ChemistryCBSE Class 11 · Equilibrium

Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle Class 11

Q: Does a large K mean a fast reaction?

No. K shows how far the reaction goes at equilibrium; the speed depends on the rate and the activation energy.

Q: Why are pure solids left out of K?

Their concentration is a constant fixed by density and molar mass, and is absorbed into the value of K.

Q: Does compression always shift the equilibrium?

Only when the number of gas moles differs on the two sides.

Q: Why does an inert gas added at constant total pressure move an equilibrium?

The volume increases, which dilutes the reacting gases and favours the side with more gas moles.

Updated 17 October 2026 · Fact-checked

At equilibrium the ratio of product to reactant concentrations, each raised to its stoichiometric coefficient, is a constant Kc at a given temperature (pure solids and liquids are left out). Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. Comparing the reaction quotient Q with K shows the direction of change: Q < K forward, Q > K backward, Q = K equilibrium. Le Chatelier's principle says a system disturbed at equilibrium shifts to oppose the disturbance, but only a change in temperature changes the value of K.

Whichever side Q lies on, the reaction moves so that Q heads toward K; only a change in temperature changes K itself.

Illustration drawn for QuestionBuddy.

Understand Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle

Dynamic equilibrium. In a closed system the forward and backward rates become equal, so the observable properties (concentration, pressure, colour) stop changing although both reactions continue.

The constant. For $a A + b B ⇌ c C + d D$ ,
$ $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} .$ $
Concentrations of pure solids and pure liquids do not appear. $K_{c}$ depends only on the temperature and on how the equation is written. For gases, $K_{p} = \frac{p _{C}^{c} p _{D}^{d}}{p _{A}^{a} p _{B}^{b}}$ and $K_{p} = K_{c} (R T)^{Δ n_{g}}$ , where $Δ n_{g}$ = gas moles of products minus gas moles of reactants.

Writing the equation differently. Reversing gives $1/ K$ ; multiplying all coefficients by $n$ gives $K^{n}$ ; adding two equations multiplies their constants. A large $K$ (well above 1) means the products dominate; a small $K$ means the reactants dominate. $K$ tells the extent, never the speed.

Reaction quotient. $Q$ has the same form as $K$ but uses the concentrations at any instant. If $Q < K$ the reaction runs forward, if $Q > K$ backward, and at $Q = K$ the system is at equilibrium.

Solving problems (ICE table). Write Initial, Change (using $x$ and the coefficients) and Equilibrium rows, substitute into $K$ , and solve for $x$ . Check that each equilibrium concentration is positive.

Le Chatelier's principle.

Adding a reactant (or removing a product) shifts the equilibrium forward; $K$ is unchanged.
Raising the pressure (by compression) shifts it towards fewer gas moles; if $Δ n_{g} = 0$ there is no shift. $K$ is unchanged.
Raising the temperature favours the endothermic direction, and $K$ changes (for an endothermic forward reaction $K$ increases).
A catalyst speeds up both directions equally; it only shortens the time to reach equilibrium.
An inert gas added at constant volume leaves all partial pressures unchanged, so there is no shift; added at constant total pressure the volume increases and the system shifts towards more gas moles.

Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle: formulas and key facts

Equilibrium constant: $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$; Leave out pure solids and liquids.
Kp and Kc: $K_{p} = K_{c} (R T)^{Δ n_{g}}$; R = 0.0831 L bar K⁻¹ mol⁻¹ when pressures are in bar.
Reversed and scaled equations: $K_{r e v} = \frac{1}{K}, K_{n \times} = K^{n}$; Adding equations multiplies the K values.
Reaction quotient: $Q = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} (at any instant)$; Compare with K for the direction.
Dissociation of A2 type: $K_{p} = \frac{4 α ^{2} P}{1 - α ^{2}}$; For A(g) ⇌ 2B(g) starting from pure A, total pressure P.

