ChemistryISC Class 11 · Chemical Equilibrium

Kc, Kp and the Reaction Quotient: Writing, Relating and Using Equilibrium Constants

Updated 17 October 2026 · Fact-checked

For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ) at equilibrium, and Kp is the same expression in partial pressures. They are related by Kp = Kc(RT)^Δn. Comparing the reaction quotient Q with K shows the direction in which a mixture will shift.

Compare Q with K: the mixture always moves toward Q = K, forward if Q is smaller, backward if Q is larger.

Illustration drawn for QuestionBuddy.

Understand Equilibrium Constants Kc, Kp and the Reaction Quotient

Law of mass action. At equilibrium the rate of the forward reaction equals the rate of the backward reaction and the concentrations stay constant. The equilibrium constant Kc = [products]^coefficients/[reactants]^coefficients is fixed at a given temperature; it depends only on T. Pure solids and pure liquids do not appear in the expression.

Manipulating K. For the reverse reaction K′ = 1/K. If the equation is multiplied by n, K becomes Kⁿ. For a sum of two reactions, K = K₁ × K₂. A large K (≫ 1) means products dominate; a small K (≪ 1) means reactants dominate. The value says nothing about speed.

Kp and Kc. For gases, Kp uses partial pressures (in atm or bar). Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants), and R = 0.0821 L atm K⁻¹ mol⁻¹ when pressure is in atm and concentration is in mol L⁻¹. If Δn = 0, Kp = Kc. Units of K follow the powers: for N₂ + 3H₂ ⇌ 2NH₃, Kc has units (mol L⁻¹)⁻².

Reaction quotient. Q has the same form as K but uses the concentrations at any instant. If Q < K the reaction goes forward; if Q > K it goes backward; if Q = K the system is at equilibrium.

Calculations. Use an ICE table (initial, change, equilibrium) with x or α. Degree of dissociation α = (moles dissociated)/(initial moles). For PCl₅ ⇌ PCl₃ + Cl₂ with initial concentration c: Kc = cα²/(1 − α). In terms of partial pressures, for N₂O₄ ⇌ 2NO₂ at total pressure P, the mole fractions are (1 − α)/(1 + α) and 2α/(1 + α).

Equilibrium Constants Kc, Kp and the Reaction Quotient: formulas and key facts

Equilibrium constant: $K c = [C]^{c} [D]^{d} / ([A]^{a} [B]^{b})$; Pure solids and liquids are omitted.
Kp and Kc: $K p = K c (R T)^{Δ} n$; R = 0.0821 L atm K⁻¹ mol⁻¹ with atm and mol L⁻¹.
Changing the equation: Reverse 1/K
Multiply by n Kⁿ
Add K₁K₂; K depends only on temperature.
Reaction quotient: $Q < K : f or w a r d ∣ Q > K : ba c k w a r d ∣ Q = K : e q u i l ib r i u m$; Q uses instantaneous concentrations.

Reverse	1/K
Multiply by n	Kⁿ
Add	K₁K₂

Equilibrium Constants Kc, Kp and the Reaction Quotient 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.

Will the mixture make more ammonia?

Suppose an engineer analyses a reactor mixture for N₂ + 3H₂ ⇌ 2NH₃ and finds 0.8 mol L⁻¹ N₂, 2.4 mol L⁻¹ H₂ and 0.4 mol L⁻¹ NH₃ at equilibrium.

Find Kc and Kp at 700 K (R = 0.0821 L atm K⁻¹ mol⁻¹).

Try it first, then show the solutionSolution

Kc = [NH₃]²/([N₂][H₂]³) = (0.4)²/(0.8 × 2.4³) = 0.16/11.06 = 0.0145 (mol L⁻¹)⁻².
Δn = 2 − 4 = −2, RT = 0.0821 × 700 = 57.5.
Kp = Kc(RT)⁻² = 0.01447/3303 = 4.4 × 10⁻⁶ atm⁻².

Answer: Kc = 0.0145; Kp = 4.4 × 10⁻⁶ atm⁻².

Memory hook: Count only gas moles in Δn.

