JAM Thermodynamics — Gibbs Energy Question Types
Gibbs energy is the single most productive concept in the IIT-JAM physical chemistry syllabus, because it is where thermodynamics, equilibrium and electrochemistry all meet. Most students know ΔG = ΔH − TΔS and stop there — and then lose marks on the questions that need the reaction quotient, the pressure dependence, or the distinction between ΔG and ΔG°. Here are the six patterns that keep reappearing, each worked in full.
The core relations
dG = V dp − S dT → (∂G/∂T)p = −S, (∂G/∂p)T = V
ΔG° = −RT ln K ; ΔG = ΔG° + RT ln Q ; ΔG° = −nFE°
Two conditions are attached to every use of ΔG as a spontaneity criterion: constant temperature and constant pressure. Under those conditions ΔG < 0 means the process is spontaneous, ΔG = 0 means equilibrium, and the magnitude |ΔG| is the maximum non-expansion work the process can deliver. That last interpretation is what makes ΔG° = −nFE° work for a cell.
Type 1 — crossover temperature
A reaction has ΔH° = +55.0 kJ mol⁻¹ and ΔS° = +175 J K⁻¹ mol⁻¹. Is it spontaneous at 298 K, and above what temperature does it become so?
At 298 K: ΔG° = 55 000 − 298(175) = 55 000 − 52 150 = +2 850 J mol⁻¹ = +2.85 kJ mol⁻¹ → not spontaneous.
Setting ΔG° = 0: T = ΔH°/ΔS° = 55 000 / 175 = 314.3 K. Above this the TΔS term overtakes ΔH and the reaction turns spontaneous.
Note the unit trap: ΔH in kJ, ΔS in J. Convert one before subtracting — this single slip is the most common cause of a wrong answer in this question type.
The sign combinations are worth having memorised: ΔH < 0 with ΔS > 0 is spontaneous at all T; ΔH > 0 with ΔS < 0 is never spontaneous; the two mixed cases have a crossover temperature, spontaneous at high T when both are positive, and at low T when both are negative.
Type 2 — equilibrium constant from ΔG°
ΔG° = −32.8 kJ mol⁻¹ at 298 K. Find K.
ln K = −ΔG°/(RT) = 32 800 / (8.314 × 298) = 32 800 / 2 477.6 = 13.24
K = e13.24 = 5.6 × 10⁵
Because K is exponential in ΔG°, small energy changes matter enormously: at 298 K a shift of just 5.71 kJ mol⁻¹ changes K by a factor of ten. Keep RT = 2.478 kJ mol⁻¹ at 298 K memorised as a working number.
Type 3 — ΔG versus ΔG°, and the reaction quotient
This is the distinction JAM tests most often, usually by giving a positive ΔG° and asking whether the reaction can still proceed.
ΔG° = +5.40 kJ mol⁻¹ at 298 K. The reaction mixture has Q = 0.010. Find ΔG and state the direction.
ΔG = ΔG° + RT ln Q = 5 400 + 2 477.6 × ln(0.010)
= 5 400 + 2 477.6 × (−4.6052) = 5 400 − 11 409 =
−6.01 kJ mol⁻¹ → spontaneous in the forward direction.
Cross-check with K: K = e−5400/2477.6 = e−2.180 = 0.113. Since Q = 0.010 is less than K = 0.113, the mixture must move forward to reach equilibrium — the same conclusion by an independent route.
The lesson: a positive ΔG° does not forbid the forward reaction. It only says that starting from all species at standard state, the reaction runs backwards. Real mixtures are rarely at standard state.
Type 4 — ΔG for compressing or expanding an ideal gas
For an isothermal change of an ideal gas, dG = V dp with V = nRT/p integrates directly:
2.00 mol of an ideal gas is compressed isothermally at 300 K from 1.00 bar to 10.0 bar. Find ΔG.
ΔG = 2.00 × 8.314 × 300 × ln(10.0/1.00) = 4 988.4 × 2.3026 = +11.49 kJ
Positive, as it must be: compression is not spontaneous, and work has to be done on the gas. Note also that ΔH = 0 and ΔU = 0 for an isothermal ideal-gas change, so all of ΔG here comes from −TΔS.
Type 5 — the electrochemical link
The Daniell cell has E° = 1.10 V with n = 2. Find ΔG°.
ΔG° = −nFE° = −2 × 96 485 × 1.10 = −212 267 J mol⁻¹ = −212.3 kJ mol⁻¹
Combining this with ΔG° = −RT ln K lets you get the equilibrium constant of a redox reaction straight from the cell potential — a two-step question JAM likes because it tests both relations at once.
The same identity is what makes reaction coupling possible: two reactions sharing a common intermediate have additive ΔG values, so a strongly negative one can drive a positive one. That is the thermodynamic basis of ATP-coupled biochemistry and of carbothermic metal extraction in Ellingham diagrams.
Type 6 — temperature dependence, Gibbs–Helmholtz and van't Hoff
Continuing the reaction above (K = 0.113 at 298 K), take ΔH° = +58.0 kJ mol⁻¹ and estimate K at 350 K.
1/350 − 1/298 = 0.0028571 − 0.0033557 = −4.986 × 10⁻⁴ K⁻¹
ΔH°/R = 58 000 / 8.314 = 6 976
ln(K₂/K₁) = −6 976 × (−4.986 × 10⁻⁴) = 3.478
K₂ = 0.113 × e3.478 = 0.113 × 32.4 = 3.66
K rises with temperature because ΔH° is positive — Le Chatelier, obtained quantitatively. The derivation assumes ΔH° is constant over the interval, which is an approximation, and a good answer says so.
Common mistakes
- Mixing J and kJ in ΔH − TΔS. Entropies are almost always tabulated in J K⁻¹ mol⁻¹ and enthalpies in kJ mol⁻¹.
- Putting ΔG into −RT ln K. Only ΔG° appears there. At equilibrium ΔG = 0, but ΔG° is generally not zero.
- Celsius in place of kelvin. Every T in these relations is absolute.
- Concluding "ΔG > 0 so it cannot happen". Under non-standard Q it very often can — see Type 3.
- Using ΔSsystem alone as a spontaneity test. The second law criterion is ΔSuniverse > 0; ΔG < 0 is the equivalent statement written entirely in system properties at constant T and p.
- Forgetting that ΔG° = −nFE° needs n as electrons per mole of reaction as written, not per mole of any one species.
Summary table
| Question asks for | Relation to use | Watch out for |
|---|---|---|
| Spontaneity at a given T | ΔG = ΔH − TΔS | Unit mismatch J vs kJ |
| Crossover temperature | T = ΔH/ΔS | Only valid if both have the same sign |
| K from energetics | ΔG° = −RT ln K | Standard state only |
| Direction of a real mixture | ΔG = ΔG° + RT ln Q | Compare Q with K |
| Ideal gas p or V change | ΔG = nRT ln(p₂/p₁) | Sign: compression is positive |
| Cell potential | ΔG° = −nFE° | F = 96 485 C mol⁻¹ |
| K at a new temperature | van't Hoff equation | ΔH° assumed constant |
Verify each step numerically. The Gibbs Free Energy calculator handles ΔG = ΔH − TΔS and the ΔG° ⇄ K conversion, so you can confirm a crossover temperature or an equilibrium constant in seconds instead of re-doing the exponentials by hand.
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