75 free MCQs on Equilibrium with worked answers and explanations. When forward and reverse reaction rates are equal, equilibrium is reached. Learn about Kc, Kp, Le Chatelier's principle, and acid-base equilibria, including pH, buffer solutions, and solubility product.
Take the timed Equilibrium chapterwise test →Below are 75 practice questions on Equilibrium, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Equilibrium notes.
As the reaction proceeds, reactant concentration falls and product concentration rises until both level off at equilibrium.
Q1.
The conjugate acid of ammonia (NH<sub>3</sub>) is:
Answer: C. NH<sub>4</sub><sup>+</sup>
Why: A conjugate acid is formed when a base gains a proton; NH<sub>3</sub> + H<sup>+</sup> gives NH<sub>4</sub><sup>+</sup>.
Q2.
At 25 °C, the sum pH + pOH for any aqueous solution equals:
Answer: B. 14
Why: Since K<sub>w</sub> = 10<sup>−14</sup> at 25 °C, pH + pOH = pK<sub>w</sub> = 14.
Q3.
According to Lewis, an acid is a species that:
Answer: A. Accepts an electron pair
Why: A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor.
Q4.
Chemical equilibrium is described as dynamic because:
Answer: D. Opposing reactions continue at equal rates
Why: At equilibrium the forward and reverse reactions continue but at equal rates, so concentrations stay constant.
Q5.
The equilibrium constant K<sub>c</sub> is dimensionless (has no units) when:
Answer: A. Δn for the reaction is zero
Why: Units of K<sub>c</sub> depend on (concentration)<sup>Δn</sup>; when Δn = 0 the units cancel and K<sub>c</sub> is dimensionless.
Q6.
Le Chatelier's principle states that when a stress is applied to a system at equilibrium, it will:
Answer: C. Oppose the change
Why: When a stress is applied, the equilibrium shifts to partially oppose and relieve that stress.
Q7.
The equilibrium constant Kc is expressed in terms of:
Answer: C. Molar concentration
Why: Kc uses molar concentrations of products and reactants raised to their stoichiometric powers.
Q8.
At chemical equilibrium, the rate of forward reaction is:
Answer: C. Equal to reverse
Why: At equilibrium, forward and reverse reaction rates are equal, so concentrations remain constant.
Q9.
A very large value of Kc indicates that the reaction:
Answer: B. Goes almost to completion
Why: Large Kc means products are highly favoured; the reaction proceeds nearly to completion.
Q10.
For N₂ + 3H₂ ⇌ 2NH₃, increasing pressure will:
Answer: B. Shift equilibrium right
Why: Increasing pressure favours the side with fewer moles of gas. 4 moles → 2 moles, so equilibrium shifts right.
Q11.
At equilibrium, which statement is correct?
Answer: C. Concentrations remain constant
Why: At equilibrium, concentrations of all species remain constant (not necessarily equal).
Q12.
The pH of a neutral solution at 25°C is:
Answer: B. 7
Why: At 25°C, a neutral solution has [H⁺] = [OH⁻] = 10⁻⁷ M, so pH = 7.
Q13.
A buffer solution resists changes in:
Answer: C. pH
Why: A buffer maintains nearly constant pH when small amounts of acid or base are added.
Q14.
The relationship between Kp and Kc is:
Answer: B. Kp = Kc(RT)^Δn
Why: Kp = Kc(RT)^Δn, where Δn is the change in moles of gas, R is gas constant, T is temperature in K.
Q15.
Adding a catalyst to a system at equilibrium:
Answer: D. Does not change the position of equilibrium
Why: A catalyst speeds up both forward and reverse reactions equally, so Kc and equilibrium position are unchanged.
Q16.
A reversible reaction reaches equilibrium when the forward and backward reaction rates become:
Answer: B. equal
Why: At equilibrium the forward and reverse rates are equal, so concentrations no longer change.
Q17.
At equilibrium, the measurable concentrations of reactants and products:
Answer: B. remain constant
Why: They stay constant with time, though not necessarily equal to one another.
Q18.
The equilibrium constant written in terms of molar concentrations is denoted:
Answer: B. Kc
Why: Kc uses equilibrium concentrations; Kp uses partial pressures for gaseous equilibria.
