A Level Chemistry - How to Solve: Entropy and Gibbs Free Energy (ΔG = ΔH − TΔS, Feasibility)

How to Solve: Entropy and Gibbs Free Energy (ΔG = ΔH − TΔS, Feasibility)

Complete Guide for GCSE/A-Level (Physics, Chemistry, Biology)

Introduction

"Mastering ΔG = ΔH − TΔS lets you predict whether a reaction will happen spontaneously—like why ice melts at room temperature or why batteries work. This formula appears in every A-Level Chemistry and Physics exam, often worth 6–10 marks in a single question. Get it right, and you’re halfway to an A."

WHAT YOU NEED TO KNOW FIRST

Before tackling Gibbs Free Energy, you must understand:
1. Enthalpy (ΔH) – Heat energy change at constant pressure (exothermic = negative, endothermic = positive).
2. Entropy (ΔS) – A measure of disorder (more disorder = positive ΔS, e.g., solid → liquid → gas).
3. Temperature (T) – Must be in Kelvin (K) (0°C = 273 K).

KEY TERMS & FORMULAS

1. Gibbs Free Energy (ΔG)

Formula: ΔG = ΔH − TΔS (MEMORISE THIS – it’s the core equation for feasibility.)

Variables: - ΔG = Gibbs Free Energy (kJ mol⁻¹) → Negative = spontaneous (feasible), Positive = non-spontaneous, Zero = equilibrium - ΔH = Enthalpy change (kJ mol⁻¹) (given on exam sheet or in question) - T = Temperature (K) (convert °C to K by +273) - ΔS = Entropy change (J K⁻¹ mol⁻¹) (given on exam sheet or in question – watch units!)

Key Notes: - ΔS must be in kJ K⁻¹ mol⁻¹ if ΔH is in kJ (divide by 1000 if given in J). - Feasibility rule: A reaction is spontaneous (feasible) if ΔG < 0.

2. Entropy (ΔS)

Definition: A measure of disorder in a system. - Increases when: - Solid → liquid → gas - More moles of gas are produced - Temperature increases - Decreases when: - Gas → liquid → solid - Fewer moles of gas are produced

Standard Entropy (S°): Entropy of 1 mole of a substance under standard conditions (J K⁻¹ mol⁻¹).

3. Enthalpy (ΔH)

Definition: Heat energy change at constant pressure. - Exothermic (ΔH < 0): Releases heat (e.g., combustion). - Endothermic (ΔH > 0): Absorbs heat (e.g., melting ice).

STEP-BY-STEP METHOD

Follow these steps for every ΔG question:

1. Identify ΔH and ΔS from the question.
2. If ΔS is in J K⁻¹ mol⁻¹, convert to kJ K⁻¹ mol⁻¹ (÷1000).

3. If temperature is in °C, convert to K (+273).

4. Write down the Gibbs equation: ΔG = ΔH − TΔS

5. Substitute the values into the equation.

6. Double-check units (ΔH in kJ, ΔS in kJ K⁻¹, T in K).

7. Calculate ΔG.

8. If ΔG < 0 → spontaneous (feasible)
9. If ΔG > 0 → non-spontaneous

10. If ΔG = 0 → equilibrium

11. If asked for feasibility at different temperatures:

12. Set ΔG = 0 and solve for T: T = ΔH / ΔS
13. This gives the minimum/maximum temperature for feasibility.

WORKED EXAMPLES

Example 1 – Basic (Direct Substitution)

Question: For a reaction, ΔH = −50 kJ mol⁻¹, ΔS = +100 J K⁻¹ mol⁻¹, T = 300 K. Calculate ΔG and state if the reaction is feasible.

Solution:
1. Convert ΔS to kJ: ΔS = +100 J K⁻¹ mol⁻¹ = +0.1 kJ K⁻¹ mol⁻¹

1. Write the equation: ΔG = ΔH − TΔS

2. Substitute values: ΔG = (−50) − (300 × 0.1) ΔG = −50 − 30 ΔG = −80 kJ mol⁻¹

3. Conclusion: ΔG < 0 → Spontaneous (feasible) at 300 K.

What we did and why: - Converted ΔS to kJ to match ΔH’s units. - Substituted directly into ΔG = ΔH − TΔS. - Negative ΔG means the reaction happens on its own.

Example 2 – Medium (Finding Feasibility Temperature)

Question: A reaction has ΔH = +60 kJ mol⁻¹ and ΔS = +150 J K⁻¹ mol⁻¹. At what temperature does the reaction become feasible?

Solution:
1. Convert ΔS to kJ: ΔS = +150 J K⁻¹ mol⁻¹ = +0.15 kJ K⁻¹ mol⁻¹

1. For feasibility, ΔG < 0: ΔG = ΔH − TΔS < 0

2. Set ΔG = 0 and solve for T: 0 = 60 − T(0.15) T(0.15) = 60 T = 60 / 0.15 T = 400 K

3. Conclusion: The reaction becomes feasible above 400 K (127°C).

What we did and why: - Used ΔG = 0 to find the minimum temperature for feasibility. - Since ΔH is positive (endothermic), the reaction only works at high temperatures.

Example 3 – Exam-Style (Disguised Question)

Question: The decomposition of calcium carbonate is: CaCO₃(s) → CaO(s) + CO₂(g) ΔH = +178 kJ mol⁻¹, ΔS = +161 J K⁻¹ mol⁻¹. At 298 K, is this reaction feasible? If not, at what temperature does it become feasible?

Solution:
1. Convert ΔS to kJ: ΔS = +161 J K⁻¹ mol⁻¹ = +0.161 kJ K⁻¹ mol⁻¹

1. Calculate ΔG at 298 K: ΔG = ΔH − TΔS ΔG = 178 − (298 × 0.161) ΔG = 178 − 47.978 ΔG = +130.022 kJ mol⁻¹

2. Conclusion at 298 K: ΔG > 0 → Not feasible at 298 K.

3. Find feasibility temperature: Set ΔG = 0: 0 = 178 − T(0.161) T = 178 / 0.161 T = 1105.6 K (832.6°C)

What we did and why: - Calculated ΔG at room temperature (298 K) and found it’s not feasible. - Found the minimum temperature (1106 K) where ΔG becomes negative. - This matches real-world lime kilns, which operate at ~900°C.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Okay, listen up—this is the one-minute crash course on Gibbs Free Energy. Memorise this equation: ΔG = ΔH − TΔS. ΔG tells you if a reaction happens on its own—negative = yes, positive = no. First, convert everything to kJ and Kelvin—ΔS is often in J, so divide by 1000, and add 273 to °C. Plug the numbers in. If ΔG is negative, it’s feasible. If you’re asked for the temperature where it becomes feasible, set ΔG = 0 and solve for T. Watch out for unit traps—examiners love giving ΔS in J and ΔH in kJ. And remember: exothermic reactions aren’t always feasible, and endothermic reactions can be feasible at high temperatures. That’s it—go ace that question!"