Chemical Equilibrium ISC Class 11: Kc, Kp and Le Chatelier's Principle

What Chemical Equilibrium is about

Weightage: Chemical Equilibrium belongs to unit 6 of the ISC Class XI Chemistry syllabus. CISCE weights units 1-6 together (Physical Chemistry) at 41 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Reversible reactions and physical equilibria

An irreversible reaction goes essentially to completion in one direction (burning magnesium). A reversible reaction can proceed in both directions under the same conditions, shown by ⇌. Physical equilibria: solid ⇌ liquid (ice and water at 0 °C and 1 atm, where the rates of melting and freezing are equal); liquid ⇌ vapour (water in a closed flask reaches a constant vapour pressure when evaporation and condensation rates are equal); solid ⇌ vapour (iodine, camphor or ammonium chloride sublimation in a closed vessel). Equilibrium is possible only in a closed system.

Worked example

Why does the vapour pressure of water in a sealed flask become constant at a given temperature?

Try it first, then show the solutionSolution

1. At first evaporation is faster than condensation, so vapour builds up
2. As vapour increases, condensation speeds up until the two rates are equal
3. Both processes continue, but the amount of vapour no longer changes

Answer: A dynamic liquid–vapour equilibrium is reached

Dynamic nature and characteristics of equilibrium

At equilibrium the rate of the forward reaction equals the rate of the backward reaction, so concentrations remain constant although molecules keep reacting (shown by isotope-labelling experiments). Characteristics: it is attained only in a closed system; it can be approached from either direction; observable properties (colour, pressure, concentration) become constant; a catalyst does not change the position, only how quickly it is reached; and the equilibrium constant depends only on temperature.

Worked example

H₂ and I₂ are mixed in one flask, and pure HI is placed in another at the same temperature. What is found after a long time?

Try it first, then show the solutionSolution

1. Both flasks reach mixtures of H₂, I₂ and HI
2. The ratio [HI]²/([H₂][I₂]) has the same value in both

Answer: The same equilibrium is reached from either direction

Law of mass action and the equilibrium constant Kc

The law of mass action: the rate of a reaction is proportional to the product of the active masses (molar concentrations) of the reactants, each raised to its coefficient. Equating forward and backward rates at equilibrium gives, for aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ). Rules: concentrations are equilibrium values in mol L⁻¹; pure solids and liquids are omitted (their 'concentration' is constant), which matters in heterogeneous equilibria; reversing a reaction gives 1/K; multiplying the equation by n gives Kⁿ; adding reactions multiplies their Ks. Units of Kc are (mol L⁻¹)^Δn.

Worked example

Write Kc for $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$ and for $CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$, with units.

Try it first, then show the solutionSolution

1. Kc = [NH₃]²/([N₂][H₂]³); Δn = 2 − 4 = −2, units (mol L⁻¹)⁻² = L² mol⁻²
2. Solids omitted: Kc = [CO₂], units mol L⁻¹ (and Kp = p_CO₂)

Answer: [NH₃]²/([N₂][H₂]³), L² mol⁻²; [CO₂], mol L⁻¹

Kp and its relation to Kc (derivation)

For gases, partial pressures can replace concentrations: Kp = p_Cᶜ p_Dᵈ/(p_Aᵃ p_Bᵇ). For an ideal gas p = (n/V)RT = [X]RT. Substituting: Kp = ([C]RT)ᶜ([D]RT)ᵈ/(([A]RT)ᵃ([B]RT)ᵇ) = Kc(RT)^((c + d) − (a + b)), so Kp = Kc(RT)^Δn, where Δn = moles of gaseous products − moles of gaseous reactants. Use R = 0.0821 L atm K⁻¹ mol⁻¹ with pressures in atm (or 0.0831 L bar K⁻¹ mol⁻¹ for bar). If Δn = 0, Kp = Kc.

Worked example

For $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$ at 500 K, Kc = 0.061 L² mol⁻². Find Kp.

