Equilibrium MCQs – 40 Practice Questions with Answers and Explanations

Master chemical equilibrium with 40 free practice MCQs, written for GCSE, IGCSE, A-Level and AP Chemistry students. The questions build up from reversible reactions and Le Chatelier’s principle to full Kc, Kp, Kw and Ksp calculations — every answer comes with a worked explanation, so you can see exactly where the marks come from.

Quick formula recap: Need the equations first? Revise with our free formula sheets:
GCSE Chemistry Formula Sheet ·
IGCSE Chemistry Formula Sheet ·
A-Level Chemistry Formula Sheet ·
AP Chemistry Formula Sheet

Reversible Reactions & Dynamic Equilibrium

Q1. Which symbol is used to show that a chemical reaction is reversible?

  • (A) →
  • (B) ⇌
  • (C) Δ
  • (D) +
Show Answer

Answer: B – The double harpoon symbol ⇌ shows a reversible reaction: the forward and reverse reactions can both occur. A single arrow → is used for reactions that go to completion.

Q2. A dynamic equilibrium can only be established in which type of system?

  • (A) An open system open to the air
  • (B) A system where all the reactants have been used up
  • (C) A closed system
  • (D) A system where concentrations of reactants and products are equal
Show Answer

Answer: C – In a closed system nothing can enter or escape, so the forward and reverse reactions can balance each other. In an open system, gases or vapours escape and equilibrium is never reached.

Q3. When a reversible reaction reaches dynamic equilibrium, which statement is correct?

  • (A) The forward rate equals the reverse rate, and both reactions keep happening
  • (B) The forward reaction stops completely
  • (C) The reverse reaction stops completely
  • (D) Both reactions stop completely
Show Answer

Answer: A – Equilibrium is dynamic: both reactions continue at the molecular level, but at equal rates, so there is no overall change. Nothing has stopped.

Q4. At equilibrium, the concentrations of reactants and products…

  • (A) stay constant, but are not necessarily equal
  • (B) are always exactly equal to each other
  • (C) keep increasing slowly
  • (D) both fall to zero
Show Answer

Answer: A – Equal rates mean amounts stop changing overall, so concentrations stay constant. They are only equal by coincidence — usually one side is present in much larger amount than the other.

Q5. Blue hydrated copper(II) sulfate crystals turn white when heated gently. This is because…

  • (A) the copper melts out of the crystals
  • (B) water of crystallisation is driven off, and adding water reverses the change
  • (C) the crystals react with oxygen in the air
  • (D) the sulfate group decomposes to sulfur dioxide
Show Answer

Answer: B – CuSO4.5H2O ⇌ CuSO4 + 5H2O. Heating drives the forward reaction (white anhydrous salt); adding water reverses it and the blue colour returns — this is the chemical test for water.

Q6. Which is the correctly balanced equation for the Haber process?

  • (A) N2 + H2 ⇌ NH3
  • (B) 2N2 + 3H2 ⇌ 2NH3
  • (C) N2 + 3H2 ⇌ 2NH3
  • (D) N2 + H2 ⇌ 2NH3
Show Answer

Answer: C – N2 + 3H2 ⇌ 2NH3: atoms balance (2 N and 6 H on each side) and the reaction is reversible, so the ⇌ symbol is used.

Q7. In the esterification reaction, ethanoic acid reacts with ethanol. The products are…

  • (A) ethyl ethanoate only
  • (B) water only
  • (C) ethane and carbon dioxide
  • (D) ethyl ethanoate and water
Show Answer

Answer: D – CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O. The reaction is reversible, which is why the ester must be separated from the mixture as it forms.

Q8. The word “dynamic” in dynamic equilibrium means that…

  • (A) the equilibrium position keeps swinging from side to side
  • (B) both the forward and reverse reactions are still occurring
  • (C) the molecules stop moving once equilibrium is reached
  • (D) the reaction alternates, forward one minute and backward the next
Show Answer

Answer: B – Forward and reverse reactions run continuously and simultaneously at the same rate. Macroscopic properties stay constant, but at particle level the reaction never stops.

Q9. Solid ammonium chloride disappears when heated in a test tube, and a white solid re-forms near the cooler top of the tube. This happens because…

  • (A) ammonium chloride melts and then freezes again
  • (B) it dissociates into ammonia and hydrogen chloride gases, which recombine on cooling
  • (C) it reacts with the glass of the test tube
  • (D) it evaporates and is lost permanently
Show Answer

Answer: B – NH4Cl(s) ⇌ NH3(g) + HCl(g). Heat drives dissociation; near the cooler glass the gases recombine to form solid ammonium chloride again — a reversible change.