Q19.
For aA + bB ⇌ cC + dD, the expression Kc = [C]ᶜ[D]ᵈ divided by:
Answer: B. [A]ᵃ[B]ᵇ
Why: Each concentration is raised to its stoichiometric coefficient: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ.
Q20.
A catalyst added to a system at equilibrium:
Answer: C. does not shift its position
Why: A catalyst speeds up both directions equally, so it reaches equilibrium faster without shifting it.
Q21.
The ionic product of water, Kw, at 25 °C is:
Answer: B. 1 × 10⁻¹⁴
Why: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.
Q22.
A solution having a pH less than 7 at 25 °C is:
Answer: A. acidic
Why: pH below 7 means [H⁺] exceeds [OH⁻], so the solution is acidic.
Q23.
The pH of a neutral aqueous solution at 25 °C is:
Answer: B. 7
Why: In a neutral solution [H⁺] = [OH⁻] = 10⁻⁷ M, giving pH = 7.
Q24.
According to Arrhenius, an acid releases which ion in aqueous solution?
Answer: B. H⁺
Why: An Arrhenius acid furnishes H⁺ (as H₃O⁺) in water; a base furnishes OH⁻.
Q25.
Le Chatelier’s principle predicts how an equilibrium responds to a change in concentration, pressure or:
Answer: C. temperature
Why: Temperature, pressure and concentration are the factors that shift equilibrium; a catalyst does not.
Q26.
For A ⇄ B, K<sub>c</sub> = 4. Starting from 1 M A and no B, the equilibrium concentration of B is:
Answer: C. 0.8 M
Why: K<sub>c</sub> = [B]/[A] = x/(1-x) = 4, so x = 0.8 M of B.
Q27.
For 2SO<sub>3</sub>(g) ⇌ 2SO<sub>2</sub>(g) + O<sub>2</sub>(g), the relation between K<sub>p</sub> and K<sub>c</sub> (Δn = +1) is:
Answer: D. K<sub>p</sub> = K<sub>c</sub>(RT)
Why: K<sub>p</sub> = K<sub>c</sub>(RT)<sup>Δn</sup>; here Δn = 3 − 2 = +1, so K<sub>p</sub> = K<sub>c</sub>(RT).
Q28.
One mole each of H<sub>2</sub> and I<sub>2</sub> are placed in a 1 L flask; at equilibrium 0.8 mol of each has reacted. K<sub>c</sub> for H<sub>2</sub> + I<sub>2</sub> ⇌ 2HI is:
Answer: C. 64
Why: At equilibrium [HI] = 1.6, [H<sub>2</sub>] = [I<sub>2</sub>] = 0.2; K<sub>c</sub> = (1.6)<sup>2</sup>/(0.2 × 0.2) = 2.56/0.04 = 64.
Q29.
The pH of a 0.005 M H<sub>2</sub>SO<sub>4</sub> solution (assume complete dissociation) is:
Answer: B. 2
Why: H<sub>2</sub>SO<sub>4</sub> gives 2H<sup>+</sup>, so [H<sup>+</sup>] = 2 × 0.005 = 0.01 M; pH = −log(0.01) = 2.
Q30.
A buffer contains 0.2 M CH<sub>3</sub>COOH and 0.2 M CH<sub>3</sub>COONa (pK<sub>a</sub> = 4.74). Its pH is:
Answer: A. 4.74
Why: pH = pK<sub>a</sub> + log([salt]/[acid]) = 4.74 + log(1) = 4.74.
Q31.
For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), increasing pressure will:
Answer: A. Favour PCl₅ formation
Why: Increasing pressure shifts equilibrium towards fewer moles of gas (left, 1 mole from 2 moles), favouring PCl₅.
Q32.
For N₂ + 3H₂ ⇌ 2NH₃, the relationship between Kp and Kc is:
Answer: B. Kp = Kc(RT)⁻²
Why: Δn = 2 - (1+3) = -2. Kp = Kc(RT)^Δn = Kc(RT)⁻². So Kp < Kc for this reaction.
Q33.
