NEET ChemistryUpdated 19 Oct 2026

Chemical Equilibrium: NEET PYQs and weightage

Chemical Equilibrium carries 2.8 of 100 Chemistry marks in NEET by past-paper share: 20 of the 707 tagged NEET Chemistry PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 19 of 26 Chemistry chapters.

Take the 25-question Chemical Equilibrium test Formula sheet ↓All NEET Chemistry chapters

2.8/100 Weight in Chemistry

Exam importance (3/5, by past-paper share)

20 NEET PYQs

385 Practice questions

364 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Chemistry PYQs in the bank (20 of 707), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Chemical Equilibrium: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Equilibrium constant Kc
$a A + b B ⇌ c C + d D$
$K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$
Use it when: you have equilibrium molar concentrations ( $mol/L$ ) and need $K$ , or the reverse
Pressure constant Kp
$K_{p} = \frac{( p _{C} ) ^{c} ( p _{D} ) ^{d}}{( p _{A} ) ^{a} ( p _{B} ) ^{b}} (pressures in bar)$
Use it when: the species are gases and partial pressures are given
Kp–Kc link
$K_{p} = K_{c} (R T)^{Δ n}$
$Δ n = (moles of gaseous products) - (moles of gaseous reactants)$
$R = 0.0831 bar L mol^{- 1} K^{- 1}$
Use it when: one constant is given and the question asks for the other ( $Δ n = 0$ means $K_{p} = K_{c}$ )
Pressure from concentration
$p = c R T (c in mol/L, p in bar)$
Use it when: switching a gas between concentration and partial pressure
Reverse reaction
$K^{'} = \frac{1}{K}$
Use it when: the equation is written backwards compared with the one whose $K$ is given
Multiplied equation
$K_{new} = K^{n} (halving the equation gives K^{1/2} = K)$
Use it when: every coefficient is scaled by the same factor $n$
Adding reactions
$K_{net} = K_{1} \times K_{2} \times \dots$
Use it when: the target equation is the sum of given equations
Reaction quotient
$Q < K \to forward$
$Q > K \to backward$
$Q = K \to at equilibrium$
$Q_{c}$ has the same form as $K_{c}$ but uses concentrations at any moment.
Use it when: asked which way a mixture will shift
Gibbs energy at any stage
$Δ G = Δ G^{\circ} + R T ln Q$
Use it when: checking spontaneity of a mixture that is not at equilibrium
Standard Gibbs energy and K
$Δ G^{\circ} = - R T ln K$
$K = e^{- Δ G^{\circ} / R T} (R = 8.314 J mol^{- 1} K^{- 1}, Δ G^{\circ} in J, not kJ)$
Use it when: converting between $Δ G^{\circ}$ and $K$ ( $Δ G^{\circ} < 0$ means $K > 1$ )
Heterogeneous equilibrium
$CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g) \Rightarrow K_{p} = p_{C O_{2}}$
Leave pure solids and liquids out of $K$ .
Use it when: any species is (s) or (l)
Extent from K
$K > 1 0^{3} (mostly products)$
$K < 1 0^{- 3} (mostly reactants)$
Values in between mean both are appreciable.
Use it when: judging the equilibrium composition from the size of $K$
Dissociation A(g) ⇌ 2B(g)
$K_{p} = \frac{4 α ^{2} p}{1 - α ^{2}}$
$α = \frac{1}{1 + 4 p / K _{p}} (p = total pressure in bar)$
Use it when: finding the degree of dissociation at a given total pressure (for example $N_{2} O_{4} ⇌ 2 N O_{2}$ )
Temperature and K
For exothermic reactions ( $Δ H < 0$ ), $K$ falls as $T$ rises; for endothermic reactions ( $Δ H > 0$ ), $K$ rises as $T$ rises. Only temperature changes $K$ . A catalyst, or an inert gas added at constant volume, leaves the equilibrium unchanged.
Use it when: a question asks how heating, a catalyst or added argon affects $K$ or the yield

Trap

counting solids, liquids or water as solvent when finding Δn or writing K. Only gaseous moles go into Δn, and pure solids and liquids never appear in K.