Q10. Which statement about a system at dynamic equilibrium is NOT true?

  • (A) The forward and reverse rates are equal
  • (B) The concentrations of reactants and products stay constant
  • (C) Both reactions are still taking place
  • (D) The concentrations of reactants and products are equal
Show Answer

Answer: D – Constant does not mean equal. At equilibrium the concentrations stop changing, but the reactant and product concentrations are usually very different.

Le Chatelier’s Principle

Q11. For the Haber process, N2 + 3H2 ⇌ 2NH3 (ΔH = −92 kJ mol−1), what happens to the yield of ammonia if the temperature is increased?

  • (A) The equilibrium shifts left, so the yield of ammonia falls
  • (B) The equilibrium shifts right, so the yield of ammonia rises
  • (C) The yield stays exactly the same
  • (D) Only the rate changes; the yield is unaffected
Show Answer

Answer: A – The forward reaction is exothermic, so raising the temperature favours the endothermic (reverse) direction to absorb the extra heat. The position shifts left and less ammonia is present at equilibrium (although it forms faster).

Q12. For the same Haber process equilibrium, what is the effect of increasing the pressure?

  • (A) The position shifts left, towards more moles of gas
  • (B) There is no effect on the position
  • (C) The position shifts right, towards fewer moles of gas
  • (D) Ammonia decomposes completely
Show Answer

Answer: C – There are 4 moles of gas on the left (1 N2 + 3 H2) but only 2 moles on the right. Higher pressure favours the side with fewer gas molecules, so the position shifts right and the ammonia yield rises.

Q13. What is the effect of adding a catalyst to a reversible reaction at equilibrium?

  • (A) It shifts the position to the right
  • (B) It has no effect on the position, but equilibrium is reached faster
  • (C) It shifts the position to the left
  • (D) It increases the equilibrium yield of product
Show Answer

Answer: B – A catalyst speeds up the forward and reverse reactions equally, so the position (and yield) is unchanged — equilibrium is simply reached sooner.

Q14. In the Contact process, 2SO2 + O2 ⇌ 2SO3. If sulfur trioxide is removed from the mixture as it forms, the position of equilibrium will…

  • (A) shift to the left
  • (B) stay exactly where it was
  • (C) cause the equilibrium constant to fall
  • (D) shift to the right to replace the SO3 removed
Show Answer

Answer: D – Removing a product disturbs the equilibrium, so the forward reaction runs faster than the reverse until balance is restored — more SO2 and O2 are converted to SO3.

Q15. Dinitrogen tetroxide dissociates on heating: N2O4(g) ⇌ 2NO2(g), ΔH = +57 kJ mol−1. NO2 is brown and N2O4 is colourless. When the mixture is heated, the brown colour deepens because…

  • (A) the equilibrium shifts right, producing more NO2
  • (B) the equilibrium shifts left, producing more NO2
  • (C) all the N2O4 evaporates
  • (D) NO2 molecules expand when heated
Show Answer

Answer: A – Dissociation is endothermic, so heating favours the forward direction to take in heat. More brown NO2 forms and the colour deepens.

Q16. The Haber process uses a compromise temperature of about 450 °C rather than a much lower temperature. Why?

  • (A) 450 °C gives the maximum possible yield of ammonia
  • (B) the iron catalyst only works above 450 °C
  • (C) a lower temperature gives a better yield but an unacceptably slow rate
  • (D) ammonia decomposes below 450 °C
Show Answer

Answer: C – Low temperature favours yield (exothermic reaction) but the rate becomes far too slow to be economic. 450 °C balances a reasonable yield against a reasonable rate.

Q17. The Contact process is run at only about 200 kPa (2 atmospheres) even though higher pressure would favour SO3 formation. Why?

  • (A) high pressure shifts the equilibrium to the left
  • (B) the conversion is already about 98%, so costly high-pressure equipment is not worth it
  • (C) the vanadium(V) oxide catalyst is poisoned by pressure
  • (D) SO3 becomes a liquid at high pressure
Show Answer

Answer: B – There are fewer gas moles on the right (3 → 2), so pressure helps a little, but with conversion already around 98% the extra gain does not justify the cost and safety demands of high-pressure plant.

Q18. For the equilibrium H2(g) + I2(g) ⇌ 2HI(g), increasing the pressure will…

  • (A) shift the position to the right
  • (B) shift the position to the left
  • (C) produce much more HI at equilibrium
  • (D) have no effect on the position of equilibrium
Show Answer

Answer: D – There are 2 moles of gas on each side of the equation, so pressure changes do not favour either direction. The position stays the same (the rates of both reactions increase equally).

