IB Chemistry HLTopic 5 — How Far? The Position of EquilibriumPaper 1 & 2HL only | Core skill~10 min read
The Equilibrium Constant and Gibbs Energy
Two ideas you have met separately turn out to be the same idea. “Is this reaction spontaneous?” and “how far does this reaction go?” are answered by one equation linking ΔG to K. Get the units right and these become some of the most predictable calculation marks in the course.
📘 What you need to know
The link is ΔG = –RT ln K, which is given to you in the data booklet.
Rearranged: ln K = –ΔG ÷ (RT), then K = e raised to that power.
ΔG negative → K > 1: products are favoured, the forward reaction is spontaneous.
ΔG = 0 → K = 1: neither side is favoured.
ΔG positive → K < 1: reactants are favoured.
The more negative ΔG is, the bigger K becomes — and it grows extremely fast.
Units trap: ΔG is in kJ mol–1 but R is in J K–1 mol–1. One of them needs a factor of 1000.
T must be in kelvin, and K itself says nothing about the rate of the reaction.
The equation, and what each part is doing
ΔG measures how strongly a reaction wants to go. K measures how far it actually gets. It should not be a surprise that the two are linked — this equation just says exactly how.
Given in section 1 of the data booklet
ΔG = –RT ln K and so ln K = –ΔGRT
Symbol
What it is
Units to use
ΔG
Standard Gibbs energy change
J mol–1 in the equation, but usually quoted in kJ mol–1
R
The gas constant, 8.31
J K–1 mol–1
T
Temperature
K — add 273 to a Celsius value
K
Equilibrium constant
No units
The one that ruins answers: R is in joules and ΔG is almost always given in kilojoules. Multiply ΔG by 1000 before it goes anywhere near the equation. An answer that is out by a factor of 1000 inside a logarithm comes out spectacularly wrong.
Reading the relationship off a graph
Because the equation has ln K in it, K responds to ΔG in a very lopsided way. A modest change in ΔG produces an enormous change in K.
The steepness is the point. Moving ΔG from –5 to –50 kJ mol–1 multiplies K by roughly 78 million. Small energy differences produce wildly different equilibrium positions.
🤔 Why is a logarithm involved at all?
Energy adds up, but equilibrium constants multiply. Combine two reactions and you add their ΔG values, yet you multiply their K values. The one function that turns multiplication into addition is the logarithm, so any equation linking the two has to have a log in it. That is also why K explodes so quickly: a linear change in ΔG is an exponential change in K.
Getting the direction right without a calculator
Many questions only want a sign or a comparison. You can answer those from the shape of the relationship, without touching the equation.
ΔG IS NEGATIVE
K is greater than 1
Equilibrium lies to the right
Products dominate at equilibrium
Forward reaction is spontaneous
The more negative ΔG gets, the bigger K gets and the more complete the reaction.
ΔG IS POSITIVE
K is less than 1
Equilibrium lies to the left
Reactants dominate at equilibrium
Backward reaction is spontaneous
The reaction still happens to a tiny extent — K is small, but never zero.
Notice that “spontaneous” does not mean fast. Diamond turning into graphite has a negative ΔG, so K favours graphite heavily — and yet nobody’s ring has ever crumbled. Thermodynamics tells you where a reaction ends up; kinetics tells you whether you will live to see it.
Worked examples
WORKED EXAMPLE
Finding K from ΔG
At 298 K, a reaction has ΔG = –12.5 kJ mol–1. Calculate K and comment on the position of equilibrium. (R = 8.31 J K–1 mol–1)
Step 1: Convert ΔG into joulesΔG = –12.5 × 1000 = –12500 J mol–1Step 2: Rearrange the equation for ln Kln K = –ΔGRT = ––125008.31 × 298Step 3: Work out the bottom, then divide8.31 × 298 = 2476.4ln K = 125002476.4 = 5.0477Step 4: Undo the natural log with exK = e5.0477 = 155.66K = 156 (3 s.f.)K is well above 1, so equilibrium lies to the right and products dominate – which matches ΔG being negative
WORKED EXAMPLE
Working backwards, from K to ΔG
A reaction has K = 4.50 × 10–3 at 500 K. Calculate ΔG in kJ mol–1 and state whether the forward reaction is spontaneous under standard conditions.
Step 1: Take the natural log of Kln (4.50 × 10–3) = –5.4037it is negative because K is less than 1 – a useful checkStep 2: Put everything into ΔG = –RT ln KΔG = –8.31 × 500 × (–5.4037)ΔG = –4155 × (–5.4037) = 22452 J mol–1Step 3: Convert back into kilojoules22452 ÷ 1000 = 22.452ΔG = +22.5 kJ mol–1 (3 s.f.)ΔG is positive, so the forward reaction is not spontaneous under standard conditions
WORKED EXAMPLE
Comparing two reactions
At 298 K, reaction P has ΔG = –5.0 kJ mol–1 and reaction Q has ΔG = –50.0 kJ mol–1. Both are spontaneous. Calculate both equilibrium constants and comment.
Reaction Pln K = 50002476.4 = 2.0191 so K = e2.0191 = 7.53Reaction Qln K = 500002476.4 = 20.191 so K = e20.191 = 5.87 × 108CommentΔG is only ten times more negative, but K is about 78 million times bigger.P gives a genuine mixture; Q goes essentially to completionboth are spontaneous, but “spontaneous” covers a huge range of outcomes
🧠 Keeping the minus signs straight
The equation has a minus sign in it, so the signs of ΔG and ln K are always opposite. Negative ΔG gives positive ln K, which gives K bigger than 1. If your answer has ΔG and K pointing the same way, you have dropped a sign somewhere.
💡 Exam tip
Convert kJ to J first. Write “× 1000” as its own line so you cannot forget it.
Use ln, not log. The equation is built on natural logarithms.
Work out RT as a single number before dividing. It cuts calculator slips in half.
Check the sign at the end: negative ΔG must give K > 1, positive ΔG must give K < 1.
Temperature in kelvin, every time. A Celsius value in this equation is always wrong.
If a question asks you to “comment on the position of equilibrium”, compare K to 1 and say which side is favoured.
Never say a large K means a fast reaction. It does not.
⚠ Common mix-up
Leaving ΔG in kilojoules. The single most common error on this equation.
Using log instead of ln. They differ by a factor of 2.303, so the answer is badly wrong.
Losing a minus sign. There is one in the equation and often one on ΔG. Two minuses make a plus.
Forgetting to take ex at the end. ln K is not K. Finish the job.
Using Celsius for T. Add 273 first.
Thinking ΔG positive means no reaction. It means K is small, not zero. There is still a little product.
Confusing spontaneous with fast. Thermodynamics and kinetics answer completely different questions.
That is the end of How Far? The Position of Equilibrium. Up next: Brønsted–Lowry Acids and Bases, where the equilibrium ideas you have just built get put straight to work on acids.
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