Tricks to remember Equilibrium Constants Kc, Kp and the Reaction Quotient

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

Products over reactants
Always write Kc as products on top and reactants below, each raised to its coefficient.
Use it when: Writing correct expressions.
Q tells the way
Q smaller than K means the reaction still has to go forward.
Use it when: Direction questions.

Equilibrium Constants Kc, Kp and the Reaction Quotient: solved examples

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

Example 1: Δn for Kp · 1 mark · MCQBoard level

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the value of Δn in Kp = Kc(RT)^Δn is:
(a) +2
(b) −2
(c) +1
(d) 0

Try it first, then show the solutionSolution

Δn = 2 − (1 + 3) = −2.

Answer: (b) −2

Example 2: Value of Kc · 3 marks · Short answerBoard level

In the reaction H₂(g) + I₂(g) ⇌ 2HI(g), the concentrations at equilibrium are [H₂] = 0.20, [I₂] = 0.20 and [HI] = 1.60 mol L⁻¹. Write Kc and calculate its value.

Try it first, then show the solutionSolution

Kc = [HI]²/([H₂][I₂]). (1 mark)
= (1.60)²/(0.20 × 0.20). (1 mark)
= 2.56/0.04 = 64 (no units, since Δn = 0). (1 mark)

Answer: Kc = 64

Example 3: Dissociation of PCl₅ · 4 marks · Short answerHarder board level

0.50 mol of PCl₅ is placed in a 2.0 L vessel at 500 K and 40% dissociates at equilibrium: PCl₅ ⇌ PCl₃ + Cl₂. Find Kc and Kp. (R = 0.0821 L atm K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

Initial [PCl₅] = 0.25 mol L⁻¹; at equilibrium [PCl₅] = 0.25 × 0.6 = 0.15 and [PCl₃] = [Cl₂] = 0.25 × 0.4 = 0.10. (1.5 marks)
Kc = (0.10 × 0.10)/0.15 = 0.0667 mol L⁻¹. (1 mark)
Δn = +1: Kp = Kc(RT) = 0.0667 × 0.0821 × 500. (1 mark)
= 2.74 atm. (0.5 mark)

Answer: Kc = 0.0667 mol L⁻¹; Kp = 2.74 atm.

Example 4: Direction of reaction · 3 marks · Short answerHarder board level

For H₂ + I₂ ⇌ 2HI, Kc = 50 at a certain temperature. Which way does the reaction go in mixtures with (a) [H₂] = [I₂] = 0.5 and [HI] = 4.0, and (b) [H₂] = [I₂] = 0.5 and [HI] = 3.0 mol L⁻¹?

Try it first, then show the solutionSolution

(a) Q = 4.0²/(0.5 × 0.5) = 64 > 50, so the reaction goes backward. (1.5 marks)
(b) Q = 3.0²/0.25 = 36 < 50, so the reaction goes forward. (1.5 marks)

Answer: (a) backward (Q = 64); (b) forward (Q = 36).

Example 5: Equilibrium composition · 5 marks · Long answerHOTS / competency

One mole each of H₂ and I₂ is heated in a 1.0 L vessel where Kc = 50 for H₂ + I₂ ⇌ 2HI. Find the equilibrium concentrations.

Try it first, then show the solutionSolution

Let x mol L⁻¹ of H₂ react: [H₂] = [I₂] = 1 − x and [HI] = 2x. (1 mark)
Kc = (2x)²/(1 − x)² = 50. (1 mark)
Taking the square root: 2x/(1 − x) = 7.071. (1 mark)
2x = 7.071 − 7.071x, so x = 7.071/9.071 = 0.780. (1 mark)
[HI] = 1.56 mol L⁻¹; [H₂] = [I₂] = 0.22 mol L⁻¹. Check: 1.559²/0.2205² = 50.0. (1 mark)

Answer: [HI] = 1.56 mol L⁻¹; [H₂] = [I₂] = 0.22 mol L⁻¹.

