Ch 4 · Gibbs free energy change, ΔG · AS & A Level Chemistry 9701
Topic 4 of 4 in Chemical energetics — AS & A Level Chemistry 9701 — 48 lessons.
Introduction
Enthalpy change alone cannot predict whether a reaction will occur, since many endothermic processes still happen spontaneously. Gibbs free energy combines enthalpy and entropy into a single quantity whose sign shows whether a reaction is thermodynamically feasible at a given temperature.
Key ideas — The Gibbs equation
The Gibbs free energy change, Δ G, of a process at temperature T (in kelvin) is given by:
🧑🏫 Teacher's explanation
Think of a chemical reaction as a business venture. For a business to be successful (feasible), it needs to satisfy two conditions: it must be profitable (low enthalpy, Δ H < 0) and it must have a high level of organizational efficiency or 'momentum' (high entropy, Δ S > 0). However, there is a trade-off. Sometimes a reaction is profitable but messy, or messy but profitable. The Gibbs free energy, Δ G, is the ultimate 'bottom line' that tells you if the reaction will actually happen spontaneously under specific conditions.
📖 Definitions
🔍 Worked example — Easy
Calculate Δ G for a reaction where Δ H = -150 kJ mol^-1 and Δ S = +0.200 kJ K^-1 mol^-1 at T = 298 K.
🔍 Worked example — Standard
For a reaction, Δ H = +80 kJ mol^-1 and Δ S = +0.300 kJ K^-1 mol^-1. Determine if the reaction is feasible at 298 K.
🔍 Worked example — Exam-style
A reaction has Δ H = +172 kJ mol^-1 and Δ S = +0.160 kJ K^-1 mol^-1. Calculate the temperature above which the reaction becomes feasible.
⚠️ Common errors
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🎯 Exam question
Key ideas — Feasibility from the sign of ΔG
A reaction or process is described as feasible (thermodynamically favourable) when Δ G is zero or negative; it is not feasible when Δ G is positive. A feasible reaction is not guaranteed to happen at an observable rate: Δ G says nothing about the activation energy or rate of a reaction, only whether it could occur given enough time, so a thermodynamically feasible reaction can still appear not to proceed because it is kinetically very slow.
🧑🏫 Teacher's explanation
Imagine you are standing at the top of a hill with a ball. The ball wants to roll down to the bottom because that is a state of lower energy. In chemistry, 'feasibility' is like asking if the ball *wants* to roll down. Even if there is a small bump (activation energy) stopping it, the ball is technically 'destined' to reach the bottom. Δ G tells us if that 'downhill' movement is possible based on both energy (enthalpy) and disorder (entropy).
📖 Definitions
🔍 Worked example — Easy
For a reaction, Δ H = -100 kJ mol^-1 and Δ S = -200 J K^-1 mol^-1. Calculate Δ G at 298 K.
🔍 Worked example — Standard
A reaction has Δ H = +50 kJ mol^-1 and Δ S = +150 J K^-1 mol^-1. Determine if the reaction is feasible at 298 K.
🔍 Worked example — Exam-style
A reaction is non-feasible at 300 K but becomes feasible at 500 K. Given Δ H = +80 kJ mol^-1 and Δ S = +300 J K^-1 mol^-1, calculate the temperature at which the reaction becomes feasible (Δ G = 0).
⚠️ Common errors
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🎯 Exam question
Key ideas — Calculating ΔG
Given Δ H, Δ S and a temperature, Δ G is found directly by substitution into Δ G = Δ H - TΔ S. The reverse calculation, finding the temperature at which a reaction becomes exactly feasible, sets Δ G = 0 and rearranges to:
🧑🏫 Teacher's explanation
Imagine you are trying to decide whether to go for a run. Your 'motivation' (enthalpy change, Δ H) is the energy you gain from the endorphins, but the 'chaos' or 'effort' (entropy change, Δ S) is the mess and exhaustion you deal with. The Gibbs free energy (Δ G) is your final decision: if the net result is negative, the run is 'feasible' or worth it.