For weak acid HA with Ka = 1.8 × 10⁻⁵, pH of 0.1 M solution is approximately:
Answer: A. 2.87
Why: [H⁺] = √(Ka × C) = √(1.8×10⁻⁵ × 0.1) ≈ 1.34×10⁻³ M; pH = -log(1.34×10⁻³) = 2.87.
Q34.
The Henderson-Hasselbalch equation (pH = pKa + log[A⁻]/[HA]) is used to calculate:
Answer: B. pH of a buffer solution
Why: This equation relates pH to pKa and the ratio of conjugate base to weak acid in a buffer.
Q35.
The common ion effect causes solubility of a sparingly soluble salt to:
Answer: B. Decrease
Why: Adding a common ion shifts the dissolution equilibrium left (Le Chatelier's principle), reducing solubility.
Q36.
For AgCl with Ksp = 1.8 × 10⁻¹⁰, its molar solubility is:
Answer: A. 1.34 × 10⁻⁵ M
Why: AgCl ⇌ Ag⁺ + Cl⁻. Ksp = s². s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M.
Q37.
For an exothermic reaction, increasing temperature will:
Answer: B. Decrease K
Why: Heat is a product in exothermic reactions. Increasing temperature shifts equilibrium left, decreasing K.
Q38.
Which of the following increases Kc for an endothermic reaction?
Answer: B. Increasing temperature
Why: For endothermic reactions, increasing temperature shifts equilibrium right and increases Kc.
Q39.
For the reaction 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) <=> 2SO<sub>3</sub>(g), how does adding an inert gas at constant volume affect the equilibrium?
Answer: D. There is no effect on the equilibrium position
Why: Adding an inert gas at constant volume does not change the partial pressures or concentrations of the reacting species, so the equilibrium position is unaffected.
Q40.
A buffer solution is prepared by mixing a weak acid with its conjugate base. Why does this mixture resist large changes in pH when a small amount of acid or base is added?
Answer: C. The weak acid neutralises added base while the conjugate base neutralises added acid, keeping their ratio steady
Why: In a buffer, the weak acid component reacts with added base and the conjugate base component reacts with added acid, so the ratio of conjugate base to weak acid changes only slightly and the pH stays nearly constant.
Q41.
For N₂ + 3H₂ ⇌ 2NH₃, Kp = Kc(RT)^Δn where Δn equals:
Answer: B. −2
Why: Δn = gaseous products − reactants = 2 − 4 = −2.
Q42.
Increasing the total pressure on the equilibrium N₂ + 3H₂ ⇌ 2NH₃ shifts it:
Answer: B. forward
Why: Higher pressure favours the side with fewer gas moles (2 vs 4), so the equilibrium shifts toward NH₃.
Q43.
For a gaseous reaction in which Δn = 0, the value of Kp is:
Answer: C. equal to Kc
Why: When Δn = 0, (RT)^Δn = 1, so Kp = Kc.
Q44.
When the reaction quotient Q is smaller than the equilibrium constant K, the reaction proceeds:
Answer: B. forward
Why: With Q < K there are too few products, so the net reaction runs forward until Q = K.
Q45.
Adding a common ion to a solution of a weak acid decreases its:
Answer: B. degree of dissociation
Why: The common-ion effect suppresses ionisation, lowering the weak acid’s degree of dissociation.
Q46.
A buffer solution is one that resists changes in:
Answer: B. pH
Why: A buffer resists pH change on adding small amounts of acid or base.
Q47.
The conjugate base of hydrochloric acid, HCl, is:
Answer: A. Cl⁻
Why: Removing a proton from HCl leaves Cl⁻, its conjugate base.
Q48.
According to the Brønsted–Lowry theory, a base is a species that:
Answer: B. accepts a proton
Why: A Brønsted–Lowry base is a proton acceptor; an acid is a proton donor.
Q49.
For the exothermic equilibrium 2SO₂ + O₂ ⇌ 2SO₃, raising the temperature shifts it:
Answer: B. backward
Why: For an exothermic reaction, added heat shifts the equilibrium in the endothermic (backward) direction.
Q50.
The pH of a 0.01 M HCl solution is:
Answer: B. 2
Why: HCl is fully dissociated, so [H⁺] = 0.01 M and pH = −log(10⁻²) = 2.
