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CBSE · Class 11 · Chemistry

Equilibrium

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
At equilibrium, the rates of the forward and backward reactions are:
  1. forward faster than backward
  2. backward faster than forward
  3. both zero
  4. equal
2.
For 2SO2(g) + O2(g) ⇌ 2SO3(g), the equilibrium concentrations are [SO2] = 0.5 M, [O2] = 0.5 M, [SO3] = 0.5 M. K_c is:
  1. 2
  2. 0.5
  3. 4
  4. 1
3.
For CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g), K_p and K_c are related by:
  1. K_p = K_c(RT)−2
  2. K_p = K_c(RT)
  3. K_p = K_c
  4. K_p = K_c(RT)2
4.
For N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92.4 kJ/mol, if a catalyst is added, the equilibrium:
  1. does not shift
  2. shifts to the right (forward)
  3. shifts to the left (backward)
  4. stops completely
5.
The pH of a 1 × 10−4 M solution of nitric acid (HNO3) at 298 K is: (assume complete ionisation)
  1. 4
  2. 3
  3. 10
  4. 5
6.
The solubility of Ag2SO4 in water is 7 × 10−2 mol/L. Its solubility product is:
  1. 1.372 × 10−4
  2. 4.9 × 10−3
  3. 1.372 × 10−2
  4. 1.372 × 10−3
7.
The conjugate acid of OH− is:
  1. H2CO3
  2. SO42−
  3. H2O
  4. H2PO4−
8.
Assertion (A): Increasing the pressure on 2SO2(g) + O2(g) ⇌ 2SO3(g) shifts the equilibrium to the left.
Reason (R): Increasing the pressure shifts an equilibrium towards the side with fewer moles of gas.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
State Le Chatelier’s principle.
10.
0.5 mol of N2O4 is heated in a 2 L vessel. At equilibrium 60% of it has dissociated (N2O4(g) ⇌ 2NO2(g)). Find K_c. (2 decimal places where needed.)
11.
The ionisation constant of a weak base BOH is 1 × 10−5. Find [OH−], the pH and the degree of ionisation of a 0.001 M solution at 298 K.
12.
A solution has [OH−] = 4 × 10−6 M at 298 K. Find its pH and pOH. (log 4 = 0.6021)
13.
At equilibrium, the partial pressures for CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) are CH4 2 bar, H2O 2 bar, CO 4 bar, H2 1 bar. Find K_p. (2 decimal places where needed.)
14.
The solubility product of CaF2 is 2.048 × 10−9. Find its solubility in mol/L.
15.
A buffer contains 0.1 M ethanoic acid (CH3COOH) and 0.05 M of its sodium salt. Find its pH. (pKa = 4.76) (log 2 = 0.3010, log 5 = 0.6990)
16.
For PCl5(g) ⇌ PCl3(g) + Cl2(g), K_c = 10. At one moment [PCl5] = 2 M, [PCl3] = 0.1 M, [Cl2] = 1 M. Find Q and say which way the reaction goes. (2 decimal places where needed.)
Section C — Word problems4 × 4 = 16 marks
17.
Explain the effect of changing concentration, pressure, temperature and adding a catalyst on the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92.4 kJ/mol.
18.
4 mol each of H2 and I2 are heated in a 2 L vessel: H2(g) + I2(g) ⇌ 2HI(g), K_c = 4. Find the equilibrium concentrations of H2, I2 and HI. (2 decimal places where needed.)
19.
A weak acid HA has Ka = 1 × 10−3. For a 0.1 M solution at 298 K find (a) [H+], (b) the pH, (c) the degree of ionisation. (d) What is the pH after the solution is diluted 10 times?
20.
The solubility product of PbI2 is 2.048 × 10−6. (a) Find its solubility in pure water. (b) Find its solubility in a solution in which [I−] = 0.05 M (from a soluble salt). (c) Explain the difference.
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CBSE · Class 11 · Chemistry

Answer key

Equilibrium · AW-11C-EQM-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) equalequilibrium is dynamic
2(a) 2K_c = [SO3]2 / ([SO2]2[O2])
3(d) K_p = K_c(RT)2K_p = K_c(RT)^Δn with Δn = 2
4(a) does not shiftLe Chatelier: it shifts a catalyst speeds up both directions equally
5(a) 4[H+] = 1 × 1 × 10−4 = 10−4 M
6(d) 1.372 × 10−3Ag2SO4 ⇌ 2Ag+ + SO42−, so K_sp = 4s3
7(c) H2Oadd one H+
8(d) A is false, but R is true.A is false: Δn = −1. R is true.
9If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, it shifts in the direction that tends to undo the change.—
10K_c = 0.9 ([N2O4] = 0.1 M, [NO2] = 0.3 M)initial 0.25 M; use an ICE table with α
11[OH−] = 1 × 10−4 M; pH = 10; α = 10%[OH−] = √(Kb × c); pOH = −log[OH−]; pH = 14 − pOH
12pOH = 5.3979; pH = 8.6021pOH = −log(4 × 10−6) = 6 − log 4; pH + pOH = 14
13K_p = 1K_p = p(CO) × p(H2)3 / (p(CH4) × p(H2O))
148 × 10−4 mol/LK_sp = 4s3, so s = ∛(K_sp/4)
15pH = 4.459pH = pKa + log([salt]/[acid]) = 4.76 + log 0.5
16Q = 0.05; it goes forward (Q < K)Q = [PCl3][Cl2] / [PCl5] with these concentrations
17Adding N2 or H2, or removing NH3, shifts it to the right. Raising the pressure shifts it to the right (4 mol of gas become 2). Raising the temperature shifts it to the left (the forward reaction is exothermic) and lowers K. A catalyst (iron) only makes equilibrium come faster. So high pressure, moderate temperature (about 700 K) and a catalyst are used in the Haber process.—
18[H2] = [I2] = 1 M; [HI] = 2 Mif x mol reacts: (2x)2/(4 − x)2 = 4, so 2x/(4 − x) = 2, x = 2
19(a) 1 × 10−2 M; (b) 2; (c) 10%; (d) 2.5[H+] = √(K × c); after dilution c is 10 times smaller
20(a) 8 × 10−3 mol/L; (b) 8.19 × 10−4 mol/L; (c) the common ion shifts PbI2(s) ⇌ Pb2+ + 2I− to the left (common ion effect), so less dissolves(a) K_sp = 4s3; (b) K_sp = s × (0.05)2