Two ideas you have always kept in separate boxes turn out to be the same idea. ΔG° tells you which way a reaction wants to go. K tells you where it ends up. One short equation joins them, and once you have it you can work out an equilibrium constant without doing a single experiment.
📚 What you need to know
ΔG° = −RT ln K, and rearranged, ln K = −ΔG° / RT. Both are in the booklet, section 1.
R = 8.31 J K−1 mol−1, so ΔG° must be in joules here, not kilojoules.
ΔG° negative → K > 1, products favoured. ΔG° positive → K < 1, reactants favoured.
ΔG° = 0 gives K = 1 exactly.
As a reaction proceeds, G falls to a minimum. That minimum is equilibrium.
Away from equilibrium, ΔG = ΔG° + RT ln Q, where Q uses the concentrations you actually have.
At equilibrium Q = K and ΔG = 0, which is exactly how the first equation is derived.
Free energy runs downhill into a valley
Up to now you have used ΔG° as a yes-or-no verdict. That is a bit of a simplification. What actually happens is that free energy falls as the reaction proceeds, and it does not fall forever — it reaches a minimum and stops. Once the mixture sits at the bottom, nothing more happens overall, because moving in either direction would mean going uphill.
That bottom of the valley is equilibrium. And where the bottom sits — near the products end or near the reactants end — is precisely what K is telling you.
Both reactions start by going downhill. The difference is where the bottom is. On the left the mixture ends up mostly products; on the right it barely gets going before it settles.
This picture clears up something that bothers a lot of students. A reaction with a positive ΔG° is not frozen — a tiny bit of product still forms, because even that small step downhill lowers the free energy. It just does not get far, which is another way of saying K is small.
The equation that joins them
Because the bottom of the valley is fixed by how negative ΔG° is, there has to be a direct link between ΔG° and K. Here it is.
Both forms are in the data booklet
ΔG° = −RT ln K — or — ln K = −ΔG° ÷ RT
Read the minus sign carefully, because it does all the work. A negative ΔG° makes −ΔG°/RTpositive, so ln K is positive, so K is bigger than 1. Products win. Flip the sign of ΔG° and every step flips with it.
Because of the logarithm, free energy changes that look modest produce enormous equilibrium constants. That is worth remembering when a question asks you to comment on the size of K.
How big is big? A feel for the numbers
All of these are worked out at 298 K, so RT = 8.31 × 298 = 2476 J mol−1.
ΔG° / kJ mol−1
K at 298 K
What the mixture looks like
−34
about 106
Essentially all products
−17
about 103
Mostly products
0
1
A real mixture of both
+17
about 10−3
Mostly reactants
+34
about 10−6
Barely any product at all
Notice the pattern. Every 17 kJ mol−1 change in ΔG° multiplies or divides K by about a thousand. So a reaction with ΔG° of −5 kJ mol−1 is genuinely reversible, while one at −50 kJ mol−1 goes to completion for all practical purposes.
🧩 Finding K from ΔG°
Convert ΔG° into joules by multiplying by 1000. R is in joules, so this is the opposite of what you did on the Gibbs page.
Check T is in kelvin.
Work out ln K = −ΔG° ÷ (RT), watching the double negative.
Take e to that power to get K. On your calculator that is the ex button, not 10x.
Comment on the value. Bigger than 1 means products favoured; smaller means reactants.
What if the mixture is not at equilibrium?
ΔG° is the standard value: it assumes everything at 1 mol dm−3 or 100 kPa. A real flask is almost never in that state, so we need the version that uses the concentrations you actually have. That is where the reaction quotient, Q, comes in. It is worked out exactly like K, but with whatever concentrations are in the flask right now.
Free energy away from equilibrium
ΔG = ΔG° + RT ln Q
Put Q = K into that equation and set ΔG = 0, because at equilibrium there is no push either way. You get 0 = ΔG° + RT ln K, which rearranges straight back to ΔG° = −RT ln K. The two equations are the same idea seen from different places on the valley slope.
