NEET Previous Year Questions (2014-2024): Equilibrium | Chemistry Class 11 PDF Download

2024

Q1: In which of the following equilibria, K_p and K_c are NOT equal?
(a)
(b)
(c)
(d)       (NEET 2024)
Ans: 

Option D:  
Reactant side moles = 2, Product side moles = 2; Δn$=2-2=0$. 
From this analysis, it is evident that K_p and K_c are not equal in Option A where Δn = 1. In all other options, since Δn = 0, K_p is equal to K_c. Thus, the correct answer is Option A.

Q2: For the reaction  2A ⇌ B + C, K_c = 4 × 10⁻³. At a given time, the composition of reaction mixture is: . [A] = [B] = [C] = 2 × 10⁻³M.  Then, which of the following is correct?
(a) Reaction is at equilibrium.
(b) Reaction has a tendency to go in forward direction.
(c) Reaction has a tendency to go in backward direction.
(d) Reaction has gone to completion in forward direction.      (NEET 2024)
Ans: (c)
To determine which option is correct regarding the reaction state and its direction, we need to calculate the reaction quotient Q_c and compare it to the equilibrium constant K_c. The reaction given is:
2A ⇌ B + C
The equilibrium constant expression K_c for this reaction is:

Given that K_c = 4 × 10⁻³ and the concentrations of A, B, and C at this time are each 2 × 10⁻³ M, we can substitute these values into the expression for K_c to calculate the reaction quotient Q_c:

Simplifying, we find:

Since Q_c > K_c (1 > 0.004), the reaction quotient is greater than the equilibrium constant. This indicates that the concentration of products (B and C) is too high relative to the concentration of reactants (A) for the system to be at equilibrium under these conditions.
This means that the reaction has a tendency to move in the backward direction to reach equilibrium, reducing the concentration of the products (B and C) and increasing the concentration of the reactant (A). Therefore, the correct answer to the given question is:
Option C: Reaction has a tendency to go in backward direction.

Q3: Consider the following reaction in a sealed vessel at equilibrium with concentrations of N₂ = 3.0 × 10⁻³M, O₂ = 4.2 × 10⁻³M and NO = 2.8 × 10⁻³M.
2NO(g) ⇌ N₂(g) + O₂(g)
If  0.1molL L⁻¹ of NO(g) is taken in a closed vessel, what will be degree of dissociation (α) of NO_(g) at equilibrium?
(a) 0.00889
(b) 0.0889
(c) 0.8889
(d) 0.717          (NEET 2024)
Ans: (d)

2022

Q2:for the above reaction at 298 K, K_C is found to be 3.0 × 10^–59. If the concentration of O₂ at equilibrium is 0.040 M then concentration of O₃ in M is       (NEET 2022 Phase 1)
(a) 2.4 × 10³¹ 
(b) 1.2 × 10²¹

(c) 4.38 × 10^–32 
(d) 1.9 × 10^–63
Ans: c

Q3:  0.01 M acetic acid solution is 1% ionised, then pH of this acetic acid solution is :      (NEET 2022 Phase 2)
(a) 1
(b) 3
(c) 2
(d) 4
Ans: (d)

Q4: K_p for the following reaction is 3.0 at 1000 K.    (NEET 2022 Phase 2)
CO₂(g) + C(s) $\rightleftharpoons$ 2CO(g)
What will be the value of K_c for the reaction at the same temperature?
(Given : R = 0.083 L bar K^$-$1 mol^$-$1)
(a) 3.6
(b) 0.36
(c) 3.6 $\times$ 10^$-$2 
(d) 3.6 $\times$ 10^$-$3 
Ans: (c)

2021

Q1: The pK_b of dimethylamine and pk_a of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate at solution is:      (NEET 2021)
(a) 7.75
(b) 6.25
(c) 8.50
(d) 5.50
Ans: (a)

Calculate the pH of dimethylammonium acetate is as follows:

pH = 7.75
Hence, option 1 is the correct answer.

