG<0\Delta G < 0 → feasible.

Solution

Worked Example: Temperature for Feasibility

Problem

Question: ΔH=+180\Delta H = +180 kJ/mol, ΔS=+120\Delta S = +120 J/K/mol. Above what temperature is this feasible?

T=180×103120=1500 KT = \frac{180 \times 10^3}{120} = 1500 \text{ K}

Feasible above 1500 K (1227°C1227°\text{C}).

Solution

Worked Example: Why Does NH₄Cl Dissolve Despite Being Endothermic?

Problem

ΔHsol=+15\Delta H_{\text{sol}} = +15 kJ/mol; ΔS=+100\Delta S = +100 J/K/mol

ΔG=15−298×0.100=15−29.8=−14.8 kJ mol−1\Delta G = 15 - 298 \times 0.100 = 15 - 29.8 = -14.8 \text{ kJ mol}^{-1}

ΔG<0\Delta G < 0 → feasible, because the large positive entropy change (solid → ions in solution) overcomes the endothermic enthalpy change.

Solution

7. Practice Questions

    1. Calculate ΔS\Delta S for: 2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s) Given: S⊖S^\ominus: Mg = 33, O₂ = 205, MgO = 27 J/K/mol
    1. Calculate ΔG\Delta G at 500 K for a reaction with ΔH=+40\Delta H = +40 kJ/mol and ΔS=+100\Delta S = +100 J/K/mol.
    1. A reaction has ΔH=−80\Delta H = -80 kJ/mol and ΔS=−200\Delta S = -200 J/K/mol. Below what temperature is it feasible?
    1. Explain why many decomposition reactions only occur at high temperatures in terms of entropy and free energy.
    1. For a reaction at equilibrium at 400 K with ΔS=+50\Delta S = +50 J/K/mol, calculate ΔH\Delta H.

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8. Exam Tips

  • Convert ΔS\Delta S from J to kJ before using ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
  • Entropy increases when gases are produced or number of gas moles increases
  • ΔG<0\Delta G < 0 means feasible but says nothing about rate
  • At T=ΔH/ΔST = \Delta H / \Delta S, the reaction is at the boundary of feasibility
  • Use ΔG=−RTln⁡K\Delta G = -RT\ln K to link free energy and equilibrium

Summary

  • Entropy (SS): measure of disorder; Sgas≫Sliquid>SsolidS_{\text{gas}} \gg S_{\text{liquid}} > S_{\text{solid}}
  • ΔSsystem=∑S(products)−∑S(reactants)\Delta S_{\text{system}} = \sum S(\text{products}) - \sum S(\text{reactants})
  • Gibbs free energy: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
  • ΔG<0\Delta G < 0 → feasible; ΔG>0\Delta G > 0 → not feasible
  • Temperature determines feasibility when ΔH\Delta H and ΔS\Delta S have the same sign
Tags:
#a-level#chemistry#thermodynamics#entropy#gibbs

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