For N₂ + 3H₂ ⇌ 2NH₃, the Kc expression is:

[NH₃]² / ([N₂][H₂]³). Products over reactants, each raised to its coefficient: [NH₃]² on top, [N₂][H₂]³ underneath.

A large value of K (≫1) means at equilibrium:

products dominate. K ≫ 1 puts products on top as the larger quantity — the reaction goes nearly to completion.

Which of these changes the VALUE of K?

temperature. Only temperature changes K; concentration, pressure and catalysts affect position or rate, not the value of K.

Adding a catalyst to a system at equilibrium:

does not change K. A catalyst only speeds up reaching equilibrium; the position and K are unchanged.

The size of K tells you about a reaction's ___, not its ___.

extent; rate. K measures how far the reaction goes (extent); how fast it gets there is kinetics.

Kp differs from Kc in that it uses:

partial pressures of gases. Kp is written with partial pressures rather than concentrations, for gaseous equilibria.

For an exothermic forward reaction, increasing the temperature will:

decrease K. Higher temperature favours the endothermic (backward) direction, lowering the yield and decreasing K.

If the reaction quotient Q is greater than K, the reaction will:

move backward. Q > K means too much product, so the reaction shifts backward until Q = K.

Chemical Equilibrium (Kc & Kp)

Putting a number on how far a reversible reaction goes

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Practise the full question bank:

The equilibrium constant

You have met dynamic equilibrium and Le Chatelier qualitatively. Now we measure it. For a reversible reaction at equilibrium, a fixed ratio of product to reactant concentrations always holds at a given temperature — the equilibrium constant, Kc.

For the general reaction aA + bB ⇌ cC + dD, each concentration is raised to its coefficient: products on top, reactants underneath.

Kc = [C]^c [D]^d / ([A]^a [B]^b)concentrations at equilibrium, each to the power of its coefficient

What the size of K tells you

The magnitude of K measures how far the reaction goes. K much greater than 1: products dominate at equilibrium — the reaction "goes nearly to completion". K much less than 1: reactants dominate — little product forms. K around 1: significant amounts of both. K says nothing about how fast equilibrium is reached — that is kinetics, a separate question.

Kp for gases

For reactions between gases it is often easier to use partial pressures instead of concentrations, giving Kp — the same expression, with each partial pressure raised to its coefficient. Kc and Kp describe the same equilibrium in different units, and are related through the ideal gas law.

What changes K, and what does not

This is the crucial, often-tested point. Changing concentration or pressure shifts the position of equilibrium (Le Chatelier) but does not change K — the ratio re-establishes itself. Adding a catalyst does not change K either; it only reaches equilibrium faster. Only temperature changes the value of K: raising it increases K for an endothermic reaction and decreases K for an exothermic one.

Only temperature changes K

Concentration, pressure and catalysts move the position of equilibrium or the speed of getting there — but the number K stays fixed. Change the temperature and K itself changes. This distinction is a classic exam discriminator.

A worked example

Worked example

For H₂ + I₂ ⇌ 2HI, write the Kc expression, and say what a value of Kc = 50 implies.

Products over reactants, each to its coefficient: Kc = [HI]² / ([H₂][I₂]).HI has coefficient 2, so its concentration is squared.
Kc = 50 is greater than 1, so at equilibrium there is more product than reactant.A K above 1 favours products.
The reaction proceeds well toward HI, though not fully to completion.K = 50 is moderately large, not enormous, so some H₂ and I₂ remain.

Kc = [HI]² / ([H₂][I₂]); Kc = 50 (>1) means products are favoured — the reaction goes substantially, but not completely, to HI.

Le Chatelier applied

Le Chatelier's principle — a system at equilibrium shifts to oppose any change — turns Kc into practical prediction. Increase a reactant's concentration and the equilibrium shifts forward to use it up; raise the pressure of a gaseous equilibrium and it shifts to the side with fewer gas molecules; raise the temperature and it shifts in the endothermic direction.

Crucially, only temperature changes the value of K. Changing concentration or pressure shifts the position of equilibrium but leaves K unchanged, and a catalyst changes neither — it only makes equilibrium arrive sooner.

Writing K, and the reaction quotient

The equilibrium expression is products over reactants, each raised to the power of its balancing number. Pure solids and pure liquids are left out, because their concentrations do not change. Comparing the reaction quotient Q (the same expression using the current, non-equilibrium concentrations) with K tells you which way a mixture will move: if Q < K it goes forward, if Q > K it goes back.

Q versus K

K describes equilibrium; Q describes the mixture right now. The system always shifts to make Q equal K — forward if Q is too small, backward if Q is too big.

Practice questions

6 questions · 21 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

4 · 9 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]

Write the equilibrium constant expression Kc for the reaction N₂ + 3H₂ ⇌ 2NH₃.

Model answer

Kc = [NH₃]² ÷ ([N₂][H₂]³).

Examiner tip. Products over reactants, each raised to the power of its balancing number.

SQ2[2 marks]

State what a large value of Kc tells you about the position of equilibrium.

Model answer

A large Kc means the equilibrium lies to the right — the mixture contains mostly products.

Examiner tip. Large Kc → products favoured; small Kc → reactants favoured.

SQ3[3 marks]

For an exothermic forward reaction at equilibrium, state and explain the effect of increasing temperature on Kc.

Model answer

Increasing temperature shifts the equilibrium in the endothermic (backward) direction, so the yield of product falls and Kc decreases.

Examiner tip. Only temperature changes the value of Kc; concentration and pressure changes do not.

SQ4[2 marks]

Write the equilibrium expression Kc for: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

Model answer

Kc = [SO₃]² ÷ ([SO₂]² [O₂]).

Examiner tip. Products over reactants, each concentration raised to the power of its balancing number.

Long questions

2 · 12 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]

Consider the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), forward reaction exothermic.

State and explain the effect of increasing the pressure on the yield of SO₃. [3]
State and explain the effect of adding a catalyst on the yield of SO₃. [2]
State the effect of increasing temperature on the value of Kc. [1]

Mark scheme

Higher pressure increases the yield of SO₃[1]
The equilibrium shifts to the side with fewer gas molecules[1]
3 gas molecules on the left, 2 on the right, so it shifts right[1]
A catalyst does not change the yield / position of equilibrium[1]
It speeds up the forward and backward reactions equally (equilibrium reached faster)[1]
Increasing temperature decreases Kc[1]

Higher pressure → more SO₃ (fewer gas moles on right); catalyst → no change in yield (faster only); higher temperature → Kc decreases.

Examiner tip. Pressure and catalyst do not change Kc; only temperature does.

LQ2[6 marks]

Consider the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), forward reaction exothermic.

State and explain the effect of increasing pressure on the yield of ammonia. [3]
State and explain the effect of increasing temperature on the value of Kc. [3]

Mark scheme

Higher pressure increases the yield of ammonia[1]
the system shifts to the side with fewer gas molecules[1]
4 gas molecules on the left, 2 on the right, so it shifts right[1]
Increasing temperature decreases Kc[1]
the forward reaction is exothermic, so higher temperature favours the endothermic reverse[1]
less product forms, so Kc falls[1]

Higher pressure → more NH₃ (fewer gas moles on the right); higher temperature → lower Kc (favours the endothermic reverse).

Examiner tip. Pressure shifts position only; temperature actually changes K.