Try it first, then show the solutionSolution

1. Δn = 2 − 4 = −2
2. RT = 0.0821 × 500 = 41.05
3. Kp = 0.061 × (41.05)⁻² = 0.061/1685 = 3.62 × 10⁻⁵ atm⁻²

Answer: 3.6 × 10⁻⁵ atm⁻²

Calculating equilibrium concentrations

Use an ICE table: Initial amounts, Change (in terms of x and the coefficients), Equilibrium amounts. Divide by volume for concentrations, substitute into Kc and solve; for symmetric cases take square roots. For dissociation reactions it is common to use the degree of dissociation α (fraction of the reactant that dissociates). Always check that the answer is physically sensible (no negative concentrations, α between 0 and 1).

Worked example

1.0 mol H₂ and 1.0 mol I₂ are heated in a 1.0 L flask; Kc for $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$ is 50. Find the equilibrium concentrations.

Try it first, then show the solutionSolution

1. Equilibrium: [H₂] = [I₂] = 1 − x, [HI] = 2x
2. (2x)²/(1 − x)² = 50 ⇒ 2x/(1 − x) = 7.07
3. x = 7.07/9.07 = 0.78; [HI] = 1.56 M, [H₂] = [I₂] = 0.22 M

Answer: [HI] = 1.56 M; [H₂] = [I₂] = 0.22 M

Reaction quotient and the meaning of K

The reaction quotient Q has the same form as K but uses the concentrations at any moment. If Q < K, the reaction proceeds forward; if Q > K, backward; if Q = K, the system is at equilibrium. The size of K shows the extent of reaction: K > 10³ products predominate; K < 10⁻³ reactants predominate; in between, appreciable amounts of both are present. K says nothing about the rate.

Worked example

For $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$, Kc = 0.50 at a certain temperature. A mixture has [N₂] = 0.20, [H₂] = 0.40, [NH₃] = 0.10 mol L⁻¹. Which way will it move?

Try it first, then show the solutionSolution

1. Q = (0.10)²/(0.20 × 0.40³) = 0.010/0.0128 = 0.78
2. Q > K, so the system has too much product

Answer: Backward (some NH₃ decomposes)

Le Chatelier's principle

If a system at equilibrium is subjected to a change, it adjusts to oppose that change. Concentration: adding a reactant or removing a product shifts the equilibrium forward. Pressure (by changing volume): an increase favours the side with fewer moles of gas; no effect if Δn = 0. Temperature: raising it favours the endothermic direction, and is the only factor that changes K (K decreases with T for exothermic reactions). Catalyst: no shift; equilibrium is reached faster. Inert gas: at constant volume, no change (partial pressures unchanged); at constant pressure, the volume increases and the equilibrium shifts towards more moles of gas.

Worked example

For $N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$ (colourless ⇌ brown, ΔH = +57 kJ), predict the colour change when (i) the mixture is heated (ii) it is compressed.

Try it first, then show the solutionSolution

1. (i) Endothermic forward: heating shifts right; more NO₂, darker brown
2. (ii) Compression favours fewer gas moles (left); after the initial darkening from compression, the colour fades as NO₂ converts to N₂O₄

Answer: (i) Darker (ii) Lighter (after an initial darkening)

Maximum yield in industry and in the laboratory

Haber process: $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$, ΔH = −92 kJ. High pressure (about 200 atm) favours fewer moles; low temperature favours the exothermic forward reaction, but is too slow, so a compromise of about 700 K with an iron catalyst is used, and ammonia is removed by cooling. Contact process: $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$, ΔH = −197 kJ; excess air, about 720 K, V₂O₅ catalyst and pressures of 1–2 atm (conversion is already high). N₂O₄ ⇌ 2NO₂ (endothermic): high temperature and low pressure favour NO₂. Ester hydrolysis: $C{H}_{3}COO{C}_2{H}_5+H_{2}O\rightleftharpoons C{H}_{3}COOH+C_2{H}_{5}OH$ has Kc of about 0.25, so excess water (or removing a product) is used to push it forward.