2020

Q1: Find out the solubility of Ni(OH)₂ in 0.1 M NaOH. Given that the ionic product of Ni(OH)₂ is 2×10^–15.     (NEET 2020)
(a) 1 × 10^-13 M
(b) 1 × 10⁸ M
(c) 2 × 10^-13 M
(d) 2 × 10^-8 M
Ans: (c)

2019

Q1:  pH of a saturated solution of Ca(OH)₂ is 9. The solubility product (K_sp) of Ca(OH)₂ is:    (NEET 2019)
(a) 0.5 × 10^-15
(b) 0.25 × 10^-10
(c) 0.125 × 10^-15
(d) 0.5 × 10^-10
Ans: (a)

pH = 9 Hence pOH = 14 - 9 = 5
[OH^-] = 10^-5 M

Thus K_sp = [Ca²⁺][OH^-]²


Q2: Conjugate base for Brönsted acids H₂O and HF are :    (NEET 2019)
(a) OH^- and H₂F⁺, respectively
(b) H₃O⁺ and F^-, respectively
(c) OH^- and F^-, respectively
(d) H₃O⁺ and H₂F⁺, respectively
Ans: (c)

HF on loss of H⁺ ion becomes F^- is the conjugate base of HF
Example :

Q3: Which will make basic buffer?    (NEET 2019)
(a) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH₃COOH
(b) 100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M NaOH
(c) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH
(d) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Ans: c

This is basic solution due to NaOH.
This is not basic buffer.

Hydrolysis of salt takes place.
This is not basic buffer.

This is basic buffer

⇒ Neutral solution

2018

Q1: The solubility of BaSO₄ in water 2.42×10³ gL^-1 at 298 K. The value of solubility product (Ksp) will be(Given molar mass of BaSO4 = 233 g mol^-1)    (NEET 2018)
(a) 1.08 × 10^-10 mol² L^-2
(b) 1.08 × 10^-12 mol² L^-2
(c) 1.08 × 10^-14 mol² L^-2
(d) 1.08 × 10^-8 mol² L^-2
Ans: (a)

Q2: Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :

pH of which one of them will be equal to 1 ?    (NEET 2018)
(1) b
(2) a
(3) d
(4) c
Ans: (d)
(A) 60 mL M/$\frac{10}{}$ HCl + 40 mL M/$\frac{10}{}$ NaOH
Mili. moles of HCl = 60 $\times$ 1/$\frac{10}{}$ = 6
Mili. moles of NaOH = 40 $\times$ 1/10 = 4
Mili. moles of HCl remaining = 6 - 4 = 2
Total volume will be 60 + 40 = 100 mL
Concentration of [H⁺] = $\frac{2}{}$2/100 = 2 $\times$ 10^-2
$\therefore$ pH = 2 - log2 = 1.7
(B) 55 mL M/$\frac{10}{}$ HCl + 45 mL M/$\frac{10}{}$ NaOH
Mili. moles of HCl = 55 $\times$ 1/10 = 5.5
Mili. moles of NaOH = 45 $\times$ $\frac{1}{}$/10 = 4.5
Mili. moles of HCl remaining = 5.5 - 4.5 = 1
Concentration of [H⁺] = $\frac{1}{100}$ = 10^-2
$\therefore$ pH = 2
(C) 75 mL M/$\frac{5}{}$ HCl + 25 mL M/$\frac{5}{}$ NaOH
Mili. moles of HCl = 75 $\times$ $\frac{1}{}$/5 = 15
Mili. moles of NaOH = 25 $\times$1/5 = 5
Mili. moles of HCl remaining = 15 - 5 = 10
Total volume will be 75 + 25 = 100 mL
Concentration of [H⁺] = 10/100 = 10^-1
$\therefore$ pH = 1
(D) 100 mL M/$\frac{10}{}$ HCl + 100 mL M/ 1$\frac{0}{}$ NaOH
Mili. moles of HCl = 100 $\times$ $\frac{1}{}$/10 = 10
Mili. moles of NaOH = 100 $\times$ $\frac{1}{}$/10 = 10
Mili. moles of HCl remaining = 10 - 10 = 0
So, it is neutral solution.
$\therefore$ pH = 7 