How to solve Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle questions

Write the balanced equation, decide whether K or Q is wanted, and then set up an ICE table. Work in concentrations (mol per litre) for Kc and in pressures for Kp.

1Write K from the balanced equation, omitting solids and liquids.
2Convert amounts to concentrations (divide by the volume).
3Compute Q if a direction is asked and compare it with K.
4Build the ICE table with the change in terms of x and the coefficients.
5Substitute the equilibrium row into K and solve for x; reject any root that makes a concentration negative.
6For Le Chatelier questions state the disturbance, the direction of shift and whether K changes.

Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A sealed fizzy-drink bottle

Suppose CO₂ dissolved in a sealed bottle is in equilibrium with CO₂ in the gap above the liquid. You open the cap and the gas escapes.

Say which way the equilibrium shifts and what you see.

Try it first, then show the solutionSolution

Opening the cap lowers the partial pressure of CO₂ in the gas phase, so Q = [CO₂(aq)]/p(CO₂) for the dissolving step becomes greater than K.
The system shifts backward to oppose the loss of gaseous CO₂, so dissolved CO₂ comes out of solution.

Answer: Dissolved CO₂ leaves the liquid, seen as bubbles.

Memory hook: Remove a product, and the system makes more of it.

Brown gas in hot water

Suppose a gas mixture in a sealed tube is brown because of NO₂ in the equilibrium N₂O₄ ⇌ 2NO₂ (colourless ⇌ brown), and the forward reaction is endothermic. The tube is placed in hot water.

Predict the colour change and the change in Kc.

Try it first, then show the solutionSolution

Heating favours the endothermic (forward) direction, so more NO₂ forms.
The forward direction is favoured, so Kc increases.

Answer: The colour deepens and Kc increases.

Memory hook: Heat pushes the endothermic side, and only heat changes K.

Tricks to remember Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle

Short memory aids. Say each one aloud once, then use it on a question the same day.

Q against K
Q small means more to make: Q < K goes forward.
Use it when: Direction of reaction.
Only temperature changes K
Concentration, pressure, inert gas and catalyst never change the value of K.
Use it when: Le Chatelier conclusions.
Kp = Kc (RT) to the Δn
Δn = gas products minus gas reactants; if it is zero, Kp = Kc.
Use it when: Converting between Kp and Kc.
Solids stay out
Pure solids and liquids have constant concentration, so they are omitted from K.
Use it when: Writing heterogeneous equilibria.

Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Kp from Kc · 2 marks · Very short answerBoard level

For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.040 mol L⁻¹ at 500 K. Find Kp at 500 K (R = 0.0831 L bar K⁻¹ mol⁻¹).

Try it first, then show the solutionSolution

Δn_g = 2 − 1 = +1, so Kp = Kc(RT). (1)
Kp = 0.040 × 0.0831 × 500 = 1.66 bar. (1)

Answer: Kp = 1.66 bar.

Example 2: ICE table for the dissociation of phosgene · 3 marks · Short answerHarder board level

1.0 mol of COCl₂ is placed in a 2.0 L flask and heated to a temperature at which Kc = 0.020 mol L⁻¹ for COCl₂(g) ⇌ CO(g) + Cl₂(g). Find the equilibrium concentrations and the degree of dissociation.

Try it first, then show the solutionSolution

Initial [COCl₂] = 1.0/2.0 = 0.50 mol L⁻¹. Let x mol L⁻¹ dissociate: [CO] = [Cl₂] = x and [COCl₂] = 0.50 − x. (1)
Kc = x²/(0.50 − x) = 0.020, so x² + 0.020x − 0.010 = 0. (1)
x = [−0.020 + √(0.0004 + 0.040)]/2 = 0.0905. [CO] = [Cl₂] = 0.0905 mol L⁻¹, [COCl₂] = 0.410 mol L⁻¹, degree of dissociation = 0.0905/0.50 = 0.18. (1)

Answer: [CO] = [Cl₂] = 0.0905 M; [COCl₂] = 0.410 M; α = 0.18.

