Equilibrium

14 previous-year questions from this chapter every option and the correct answer, free · where the marks are

1. Law of mass action, Kc and Kp, Le Chatelier

Exam focus: Kp and Kc are related by a factor of (RT) raised to the power Δn — moles of gaseous product minus reactant — and getting that exponent's sign backwards inverts the whole conversion. Separately, adding an inert gas at constant volume leaves equilibrium untouched, but adding it at constant pressure shifts it toward the side with more gas moles, because the container must expand to keep pressure fixed.

Chemical Equilibrium: Kc, Kp and Le Chatelier's Principle

Competition Wallah

Full-chapter NCERT one-shot covering the law of mass action, Kc/Kp, and Le Chatelier's principle with PYQs.

ChemistryNEET 2025show ▾

For the reaction

A (g) ⇌ 2B (g)

, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 at 1000 K. [Given

R = 0.0831 L atm mo l^{- 1} K^{- 1}

K_{p}

for the reaction at 1000 K is:

A.83.1
B. $2.077 \times 1 0^{5}$
C.0.033✓
D.0.021

Solution

$K_{c} = \frac{k _{f}}{k _{b}} = \frac{1}{2500} = 4 \times 1 0^{- 4}$ . $Δ n_{g} = 2 - 1 = 1$ , so $K_{p} = K_{c} (R T) = 4 \times 1 0^{- 4} \times 0.0831 \times 1000 \approx 0.033$ .

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ChemistryNEET 2025show ▾

A higher yield of NO in

N_{2} (g) + O_{2} (g) ⇌ 2NO (g)

can be obtained at [

Δ H

of the reaction

= + 180.7 kJ mo l^{- 1}

]: A. Higher temperature · B. Lower temperature · C. Higher concentration of

N_{2}

· D. Higher concentration of

O_{2}

Choose the correct answer from the options given below:

A.A, D only
B.B, C only
C.B, C, D only
D.A, C, D only✓

Solution

The reaction is endothermic, so higher temperature shifts it forward (A). Adding either reactant also pushes it forward (C, D). Lower temperature would reduce the yield. Hence A, C and D.

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ChemistryNEET 2024show ▾

In which of the following equilibria are

K_{p}

and

K_{c}

not equal?

A. $H_{2} (g) + I_{2} (g) ⇌ 2HI (g)$
B. $CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$
C. $2BrCl (g) ⇌ B r_{2} (g) + C l_{2} (g)$
D. $PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$ ✓

Solution

$K_{p} = K_{c} (R T)^{Δ n_{g}}$ ; they are equal only when $Δ n_{g} = 0$ . For $PC l_{5} ⇌ PC l_{3} + C l_{2}$ , $Δ n_{g} = 2 - 1 = 1$ , so $K_{p} \neq = K_{c}$ . The other three have equal moles of gas on both sides.

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ChemistryNEET 2024show ▾

For the reaction

2A ⇌ B + C

K_{c} = 4 \times 1 0^{- 3}

. At a given time the composition of the reaction mixture is

[A] = [B] = [C] = 2 \times 1 0^{- 3}

M. Then which of the following is correct?

A.The reaction has a tendency to go in the forward direction.
B.The reaction has a tendency to go in the backward direction.✓
C.The reaction has gone to completion in the forward direction.
D.The reaction is at equilibrium.

Solution

$Q_{c} = \frac{[ B ] [ C ]}{[ A ] ^{2}} = \frac{( 2 \times 1 0 ^{- 3} ) ^{2}}{( 2 \times 1 0 ^{- 3} ) ^{2}} = 1$ , which is greater than $K_{c} = 4 \times 1 0^{- 3}$ . The reaction proceeds backward to reduce $Q_{c}$ .

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ChemistryNEET 2024show ▾

Consider the following reaction in a sealed vessel at equilibrium with concentrations

N_{2} = 3.0 \times 1 0^{- 3}

O_{2} = 4.2 \times 1 0^{- 3}

M and

NO = 2.8 \times 1 0^{- 3}

2NO (g) ⇌ N_{2} (g) + O_{2} (g)

0.1 mol L^{- 1}

of NO(g) is taken in a closed vessel, what will be the degree of dissociation (

α

) of NO(g) at equilibrium?

A.0.0889
B.0.8889
C.0.717✓
D.0.00889

Solution

$K_{c} = \frac{[ N _{2} ] [ O _{2} ]}{[ NO ] ^{2}} = \frac{3.0 \times 1 0 ^{- 3} \times 4.2 \times 1 0 ^{- 3}}{( 2.8 \times 1 0 ^{- 3} ) ^{2}} \approx 1.607$ . Starting from $0.1$ M NO: $K_{c} = \frac{( 0.05 α ) ^{2}}{( 0.1 ( 1 - α ) ) ^{2}} = \frac{α ^{2}}{4 ( 1 - α ) ^{2}}$ , so $\frac{α}{2 ( 1 - α )} = 1.268 \Rightarrow α \approx 0.717$ .

