GATE Chemistry Thermodynamics — The Question Types That Repeat
Thermodynamics in GATE Chemistry is not a large syllabus, but it is a deep one, and the same half-dozen question shapes keep reappearing dressed in different chemistry. If you can recognise the shape in the first ten seconds, you have already saved the two minutes that decide whether you finish the paper. This article names those shapes, gives the governing relation for each, and works four numericals end to end — deliberately chosen so that the answers cross-check one another.
Shape 1 — Spontaneity and the crossover temperature
You are given ΔH° and ΔS° and asked either for ΔG° at some temperature, or for the temperature at which the sign of ΔG° flips.
ΔG° is the standard Gibbs free-energy change (J mol⁻¹), ΔH° the standard enthalpy change (J mol⁻¹), ΔS° the standard entropy change (J K⁻¹ mol⁻¹) and T the absolute temperature in kelvin. The single most common slip is mixing kJ with J inside the same subtraction.
Worked example 1. For a reaction, ΔH° = +40.0 kJ mol⁻¹ and ΔS° = +120 J K⁻¹ mol⁻¹, both assumed independent of temperature. Find (a) ΔG° at 298 K and (b) the temperature above which the reaction becomes spontaneous.
(a) Convert first: ΔH° = 40 000 J mol⁻¹.
ΔG°(298) = 40 000 − (298 × 120) = 40 000 − 35 760 = +4240 J mol⁻¹ = +4.24 kJ mol⁻¹
Positive, so the reaction is not spontaneous under standard conditions at 298 K.
(b) T = ΔH° / ΔS° = 40 000 / 120 = 333.3 K. Above this temperature the −TΔS° term outweighs ΔH° and ΔG° turns negative.
Shape 2 — Converting between ΔG° and K
This is the highest-frequency single relation in the whole topic, and it runs in both directions.
K is the thermodynamic equilibrium constant (dimensionless, referenced to the standard state), Q the reaction quotient at the actual composition, R = 8.314 J K⁻¹ mol⁻¹. The second form is what tells you the direction of a reaction that is not at equilibrium: if Q < K then ΔG < 0 and the reaction runs forward.
Worked example 2. Using ΔG°(298) = +4240 J mol⁻¹ from example 1, find K at 298 K.
ln K = −ΔG° / RT = −4240 / (8.314 × 298) = −4240 / 2477.6 = −1.7113
K = e−1.7113 = 0.181
Sanity check on the sign: ΔG° is positive, so K must be less than 1. It is.
Shape 3 — Temperature dependence of K (van't Hoff)
Read the sign structure rather than memorising it: for an endothermic reaction (ΔH° > 0), raising T raises K. A plot of ln K against 1/T is a straight line of slope −ΔH°/R, and GATE likes asking you to extract ΔH° from that slope.
Worked example 3. For the same reaction, find K at 350 K.
1/T₂ − 1/T₁ = 1/350 − 1/298 = 0.00285714 − 0.00335570 = −0.00049856 K⁻¹
ln(K₂/K₁) = −(40 000 / 8.314) × (−0.00049856) = −(4811.2) × (−0.00049856) = +2.3987
K₂ = 0.181 × e2.3987 = 0.181 × 11.01 = 1.99
Cross-check with shape 1. ΔG°(350) = 40 000 − (350 × 120) = −2000 J mol⁻¹.
Then K = e−ΔG°/RT = e2000/(8.314 × 350) = e0.6873 = 1.99. ✔
The two routes agree, which is exactly the kind of internal check you should build into
your practice.
Shape 4 — Maxwell relations and the thermodynamic equation of state
GATE regularly asks you to prove or apply a partial-derivative identity. You do not need to memorise all four Maxwell relations if you can write the four fundamental equations and read the cross-derivatives off them.
From dG follow the two most-used results directly: (∂G/∂T)P = −S and (∂G/∂P)T = V. The internal-pressure identity is the classic application:
For an ideal gas, P = RT/Vm, so T(∂P/∂T)V = RT/Vm = P and the internal pressure is zero — the formal statement that ideal-gas internal energy depends only on temperature. For a van der Waals gas the same substitution gives (∂U/∂V)T = a/Vm², which is a standard one-line GATE answer.
Shape 5 — Entropy of mixing and ideal solutions
Worked example 4. One mole of ideal gas A is mixed with one mole of ideal gas B at constant T and P. Find ΔSmix.
n = 2 mol, xA = xB = 0.5, ln 0.5 = −0.6931
ΔSmix = −(2)(8.314)[(0.5)(−0.6931) + (0.5)(−0.6931)] = −(16.628)(−0.6931) = +11.53 J K⁻¹
Since ΔHmix = 0 for an ideal mixture, ΔGmix = −TΔSmix = −(298)(11.53) = −3436 J — negative at every temperature, which is why ideal gases always mix spontaneously.
The relations worth having on instant recall
| Question shape | Governing relation | Watch out for |
|---|---|---|
| Spontaneity / crossover T | ΔG° = ΔH° − TΔS° | kJ vs J mismatch |
| Equilibrium constant | ΔG° = −RT ln K | ln, not log; K is dimensionless |
| Direction of a non-equilibrium mixture | ΔG = ΔG° + RT ln Q | Q uses actual, not standard, activities |
| K at a second temperature | van't Hoff | Sign of (1/T₂ − 1/T₁) |
| ΔG at a second temperature | Gibbs–Helmholtz | Holds ΔH° constant — say so if asked |
| Vapour pressure vs temperature | Clausius–Clapeyron | Uses ΔHvap, not ΔHrxn |
| Partial derivative identity | Maxwell relations | Note which variable is held constant |
| Mixing | ΔSmix = −nR Σ x ln x | n is total moles; x are mole fractions |
Errors that repeatedly cost marks in this topic
- Mixing units inside one subtraction. ΔH° in kJ and TΔS° in J is the single most common wrong answer in the whole topic. Convert everything to joules before you subtract, then convert the answer once.
- Using log where the formula wants ln. ΔG° = −RT ln K. If you use log₁₀, you must carry the factor 2.303, and half the candidates who do this forget it.
- Confusing ΔG with ΔG°. ΔG° = 0 means K = 1. ΔG = 0 means the system is at equilibrium. These are different statements and GATE tests the difference.
- Assuming ΔH° and ΔS° are temperature-independent without saying so. That assumption is usually intended, but if a question gives you Cp data it is telling you not to make it.
- Getting the van't Hoff sign wrong by writing (1/T₁ − 1/T₂) instead of (1/T₂ − 1/T₁). Test your version against a case you know: endothermic reactions must give larger K at higher T.
Check your spontaneity numericals in one step. The Gibbs Free Energy calculator takes ΔH, ΔS and T, returns ΔG with the spontaneity verdict, and shows the substitution so you can see exactly where a hand calculation diverged.
Open the Gibbs Free Energy Calculator →Preparing for GATE, IIT-JAM, CSIR-NET or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches, available online across India as well as at the coaching centre — details at abcchemistry.in.