Example 3: Direction of change and the new equilibrium · 4 marks · Long answerHarder board level

For CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), Kc = 4.0 at a certain temperature. A 1.0 L vessel contains 0.10 mol CO, 0.40 mol H₂O, 0.20 mol CO₂ and 0.20 mol H₂. (a) Which way does the reaction proceed? (b) Find the equilibrium amount of CO.

Try it first, then show the solutionSolution

(a) Q = (0.20 × 0.20)/(0.10 × 0.40) = 1.0, which is less than Kc = 4.0, so the reaction proceeds forward. (1)
(b) Let x mol of CO react: Kc = (0.20 + x)²/[(0.10 − x)(0.40 − x)] = 4.0. (1)
Expanding: 0.04 + 0.4x + x² = 0.16 − 2x + 4x², so 3x² − 2.4x + 0.12 = 0. (1)
x = [2.4 − √(5.76 − 1.44)]/6 = 0.0536 (the other root, 0.746, would make [CO] negative). [CO] at equilibrium = 0.10 − 0.0536 = 0.046 mol L⁻¹. (1)

Answer: Forward; 0.046 mol of CO remains.

Example 4: Combining and reversing equilibrium constants · 4 marks · Long answerHarder board level

For A + B ⇌ C, K₁ = 2.0 and for C + B ⇌ D, K₂ = 5.0. Find K for (a) A + 2B ⇌ D, (b) D ⇌ A + 2B, (c) 2A + 4B ⇌ 2D and (d) ½A + B ⇌ ½D.

Try it first, then show the solutionSolution

(a) Adding the two equations gives A + 2B ⇌ D, so K = K₁K₂ = 10. (1)
(b) The reverse gives K = 1/10 = 0.10. (1)
(c) Doubling all coefficients squares K: 10² = 100. (1)
(d) Halving takes the square root: √10 = 3.16. (1)

Answer: (a) 10; (b) 0.10; (c) 100; (d) 3.16.

Example 5: Pressure effect on dissociation · 4 marks · Long answerHOTS / competency

N₂O₄(g) ⇌ 2NO₂(g) is 20% dissociated at a total pressure of 1 bar at 350 K. (a) Find Kp. (b) Find the degree of dissociation at the same temperature when the total pressure is raised to 4 bar. (c) Explain the trend with Le Chatelier's principle.

Try it first, then show the solutionSolution

(a) Kp = 4α²P/(1 − α²) = 4 × 0.04 × 1/0.96 = 0.167 bar. (1)
(b) With P = 4 bar: 16α²/(1 − α²) = 0.1667, so α² = 0.1667/16.1667 = 0.0103 and α = 0.101. (2)
(c) The forward reaction increases the moles of gas (1 to 2). Higher pressure shifts the equilibrium to the side with fewer moles, so dissociation falls from 20% to about 10%; Kp is unchanged. (1)

Answer: (a) 0.167 bar; (b) α ≈ 0.10 (about 10%); (c) pressure favours the side with fewer gas moles.

Where students lose marks in Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle

The slips examiners see most often, and what to write instead.

Mistake: Including solid CaCO₃ or CaO in the Kc expression.
Treating all species alike.
Do this: Leave out pure solids and liquids; write Kc = [CO₂] for the decomposition of calcium carbonate.
Usually costs: The expression mark.
Mistake: Using moles instead of concentrations when the species have different coefficients.
The volume does not cancel.
Do this: Divide by the volume before substituting, unless Δn_g = 0.
Usually costs: The numerical.
Mistake: Saying that a catalyst increases the yield, or that pressure changes the value of K.
Confusing rate, position and constant.
Do this: A catalyst only saves time; pressure and concentration shift the position but K stays fixed; only temperature changes K.
Usually costs: Concept marks.
Mistake: Using R = 8.314 with Kc in L mol⁻¹ and pressures in bar.
Wrong units for R.
Do this: Use R = 0.0831 L bar K⁻¹ mol⁻¹ when pressures are in bar.
Usually costs: The value of Kp.

