Van't Hoff Equation — How the Equilibrium Constant Changes with Temperature
Le Chatelier's principle tells you the direction an equilibrium shifts when you heat it. The van't Hoff equation tells you by how much. It is the quantitative version of the same idea, and it turns a one-line qualitative statement into a number you can calculate. It appears in Class 12 equilibrium, in IIT-JAM and GATE physical chemistry, and in CSIR-NET thermodynamics.
First, clear up a name clash
Two different things in your syllabus are called "van't Hoff". They are not related and mixing them up is a guaranteed lost mark:
| Van't Hoff equation (this article) | Van't Hoff factor i | |
|---|---|---|
| Topic | Chemical equilibrium and thermodynamics | Colligative properties of solutions |
| Question it answers | How does K change when T changes? | How many particles does one formula unit give in solution? |
| Typical form | ln(K₂/K₁) = (ΔH°/R)(1/T₁ − 1/T₂) | ΔT_f = i K_f m |
| Example value | K rises from 0.113 to 3.26 | i ≈ 2 for NaCl |
The formula
Both forms in that box are the same equation — the second is just the first with the bracket reversed and the minus sign absorbed. Pick one and always write it the same way; that alone removes most sign errors. The differential form, which is where the graph comes from, is:
What each symbol means
| Symbol | Meaning | Unit |
|---|---|---|
| K₁, K₂ | Equilibrium constants at the two temperatures | none (thermodynamic K is dimensionless) |
| T₁, T₂ | The two absolute temperatures | kelvin (K) — never °C |
| ΔH° | Standard enthalpy change of the reaction | J mol⁻¹ if you use R in J |
| R | Gas constant | 8.314 J K⁻¹ mol⁻¹ |
The equation contains ΔH° and nothing else about the reaction. That is the deep point: only the enthalpy decides how sensitive K is to temperature. ΔS° cancels out between the two temperatures, which is why it does not appear.
Reading the sign before you calculate
- Endothermic (ΔH° positive): raising T increases K. Heat behaves like a reactant, so adding heat pushes the reaction forward.
- Exothermic (ΔH° negative): raising T decreases K. Heat behaves like a product, so adding heat pushes the reaction back.
- ΔH° ≈ 0: K barely changes with temperature at all.
Decide this first, in one line, then calculate. If your arithmetic disagrees with your prediction, you have made a sign error — not discovered new chemistry.
Worked example 1 — endothermic reaction, K rises
Question: A reaction has ΔH° = +58.0 kJ mol⁻¹ and K = 0.113 at 298 K. Estimate K at 348 K.
Prediction first: endothermic, temperature raised → K must increase.
Step 1 — put ΔH° in joules: ΔH° = 58 000 J mol⁻¹
ΔH°/R = 58 000 / 8.314 = 6976.2
Step 2 — the reciprocal temperatures:
1/T₁ = 1/298 = 3.355705 × 10⁻³
1/T₂ = 1/348 = 2.873563 × 10⁻³
(1/T₁ − 1/T₂) = 4.82142 × 10⁻⁴
Step 3 — multiply:
ln(K₂/K₁) = 6976.2 × 4.82142 × 10⁻⁴ = 3.3635
Step 4 — undo the logarithm:
K₂/K₁ = e³·³⁶³⁵ = 28.89
K₂ = 0.113 × 28.89 = 3.2646
K at 348 K ≈ 3.26 — it rose, as predicted, by a factor of about 29 for a 50 K rise.
Worked example 2 — exothermic reaction, K falls
Question: An exothermic reaction has ΔH° = −92.4 kJ mol⁻¹ and K = 4.0 × 10² at 500 K. Find K at 700 K.
Prediction first: exothermic, temperature raised → K must decrease.
ΔH°/R = −92 400 / 8.314 = −11 113.8
1/T₁ − 1/T₂ = 1/500 − 1/700 = 2.000000 × 10⁻³ − 1.428571 × 10⁻³ = 5.71429 × 10⁻⁴
ln(K₂/K₁) = (−11 113.8) × (5.71429 × 10⁻⁴) = −6.3507
K₂/K₁ = e⁻⁶·³⁵⁰⁷ = 1.7455 × 10⁻³
K₂ = 400 × 1.7455 × 10⁻³ = 0.6982
K at 700 K ≈ 0.698 — a fall from 400 to below 1. This is exactly why industrial exothermic syntheses face a real conflict: a higher temperature makes the reaction faster but the yield at equilibrium smaller.