Q19. Extra reactant is added to a mixture that is already at equilibrium. The position of equilibrium will…

  • (A) shift to the left, making more of the other reactant
  • (B) shift to use up some of the added reactant
  • (C) not change at all
  • (D) change so that the equilibrium constant increases
Show Answer

Answer: B – By Le Chatelier’s principle, the system opposes the change: the forward reaction speeds up relative to the reverse, consuming some of the added reactant until a new equilibrium is reached. Kc itself is unchanged (temperature is constant).

Q20. For CO(g) + H2O(g) ⇌ CO2(g) + H2(g), ΔH = −41 kJ mol−1. Raising the temperature will…

  • (A) increase Kc and produce more CO2
  • (B) produce more CO2 but leave Kc unchanged
  • (C) decrease Kc and shift the position to the left
  • (D) leave both Kc and the position unchanged
Show Answer

Answer: C – The forward reaction is exothermic, so heating shifts the position left (towards CO and H2O). Because there are fewer products at equilibrium, Kc decreases. Temperature is the only change that alters K.


Kc & Kp Calculations

Q21. What is the correct Kc expression for N2(g) + 3H2(g) ⇌ 2NH3(g)?

  • (A) [N2][H2]3 / [NH3]2
  • (B) [NH3]2 / ([N2][H2]3)
  • (C) [NH3] / ([N2][H2])
  • (D) 2[NH3] / ([N2] + 3[H2])
Show Answer

Answer: B – Kc = [products] / [reactants], with each concentration raised to the power of its balancing number: [NH3]2 divided by [N2]1[H2]3.

Q22. At equilibrium for H2 + I2 ⇌ 2HI, the concentrations are [H2] = 0.10, [I2] = 0.10 and [HI] = 0.80 mol dm−3. What is Kc?

  • (A) 64
  • (B) 8.0
  • (C) 0.64
  • (D) 80
Show Answer

Answer: A – Kc = [HI]2 / ([H2][I2]) = (0.80)2 / (0.10 × 0.10) = 0.64 / 0.010 = 64. The units cancel (2 powers top and bottom), so Kc has no units here.

Q23. A 1.0 dm3 flask at equilibrium for CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O contains 0.20 mol acid, 0.20 mol ethanol, 0.80 mol ester and 0.80 mol water. Kc is…

  • (A) 16
  • (B) 4.0
  • (C) 0.25
  • (D) 64
Show Answer

Answer: A – In 1.0 dm3, moles = concentrations. Kc = (0.80 × 0.80) / (0.20 × 0.20) = 0.64 / 0.040 = 16 (no units, as the powers cancel).

Q24. For 2A ⇌ B, equilibrium concentrations are [A] = 0.50 and [B] = 0.25 mol dm−3. What is Kc including its units?

  • (A) 0.50 mol dm−3
  • (B) 2.0 dm3 mol−1
  • (C) 1.0 dm3 mol−1
  • (D) 1.0 (no units)
Show Answer

Answer: C – Kc = [B] / [A]2 = 0.25 / (0.50)2 = 0.25 / 0.25 = 1.0. Units: (mol dm−3) / (mol dm−3)2 = dm3 mol−1.

Q25. 1.00 mol of N2O4 is placed in a 1.0 dm3 flask: N2O4 ⇌ 2NO2. At equilibrium the flask contains 0.40 mol of NO2. What is Kc?

  • (A) 0.40 mol dm−3
  • (B) 0.20 mol dm−3
  • (C) 0.10 mol dm−3
  • (D) 0.80 mol dm−3
Show Answer

Answer: B – Forming 0.40 mol NO2 uses up 0.20 mol N2O4, so [N2O4] = 0.80 and [NO2] = 0.40 mol dm−3. Kc = (0.40)2 / 0.80 = 0.16 / 0.80 = 0.20 mol dm−3. Using the initial moles (1.00) instead of equilibrium moles is the classic mistake.

Q26. What are the units of Kc for N2 + 3H2 ⇌ 2NH3?

  • (A) mol dm−3
  • (B) dm3 mol−1
  • (C) mol2 dm−6
  • (D) dm6 mol−2
Show Answer

Answer: D – Units = (concentration)2 / (concentration)4 = (concentration)−2 = (mol dm−3)−2 = dm6 mol−2.