Exam tips

Write the balanced equation and the expression for K before any numbers.
Show the ICE table in three rows; it earns the method marks.
Reject the root that gives a negative concentration and say so.
In Le Chatelier answers state the disturbance, the shift and whether K changes.
Compare Q with K in words before declaring a direction.

Practice questions on Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For which of the following is Kp equal to Kc(RT)⁻¹?
(a) H₂(g) + I₂(g) ⇌ 2HI(g) (b) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (c) N₂O₄(g) ⇌ 2NO₂(g) (d) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Hint
Find Δn_g = −1.
Answer
(b): Δn_g = 2 − 3 = −1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The value of Kc for an endothermic reaction increases when the temperature is raised.
Reason (R): A rise in temperature favours the endothermic direction, so more product is present at the new equilibrium.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Temperature is the one factor that changes K.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Write the expression for Kc for 2NO(g) + O₂(g) ⇌ 2NO₂(g) and state its units in mol L⁻¹.
Hint
Count the powers on top and bottom.
Answer
Kc = [NO₂]²/([NO]²[O₂]); the units are (mol L⁻¹)² / (mol L⁻¹)³ = L mol⁻¹.
Q4 · 3 marks · Short answerHarder board level
For H₂(g) + I₂(g) ⇌ 2HI(g) at a certain temperature Kc = 50. A 1.0 L vessel contains 0.50 mol H₂, 0.50 mol I₂ and 2.0 mol HI. Is the mixture at equilibrium? If not, say in which direction it moves and justify.
Hint
Find Q first.
Answer
Q = 2.0²/(0.50 × 0.50) = 16, which is less than 50. The mixture is not at equilibrium; it moves forward, forming more HI.
Q5 · 4 marks · Long answerHOTS / competency
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.50 L² mol⁻² at 400 K (R = 0.0831 L bar K⁻¹ mol⁻¹). (a) Find Kp. (b) Find Kc and Kp for ½N₂(g) + 3/2 H₂(g) ⇌ NH₃(g). (c) State, with a reason, the effect of increasing the pressure on the yield of NH₃.
Hint
Δn_g = −2 for the first equation and −1 for the second.
Answer
(a) Kp = 0.50/(0.0831 × 400)² = 4.5 × 10⁻⁴ bar⁻². (b) Kc = √0.50 = 0.71 L mol⁻¹; Kp = 0.71/(0.0831 × 400) = 2.1 × 10⁻² bar⁻¹. (c) The yield increases: higher pressure favours the side with fewer gas moles (4 moles to 2), while K does not change.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Relation between Kp and Kc?Show answer
Kp = Kc(RT)^Δn_g.
2.K for the reverse reaction?Show answer
1/K.
3.What does Q < K mean?Show answer
The reaction proceeds forward.
4.Which factor alone changes K?Show answer
Temperature.
5.Effect of a catalyst on the equilibrium?Show answer
None on its position; it only speeds up reaching it.
6.Units of Kc for N₂ + 3H₂ ⇌ 2NH₃?Show answer
L² mol⁻².

Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle: frequently asked questions

Does a large K mean a fast reaction?

No. K shows how far the reaction goes at equilibrium; the speed depends on the rate and the activation energy.

Why are pure solids left out of K?

Their concentration is a constant fixed by density and molar mass, and is absorbed into the value of K.

What happens to K if the equation is multiplied by 2?

K is squared.

Does compression always shift the equilibrium?

Only when the number of gas moles differs on the two sides.

Why does an inert gas added at constant total pressure move an equilibrium?

The volume increases, which dilutes the reacting gases and favours the side with more gas moles.