Chemical Equilibrium — Kc, Kp & Le Chatelier

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A reaction at equilibrium has not stopped: the forward and backward reactions run at the same rate, so nothing seems to change. One number, K, describes where that balance sits, and almost every question in this chapter asks you to calculate it, compare with it, or predict how it moves.

What you will be able to do

• Write Kc and Kp for any reaction, leaving out pure solids and liquids, and convert between them
• Manipulate K when an equation is reversed, multiplied or added
• Use Q to predict direction, and solve ICE-table and degree-of-dissociation numericals
• Apply Le Chatelier’s principle to concentration, pressure, inert gas, temperature and catalysts
• Link K to ΔG° and use the van’t Hoff equation for K at a new temperature

The big picture

Equilibrium is a balance of rates, not a stop.

In one line: At equilibrium the forward and backward rates are equal; K fixes the ratio of products to reactants at that point, changes only with temperature, and Le Chatelier tells you which way the system shifts when disturbed.

• Dynamic: both reactions continue at equal rates. Mixing H₂ + I₂ or starting from pure HI reaches the same equilibrium.
• Closed system: equilibrium needs a closed vessel (CaCO₃ in an open container decomposes completely as CO₂ escapes).
• Physical equilibria follow the same idea: ice ⇌ water at 273 K, liquid ⇌ vapour in a sealed bottle, a saturated solution.

Think of it like this: Picture an escalator going up and a crowd walking down it at the same speed. People are moving all the time, but the number on each floor stays the same. That is dynamic equilibrium — constant amounts, not zero activity.

The equilibrium constant

The law of mass action, written for any reaction.

aA + bB ⇌ cC + dD
Kc = [C]ᶜ [D]ᵈ / ([A]ᵃ [B]ᵇ)

Concentrations at equilibrium, in mol L⁻¹, each raised to its stoichiometric coefficient

• Leave out pure solids and pure liquids — their concentration is constant. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂] and Kp = p(CO₂).
• Water is left out when it is the solvent, but included when it is a gas or one reactant among others in a non-aqueous mixture (e.g. esterification).
• K depends only on temperature — not on starting amounts, pressure, volume or a catalyst.
• Size of K: K > 10³ → products dominate; K < 10⁻³ → reactants dominate; in between, both are present in significant amounts.

Solved example 01 — Calculating Kc

For H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium concentrations are [H₂] = [I₂] = 0.10 M and [HI] = 0.70 M. Find Kc.

1. Kc = [HI]² / ([H₂][I₂]).
2. = (0.70)² / (0.10 × 0.10) = 0.49 / 0.010.

Answer: Kc = 49 (no units, since Δn = 0).

Kp and Kc

Partial pressures for gases, concentrations for solutions.

Kp = Kc (RT)^Δn

Δn = moles of gaseous products − moles of gaseous reactants; R = 0.0821 L atm K⁻¹ mol⁻¹ (or 0.0831 L bar K⁻¹ mol⁻¹)

Δn	Relation	Example
0	Kp = Kc	H₂ + I₂ ⇌ 2HI; N₂ + O₂ ⇌ 2NO
> 0	Kp > Kc (when RT > 1)	PCl₅ ⇌ PCl₃ + Cl₂ (Δn = +1)
< 0	Kp < Kc (when RT > 1)	N₂ + 3H₂ ⇌ 2NH₃ (Δn = −2)

Key idea: Match R to the pressure unit. With pressures in atm use 0.0821; with bar use 0.0831. Using 8.314 here is the commonest error — that value belongs with SI pressure (Pa) and volume (m³).

Solved example 02 — Kp from Kc

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 6.0 × 10⁻² at 500 K. Find Kp (atm).

1. Δn = 2 − 4 = −2.
2. RT = 0.0821 × 500 = 41.05; (RT)² = 1685.
3. Kp = Kc / (RT)² = 6.0 × 10⁻² / 1685.

Answer: Kp ≈ 3.6 × 10⁻⁵ atm⁻².

Manipulating K

What happens to K when you change the equation.