Compare Q with K
Sign of ΔG
What happens next
Q < K
Negative
Forward reaction runs, making more product
Q = K
Zero
At equilibrium, no net change
Q > K
Positive
Reverse reaction runs, remaking reactants
Worked examples
WORKED EXAMPLE
For N2(g) + 3H2(g) → 2NH3(g), ΔG° = −32.7 kJ mol−1 at 298 K. Calculate K and comment on the position of equilibrium. (R = 8.31 J K−1 mol−1)
Step 1: get ΔG° into joules−32.7 × 1000 = −32700 J mol⁻¹Step 2: use ln K = −ΔG° ÷ RTln K = −(−32700) ÷ (8.31 × 298) = 32700 ÷ 2476Step 3: work out ln K, then Kln K = 13.2, so K = e13.2 = 5.4 × 10⁵K ≈ 5.4 × 10⁵, so equilibrium lies far to the products sideWhich raises an obvious question: if K is that big at room temperature, why is the Haber process run at 450 °C? Because K says nothing about rate. At 298 K you would wait forever.
WORKED EXAMPLE
A reaction has K = 4.5 × 10−3 at 500 K. Calculate ΔG° at this temperature.
Step 1: pick the right form
We have K and want ΔG°, so use ΔG° = −RT ln KStep 2: find ln Kln (4.5 × 10⁻³) = −5.40Step 3: substituteΔG° = −(8.31 × 500 × −5.40) = +22452 J mol⁻¹ΔG° = +22.5 kJ mol⁻¹K smaller than 1 always gives a positive ΔG°. If your signs disagree with that, you have lost a minus sign somewhere.
WORKED EXAMPLE
For H2(g) + I2(g) ⇌ 2HI(g), ΔG° = −8.0 kJ mol−1 at 700 K. A flask contains [H2] = 0.20, [I2] = 0.10 and [HI] = 0.20 mol dm−3. Find ΔG and say which way the reaction goes.
Step 1: write and work out QQ = [HI]² ÷ ([H₂][I₂]) = 0.20² ÷ (0.20 × 0.10)Q = 0.040 ÷ 0.020 = 2.0Step 2: use ΔG = ΔG° + RT ln Q, in kJRT ln Q = (8.31 × 700 × ln 2.0) ÷ 1000 = +4.03 kJ mol⁻¹Step 3: add themΔG = −8.0 + 4.03 = −3.97 kJ mol⁻¹ΔG is negative, so the forward reaction keeps goingDivide the RT ln Q term by 1000 because R is in joules while ΔG° is in kilojoules. Same trap as the Gibbs page, opposite direction.
💡 Exam tip
Decide which unit you are working in first. Either put everything in joules or divide the RT term by 1000. Pick one and label it.
Use ex, not 10x. The equation has a natural log, so the reverse step is the exponential.
Write the minus sign as a bracket: ln K = −(−32700) ÷ 2476. It makes the double negative obvious.
Quote K in standard form when it is huge or tiny, and give the comment sentence too.
ΔG° and ΔG are different quantities. The one with the ° is fixed for a given temperature; the one without depends on what is in the flask.
Remember the ΔG° = 0 case. It gives ln K = 0, so K = 1, not K = 0.
⚠ Common mix-up
Leaving ΔG° in kJ while R is in J. Your ln K comes out a thousand times too small and K looks like 1.
Losing the minus sign in ln K = −ΔG°/RT, which turns a huge K into a tiny one.
Saying ΔG° = 0 means K = 0. It means K = 1, because e0 = 1.
Thinking a positive ΔG° means no reaction at all. A small amount of product still forms — K is small, not zero.
Confusing Q and K. Same expression, but K only applies at equilibrium.
Using ΔG° from 298 K at a different temperature. Both ΔG° and K change when T changes, so use the value for the temperature in the question.
That closes the loop on this topic: entropy explains why things spread out, Gibbs free energy weighs that against enthalpy, and K tells you where the balance lands. Up next: Features of Dynamic Equilibrium, where we look at that balance point from the kinetics side.
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