IB Chemistry HL Topic 5 — How Far? The Position of Equilibrium Paper 1 & 2 Core skill ~10 min read

The Equilibrium Constant, Kc

The last page taught you how to write the expression. This page is about the number that comes out of it. That single number tells you how far a reaction goes before it gives up — and once you can read it at a glance, a lot of exam questions become one-liners.

📘 What you need to know

Reading the size of K

Think of K as the score at the end of a tug of war between the forward and backward reactions. A huge number means the forward side won convincingly. A tiny number means the backward side barely let anything through. A number near 1 means it ended up roughly even.

What K is comparing K = products at equilibriumreactants at equilibrium
What the size of K is telling you K values span a huge range, so the scale below rises by a factor of 10000 each step mostly reactants mostly products K = 1 10⁻¹² 10⁻⁸ 10⁻⁴ 1 10⁴ 10⁸ 10¹² K ≪ 1 K ≈ 1 K ≫ 1 barely reacts at all plenty of both nearly complete reactants products Both reactants and products are always present, even when K looks extreme.
The bars are the useful part. Even at K = 10–12 there is a trace of product, and at K = 1012 there is a trace of reactant left. Equilibrium never quite reaches zero.
Rough working values: K bigger than about 103 and chemists call the reaction essentially complete. K smaller than about 10–3 and they say it hardly happens. Between those two you have a genuine mixture to talk about.

Only temperature moves K

This is the sentence that unlocks half of the Le Chatelier questions later on. You can pump up the pressure, add more reactant, or throw in a catalyst — the concentrations will all shuffle around, but they shuffle in exactly the way that keeps the ratio the same. Temperature is the one thing that changes the ratio itself.

CHANGES K

  • Temperature, and nothing else

Heating an endothermic reaction increases K. Heating an exothermic reaction decreases K. Any K value you are given is only valid at the temperature stated with it.

DOES NOT CHANGE K

  • Adding or removing reactant or product
  • Changing the pressure or the volume
  • Adding a catalyst
  • Adding more of a solid

These shift the position of equilibrium but the concentrations always settle back to the same ratio.

If a question ever asks “what happens to K when you increase the pressure?”, the answer is a single word: nothing. Students write a paragraph about shifting to the side with fewer moles, which is a fine answer to a different question.

Calculating Kc from data

Most Kc questions hand you either concentrations directly, or moles plus a volume. If you get moles, divide by the volume in dm3 first. Everything else is substitution.

🧩 The method

  1. Write the balanced equation and the K expression from it.
  2. Convert moles to concentrations: divide each amount by the total volume in dm3. Watch for cm3, which you divide by 1000 first.
  3. Substitute the equilibrium values into the expression, powers included.
  4. Work out the number and give it to 3 significant figures unless told otherwise.
  5. Sanity check the size. If K came out huge, the mixture should be mostly product. Does that match the data you were given?

Worked examples

WORKED EXAMPLE

Calculating Kc from moles and volume

A 2.00 dm3 sealed flask is left to reach equilibrium. It is found to contain 0.400 mol H2, 0.400 mol I2 and 3.00 mol HI.
H2(g) + I2(g) ⇌ 2HI(g)
Calculate Kc and comment on the position of equilibrium.

Step 1: Turn moles into concentrations [H2] = 0.400 ÷ 2.00 = 0.200 mol dm–3 [I2] = 0.400 ÷ 2.00 = 0.200 mol dm–3 [HI] = 3.00 ÷ 2.00 = 1.50 mol dm–3 Step 2: Write the expression Kc = [HI]2[H2][I2] Step 3: Substitute Kc = 1.5020.200 × 0.200 = 2.250.0400 Kc = 56.3 (3 s.f.) K is well above 1, so the equilibrium lies to the right and the mixture is mainly HI
WORKED EXAMPLE

A reaction with a power that changes the answer

At a certain temperature, a 0.500 dm3 flask at equilibrium contains 0.0600 mol N2O4 and 0.0400 mol NO2.
N2O4(g) ⇌ 2NO2(g)
Calculate Kc.

Step 1: Concentrations first [N2O4] = 0.0600 ÷ 0.500 = 0.120 mol dm–3 [NO2] = 0.0400 ÷ 0.500 = 0.0800 mol dm–3 Step 2: The 2 in front of NO2 becomes a square Kc = [NO2]2[N2O4] = 0.080020.120 = 0.006400.120 = 0.05333… Kc = 0.0533 (3 s.f.) forget the square and you get 0.667 – twelve times too big
WORKED EXAMPLE

Reading three K values

State whether each equilibrium mixture contains mostly reactants, mostly products, or a lot of both.
(a) K = 3.2 × 10–14   (b) K = 1.7 × 108   (c) K = 2.4

(a) K is far smaller than 1 The bottom of the fraction must be much bigger than the top, so reactant concentration greatly exceeds product concentration. Mostly reactants — equilibrium far to the left (b) K is far bigger than 1 The top of the fraction dominates, so product concentration greatly exceeds reactant concentration. Mostly products — the reaction goes almost to completion (c) K is close to 1 Significant amounts of both, tipped slightly towards products to earn the mark, always say why: name which part of the fraction is bigger

🧠 Which way is “right”?

Products are written on the right of the equation, and products go on the top of the K expression. So a big top means a big K means the equilibrium lies to the right. Big, top, right — three words that always travel together.

💡 Exam tip

⚠ Common mix-up

Up next: Le Chatelier’s Principle — how to predict which way an equilibrium shifts when you disturb it, and why only temperature touches K.

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