How does NEET ask Chemical Equilibrium?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1ChemistryNEET UG PYQApplications of Equilibrium Constant

Try right here · No sign-in

For the reaction 2A ⇌ B + C, K

_{c}

= 4 × 10–3. At a given time, the composition of reaction mixture is : [A] = [B] = [C] = 4 × 10–3 Then, which of the following is correct?

Answer key (B) & worked solution

Correct answer: Option B

K_{C} = \frac{[ B ] [ C ]}{[ A ] ^{2}}

Q_{C} = \frac{[ 2 \times 10 ^{- 3} ] [ 2 \times 10 ^{- 3} ]}{[ 2 \times 10 ^{- 3} ] ^{2}} = 1

Q_{C} > K_{C}

Q2ChemistryNEET UG PYQEquilibrium Constants (Kp and Kc) and Their Properties

A 20 litre container at 400 K contains CO

_{2}

(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 CO

_{2}

attains its maximum value will be.(Given that : SrCO

_{3}

(s) = SrO(s) + CO

_{2}

(g), Kp = 1.6 atm)

A4 litre

B2 litre

C5 litre

D10 litre

Answer key (C) & worked solution

Correct answer: Option C

Max. pressure of CO

_{2}

= Pressure of CO

_{2}

at equilibrium For reaction, SrCO

_{3}

(s) ⇌ SrO(s) + CO

_{2}

Kp = P

_{C O 2}

= 1.6 atm = maximum pressure of CO

_{2}

Volume of container at this stage,

V = \frac{n R T}{P}

… (i) Since container is sealed and reaction was not earlier at equilibrium ∴ n is constant

n = \frac{P V}{R T} = \frac{0.4 \times 20}{R T}

… (ii) Put equation (ii) in equation (i)

V = [\frac{0.4 \times 20}{R T}] \frac{R T}{1.6} = 5 L

Q3ChemistryNEET UG PYQThermodynamics of Equilibrium

Which of the following statements is correct for a reversible process in a state of equilibrium?

AΔG

^{0}

= 2.303 RTlogK

BΔG = − 2.303 RTlogK

CΔG = 2.303 RTlogK

DΔG

^{o}

= − 2.303 RTlogK

Answer key (D) & worked solution

Correct answer: Option D

ΔG = ΔG

^{0}

+ 2.303 RTlogQ 0 = ΔG

^{0}

+ 2.303 RTlogK ΔG

^{0}

= − 2.303 RTlogK

Q4ChemistryNEET UG PYQEquilibrium Constants (Kp and Kc) and Their Properties

In which of the following equilibria, K

_{p}

and K

_{c}

are NOT equal?

_{2}

(g) + I

_{2}

(g) ⇌ 2HI(g)

BCO

_{2}

(g) + H

_{2}

O(g) ⇌ CO

_{2}

(g) + H

_{2}

(g)

C2BrCl(g) ⇌ Br

_{2}

(g) + Cl

_{2}

(g)

DPCl

_{5}

(g) ⇌ PCl

_{3}

(g) + Cl

_{2}

(g)

Answer key (D) & worked solution

Correct answer: Option D

Δ K_{p} = K_{c} (R T)^{Δ n_{g}}

In option 4 PCl

_{5}

(g) ⇌ PCl

_{3}

(g) + Cl

_{2}

(g) ∆n

_{g}

= 1 K

_{p}

= K

_{c}

RT In rest option 1,2,3 ∆n

_{g}

= 0 So, in that K

_{p}

= K

_{c}

Q5ChemistryNEET UG PYQLe Chatelier's Principle

KMnO

_{4}

can be prepared from K

_{2}

MnO

_{4}

as per the reaction

3 M n O_{4}^{2 -} + 2 H_{2} O \to 2 M n O_{4}^{-} + M n O_{2} + 4 O H^{-}

The reaction can go completion by removing OH

^{-}

ions by adding:

AHCI

BKOH

CCO

_{2}

DSO

_{2}

Answer key (C) & worked solution

Correct answer: Option C

The reaction can go to completion by removing OH

^{-}

ions by adding CO

_{2}

Q6ChemistryNEET UG PYQLe Chatelier's Principle

For the reversible reaction:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) + heat

The equilibrium shifts in forward direction:

ABy decreasing the concentration of N

_{2}

(g) and H

_{2}

(g)

BBy increasing pressure and decreasing temperature

CBy increasing the concentration of NH

_{3}

(g)

DBy decreasing the pressure

Answer key (B) & worked solution

Correct answer: Option B

∵ The forward reaction is exothermic, it will be favoured by decrease in Temperature. Also on increasing P, direction having lesser no. of gaesous moles is favoured that is forward direction in this case.