Q3: Which one of the following conditions will favour maximum formation of the product in the reaction
A₂(g) + B₂(g) ⇌ X₂(g) Δ_rH = -X kJ ?    (NEET 2018)
(a) Low temperature and high pressure
(b) Low temperature and low pressure
(c) High temperature and high pressure
(d) High temperature and low pressure
Ans: (a)
For reaction ΔH = - ve and Δn_g = - ve
∴ High P, Low T, favour product formation.

2017

Q1: Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10^-4 mol L^-1 Solubility product of Ag₂C₂O₄ is :-    (NEET 2017)
(a) 2.66 × 10^-12
(b) 4.5 × 10^-11
(c) 5.3 × 10^-12
(d) 2.42 × 10^-8
Ans: c

Q2: The equilibrium constant of the following are :    (NEET 2017)

The equilibrium constant (K) of the reaction:

(a)
(b)
(c)
(d)
Ans: a



Q3: A 20 litre container at 400 K contains CO₂(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO₂ attains its maximum value, will be :-    (NEET 2017)

(a) 10 litre
(b) 4 litre
(c) 2 litre
(d) 5 litre
Ans:(d)

2016

Q1: MY and NY₃, two nearly insoluble salts, have the same Ksp values of 6.2 x 10^-13 at room temperature, which statements would be true in regard to MY and NY₃?    (NEET 2016 Phase 1)
(a) The addition of the salt of KY to solution of MY and NY₃ will have no effect on their solubilities.
(b) The molar solubilities of MY and NY₃ in water are identical.
(c) The molar solubility of MY in water is less than that of NY₃.
(d) The salts MY and NY₃ are more soluble in 0.5 M KY than in pure water.
Ans: (c)
For MY,

Q2: Consider the following liquid-vapour equilibrium.   (NEET 2016 Phase 1)
Liquid $\rightleftharpoons$ Vapour
Which of the following relations is correct ? 
(a)

(b)

(c)

(d)

Ans: (b)
This is Clausius-Clapeyron equation.

Q3: The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅N⁺H) ina 0.10 M aqueous pyridine solution (Kb for C₅H5N = 1.7 × 10⁻⁹) is
(a) 0.0060%
(b) 0.013%
(c) 0.77%
(d) 1.6%     (NEET 2016 Phase 2)
Ans: (b)
C₅H₅N + H₂O → C₅H₅N⁺H + OH^–
So, the amount of pyridine that forms pyridinium ion is α. 

So, percentage of pyridine that forms pyridinium ion
= 1.30 × 10^–4 × 100
= 1.30 × 10^–2 = 0.013% 

Q4: Which of the following fluro-compounds is most likely to behave as a Lewis base ?
(a) BF₃
(b) PF₃
(c) CF₄
(d) SiF₄     (NEET 2016 Phase 2)
Ans: (b)
PF₃ is lewis base because on P atom there is lone pair.

Q5: The solubility of AgCl(s) with solubility product 1.6 × 10⁻¹⁰ in 0.1 M NaCl solution would be
(a) 1.26 $\times$ 10^$-$5 M 
(b) 1.6 $\times$ 10^$-$9 M 
(c) 1.6 $\times$ 10^$-$11 M 
(d) Zero     (NEET 2016 Phase 2)
Ans: (b)

Concentration of Cl^– is (s + 0.1) mol L^–1 because s mol L^–1 from ionization of AgCl and 0.1 mol L^–1 from ionization of 0.1 M NaCl.
Now, K_sp = [Ag⁺][Cl^– ]
$\Rightarrow$ 1.6 × 10^–10 = s (s + 0.1)
$\Rightarrow$ 1.6 × 10^–10 = s (0.1) {$\because$ s << 0.1}
$\Rightarrow$ s = 1.6 × 10^–9 M 