Where students lose marks in Equilibrium Constants Kc, Kp and the Reaction Quotient

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

Mistake: Including solids in the expression
Writing [CaCO₃] in Kc.
Do this: Pure solids and liquids have constant activity and are omitted.
Usually costs: 1 to 2 marks.
Mistake: Wrong Δn
Counting liquids or solids in Δn.
Do this: Δn is the change in the number of gas moles only.
Usually costs: 2 marks.
Mistake: Using initial amounts in K
Putting the initial concentrations in the Kc expression.
Do this: Use equilibrium concentrations; initial ones give Q.
Usually costs: 2 marks.

Exam tips

Write the expression with coefficients as powers before substituting.
Use an ICE table in every calculation.
State units of K or say dimensionless when Δn = 0.
Check the answer by substituting back into Kc.

Practice questions on Equilibrium Constants Kc, Kp and the Reaction Quotient

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
The units of Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) are:
(a) mol L⁻¹
(b) L mol⁻¹
(c) mol² L⁻²
(d) none (dimensionless)
Hint
Δn = −1.
Answer
(b) L mol⁻¹
Q2 · 2 marks · Short answerBoard level
If Kc = 4 for A ⇌ B, find Kc for B ⇌ A and for 2A ⇌ 2B.
Hint
Reverse: 1/K; double: K².
Answer
B ⇌ A: 1/4 = 0.25. 2A ⇌ 2B: 4² = 16.
Q3 · 3 marks · Short answerHarder board level
N₂O₄(g) ⇌ 2NO₂(g) is 20% dissociated at a total pressure of 1 atm. Find the partial pressures and Kp.
Hint
Use mole fractions (1 − α)/(1 + α) and 2α/(1 + α).
Answer
p(N₂O₄) = 0.8/1.2 = 0.667 atm; p(NO₂) = 0.4/1.2 = 0.333 atm. Kp = 0.333²/0.667 = 0.167 atm.
Q4 · 3 marks · Short answerHarder board level
For 2SO₂ + O₂ ⇌ 2SO₃, Kc = 280 L mol⁻¹ at 1000 K. Find Kp. (R = 0.0821 L atm K⁻¹ mol⁻¹)
Hint
Δn = −1.
Answer
Kp = Kc(RT)⁻¹ = 280/(0.0821 × 1000) = 280/82.1 = 3.41 atm⁻¹.
Q5 · 3 marks · Short answerHarder board level
PCl₅ ⇌ PCl₃ + Cl₂ has Kc = 0.05 mol L⁻¹ at some temperature. If 1 mol of PCl₅ is placed in a 1 L vessel, find the degree of dissociation.
Hint
Kc = cα²/(1 − α) with c = 1.
Answer
α² + 0.05α − 0.05 = 0 → α = [−0.05 + √(0.0025 + 0.2)]/2 = (−0.05 + 0.45)/2 = 0.20, so 20%.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose an ammonia reactor starts with 1 mol N₂ and 3 mol H₂ in a 1 L vessel, and at 700 K 0.4 mol of NH₃ forms at equilibrium (R = 0.0821 L atm K⁻¹ mol⁻¹).
(i) Find the equilibrium concentrations of all three gases. (1 mark)
(ii) Find Kc. (1 mark)
(iii) Find Kp. (1 mark)
(iv) Find the percentage of the maximum possible NH₃ that is formed. (1 mark)
Hint
0.4 mol NH₃ uses 0.2 mol N₂ and 0.6 mol H₂.
Answer
(i) [NH₃] = 0.4, [N₂] = 0.8, [H₂] = 2.4 mol L⁻¹. (ii) Kc = 0.16/(0.8 × 13.82) = 0.0145 (mol L⁻¹)⁻². (iii) Kp = 0.01447/(57.47)² = 4.4 × 10⁻⁶ atm⁻². (iv) The maximum is 2 mol; 0.4/2 = 20%.

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.
2.K for the reverse reaction?Show answer
1/K.
3.What does Q > K mean?Show answer
The reaction proceeds backward.
4.Does K depend on concentration?Show answer
No, only on temperature.

Equilibrium Constants Kc, Kp and the Reaction Quotient: frequently asked questions

Does a large K mean a fast reaction?

No. K tells the extent of reaction at equilibrium, not the rate.

Why are pure solids left out of K?

Their concentration is a constant, which is absorbed into the value of K.

Can K change with pressure?

No. Changing the pressure can change the composition but not K at a fixed temperature.