Worked example 3 — finding ΔH° from two measured K values
This is the direction that appears most often in JAM and GATE, because it is what an experimenter actually does.
Question: K = 1.80 × 10⁻³ at 310 K and 9.00 × 10⁻³ at 340 K. Find ΔH°.
Rearranged: ΔH° = R × ln(K₂/K₁) ÷ (1/T₁ − 1/T₂)
Step 1: K₂/K₁ = 9.00 × 10⁻³ / 1.80 × 10⁻³ = 5.00, so ln(K₂/K₁) = ln 5.00 = 1.60944
Step 2: 1/310 = 3.225806 × 10⁻³, 1/340 = 2.941176 × 10⁻³, difference = 2.84630 × 10⁻⁴
Step 3: ΔH° = 8.314 × 1.60944 ÷ (2.84630 × 10⁻⁴)
numerator = 13.3809
ΔH° = 13.3809 / 2.84630 × 10⁻⁴ = 47 011 J mol⁻¹
ΔH° ≈ +47.0 kJ mol⁻¹
Sanity check: K went up with temperature, so ΔH° must come out positive. It did ✓
The graph: ln K against 1/T
Rearranging the integrated equation into the form of a straight line gives:
So plotting ln K on the y-axis against 1/T on the x-axis gives a straight line — the van't Hoff plot. Reading it:
| Feature of the plot | What it gives you |
|---|---|
| Slope | −ΔH°/R, so ΔH° = −R × slope |
| Intercept on the ln K axis | ΔS°/R, so ΔS° = R × intercept |
| Line slopes downward (negative slope) | ΔH° is positive — endothermic |
| Line slopes upward (positive slope) | ΔH° is negative — exothermic |
| Line is curved, not straight | ΔH° is not constant over that temperature range |
Note the intercept is only reachable by extrapolating to 1/T = 0, i.e. infinite temperature — so ΔS° from a van't Hoff plot is always a long extrapolation and is less reliable than ΔH° from the slope. Say so if an exam asks you to comment on the accuracy.
The assumption you must state
The two-point form of the equation is obtained by integrating and treating ΔH° as constant between T₁ and T₂. Over a 20–50 K window that is usually a good approximation. Over hundreds of kelvin it is not, because ΔH° itself varies with temperature (through the heat capacities). Examiners give credit for writing "assuming ΔH° is independent of temperature over this range" — it costs one line and shows you understand the limit of the tool.
Common mistakes that cost marks
- Temperature in °C. Every T in this equation is absolute. 25 °C is 298.15 K.
- Mixing kJ with R in J. If R = 8.314 J K⁻¹ mol⁻¹, ΔH° must be in J mol⁻¹. Writing 58 instead of 58 000 makes the answer wrong by a factor of a thousand inside a logarithm — the result is not just "a bit off", it is meaningless.
- Flipping the bracket. (1/T₁ − 1/T₂) goes with +ΔH°/R; (1/T₂ − 1/T₁) goes with −ΔH°/R. Fix one form and never switch mid-question.
- Forgetting to exponentiate. ln(K₂/K₁) = 3.36 does not mean K₂/K₁ = 3.36. It means K₂/K₁ = e³·³⁶ = 28.9.
- Confusing K₂/K₁ with K₂. The equation gives the ratio. Multiply by K₁ to get K₂.
- Thinking a catalyst changes K. A catalyst speeds up both directions equally and changes neither K nor ΔH°. Only temperature changes K.
- Using it for the van't Hoff factor. Different topic entirely — see the table at the top of this page.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 12 | Effect of temperature on K, qualitatively and with simple numbers |
| IIT-JAM / CUET-PG | Two-point calculations both ways; sign reasoning from ΔH° |
| GATE Chemistry | Van't Hoff plots — reading ΔH° from a slope, ΔS° from an intercept |
| CSIR-NET | Combined with ΔG° = −RT ln K and ΔG° = ΔH° − TΔS° in multi-step problems |
Check your answer, including the sign. The van't Hoff calculator takes ΔH°, T₁, T₂ and either K value and returns the other — or works backwards from two K values to ΔH°, which is the version most exam questions actually ask.
Open the Van't Hoff Equation Calculator →Thermodynamics numericals reward method, not memory — the same four steps solve nearly every one. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and through online classes across India: abcchemistry.in.