2015

Q1: The Ksp of Ag₂CrO₄, AgCl, AgBr and AgI are respectively, 1.1 x 10^-12, 1.8 x 10^-10, 5.0 x 10^-13, 8.3 x 10^-17. Which one of the following salts will precipitate last if AgNO₃ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na₂CrO₄ ?    (NEET / AIPMT Cancelled Paper 2015)
(a) Ag₂CrO₄
(b) AgI
(c) AgCl
(d) AgBr
Ans: (a)



Q2: If the value of an equilibrium constant for a particular reaction is 1.6 x 10¹², then at equilibrium the system will contain :
(a) similar amounts of reactants and products
(b) all reactants
(c) mostly reactants
(d) mostly products   (NEET / AIPMT Cancelled Paper 2015)
Ans: (d)

Q3:What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
(a) 2.0
(b) 7.0
(c) 1.04
(d) 12.65                   (NEET / AIPMT 2015)
Ans: (d)

As equal volumes of HCl and NaOH are added so the volume of resulting solution becomes double and the concentration of the solution becomes half.
$\therefore$ [OH^-] = 0.09/2 = 0.045 M
$\therefore$ pOH = – log[OH^– ] = –log [0.045] = 1.35
$\therefore$ pH = 14 – pOH = 14 – 1.35 = 12.65 

Q4: Which one of the following pairs of solution is not an acidic buffer ?
(a) CH₃COOH and CH₃COONa 
(b) H₂CO₃ and Na₂CO₃ 
(c) H₃PO₄ and Na₃PO₄ 
(d) HClO₄ and NaClO₄         (NEET / AIPMT 2015)
Ans: (d)
An acidic buffer is a mixture of a weak acid and its salt with a strong base. Among CH₃COOH, H₂CO₃, H₃PO₄ and HClO₄, the HClO₄ is a strong acid while all other are weak acid thus, HClO₄ and NaClO₄ does not constitute to form an acidic buffer. 

2014

Q1: For the reversible reaction :
N₂(g) + 3H₂(g) ⇔ 2NH₃ (g) + heat
The equilibrium shifts in forward direction :    (NEET / AIPMT 2014)
(a) by decreasing the concentrations of N₂(g) and H₂(g)
(b) by increasing pressure and decreasing temperature
(c) by increasing the concentration of NH₃(g)
(d) by decreasing the pressure
Ans: (b)
As the forward reaction is exothermic and leads to lowering of pressure (produces lesser number of gaseous moles). Hence, According to Le Chatelier's principle, at high pressure and low temperature, the given reversible reaction will shift in forward direction to form more product.

Q2: Using the Gibbs energy change, ΔG° = +63.3 kJ, for the following reaction,

    (NEET / AIPMT 2014)
(a) 2.9 × 10^-3
(b) 7.9 × 10^-2
(c) 3.2 × 10^-26
(d) 8.0 × 10^-12
Ans: (d)


Q3: For a given exothermic reaction, K_p and K'_p are the equilibrium constants at temperatures T₁ and T₂ , respectively. Assuming that heat of reaction is constant in temperature range between T₁ and T₂ , it is readily observed that :  (NEET / AIPMT 2014)
(a)
(b)
(c)
(d)
Ans: (c)
For exothermic reactions, on increasing the temperature the value of equilibrium constant decreases.

Q4: Which of the following salts will give highest pH in water ?    (NEET / AIPMT 2014)
(a) Na₂CO₃
(b) CuSO₄
(c) KCl
(d) NaCl
Ans: (a)
The highest pH refers to the basic solution containing OH^- ions. Therefore, the basic salt releasing OH^- ions on hydrolysis will give highest pH in water.
Only the salt of a strong base and weak acid would release OH^- ion on hydrolysis. Among the given salts, Na₂CO₃ corresponds to the basic salt as it is formed by the neutralisation of NaOH [strong base] and H₂CO₃ [weak acid]