Q27. At a total pressure of 200 atm, an equilibrium Haber mixture has mole fractions: N2 = 0.20, H2 = 0.60, NH3 = 0.20. What is Kp for N2 + 3H2 ⇌ 2NH3?

  • (A) 2.3 × 10−5 atm−2
  • (B) 0.93 atm−2
  • (C) 4.3 × 104 atm2
  • (D) 5.8 × 10−3 atm−2
Show Answer

Answer: A – Partial pressures: p(N2) = 0.20 × 200 = 40, p(H2) = 0.60 × 200 = 120, p(NH3) = 40 atm. Kp = (40)2 / (40 × 1203) = 1600 / 6.912 × 107 = 2.3 × 10−5 atm−2. Using mole fractions directly (option B) ignores the total pressure.

Q28. For a reaction with Kc = 4.0, a mixture is analysed and the reaction quotient Q is found to be 16. What will happen?

  • (A) Nothing — the mixture is already at equilibrium
  • (B) The forward reaction dominates and more product forms
  • (C) The reverse reaction dominates and the mixture moves back towards reactants
  • (D) Kc will increase to 16
Show Answer

Answer: C – Q > K means there is too much product compared with equilibrium, so the reverse reaction runs faster until Q falls to 4.0. K itself never changes at constant temperature.

Q29. What is the correct Kp expression for 2SO2(g) + O2(g) ⇌ 2SO3(g)?

  • (A) p(SO2)2 × p(O2) / p(SO3)2
  • (B) p(SO3)2 / (p(SO2)2 × p(O2))
  • (C) p(SO3) / (p(SO2) × p(O2))
  • (D) [SO3]2 / ([SO2]2[O2])
Show Answer

Answer: B – Kp uses partial pressures, products over reactants, each raised to its balancing number: p(SO3)2 divided by p(SO2)2 × p(O2). Option D is the Kc expression, not Kp.

Q30. A + B ⇌ C is studied in a 1.0 dm3 flask. Initially there are 2.0 mol of A and 2.0 mol of B and no C. At equilibrium, 1.0 mol of C has formed. What is Kc?

  • (A) 0.25 dm3 mol−1
  • (B) 0.50 dm3 mol−1
  • (C) 4.0 dm3 mol−1
  • (D) 1.0 dm3 mol−1
Show Answer

Answer: D – Forming 1.0 mol C consumes 1.0 mol each of A and B, so equilibrium amounts are A = B = C = 1.0 mol, i.e. 1.0 mol dm−3 each. Kc = 1.0 / (1.0 × 1.0) = 1.0 dm3 mol−1. Option A uses the initial concentrations — a common error.

Equilibrium in Solution (A-Level / AP Stretch)

Q31. The ionisation of water is endothermic, so Kw increases as temperature rises. What happens to the pH of pure water when it is heated?

  • (A) It rises above 7 and the water becomes alkaline
  • (B) It stays at exactly 7.00 at all temperatures
  • (C) It falls below 7, but the water is still neutral because [H+] = [OH−]
  • (D) It falls below 7 and the water becomes acidic
Show Answer

Answer: C – A larger Kw means more ionisation, so [H+] rises and pH falls below 7. But [H+] still equals [OH−], so the water remains neutral — pH 7 is only the neutral point at 25 °C.

Q32. At 50 °C, Kw = 5.5 × 10−14 mol2 dm−6. What is the pH of pure water at this temperature?

  • (A) 7.00
  • (B) 6.63
  • (C) 6.13
  • (D) 7.37
Show Answer

Answer: B – In pure water [H+] = √Kw = √(5.5 × 10−14) = 2.35 × 10−7 mol dm−3. pH = −log(2.35 × 10−7) = 6.63.

Q33. What is the correct Ksp expression for calcium fluoride, CaF2?

  • (A) [Ca2+][F−]
  • (B) [Ca2+]2[F−]
  • (C) [CaF2] / ([Ca2+][F−])
  • (D) [Ca2+][F−]2
Show Answer

Answer: D – CaF2(s) ⇌ Ca2+(aq) + 2F−(aq), so Ksp = [Ca2+][F−]2 — the fluoride concentration is squared because 2 mol of F− are produced. The solid itself does not appear in the expression.

Q34. Silver chloride has Ksp = 1.6 × 10−10 mol2 dm−6. Its molar solubility in pure water is…

  • (A) 1.3 × 10−5 mol dm−3
  • (B) 1.6 × 10−10 mol dm−3
  • (C) 8.0 × 10−11 mol dm−3
  • (D) 2.6 × 10−20 mol dm−3
Show Answer

Answer: A – For a 1:1 salt, if solubility = s, then [Ag+] = [Cl−] = s and Ksp = s2. So s = √(1.6 × 10−10) = 1.3 × 10−5 mol dm−3.