Change to the equation	New constant	Example
Reverse it	1/K	2HI ⇌ H₂ + I₂: K = 1/49
Multiply by n	Kn	2H₂ + 2I₂ ⇌ 4HI: K = 49² = 2401
Divide by n (e.g. halve)	K1/n (the nth root)	½H₂ + ½I₂ ⇌ HI: K = √49 = 7
Add two equations	K₁ × K₂	Overall K is the product

Solved example 03 — Reversing and halving

For N₂ + 3H₂ ⇌ 2NH₃ the constant is K. Write the constant for NH₃ ⇌ ½N₂ + 3/2 H₂.

1. Reverse the equation → 1/K.
2. Halve it → square root.

Answer: K′ = 1/√K.

Check yourself

• C1. Write Kp for C(s) + H₂O(g) ⇌ CO(g) + H₂(g).
• C2. If K = 4 for A ⇌ 2B, what is K for 2B ⇌ A?

Answers at the end.

Q and the direction of change

Compare the present mixture with equilibrium.

The reaction quotient Q has the same form as K but uses the concentrations right now, not at equilibrium.

Comparison	What the system does
Q < K	Too few products: reaction goes forward
Q = K	At equilibrium: no net change
Q > K	Too many products: reaction goes backward

Solved example 04 — Which way?

For H₂ + I₂ ⇌ 2HI, K = 49. A mixture has [HI] = 0.50 M and [H₂] = [I₂] = 0.10 M. Which way does it move?

1. Q = (0.50)² / (0.10 × 0.10) = 0.25 / 0.010 = 25.
2. Q (25) < K (49).

Answer: The reaction moves forward, forming more HI.

ICE tables

Initial, Change, Equilibrium — the method for every “find the amounts” numerical.

1. Write the balanced equation.
2. I: initial amounts (mol L⁻¹, or moles if Δn = 0).
3. C: changes in terms of x, in the ratio of the coefficients (reactants −, products +).
4. E: initial + change. Substitute into K and solve for x.

Solved example 05 — Solving an ICE table

1.0 mol each of H₂ and I₂ are heated in a 1 L flask. Kc = 64. Find the equilibrium concentrations.

1. I: H₂ 1.0, I₂ 1.0, HI 0. C: −x, −x, +2x. E: 1 − x, 1 − x, 2x.
2. Kc = (2x)² / (1 − x)² = 64 → take square roots: 2x / (1 − x) = 8.
3. 2x = 8 − 8x → 10x = 8 → x = 0.8.

Answer: [HI] = 1.6 M; [H₂] = [I₂] = 0.2 M.

Key idea: Perfect squares save time. When both sides are squares (as above), take the square root instead of solving a quadratic. When K is very small, assume x is negligible beside the initial amount — then check that x is under about 5% of it.

Degree of dissociation

How much of one substance falls apart.

For a dissociation, α is the fraction of one mole that dissociates. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) starting with 1 mol: at equilibrium PCl₅ = 1 − α, PCl₃ = α, Cl₂ = α, total = 1 + α.

Kp = α² P / (1 − α²)

P = total pressure at equilibrium. If α is small: α ≈ √(Kp / P)

α = (D − d) / [(n − 1) d]

Vapour density method: D = vapour density before dissociation, d = observed (mixture) vapour density, n = moles of products from one mole

Solved example 06 — PCl₅ dissociation

PCl₅ is 50% dissociated at 1 atm total pressure. Find Kp.

1. α = 0.5.
2. Kp = (0.5)² × 1 / (1 − 0.25) = 0.25 / 0.75.

Answer: Kp ≈ 0.33 atm.

Key idea: Lowering the total pressure increases α for any dissociation that increases the number of gas moles: from α ≈ √(Kp/P), α rises as P falls. K itself does not change.

Check yourself

• C3. Q = 0.5 and K = 2.0. Which direction does the reaction go?
• C4. Why does adding more CaCO₃(s) not shift CaCO₃ ⇌ CaO + CO₂?

Answers at the end.

Le Chatelier’s principle

Disturb an equilibrium and it shifts to partly undo the disturbance.

Change	Shift	Does K change?
Add a reactant / remove a product	Forward	No
Add a product / remove a reactant	Backward	No
Increase pressure (decrease volume)	Toward fewer gas moles	No
Decrease pressure	Toward more gas moles	No
Add inert gas at constant volume	No shift (partial pressures unchanged)	No
Add inert gas at constant pressure	Toward more gas moles (volume grows)	No
Raise temperature	In the endothermic direction	Yes — K rises for endothermic, falls for exothermic
Add a catalyst	No shift; equilibrium reached sooner	No

Key idea: Temperature is the only thing that changes K. Everything else changes Q and the system moves back to the same K.

