Equilibrium — NEET Chemistry MCQs with answers

Equilibrium is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 1,098 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Chemical equilibrium: K, Q and Le Chatelier · Chemical equilibrium · Ionic equilibrium: acids, bases and pH · Buffers and salt hydrolysis · Ionic equilibrium · Solubility equilibrium · Equilibrium constant · Kp and Kc relation · Titration curve · ICE table setup · Factors affecting equilibrium · Weak acid ionization · Solubility experiment · Percent dissociation · Buffer capacity · Temperature effect on equilibrium · Le Chatelier prediction · PCl5 dissociation · Relative acid strength · Molar solubility from Ksp · Catalyst and equilibrium constant · Strong electrolyte dissociation · pH of weak base · Equilibrium perturbation · Factors not affecting Kc · Ka expression setup · Acid-base equilibrium · Acid-base theories · Law of mass action · Relationship between Kp and Kc · Degree of dissociation · pH calculation · pOH and Kw · Weak acid dissociation · Common ion effect · Solubility product · Solubility product for 1:2 electrolyte · Le Chatelier's principle - temperature effect · Reaction quotient Q · Hydrolysis of salts

8 free Equilibrium practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Buffers and salt hydrolysis

   Which statement about buffer capacity is correct?

   • A.Buffer capacity is maximum when the pH is far from the pKa of the acid used
   • B.Buffer capacity depends only on the pKa value and not on the total concentration of the components
   • C.Buffer capacity is maximum when [acid] = [salt] (pH = pKa), and increases with higher total concentration of the buffer components
   • D.Buffer capacity is independent of the total concentration of the buffer components
   Show answer and explanation

   Answer: C. Buffer capacity is maximum when [acid] = [salt] (pH = pKa), and increases with higher total concentration of the buffer components

   A buffer resists pH change best when acid and conjugate base are present in equal, appreciable amounts, and greater total concentration means more moles of each are available to neutralise added acid or base.

   Buffer capacity measures how much strong acid/base a buffer can absorb per unit pH change. It is maximum at pH = pKa (equal concentrations of acid and conjugate base) because both components are then present in the largest simultaneous amounts to react with either added H+ or OH-. It also scales with the total concentration of the buffer components — a more concentrated buffer has more 'reserve' of each species and thus higher capacity, all else equal.

   Common mistake: Believing buffer capacity is unrelated to total concentration, or is maximum away from pKa

2. Question 2 · hard · Degree of dissociation

   1 mol of PCl5 is placed in a 10 L vessel and dissociates as PCl5 <-> PCl3 + Cl2. At equilibrium, the degree of dissociation is 0.4. The value of Kc is:

   • A.0.0267 mol/L
   • B.0.16 mol/L
   • C.0.4 mol/L
   • D.0.0667 mol/L
   Show answer and explanation

   Answer: A. 0.0267 mol/L

   C=0.1M; [PCl5]=0.06, [PCl3]=[Cl2]=0.04; Kc=0.04x0.04/0.06=0.0267

   Initial concentration C = 1/10 = 0.1 M. At equilibrium: [PCl5] = C(1-α) = 0.1x0.6 = 0.06 M; [PCl3]=[Cl2] = Cα = 0.1x0.4 = 0.04 M. Kc = (0.04x0.04)/0.06 = 0.0016/0.06 = 0.02667 mol/L.

   Common mistake: Not converting initial moles to concentration using the vessel volume

3. Question 3 · hard · Degree of dissociation

   1 mol of N2O4 dissociates as N2O4 <-> 2NO2 with degree of dissociation 0.5 at a total pressure of 2 atm. The value of Kp is:

   • A.2.67 atm
   • B.1.33 atm
   • C.0.67 atm
   • D.4 atm
   Show answer and explanation

   Answer: A. 2.67 atm

   At eq: N2O4=0.5,NO2=1,total=1.5; P(N2O4)=0.667atm,P(NO2)=1.333atm; Kp=(1.333)^2/0.667=2.67

   Moles at equilibrium: N2O4 = 1-0.5=0.5, NO2=2x0.5=1; total moles=1.5. Mole fractions: N2O4=0.5/1.5=0.333, NO2=1/1.5=0.667. Partial pressures: P(N2O4)=0.333x2=0.667atm, P(NO2)=0.667x2=1.333atm. Kp=P(NO2)^2/P(N2O4)=(1.333)^2/0.667=1.778/0.667=2.665 ≈ 2.67 atm.

