Key Learning Objectives
- Calculate standard entropy changes (ΔS) from absolute entropy data.
- Apply the Gibbs Free Energy equation: ΔG = ΔH - TΔS to determine reaction feasibility.
- Calculate the threshold temperature at which a non-spontaneous reaction becomes feasible (ΔG ≤ 0).
Topic Summary & Core Notes
Master thermodynamics and entropy calculations. Solve ΔS = ΣS(products) - ΣS(reactants), handle J/K/mol to kJ/mol conversions, and calculate feasibility with ΔG = ΔH - TΔS.
Key Rules, Formulas & Definitions
Standard Entropy Change
ΔS = Σ S°(products) - Σ S°(reactants) [units: J K⁻¹ mol⁻¹]
Gibbs Free Energy
ΔG = ΔH - TΔS [Feasible when ΔG ≤ 0; convert ΔS from J to kJ by dividing by 1,000]
Feasibility Temperature
T = ΔH ÷ ΔS (when ΔG = 0)
2 Step-by-Step Worked Examples
Worked Example 1
Worked Example 1: Calculating Feasibility Temperature
Question: For a reaction, ΔH = +178 kJ/mol and ΔS = +161 J K⁻¹ mol⁻¹. Calculate the temperature above which the reaction becomes feasible.
Step-by-Step Solution:
- Step 1: Convert units: ΔS = +161 J K⁻¹ mol⁻¹ ÷ 1,000 = +0.161 kJ K⁻¹ mol⁻¹.
- Step 2: Set ΔG = 0: ΔG = ΔH - TΔS = 0 → T = ΔH ÷ ΔS.
- Step 3: Calculate temperature: T = 178 ÷ 0.161 = 1,105.6 K.
Final Answer: 1,106 K
Senior Examiner Insight: The number 1 student error in thermodynamics is forgetting to divide ΔS by 1,000 to convert J to kJ.
Worked Example 2
Worked Example 2: Qualitative Entropy Predictions
Question: Explain why the decomposition reaction CaCO3(s) → CaO(s) + CO2(g) has a large positive entropy change (ΔS > 0).
Step-by-Step Solution:
- Step 1: State changes: 1 mole of solid reactant produces 1 mole of solid and 1 mole of gas.
- Step 2: Particle disorder: Gas molecules have far greater translational freedom, random motion, and dispersal of energy than ordered crystalline solids.
- Step 3: Conclude: Producing a gas from a solid substantially increases system disorder (entropy).
Final Answer: A gas (CO2) is produced from a solid. Gas particles are much more disordered with higher dispersal of energy quanta.
Senior Examiner Insight: Always mention the state change from solid to gas and the dramatic increase in disorder / dispersal of energy.
Common Pitfalls & Examiner Warnings
Common Mistake: Unit mismatch between ΔH (kJ) and ΔS (J)
ΔH is measured in kJ/mol while ΔS is in J K⁻¹ mol⁻¹. You MUST divide ΔS by 1,000 before substituting into ΔG = ΔH - TΔS.
Common Mistake: Assuming feasible reactions occur instantly
ΔG < 0 means thermodynamically feasible, but high activation energy may make the reaction kinetically inert.
Interactive Self-Check Quiz (3 Questions)
Question 1 of 3
What is the critical condition required for a chemical reaction to be thermodynamically feasible?
Click to Reveal Answer & Mark Scheme
Correct Answer: A) ΔG ≤ 0
Detailed Explanation: A reaction is thermodynamically feasible when the Gibbs free energy change ΔG is zero or negative.
Mark Scheme & Scoring Points: 1 mark for ΔG ≤ 0.
Question 2 of 3
If a reaction has ΔH > 0 (endothermic) and ΔS < 0 (decreased disorder), at what temperatures is it feasible?
Click to Reveal Answer & Mark Scheme
Correct Answer: A) Never feasible at any temperature
Detailed Explanation: If ΔH is positive and -TΔS is positive (since ΔS < 0), ΔG will ALWAYS be positive regardless of temperature.
Mark Scheme & Scoring Points: 1 mark for Never feasible at any temperature.
Question 3 of 3
Which substance in its standard state has an absolute entropy S° equal to ZERO J K⁻¹ mol⁻¹?
Click to Reveal Answer & Mark Scheme
Correct Answer: A) A perfect crystal at absolute zero (0 Kelvin)
Detailed Explanation: By the Third Law of Thermodynamics, only a perfect crystal at 0 K has an entropy of zero.
Mark Scheme & Scoring Points: 1 mark for a perfect crystal at absolute zero (0 K).
Created by Fiaraz Iqbal
Former Headteacher & Senior Science / Maths Examiner
Fiaraz produces structured video walkthroughs, Tier 3 literacy packs, and exam mark scheme breakdowns to help secondary students achieve top grades.
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