Worked example

Give two conditions that increase the yield of SO₃ in the contact process and explain each.

Try it first, then show the solutionSolution

1. Excess oxygen (air): increases a reactant's concentration, so the equilibrium shifts forward
2. Moderately low temperature (about 720 K): the forward reaction is exothermic, so lower T favours SO₃, while V₂O₅ keeps the rate acceptable

Answer: Excess air; moderately low temperature with a catalyst

Chemistry in real life: Chemical Equilibrium

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.

Topics in this chapter

• Equilibrium Constants Kc, Kp and the Reaction Quotient
• Le Chatelier's Principle and Industrial Equilibria

Chemical Equilibrium formula sheet

Equilibrium constant

$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$

Omit pure solids and liquids.

Kp and Kc

$K_p=K_c(RT{)}^{\Delta n}$

Δn = gaseous product moles − gaseous reactant moles.

Manipulating K

$K_{\text{reverse}}=\frac{1}{K};K_{(n\times \text{eqn})}=K^n;K_{\text{sum}}=K_1{K}_2$

Halving an equation gives √K.

Direction of change

$Q<K:\text{forward};Q>K:\text{backward};Q=K:\text{equilibrium}$

Q uses current concentrations.

Dissociation of PCl₅ type

$K_p=\frac{{\alpha }^{2}P}{1-{\alpha }^2}$

For A ⇌ B + C at total pressure P.

Dissociation of N₂O₄ type

$K_p=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$

For A ⇌ 2B at total pressure P.

Chemical Equilibrium: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Chemical Equilibrium

The question patterns to expect, and the method that works for each.

Tricks to remember Chemical Equilibrium

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

• Products on top

  K = products over reactants, powers from coefficients, solids and liquids out.

  Use it when: Writing K.

• Delta n on RT

  Kp = Kc(RT)^Δn; Δn counts only gases.

  Use it when: Kp–Kc conversion.

• Q chases K

  Q smaller than K: go forward; bigger: go back.

  Use it when: Direction of reaction.

• Squeeze to fewer

  Higher pressure favours fewer gas molecules.

  Use it when: Pressure changes.

• Only heat changes K

  Concentration and pressure shift the position; temperature changes K itself.

  Use it when: Le Chatelier questions.

Chemical Equilibrium: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Chemical Equilibrium

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

• ✗Mistake: Including solids or pure liquids in K

  Students copy every species.

  Wrong working

  Kc = [CaO][CO₂]/[CaCO₃].

  Right working

  Kc = [CO₂].

  ✓Do this: Cross out (s) and (l) before writing K.

  Usually costs: 1 mark.

• ✗Mistake: Wrong Δn in Kp = Kc(RT)^Δn

  Students subtract the wrong way or count non-gases.

  Wrong working

  N₂ + 3H₂ ⇌ 2NH₃: Δn = +2.

  Right working

  Δn = 2 − 4 = −2.

  ✓Do this: Products minus reactants, gases only.

  Usually costs: 1-2 marks.

• ✗Mistake: Saying pressure or concentration changes K

  Shifts of position are confused with changes of the constant.

  Wrong working

  Adding N₂ increases K.

  Right working

  Q changes; the system shifts until Q = K again; K is unchanged.

  ✓Do this: Only temperature changes K.

  Usually costs: 1 mark.

• ✗Mistake: Using moles instead of concentrations

  Students forget to divide by volume when Δn ≠ 0.

  Wrong working

  Kc = 0.20 × 0.20/0.30 in a 2.0 L flask.

  Right working

  Use 0.10 × 0.10/0.15 = 0.067 mol L⁻¹.

  ✓Do this: Convert to mol L⁻¹ unless volumes cancel (Δn = 0).

  Usually costs: 1-2 marks.

• ✗Mistake: Claiming an inert gas always shifts equilibrium

  The constant-volume case is overlooked.