Q35. A salt MX2 has Ksp = 4.0 × 10−9 mol3 dm−9. Its molar solubility is…

  • (A) 6.3 × 10−5 mol dm−3
  • (B) 1.6 × 10−3 mol dm−3
  • (C) 1.0 × 10−3 mol dm−3
  • (D) 1.0 × 10−9 mol dm−3
Show Answer

Answer: C – If solubility = s, then [M2+] = s and [X−] = 2s, so Ksp = s × (2s)2 = 4s3. s = (4.0 × 10−9 / 4)1/3 = (1.0 × 10−9)1/3 = 1.0 × 10−3 mol dm−3. Option A wrongly takes a square root; option B forgets to divide by 4.

Q36. What is the solubility of AgCl (Ksp = 1.6 × 10−10) in 0.10 mol dm−3 sodium chloride solution?

  • (A) 1.3 × 10−5 mol dm−3 (the same as in water)
  • (B) 1.6 × 10−9 mol dm−3
  • (C) 1.6 × 10−11 mol dm−3
  • (D) 1.6 × 10−10 mol dm−3
Show Answer

Answer: B – This is the common ion effect: [Cl−] is fixed at about 0.10 by the NaCl, so solubility s = [Ag+] = Ksp / [Cl−] = 1.6 × 10−10 / 0.10 = 1.6 × 10−9 mol dm−3 — roughly 10,000 times less soluble than in pure water.

Q37. For the weak acid equilibrium HA ⇌ H+ + A−, a larger value of Ka means…

  • (A) the acid is weaker
  • (B) fewer H+ ions are present at equilibrium
  • (C) the value of pKa is larger
  • (D) the ionisation equilibrium lies further to the right, so the acid is stronger
Show Answer

Answer: D – Ka is just an equilibrium constant for acid ionisation. A larger Ka means the equilibrium lies further right, more H+ is produced, and the acid is stronger (and its pKa is smaller).

Q38. A 0.10 mol dm−3 solution of a weak acid has [H+] = 1.3 × 10−3 mol dm−3 at equilibrium. What is Ka for the acid?

  • (A) 1.7 × 10−5 mol dm−3
  • (B) 1.3 × 10−3 mol dm−3
  • (C) 1.7 × 10−6 mol dm−3
  • (D) 1.7 × 10−3 mol dm−3
Show Answer

Answer: A – Ka = [H+][A−] / [HA] = (1.3 × 10−3)2 / 0.10 = 1.69 × 10−6 / 0.10 = 1.7 × 10−5 mol dm−3. (Option C is the value before dividing by the acid concentration.)

Q39. A solute distributes itself between two immiscible solvents, water and an organic solvent, with distribution (partition) coefficient K = [solute]organic / [solute]water = 20. This means that at equilibrium…

  • (A) the solute has equal concentration in both solvents
  • (B) the concentration in the organic layer is 20 times that in the water layer
  • (C) the solute is completely insoluble in water
  • (D) 20 g of solute dissolves in the organic layer
Show Answer

Answer: B – The distribution coefficient is the ratio of the equilibrium concentrations in the two layers. K = 20 means the solute strongly prefers the organic layer: its concentration there is 20× the concentration left in the water, which is why repeated extraction with small portions of solvent works so well.

Q40. The molar solubility of calcium hydroxide is 0.010 mol dm−3. What is its Ksp?

  • (A) 1.0 × 10−4 mol3 dm−9
  • (B) 2.0 × 10−4 mol3 dm−9
  • (C) 1.0 × 10−6 mol3 dm−9
  • (D) 4.0 × 10−6 mol3 dm−9
Show Answer

Answer: D – Ca(OH)2 gives [Ca2+] = 0.010 and [OH−] = 2 × 0.010 = 0.020 mol dm−3. Ksp = 0.010 × (0.020)2 = 0.010 × 4.0 × 10−4 = 4.0 × 10−6 mol3 dm−9 (equivalently 4s3). Option A forgets both the doubling and the square.

Want more practice? Try our full test series — 1,200 more free MCQs in the full practice tests:

A-Level Chemistry Practice Tests · AP Chemistry Practice Tests

Still stuck on equilibrium calculations? Book a 1-on-1 online chemistry lesson ($10/hour) and work through them live with a tutor.

All questions on this page are original practice questions written for QuaidianPoint.