The shape of the lesson
By the end of the lesson, learners can
- The Gibbs equation: Work through the decomposition of calcium carbonate at two temperatures.
- Temperature decides: Slide the temperature in the simulator for the decomposition of calcium carbonate.
- ΔG = ΔH − TΔS, T in K.
- ΔS ÷ 1000 to get kJ K−1 mol−1.
- Feasible when ΔG ≤ 0.
- Switch temperature: T = ΔH / ΔS; say above or below.
One picture every idea returns to
The Gibbs free energy change combines enthalpy and entropy: ΔG° = ΔH° − TΔS°, with T in kelvin. ΔH is in kJ mol−1 but ΔS is in J K−1 mol−1, so divide ΔS by 1000 first. A reaction is feasible when ΔG ≤ 0. When ΔH and ΔS have the same sign, temperature decides: the switch-over happens where ΔG = 0, at T = ΔH / ΔS. For CaCO3 → CaO + CO2 (ΔH +178 kJ mol−1, ΔS +161 J K−1 mol−1), T = 178 / 0.161 = 1106 K. A feasible reaction may still be very slow if its activation energy is high.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Will it go? Ask ΔG.. Enthalpy alone cannot say whether a reaction is feasible: some endothermic reactions happen on their own. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: The Gibbs equation | Work through the decomposition of calcium carbonate at two temperatures. Project tab 6 of the lab (Chemical energetics (A Level) lab, Gibbs and temperature) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Temperature decides | Slide the temperature in the simulator for the decomposition of calcium carbonate. Project tab 6 of the lab (Chemical energetics (A Level) lab, Gibbs and temperature) and let a student drive it while the class predicts. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Temperature decides, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The classic slip is forgetting to divide Δ S by 1000: 178 − 298 × 161 gives a nonsense −47 800 kJ mol −1. | Temperature must be in kelvin. Δ G ≤ 0 means feasible, not fast. When Δ H and Δ S have the same sign, always state whether the reaction is feasible above or below the calculated temperature. |
Gibbs free energy change: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
For CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ mol−1 and ΔS = +161 J K−1 mol−1. Calculate ΔG at 298 K and state whether the reaction is feasible.
Calculate the minimum temperature at which calcium carbonate decomposes, using the data in the previous question.
For N2(g) + 3H2(g) → 2NH3(g), ΔH = −92 kJ mol−1 and ΔS = −199 J K−1 mol−1. Calculate ΔG at 700 K and comment on feasibility.
For N2O4(g) → 2NO2(g), ΔH = +57 kJ mol−1 and ΔS = +176 J K−1 mol−1. Calculate ΔG at 298 K and the temperature above which the reaction is feasible.
Ammonium nitrate dissolves in water spontaneously even though the process is endothermic. Explain using the Gibbs equation.
A reaction has ΔH negative and ΔS negative. Explain how its feasibility depends on temperature, and why a feasible reaction might still not be observed.
Answer key and marking notes
- ΔS = 0.161 kJ K−1 mol−1.
- ΔG = 178 − 298 × 0.161 = 178 − 48.0 = +130 kJ mol−1.
- ΔG is positive, so it is not feasible at 298 K.
- At the switch-over ΔG = 0, so T = ΔH / ΔS.
- T = 178 / 0.161 = 1106 K (1110 K to 3 s.f.); feasible above this temperature.
- ΔG = −92 − 700 × (−0.199) = −92 + 139.3.
- ΔG = +47.3 kJ mol−1.
- Positive, so not feasible at 700 K (it is feasible only below 462 K); the industrial process relies on high pressure and the equilibrium being reached quickly with a catalyst.
- ΔG = 57 − 298 × 0.176 = 57 − 52.4 = +4.55 kJ mol−1: not feasible at 298 K.
- T = 57 / 0.176 = 324 K; feasible above 324 K.
- Dissolving the solid gives ions dispersed through the solution, so ΔS is positive.
- At room temperature TΔS is larger than ΔH, so ΔG = ΔH − TΔS is negative and dissolving is feasible.
- At low T, −TΔS is small, so the negative ΔH makes ΔG negative: feasible.
- At high T, −TΔS becomes large and positive, so ΔG becomes positive above T = ΔH/ΔS.
- ΔG gives no information about rate; a high activation energy can make a feasible reaction too slow to observe.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A reaction has ΔH positive and ΔS negative. When is it feasible?
- Aat no temperature
- Bat all temperatures
- Conly at high temperatures
- Donly at low temperatures
For CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ mol−1 and ΔS = +161 J K−1 mol−1. What is ΔG at 298 K?