  Wrong working

  Adding argon at constant volume shifts N₂O₄ ⇌ 2NO₂ right.

  Right working

  At constant volume partial pressures are unchanged: no shift.

  ✓Do this: State whether volume or pressure is held constant.

  Usually costs: 1 mark.

Practice questions on Chemical Equilibrium

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

1. Q1 · 1 mark · MCQExam level

   For which reaction is Kp = Kc?
   (a) N₂ + 3H₂ ⇌ 2NH₃
   (b) H₂ + I₂ ⇌ 2HI
   (c) PCl₅ ⇌ PCl₃ + Cl₂
   (d) 2SO₂ + O₂ ⇌ 2SO₃

   Hint

   Δn = 0.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   Increasing the pressure on N₂ + 3H₂ ⇌ 2NH₃ at equilibrium:
   (a) shifts it forward
   (b) shifts it backward
   (c) has no effect
   (d) changes K

   Hint

   Fewer gas moles on the right.

   Answer

   (a)

3. Q3 · 1 mark · Very short answerExam level

   Write Kc for $C(s)+C{O}_2(g)\rightleftharpoons 2CO(g)$.

   Hint

   Omit carbon.

   Answer

   Kc = [CO]²/[CO₂]

4. Q4 · 1 mark · Very short answerExam level

   Give one example of a solid–vapour equilibrium.

   Hint

   Sublimation in a closed vessel.

   Answer

   Iodine (or camphor) in a closed jar: I₂(s) ⇌ I₂(g)

5. Q5 · 2 marks · Short answerExam level

   A reaction has Kc = 1 × 10⁻¹⁰ at 298 K. What does this tell you, and what does it not tell you?

   Hint

   Extent vs rate.

   Answer

   At equilibrium almost no products are present (reactants predominate). It says nothing about how fast equilibrium is reached.

6. Q6 · 2 marks · Short answerHarder exam level

   Kc for H₂ + I₂ ⇌ 2HI is 50. Find Kc for (i) 2HI ⇌ H₂ + I₂ (ii) HI ⇌ ½H₂ + ½I₂.

   Hint

   Reverse; then square root.

   Answer

   (i) 1/50 = 0.020 (ii) √0.020 = 0.141

7. Q7 · 2 marks · Short answerHarder exam level

   For CO(g) + 2H₂(g) ⇌ CH₃OH(g), Kc = 300 L² mol⁻² at 500 K. Find Kp.

   Hint

   Δn = −2.

   Answer

   Kp = 300 × (0.0821 × 500)⁻² = 300/1685 = 0.178 atm⁻²

8. Q8 · 3 marks · Short answerHarder exam level

   For A + B ⇌ C + D, Kc = 9.0. 1.0 mol each of A and B are mixed in 1.0 L. Find the equilibrium concentration of C.

   Hint

   x²/(1 − x)² = 9.

   Answer

   x/(1 − x) = 3 ⇒ x = 0.75 mol L⁻¹

9. Q9 · 3 marks · Short answerHarder exam level

   For $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$, ΔH < 0, state the effect on the amount of SO₃ of (i) adding O₂ (ii) raising the temperature (iii) increasing the pressure (iv) adding V₂O₅ (v) adding argon at constant volume.

   Hint

   Le Chatelier for each.

   Answer

   (i) increases (ii) decreases (iii) increases (iv) no change (faster) (v) no change

10. Q10 · 3 marks · Short answerHarder exam level

    Kc = 0.061 for N₂ + 3H₂ ⇌ 2NH₃ at 500 K. A mixture has [N₂] = 3.0, [H₂] = 2.0, [NH₃] = 0.50 mol L⁻¹. Is it at equilibrium? If not, which way will it move?

    Hint

    Find Q.

    Answer

    Q = 0.25/(3.0 × 8.0) = 0.0104 < 0.061: not at equilibrium; it moves forward, forming more NH₃.