Q51.
A weak acid is 1% dissociated (alpha = 0.01) in a 0.1 M solution. Its dissociation constant K<sub>a</sub> is about:
Answer: B. 1 x 10<sup>-5</sup>
Why: K<sub>a</sub> = C x alpha<sup>2</sup> = 0.1 x (0.01)<sup>2</sup> = 1 x 10<sup>-5</sup>.
Q52.
A buffer is made from 0.1 mol CH<sub>3</sub>COOH and 0.05 mol CH<sub>3</sub>COO<sup>−</sup> (pK<sub>a</sub> = 4.74). Its pH is about:
Answer: B. 4.44
Why: pH = pK<sub>a</sub> + log([A<sup>−</sup>]/[HA]) = 4.74 + log(0.05/0.1) = 4.74 − 0.30 = 4.44.
Q53.
If K<sub>c</sub> for N<sub>2</sub> + 3H<sub>2</sub> ⇌ 2NH<sub>3</sub> equals K, then K<sub>c</sub> for NH<sub>3</sub> ⇌ ½N<sub>2</sub> + 3/2 H<sub>2</sub> is:
Answer: C. K<sup>−1/2</sup>
Why: Halving the equation gives power ½ and reversing inverts it: K becomes (1/K)<sup>1/2</sup> = K<sup>−1/2</sup>.
Q54.
Equal volumes of 2 × 10<sup>−3</sup> M BaCl<sub>2</sub> and 2 × 10<sup>−3</sup> M Na<sub>2</sub>SO<sub>4</sub> are mixed. Given K<sub>sp</sub>(BaSO<sub>4</sub>) = 1 × 10<sup>−10</sup>, will BaSO<sub>4</sub> precipitate?
Answer: C. Yes, because Q > K<sub>sp</sub>
Why: After mixing each ion is halved to 1 × 10<sup>−3</sup> M; Q = (10<sup>−3</sup>)(10<sup>−3</sup>) = 10<sup>−6</sup> > K<sub>sp</sub>, so it precipitates.
Q55.
Assertion (A): For an endothermic reaction, K<sub>c</sub> increases as temperature rises. Reason (R): Raising the temperature favours the forward (heat-absorbing) direction.
Answer: A. Both A and R true; R correctly explains A
Why: By Le Chatelier, heating an endothermic reaction shifts it forward, raising K<sub>c</sub>; R correctly explains A.
Q56.
For H₂ + I₂ ⇌ 2HI, Kc = 45.9 at 490°C. If [H₂] = [I₂] = 0.1 M at equilibrium, [HI] is:
Answer: A. 0.677 M
Why: Kc = [HI]²/([H₂][I₂]) = x²/(0.01) = 45.9 → x² = 0.459 → x = 0.677 M.
Q57.
If Kc for A ⇌ B is 4, the Kc for 2B ⇌ 2A is:
Answer: B. 1/16
Why: Reversing gives Kc = 1/4. Multiplying the equation by 2 squares the constant: Kc = (1/4)² = 1/16.
Q58.
For PCl₅ ⇌ PCl₃ + Cl₂ at pressure P with small degree of dissociation α, Kp is approximately:
Answer: A. α²P
Why: Kp = α²P/(1-α²) ≈ α²P for small α. So α ≈ √(Kp/P).
Q59.
Ksp of Ag₂CrO₄ = 1.12 × 10⁻¹². Its molar solubility is:
Answer: A. 6.54 × 10⁻⁵ M
Why: Ag₂CrO₄ → 2Ag⁺ + CrO₄²⁻. Ksp = (2s)²s = 4s³. s = (Ksp/4)<sup>1/3</sup> = (2.8×10⁻¹³)<sup>1/3</sup> = 6.54×10⁻⁵ M.
Q60.
Which salt hydrolyses to give an alkaline solution?
Answer: C. CH₃COONa
Why: CH₃COONa is a salt of weak acid and strong base. The acetate ion hydrolyses to give OH⁻, making the solution alkaline.
Q61.
The ionic product of water Kw increases with temperature because:
Answer: B. Water ionisation is endothermic
Why: The ionisation of water is endothermic. Increasing temperature shifts equilibrium right, increasing Kw above 10⁻¹⁴.