- A+130 kJ mol−1
- B−47 800 kJ mol−1
- C+226 kJ mol−1
- D+178 kJ mol−1
Why must ΔS usually be divided by 1000 before it is used in ΔG = ΔH − TΔS?
- AΔS is given in J K−1 mol−1 but ΔH is in kJ mol−1
- Btemperature must be converted from °C to K
- CΔS is given per 1000 mol
- DΔG must always be smaller than ΔH
Which combination makes a reaction feasible at ALL temperatures?
- AΔH negative, ΔS positive
- BΔH negative, ΔS negative
- CΔH positive, ΔS positive
- DΔH positive, ΔS negative
For MgCO3(s) → MgO(s) + CO2(g), ΔH = +117 kJ mol−1 and ΔS = +175 J K−1 mol−1. What is ΔG at 298 K, and is the reaction feasible?
- A+64.9 kJ mol−1; not feasible
- B+64.9 kJ mol−1; feasible
- C−52 000 kJ mol−1; feasible
- D+169 kJ mol−1; not feasible
The graph shows ΔG against temperature for N2O4(g) → 2NO2(g). Use it to find ΔS for the reaction.
- A+176 J K−1 mol−1
- B−176 J K−1 mol−1
- C+57 J K−1 mol−1
- D+0.176 J K−1 mol−1
Using ΔH = +117 kJ mol−1 and ΔS = +175 J K−1 mol−1 for the decomposition of MgCO3, above what temperature is the reaction feasible?
- A669 K
- B0.669 K
- C396 K
- D1500 K
For 2Mg(s) + O2(g) → 2MgO(s), ΔH = −1204 kJ mol−1 and ΔS = −217 J K−1 mol−1. Which statement is correct?
- Ait is feasible below about 5550 K, and a lit Mg ribbon is still needed because the activation energy is high
- Bit is feasible above about 5550 K only
- Cit is never feasible because ΔS is negative
- Dit is feasible at all temperatures because ΔH is negative
Quiz key
ΔH is positive and −TΔS is also positive (minus a negative), so ΔG is positive at every temperature: never feasible. ΔH negative with ΔS positive is feasible at all temperatures. ΔH and ΔS both positive is feasible only at high temperature. Both negative is feasible only at low temperature.
ΔG = 178 − 298 × 0.161 = 178 − 48.0 = +130 kJ mol−1, so the decomposition is not feasible at room temperature. −47 800 forgets to divide ΔS by 1000. +226 adds TΔS instead of subtracting it. +178 ignores the entropy term altogether.
Entropies are tabulated in joules but enthalpy changes in kilojoules, so ΔS is divided by 1000 to give kJ K−1 mol−1 before the subtraction. Converting °C to K means adding 273, which is a separate step. Entropy is quoted per mole, not per 1000 mol. ΔG can be bigger or smaller than ΔH, depending on the sign of ΔS.
A negative ΔH and a negative −TΔS term (from positive ΔS) make ΔG negative at every temperature. With both negative, ΔG becomes positive at high temperature. With both positive, it is feasible only above T = ΔH/ΔS. With ΔH positive and ΔS negative it is never feasible.
ΔG = 117 − 298 × 0.175 = 117 − 52.15 = +64.9 kJ mol−1. A positive ΔG means the reaction is not feasible at 298 K, so the second option reads the sign wrongly. −52 000 forgets to convert J to kJ. +169 adds TΔS instead of subtracting it.
The intercept is ΔH = +57 kJ mol−1 and the line falls to −48.6 at 600 K, so the gradient = (−48.6 − 57) ÷ 600 = −0.176 kJ K−1 mol−1. Since the gradient is −ΔS, ΔS = +0.176 kJ K−1 mol−1 = +176 J K−1 mol−1; positive, as 1 mol of gas becomes 2 mol. −176 forgets the minus sign in the gradient. +57 is the intercept, which is ΔH. +0.176 is the value in kJ written with J units.
T = 117 ÷ 0.175 = 669 K; above this −TΔS outweighs the positive ΔH. 0.669 K forgets to convert ΔS to kJ. 396 is 669 K converted to °C, wrongly labelled K. 1500 K inverts the division (175 ÷ 117 × 1000).
ΔG = 0 at T = 1204 ÷ 0.217 = 5548 K. Below this, the large negative ΔH outweighs −TΔS, so ΔG is negative; but ΔG says nothing about rate, and magnesium must be heated to overcome the activation energy. Above 5548 K the entropy term wins, so 'above only' is backwards. A negative ΔS does not by itself stop a reaction. A negative ΔH is not enough at very high temperatures, because −TΔS keeps growing.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.