11. Q11 · 3 marks · Short answerHarder exam level

    For N₂O₄ ⇌ 2NO₂ at 298 K, Kp = 0.15 atm. Find the degree of dissociation at a total pressure of 1.0 atm.

    Hint

    Kp = 4α²P/(1 − α²).

    Answer

    4α²/(1 − α²) = 0.15 ⇒ α² = 0.15/4.15 = 0.0361 ⇒ α = 0.19

12. Q12 · 2 marks · Short answerHarder exam level

    Explain, in terms of rates, why ice and water can coexist indefinitely at 0 °C and 1 atm.

    Hint

    Two opposing processes.

    Answer

    At 0 °C the rate at which ice melts equals the rate at which water freezes, so the amounts of each stay constant: a dynamic solid–liquid equilibrium.

13. Q13 · 5 marks · Long answerHarder exam level

    For $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$, Kc = 49 at 700 K.
    (a) Find Kc for the decomposition of HI. [1]
    (b) 2.0 mol of HI is heated in a 1.0 L vessel at 700 K. Find the equilibrium concentrations. [3]
    (c) What is the effect of increasing the pressure? [1]

    Hint

    Let [H₂] = [I₂] = x.

    Answer

    (a) 1/49 = 0.0204. (b) x²/(2 − 2x)² = 1/49 ⇒ x/(2 − 2x) = 1/7 ⇒ x = 2/9 = 0.22 M; [HI] = 2 − 4/9 = 1.56 M. (c) None, because Δn = 0.

14. Q14 · 3 marks · Short answerHOTS / competency

    For PCl₅ ⇌ PCl₃ + Cl₂, show that α = √(Kp/(Kp + P)) and find α at P = 4.0 atm when Kp = 1.78 atm. Comment on the effect of pressure.

    Hint

    Rearrange Kp = α²P/(1 − α²).

    Answer

    Kp(1 − α²) = α²P ⇒ α² = Kp/(Kp + P). At 4.0 atm: α² = 1.78/5.78 = 0.308, α = 0.555 (less than 0.80 at 1 atm): higher pressure suppresses dissociation, as Le Chatelier predicts.

15. Q15 · 3 marks · Short answerHOTS / competency

    For esterification, CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, Kc = 4.0. If 1.0 mol of acid reacts with 2.0 mol of alcohol, how much ester forms, and what fraction of the acid is converted?

    Hint

    Volumes cancel; solve the quadratic and reject the root above 1.

    Answer

    x²/[(1.0 − x)(2.0 − x)] = 4 ⇒ 3x² − 12x + 8 = 0 ⇒ x = (12 − √48)/6 = 0.845 mol (the other root, 3.15, is impossible). About 85% of the acid is converted, against 67% with equal amounts.

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. 1.What is dynamic equilibrium?Show answer

   Forward and backward rates are equal, so concentrations stay constant while reactions continue.

2. 2.Law of mass action?Show answer

   Rate ∝ product of active masses of reactants raised to their coefficients.

3. 3.Kp and Kc relation?Show answer

   Kp = Kc(RT)^Δn.

4. 4.K for the reversed reaction?Show answer

   1/K.

5. 5.Q > K means?Show answer

   Reaction moves backward.

6. 6.Which factor changes K?Show answer

   Temperature only.

7. 7.Effect of a catalyst?Show answer

   Equilibrium reached faster; no change in position.

8. 8.Inert gas at constant pressure?Show answer

   Shifts towards more moles of gas.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you give one example of each physical equilibrium?
• ☐Can you list four characteristics of chemical equilibrium?
• ☐Can you write Kc and Kp with units for any reaction?
• ☐Can you derive Kp = Kc(RT)^Δn?
• ☐Can you find equilibrium concentrations from Kc?
• ☐Can you use Q to predict the direction of change?
• ☐Can you predict the effect of each factor, including an inert gas?
• ☐Can you justify the conditions used in the Haber and contact processes?

Chemical Equilibrium: frequently asked questions