Q7ChemistryNEET UG PYQThermodynamics of Equilibrium

Using the Gibbs energy change, ΔG

^{0}

= + 63.3KJ, for the following reaction,

A g_{2} C O_{3} (s) ⇌ 2 A g^{+} (a q) + C O_{3}^{2 -} (a q)

The K

_{s p}

of Ag

_{2}

_{3}

(s) in water at 25

^{o}

C is: (R = 8.314 J.K

^{- 1}

mol

^{- 1}

)

A2.9 × 10− 3

B7.9 × 10− 2

C3.2 × 10− 26

D8.0 × 10− 12

Answer key (D) & worked solution

Correct answer: Option D

ΔG

^{0}

= − 2.303 RT log K

_{s p}

63.3 × 1000 = − 2.303 × 8.314 × 298 log K

_{s p}

log K

_{s p}

= -11.09 K

_{s p}

= 8.0 × 10− 12

Q8ChemistryNEET UG PYQEquilibrium Constants (Kp and Kc) and Their Properties

The equilibrium constants of the following are: N

_{2}

+ 3H

_{2}

⇌

2NH

_{3}

_{1}

_{2}

+ O

_{2}

⇌

2NO K

_{2}

H_{2} + \frac{1}{2} O_{2} \to H_{2} O

_{3}

The equilibrium constant (K) of the reaction

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

will be

K_{2} K_{3} / K_{1}

K_{2}^{3} K_{3} / K_{1}

K_{1} K_{3}^{3} / K_{2}

K_{2} K_{3}^{3} / K_{1}

Answer key (D) & worked solution

Correct answer: Option D

(I)

N_{2} + 3 H_{2} ⇌ 2 N H_{3}; K_{1} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

(II)

N_{2} + O_{2} ⇌ 2 N O; K_{2} = \frac{[ N O ] ^{2}}{[ N _{2} ] [ O _{2} ]}

(III)

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O; K_{3} = \frac{[ H _{2} O ]}{[ H _{2} ] [ O _{2} ] ^{1/2}}

(II + 3 × III − I) will give

2 N H_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

∴ K = K_{2} \times K_{3}^{3} / K_{1}

Which topics in Chemical Equilibrium carry the most questions?

All 12 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Equilibrium Constants (Kp and Kc) and Their Properties	7	166
Thermodynamics of Equilibrium	6	14
Le Chatelier's Principle	4	92
Applications of Equilibrium Constant	2	30
Degree of Dissociation	1	26
Introduction to Equilibrium	0	15
Phase Equilibria	0	12
Graphical Representation and Characteristics of Chemical Equilibrium	0	8
Law of Mass Action	0	8
Homogeneous Equilibria	0	7
Heterogeneous Equilibria	0	6
Miscellaneous Examples	0	1

Why does Chemical Equilibrium show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Chemical Equilibrium, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Chemical Equilibrium?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Chemical Equilibrium test

Chapters either side of Chemical Equilibrium by weight

Classification of Elements and Periodicity Weight 3.1/100 Organic Chemistry: Some Basic Principles and Techniques Weight 3.0/100 Redox Reactions Weight 2.7/100 p-Block Elements (Group 13 and 14) Weight 2.7/100

Chemical Equilibrium: NEET PYQs and weightage

Chemical Equilibrium: formula sheet for NEET

Equilibrium constant Kc

Pressure constant Kp

Kp–Kc link

Pressure from concentration

Reverse reaction

Multiplied equation

Adding reactions

Reaction quotient

Gibbs energy at any stage

Standard Gibbs energy and K

Heterogeneous equilibrium

Extent from K

Dissociation A(g) ⇌ 2B(g)

Temperature and K

How does NEET ask Chemical Equilibrium?

Which topics in Chemical Equilibrium carry the most questions?

Why does Chemical Equilibrium show "–" until you answer 3 questions?

What happens when you test Chemical Equilibrium?

Chapters either side of Chemical Equilibrium by weight