The Thermodynamics of Protein Folding — ΔG = ΔH − TΔS, Qualitatively
A protein chain has an astronomical number of possible shapes, yet in a cell it settles into one. Students often imagine this needs some special biological principle. It does not. Folding is an ordinary chemical equilibrium between two states, and the equation that governs it is the one you already use for every reaction: ΔG = ΔH − TΔS. What makes proteins interesting is not a new law but the size of the numbers — two enormous, opposing terms that nearly cancel, leaving a stability so small that a single hot afternoon can undo it. This article computes that balance, then draws the line where the simple two-state picture stops working.
The formula you already know
For the two-state equilibrium U ⇌ F: K = [F] / [U] funfolded = 1 / (1 + K) ffolded = K / (1 + K)
At the melting temperature, ΔG° = 0, so Tm = ΔH° / ΔS°
What each term means here
| Term | What it physically is for a protein | Typical sign |
|---|---|---|
| U, F | The unfolded (random-coil) and folded (native) states of the same chain | — |
| ΔH° | Enthalpy change of folding: hydrogen bonds, van der Waals contacts and salt bridges formed inside the folded core | Negative (favourable) near room temperature |
| ΔS° | Entropy change of folding for the system as a whole — the chain loses freedom, but ordered water released from around exposed non-polar groups gains it | Negative overall in most cases |
| ΔG° | Net stability of the folded state | Small and negative — often only tens of kJ/mol |
| Tm | Temperature at which half the molecules are unfolded | K (report in K, convert for reading) |
| R | Gas constant, 8.314 J mol⁻¹ K⁻¹ | — |
Note the trap hidden in row three. The chain itself certainly becomes more ordered, so its own conformational entropy falls sharply. But when non-polar side chains bury themselves in the core, the cage-like water that was forced to order around them is released into bulk solvent and its entropy rises. That release — the hydrophobic effect — is the main driving force of folding, and it is an entropy term belonging to the solvent. Any analysis that looks only at the chain gets the sign of the argument backwards.
Worked example 1 — how stable is "stable"?
For a small single-domain protein, take ΔH° = −250 kJ/mol and ΔS° = −0.750 kJ mol⁻¹ K⁻¹ for folding. Find ΔG° at 300 K, the equilibrium constant, and the fraction unfolded.
ΔG° = ΔH° − TΔS° = −250 − (300 × −0.750) = −250 + 225 = −25 kJ/mol
K = e−ΔG°/RT, with RT = 8.314 × 300 = 2494.2 J/mol
−ΔG°/RT = 25 000 ÷ 2494.2 = 10.023
K = e10.023 = 2.25 × 10⁴
funfolded = 1 ÷ (1 + 22 545) = 4.44 × 10⁻⁵ → about 0.004% unfolded
Look at the two numbers together. Bonds worth 250 kJ/mol are formed and almost all of that is paid back to entropy; what survives is 25 kJ/mol — comparable to a handful of hydrogen bonds. A protein is a large molecule held together by a margin the size of a small one. Yet because that margin sits inside an exponential, it is still enough to keep 99.996% of the molecules folded at any instant.
Worked example 2 — the melting temperature
At Tm, ΔG° = 0, so ΔH° = TmΔS°
Tm = ΔH° ÷ ΔS° = (−250) ÷ (−0.750) = 333.3 K = 60.2 °C
Check: substitute back — ΔG° = −250 − (333.3 × −0.750) = −250 + 250.0 = 0 ✔, and K = e⁰ = 1, so exactly half the molecules are folded, which is the definition of Tm. ✔
Worked example 3 — why a few degrees matter so much
Take the same protein at 310 K (body temperature) instead of 300 K.
ΔG° = −250 + (310 × 0.750) = −250 + 232.5 = −17.5 kJ/mol
RT = 8.314 × 310 = 2577.3 J/mol · 17 500 ÷ 2577.3 = 6.790
K = e6.790 = 889
funfolded = 1 ÷ 890 = 1.12 × 10⁻³ → 0.11% unfolded
Compare with Example 1: 0.0044% → 0.11%, a 25-fold rise in unfolded molecules for a 10 K rise in temperature. Nothing dramatic happened to ΔG — it moved by 7.5 kJ/mol — but the exponential turned that into a large change in population.