Haber process: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ

• High pressure (about 200 atm) favours the side with fewer moles (2 vs 4) → more NH₃.
• Low temperature favours the exothermic forward reaction, but is too slow — so a compromise of about 700 K is used.
• Iron catalyst with molybdenum promoter speeds up the approach to equilibrium; it does not change the yield at equilibrium.
• NH₃ is liquefied and removed, pulling the equilibrium forward.

Solved example 07 — Predicting shifts

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0, what happens to the amount of SO₃ when (a) pressure is increased, (b) temperature is increased, (c) argon is added at constant volume?

1. (a) 3 gas moles → 2: higher pressure favours SO₃.
2. (b) Forward is exothermic: heating favours the backward reaction, and K falls.
3. (c) Partial pressures of the reacting gases do not change.

Answer: (a) More SO₃. (b) Less SO₃. (c) No change.

K, ΔG° and temperature

The thermodynamics behind K.

ΔG° = −2.303 RT log K

K > 1 ⇔ ΔG° < 0; K = 1 ⇔ ΔG° = 0; K < 1 ⇔ ΔG° > 0

log (K₂ / K₁) = [ΔH° / (2.303 R)] × [(T₂ − T₁) / (T₁ T₂)]

van’t Hoff equation; R = 8.314 J K⁻¹ mol⁻¹ here, ΔH° in J

Solved example 08 — van’t Hoff

The equilibrium constant of a reaction doubles when the temperature rises from 300 K to 310 K. Find ΔH°. (log 2 = 0.301)

1. log 2 = [ΔH° / (2.303 × 8.314)] × [10 / (300 × 310)].
2. ΔH° = 0.301 × 19.147 × 9300 J.

Answer: ΔH° ≈ +53.6 kJ mol⁻¹ — positive, as expected when K rises with temperature.

Exam trap gallery

Where marks are lost in equilibrium.

#	What students write	What is actually correct	Why
1	At equilibrium the reactions stop.	Both continue at equal rates (dynamic equilibrium).	Isotope-labelling experiments show exchange continues.
2	Include CaCO₃ and CaO in Kc.	Pure solids and liquids are left out: Kc = [CO₂].	Their concentrations are constant.
3	Use R = 8.314 in Kp = Kc(RT)Δn with atm.	Use 0.0821 L atm K⁻¹ mol⁻¹ (or 0.0831 for bar).	R must match the pressure and volume units.
4	Increasing pressure changes K.	Only temperature changes K; pressure changes Q.	The system shifts back to the same K.
5	A catalyst increases the yield at equilibrium.	A catalyst only gets there faster.	It speeds forward and backward reactions equally.
6	Adding inert gas always shifts the equilibrium.	At constant volume: no shift. At constant pressure: shifts toward more gas moles.	Only partial pressures of reacting gases matter.
7	Doubling an equation doubles K.	K is squared.	Concentrations are raised to the new coefficients.
8	Q > K means the reaction goes forward.	Q > K: backward. Q < K: forward.	Too many products already present.

Memory hook: “Solids and liquids sit out” — never in K. · “Kp = Kc (RT) to the Δn” — gas moles products minus reactants. · “Only heat moves K.” · “Q small, go forward; Q big, go back.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. At equilibrium: (Easy)

• (a) both reactions stop
• (b) forward rate = backward rate
• (c) concentrations of reactants and products are equal
• (d) K = 1

Q02. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp equals: (Easy)

• (a) p(CO₂)
• (b) p(CO₂)/p(CaCO₃)
• (c) 1/p(CO₂)
• (d) p(CaO) × p(CO₂)

Q03. Kp = Kc for: (Easy)

• (a) PCl₅ ⇌ PCl₃ + Cl₂
• (b) N₂ + 3H₂ ⇌ 2NH₃
• (c) H₂ + I₂ ⇌ 2HI
• (d) 2SO₂ + O₂ ⇌ 2SO₃

Q04. The value of K changes with: (Easy)

• (a) pressure
• (b) concentration
• (c) catalyst
• (d) temperature

Q05. If K for A ⇌ B is 10, K for B ⇌ A is: (Easy)