   Common mistake: Not squaring the partial pressure of the product with coefficient 2

4. Question 4 · hard · Solubility Product

   The Ksp of Mg(OH)2 is 1.2 x 10^-11. Its molar solubility in water is: [Mg(OH)2 -> Mg2+ + 2OH-]

   • A.1.44 x 10^-4 mol/L
   • B.3.46 x 10^-6 mol/L
   • C.6 x 10^-12 mol/L
   • D.1.2 x 10^-11 mol/L
   Show answer and explanation

   Answer: A. 1.44 x 10^-4 mol/L

   Ksp=4s^3; s=(Ksp/4)^(1/3)=(3x10^-12)^(1/3)≈1.44x10^-4

   For Mg(OH)2, Ksp = [Mg2+][OH-]^2 = s(2s)^2 = 4s^3. s^3 = Ksp/4 = 1.2x10^-11/4 = 3x10^-12. s = (3x10^-12)^(1/3) = (3)^(1/3) x 10^-4 = 1.4422x10^-4 mol/L.

   Common mistake: Using the 1:1 electrolyte formula Ksp=s^2 for a 1:2 salt

5. Question 5 · hard · Van't Hoff Equation

   For a reaction, K1=1 at T1=300 K and Delta H = +50000 J/mol. Using the van't Hoff equation, the value of K2 at T2=350 K is approximately: (R=8.314 J/mol K)

   • A.17.5
   • B.1.0
   • C.2.86
   • D.0.057
   Show answer and explanation

   Answer: A. 17.5

   ln(K2/K1)=-ΔH/R(1/T2-1/T1)=2.864; K2=e^2.864≈17.5

   ln(K2/K1) = -ΔH/R x (1/T2 - 1/T1) = -50000/8.314 x (1/350-1/300) = -6013.9 x (-4.7619x10^-4) = 2.864. K2 = K1 x e^2.864 = 1 x 17.54 ≈ 17.5.

   Common mistake: Sign errors when handling the (1/T2-1/T1) term

6. Question 6 · medium · Reaction Quotient

   For a reaction, the reaction quotient Q = 2 and the equilibrium constant K = 5. The reaction will proceed in which direction to reach equilibrium?

   • A.Forward direction, since Q < K
   • B.Reverse direction, since Q < K
   • C.No net reaction, since Q and K are both positive
   • D.Reverse direction, since Q > K
   Show answer and explanation

   Answer: A. Forward direction, since Q < K

   Since Q(2) < K(5), the reaction proceeds forward to increase products until Q=K

   When Q < K, the ratio of products to reactants is less than at equilibrium, so the reaction proceeds in the forward direction (toward products) until Q equals K.

   Common mistake: Confusing the direction rule when Q<K versus Q>K

7. Question 7 · medium · Ionic equilibrium

   A weak acid HA (0.1 M) is 1% dissociated. Its Ka is nearest to:

   • A.1.01 x 10^-5
   • B.1 x 10^-3
   • C.1 x 10^-2
   • D.1 x 10^-4
   Show answer and explanation

   Answer: A. 1.01 x 10^-5

   Ka=C(alpha)^2=0.1x(0.01)^2=0.1x0.0001=1x10^-5 (approx 1.01x10^-5 with 1-alpha correction)

   Ka = C(alpha)^2/(1-alpha) ≈ C(alpha)^2 for small alpha = 0.1x(0.01)^2 = 0.1x0.0001 = 1x10^-5, with slight correction for (1-alpha)=0.99, Ka=1x10^-5/0.99≈1.01x10^-5.

   Common mistake: Forgetting to square the degree of dissociation

8. Question 8 · medium · Le Chatelier's principle

   For A <-> B, Kc=3 at a fixed temperature. If [A] is changed but temperature stays constant, the new value of Kc at equilibrium is:

   • A.3 (unchanged)
   • B.6 (doubled)
   • C.1.5 (halved)
   • D.Cannot be determined
   Show answer and explanation

   Answer: A. 3 (unchanged)

   Kc depends only on temperature, so it remains 3

   Equilibrium constant Kc is a function of temperature only and is unaffected by concentration changes. Since temperature is constant, Kc remains 3.

   Common mistake: Believing Kc changes when concentrations are changed

Questions about Equilibrium for NEET

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