📖 Definitions
| Term | Cambridge wording | | --- | --- | | Gibbs free energy change (Δ G) | the change in Gibbs free energy of a system during a chemical reaction | | Feasible reaction | a reaction that can occur spontaneously under a given set of conditions | | Spontaneity | the tendency of a process to occur without continuous external intervention |
🔍 Worked example — Easy
Calculate Δ G for a reaction where Δ H = -150 kJ mol^-1, Δ S = +120 J K^-1 mol^-1 at T = 298 K.
🔍 Worked example — Standard
A reaction has Δ H = +80 kJ mol^-1 and Δ S = +0.25 kJ K^-1 mol^-1. Calculate the temperature at which the reaction becomes feasible.
🔍 Worked example — Exam-style
For the reaction N_2(g) + 3H_2(g) leftharpoons 2NH_3(g), Δ H = -92 kJ mol^-1 and Δ S = -198 J K^-1 mol^-1. Determine if the reaction is feasible at 500 K.
⚠️ Common errors
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🎯 Exam question
(a) Calculate the temperature (in Kelvin) at which the decomposition becomes feasible. [2] (b) Explain why the reaction is non-feasible at 298 K. [2] [4 marks]
Key ideas — Effect of temperature on feasibility
Since Δ G = Δ H - TΔ S, the effect of temperature on feasibility depends on the signs of Δ H and Δ S. If Δ H is negative and Δ S is positive, Δ G is negative at every temperature, so the reaction is always feasible. If Δ H is positive and Δ S is negative, Δ G is always positive, so the reaction is never feasible. If Δ H is negative and Δ S is negative, Δ G becomes positive above a certain temperature, so the reaction is only feasible below that temperature. If Δ H is positive and Δ S is positive, as in most thermal decompositions, Δ G becomes negative above a certain temperature, so the reaction only becomes feasible once heated above it.
🧑🏫 Teacher's explanation
Think of Gibbs free energy (Δ G) as the 'budget' for a chemical reaction. For a reaction to happen spontaneously (feasibly), the universe wants to 'save energy' and 'increase chaos'. If the total budget Δ G is negative, the reaction is 'on sale' and will happen.
📖 Definitions
🔍 Worked example — Easy
Calculate Δ G for a reaction at 298 K where Δ H = -150 kJ mol^-1 and Δ S = +200 J K^-1 mol^-1.
🔍 Worked example — Standard
A reaction has Δ H = +80 kJ mol^-1 and Δ S = +150 J K^-1 mol^-1. Calculate the temperature (in Kelvin) at which the reaction becomes feasible.
🔍 Worked example — Exam-style
For the decomposition of calcium carbonate, Δ H = +178 kJ mol^-1 and Δ S = +160 J K^-1 mol^-1. Determine if the reaction is feasible at 1000 K.
⚠️ Common errors
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🎯 Exam question
Worked example
Question. For the decomposition CaCO_3(s) arrow CaO(s) + CO_2(g), Δ H^ominus = +178\ mathrmkJ\ mol^-1 and Δ S^ominus = +160.4\ mathrmJ\ K^-1\ mol^-1. (a) Calculate Δ G^ominus at 298 K and state whether the reaction is feasible at this temperature. [3] (b) Calculate the minimum temperature at which the reaction becomes feasible. [2]
🔍 Worked example — Easy
Calculate Δ G for a reaction where Δ H = -120 kJ mol^-1, Δ S = -150 J K^-1 mol^-1 at T = 300 K.
🔍 Worked example — Standard
For the reaction CaCO_3(s) arrow CaO(s) + CO_2(g), Δ H = +178 kJ mol^-1 and Δ S = +160.4 J K^-1 mol^-1. Calculate Δ G at 298 K and state if it is feasible.
🔍 Worked example — Exam-style
At what temperature (in K) does the reaction in the previous example become feasible?
⚠️ Common errors
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