• (a) 10
• (b) 0.1
• (c) −10
• (d) 100

Q06. For N₂ + 3H₂ ⇌ 2NH₃, Kp/Kc equals: (Moderate)

• (a) (RT)²
• (b) (RT)⁻²
• (c) RT
• (d) 1

Q07. If Q < K, the reaction: (Moderate)

• (a) is at equilibrium
• (b) goes backward
• (c) goes forward
• (d) stops

Q08. Increasing pressure on 2NO₂ ⇌ N₂O₄ will: (Moderate)

• (a) increase NO₂
• (b) increase N₂O₄
• (c) have no effect
• (d) change K

Q09. For an exothermic reaction, raising temperature: (Moderate)

• (a) increases K
• (b) decreases K
• (c) does not affect K
• (d) makes K = 1

Q10. Adding argon at constant volume to N₂ + 3H₂ ⇌ 2NH₃: (Moderate)

• (a) shifts forward
• (b) shifts backward
• (c) no change
• (d) increases K

Q11. If K for ½A ⇌ B is 4, K for A ⇌ 2B is: (Moderate)

• (a) 2
• (b) 8
• (c) 16
• (d) 1/16

Q12. ΔG° for a reaction with K = 1 is: (Moderate)

• (a) negative
• (b) positive
• (c) zero
• (d) infinite

Q13. In PCl₅ ⇌ PCl₃ + Cl₂, lowering the total pressure: (Tough)

• (a) decreases α
• (b) increases α
• (c) does not change α
• (d) changes Kp

Q14. For H₂ + I₂ ⇌ 2HI with Kc = 64, starting with 1 mol each in 1 L, [HI] at equilibrium is: (Tough)

• (a) 0.8 M
• (b) 1.6 M
• (c) 0.2 M
• (d) 1.0 M

Q15. Adding inert gas at constant pressure to PCl₅ ⇌ PCl₃ + Cl₂: (Tough)

• (a) shifts forward
• (b) shifts backward
• (c) no change
• (d) increases Kp

Answer key with explanations

1. (b) forward rate = backward rate — Dynamic balance.
2. (a) p(CO₂) — Solids are left out.
3. (c) H₂ + I₂ ⇌ 2HI — Δn = 0.
4. (d) temperature — Only temperature.
5. (b) 0.1 — Reversed: 1/K.
6. (b) (RT)⁻² — Δn = −2.
7. (c) goes forward — Needs more products.
8. (b) increase N₂O₄ — Fewer moles on the right.
9. (b) decreases K — Backward (endothermic) direction favoured.
10. (c) no change — Partial pressures unchanged.
11. (c) 16 — Doubled equation: 4².
12. (c) zero — log 1 = 0.
13. (b) increases α — α ≈ √(Kp/P).
14. (b) 1.6 M — x = 0.8, [HI] = 2x.
15. (a) shifts forward — Volume grows → toward more moles.

Check-yourself answers

• C1. Kp = p(CO) × p(H₂) / p(H₂O). Carbon is a solid.
• C2. K = 1/4 = 0.25.
• C3. Forward (Q < K).
• C4. CaCO₃ is a solid; its “concentration” is constant and not in K.

One-page revision

Read this page the night before the exam.

The constant

• Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
• Leave out pure solids and liquids
• K changes only with T
• K > 10³ products; K < 10⁻³ reactants

Kp and Kc

• Kp = Kc (RT)^Δn
• Δn = gas moles products − reactants
• R = 0.0821 (atm) or 0.0831 (bar)
• Δn = 0 → Kp = Kc

Manipulating K

• Reverse → 1/K
• × n → K to the n
• ÷ 2 → √K
• Add equations → multiply K

Q and ICE

• Q < K forward; Q > K backward
• I, C (in ratio), E → into K
• Perfect squares: take roots
• Small K: neglect x, then check

Le Chatelier

• P up → fewer gas moles
• Inert gas: const V none; const P → more moles
• T up → endothermic side; K changes
• Catalyst: faster, no shift

Dissociation and ΔG°

• PCl₅: Kp = α²P/(1 − α²)
• α = (D − d)/[(n − 1)d]
• ΔG° = −2.303RT log K
• van’t Hoff: log(K₂/K₁) = ΔH°/2.303R × (T₂−T₁)/T₁T₂