This is the single most useful intuition in the whole topic: a linear change in ΔG becomes an exponential change in what you can measure. It is also why the unfolding of many small proteins looks so sharp when plotted against temperature, giving a curve that goes from almost fully folded to almost fully unfolded over a narrow window.
Where this is actually used
| Where | What the thermodynamics is doing |
|---|---|
| Protein-based medicines | Formulation, storage temperature and cold-chain limits are set by how far Tm sits above the storage temperature |
| Industrial enzymes | Detergent, food and textile processes need enzymes that stay folded at the process temperature and pH |
| Protein engineering | Mutations are scored by their change in ΔG of unfolding — a redesign is judged on stability as well as activity |
| Screening in the laboratory | Thermal-shift and calorimetric methods measure Tm and its shift when a ligand binds — a bound ligand stabilises the folded state |
| Food processing | Cooking, pasteurisation and gel formation are all controlled protein unfolding followed by aggregation |
| Misfolding diseases | Understanding why a chain sometimes settles into an ordered aggregate rather than the native fold |
The same reasoning explains familiar denaturants without any new equation. Urea and guanidinium salts stabilise the unfolded chain by interacting favourably with exposed groups, which raises K's denominator. Extremes of pH give like charges on the surface that repel each other. Heat works through the TΔS term above. Each one attacks a different part of ΔG.
Where the simple formula stops being valid
- ΔH° and ΔS° are not constant with temperature. Everything above assumed they are, which is exactly what you are taught to assume in Class 12 — and for proteins it is the weakest assumption in the calculation. Burying non-polar surface gives folding a large heat-capacity change (ΔCp), so ΔG plotted against T is a downward curve, not a straight line. A striking consequence follows: the curve crosses zero twice, so proteins can unfold on cooling as well as heating. A linear extrapolation far from the measured range is not trustworthy.
- The two-state model is an approximation. It holds reasonably for small single-domain proteins. Larger multi-domain proteins unfold through intermediates, each domain with its own transition, and a single K then describes nothing real. Fitting a two-state curve to such data returns numbers that look precise and mean little.
- ΔG is a small difference between huge numbers. Both ΔH and TΔS run into hundreds of kJ/mol while their difference is tens. A 2% error in either term can swamp the answer, so quoting protein stabilities to three significant figures is usually false precision.
- Real denaturation is often irreversible. Unfolded chains expose sticky hydrophobic surface and aggregate. Once that happens the reaction is not U ⇌ F any more, the equilibrium treatment is invalid, and the "Tm" measured is a kinetic, scan-rate dependent number rather than a thermodynamic one. Boiling an egg is the everyday demonstration — you cannot cool it back.
- Thermodynamics says nothing about how fast, or by what route. ΔG gives the destination, not the path or the time taken. Folding rates, intermediates and misfolding traps are kinetic questions and need a different treatment entirely.
- A van 't Hoff ΔH and a calorimetric ΔH need not agree. When they differ, that disagreement is itself the evidence that the two-state assumption has failed — it is information, not experimental error.
- Common exam slip: mixing units. ΔH in kJ/mol with R in J mol⁻¹ K⁻¹ produces an answer wrong by a factor of 1000. Convert to joules before the exponential, every time.
Why this matters for JAM, GATE, NET and CUET-PG
| Exam area | What is typically asked |
|---|---|
| Chemical thermodynamics | ΔG = ΔH − TΔS; sign analysis; spontaneity as a function of temperature |
| Chemical equilibrium | ΔG° = −RT ln K; computing K and the population of each state |
| Temperature dependence | van 't Hoff equation; obtaining ΔH from the slope of ln K against 1/T |
| Biophysical / applied chemistry | Hydrophobic effect, denaturation, heat-capacity change on unfolding |
| Assertion–reason questions | Why a large negative ΔH does not by itself guarantee a stable folded state |
Do the ΔG = ΔH − TΔS step without arithmetic slips. The Gibbs Free Energy calculator takes ΔH, ΔS and T, returns ΔG with the units handled, and reports spontaneity — exactly the step where the kJ/J mix-up above costs marks.
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