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ChemistryNEET 2022show ▾

3 O_{2} (g) ⇌ 2 O_{3} (g)

For the above reaction at 298 K,

K_{c}

is found to be

3.0 \times 1 0^{- 59}

. If the concentration of

O_{2}

at equilibrium is 0.040 M, then the concentration of

O_{3}

in M is:

A. $1.2 \times 1 0^{21}$
B. $4.38 \times 1 0^{- 32}$ ✓
C. $1.9 \times 1 0^{- 63}$
D. $2.4 \times 1 0^{31}$

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2. Ionic equilibrium: acids, bases, pH, buffers

Exam focus: For a strong acid diluted to around 10⁻⁷–10⁻⁸ M, pH = −log(C) alone gives a value above 7 — impossible for an acid — because it ignores water's own contribution of H⁺ at that concentration. This "pH of an extremely dilute strong acid" question is a NEET standard precisely because the naive formula gives a physically impossible answer.

Ionic Equilibrium: Acids, Bases, pH, Buffers

Competition Wallah

Full-chapter one-shot covering acid-base theories, pH, hydrolysis and buffers; also includes solubility-product/common-ion content.

ChemistryNEET 2025show ▾

Phosphoric acid ionises in three steps with ionisation constant values

K_{a_{1}}

K_{a_{2}}

and

K_{a_{3}}

respectively, while

K

is the overall ionisation constant. Which of the following statements are true? A.

lo g K = lo g K_{a_{1}} + lo g K_{a_{2}} + lo g K_{a_{3}}

H_{3} P O_{4}

is a stronger acid than

H_{2} P O_{4}^{-}

and

HP O_{4}^{2 -}

K_{a_{1}} > K_{a_{2}} > K_{a_{3}}

\frac{K _{a_{2}} + K _{a_{3}}}{K _{a_{1}}} = 2

Choose the correct answer from the options given below:

A.A and B only
B.A and C only
C.B, C and D only
D.A, B and C only✓

Solution

The overall constant is the product $K = K_{a_{1}} K_{a_{2}} K_{a_{3}}$ , so the logs add (A). Removing successive protons from an increasingly negative ion gets harder, so $K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$ (C) and $H_{3} P O_{4}$ is the strongest acid of the three (B). D is baseless. Hence A, B and C.

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ChemistryNEET 2023show ▾

Amongst the given options, which of the following molecules/ions acts as a Lewis acid?

A. $N H_{3}$
B. $H_{2} O$
C. $B F_{3}$ ✓
D. $O H^{-}$

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ChemistryNEET 2022show ▾

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is (given

p K_{a}

C H_{3} COOH

= 4.57):

A.2.57
B.5.57✓
C.3.57
D.4.57

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ChemistryNEET 2021show ▾

The

p K_{b}

of dimethylamine and

p K_{a}

of acetic acid are 3.27 and 4.77 respectively at

T

(K). The correct option for the pH of dimethylammonium acetate solution is:

A.8.50
B.5.50
C.7.75✓
D.6.25

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ChemistryNEET 2019show ▾

The conjugate bases for the Brønsted acids

H_{2} O

and

HF

are, respectively:

A. $O H^{-}$ and $H_{2} F^{+}$
B. $H_{3} O^{+}$ and $F^{-}$
C. $O H^{-}$ and $F^{-}$ ✓
D. $H_{3} O^{+}$ and $H_{2} F^{+}$

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ChemistryNEET 2019show ▾

Which of the following mixtures will make a basic buffer?

A.50 mL of 0.1 M NaOH + 25 mL of 0.1 M $C H_{3} COOH$
B.100 mL of 0.1 M $C H_{3} COOH$ + 100 mL of 0.1 M NaOH
C.100 mL of 0.1 M HCl + 200 mL of 0.1 M $N H_{4} OH$ ✓
D.100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH

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3. Solubility product and common-ion effect

Exam focus: The Ksp expression's exponents come straight from dissolution stoichiometry — Ag₂CrO₄ gives [Ag⁺]²[CrO₄²⁻], and dropping that square is the most common setup error. A common ion lowers molar solubility but never changes Ksp itself, which stays constant at a given temperature — a distinction NEET tests by asking what actually changes when a common ion is added.

Ionic Equilibrium — Full Revision (incl. Solubility Product and Common-Ion Effect)

Sri Chaitanya Academy NEET

Bundled within a full Ionic Equilibrium chapter revision rather than a video dedicated solely to Ksp/common-ion effect.

1. Law of mass action, Kc and Kp, Le Chatelier

Law of mass action, Kc and Kp, Le Chatelier's principle

2. Ionic equilibrium: acids, bases, pH, buffers

Ionic equilibrium: acids, bases, pH and buffers

3. Solubility product and common-ion effect

Solubility product and common-ion effect