Q62.
AgCl (Ksp = 10⁻¹⁰) and AgBr (Ksp = 10⁻¹³) are both dissolved. When Ag⁺ is slowly added, which precipitates first?
Answer: B. Br⁻ precipitates first
Why: AgBr has lower Ksp, so it needs less Ag⁺ to initiate precipitation. Br⁻ precipitates first.
Q63.
For an exothermic reaction with Kc = 0.04 at 1000 K, as temperature increases Kc will:
Answer: B. Decrease
Why: For exothermic reactions (ΔH < 0), increasing temperature shifts equilibrium left (Le Chatelier's), decreasing Kc.
Q64.
A 0.1 M solution of a weak base BOH has a degree of dissociation of 0.02. What is the value of Kb for this base?
Answer: A. 4 x 10<sup>-5</sup>
Why: Kb = c x alpha<sup>2</sup> = 0.1 x (0.02)<sup>2</sup> = 0.1 x 0.0004 = 4 x 10<sup>-5.</sup>
Q65.
For the equilibrium N<sub>2</sub>O<sub>4</sub>(g) <=> 2NO<sub>2</sub>(g), the total pressure at equilibrium is P and the degree of dissociation of N<sub>2</sub>O<sub>4</sub> is alpha. Which expression correctly gives Kp in terms of alpha and P?
Answer: A. Kp = 4 alpha<sup>2</sup> P / (1 - alpha<sup>2</sup>)
Why: For N<sub>2</sub>O<sub>4</sub> <=> 2NO<sub>2</sub> with initial 1 mole and degree of dissociation alpha, total moles become (1+alpha), and Kp works out to 4 alpha<sup>2</sup> P / (1 - alpha<sup>2</sup>) after substituting mole fractions and total pressure.
Q66.
For a sparingly soluble salt AB with molar solubility S, its solubility product is:
Answer: B. Ksp = S²
Why: AB ⇌ A⁺ + B⁻ gives [A⁺] = [B⁻] = S, so Ksp = S².
Q67.
By Ostwald’s dilution law for a weak acid, Ka in terms of concentration C and degree of dissociation α is:
Answer: B. α² C/(1 − α)
Why: Ka = Cα²/(1 − α); for small α this simplifies to Ka ≈ Cα².
Q68.
In the Henderson–Hasselbalch equation, the pH of an acidic buffer is pH = pKa +:
Answer: B. log([salt]/[acid])
Why: pH = pKa + log([salt]/[acid]); equal salt and acid concentrations give pH = pKa.
Q69.
The degree of hydrolysis of the salt of a weak acid and a strong base increases on:
Answer: B. increasing the dilution
Why: Dilution promotes hydrolysis, so the degree of hydrolysis increases as the solution is diluted.
Q70.
If the Ka of acetic acid is 1.8 × 10⁻⁵, its pKa is approximately:
Answer: A. 4.74
Why: pKa = −log(1.8 × 10⁻⁵) ≈ 4.74.
Q71.
A large equilibrium constant (K ≫ 1) indicates that at equilibrium:
Answer: B. products are favoured
Why: A large K means the numerator (products) dominates, so the equilibrium lies well toward the products.
Q72.
The pH of a 0.001 M NaOH solution at 25 °C is:
Answer: B. 11
Why: [OH⁻] = 10⁻³ so pOH = 3, and pH = 14 − 3 = 11.
Q73.
Adding an inert gas to a gaseous equilibrium at constant volume:
Answer: C. leaves the position unchanged
Why: At constant volume the partial pressures of the reacting gases are unchanged, so the equilibrium does not shift.
Q74.
Given ΔG° = −2.303 RT log K, a positive value of ΔG° means the equilibrium constant K is:
Answer: B. less than 1
Why: A positive ΔG° forces log K to be negative, so K < 1 and reactants are favoured.
Q75.
A salt formed from a strong acid and a weak base, when dissolved in water, gives a solution that is:
Answer: A. acidic
Why: The cation of the weak base hydrolyses to release H⁺, making the solution acidic (e.g. NH₄Cl).