How much of JEE Main comes from Equilibrium?

JEEGenie estimates about 5% of JEE Main marks come from Equilibrium, based on recent papers. The share varies from year to year.

Which formulas matter most in Equilibrium?

The Equilibrium sheet on this page has 17 formula entries, starting with Law of Mass Action, Equilibrium Constant K_c and K_p, Reaction Quotient Q, Effect of Temperature on K, Bronsted-Lowry Acid and Base. Each one comes with when to use it and the trap to avoid.

What topics does Equilibrium cover for JEE?

Equilibrium covers Law of Mass Action, Le Chatelier's Principle, Ionic Equilibrium, pH and Buffers, Solubility Product, Acids and Bases.

How should I revise Equilibrium before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Equilibrium questions on JEEGenie.

How many previous-year questions are there on Equilibrium?

JEEGenie has 489 previous-year questions on Equilibrium, and this page shows a few with worked explanations.

Is practice on Equilibrium free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Equilibrium for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 5% of JEE Main marks come from Equilibrium. This page covers 17 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Equilibrium is dynamic, not static; reactions continue at equal rates.

Key points to know

Chemical Equilibrium. A state in a reversible reaction where the rates of forward and backward reactions are equal, and concentrations of reactants and products remain constant.
Law of Mass Action. At equilibrium, the rate of a reaction is proportional to the product of active masses (concentrations) of reactants, each raised to their stoichiometric coefficients.
Le Chatelier's Principle. If a system at equilibrium is disturbed, it shifts to counteract the disturbance and restore equilibrium.
Effect of Temperature on $K$ . For exothermic reactions ( $Δ H < 0$ ), increasing temperature decreases $K$ ; for endothermic ( $Δ H > 0$ ), increasing temperature increases $K$ .
Acid and Base (Arrhenius). Arrhenius acid: substance that gives H⁺ in water. Arrhenius base: gives OH⁻ in water.
Bronsted-Lowry Acid and Base. Acid donates a proton (H⁺); base accepts a proton. Conjugate acid-base pairs differ by one H⁺.
Lewis Acid and Base. Lewis acid accepts an electron pair; Lewis base donates an electron pair.
Buffer Solution. A solution that resists pH change upon addition of small amounts of acid or base. Contains a weak acid and its conjugate base (or weak base and its conjugate acid).
Degree of Ionization ( $α$ ). Fraction of initial molecules that ionize. For weak acid: $α = \sqrt{K_{a} / C}$ (if $α < 0.05$ ).
Hydrolysis of Salts. Reaction of salt ions with water to produce acidic or basic solutions. Salt of weak acid + strong base gives basic solution; weak base + strong acid gives acidic; weak acid + weak base depends on $K_{a}$ and $K_{b}$ .
Relation Between $K_{p}$ and $K_{c}$ for Gaseous Reactions. $K_{p}$ and $K_{c}$ are related by $K_{p} = K_{c} (R T)^{Δ n_{g}}$ , where $Δ n_{g}$ is change in moles of gas.

Formula sheet

Name	Formula	When to use
Law of Mass Action	For $a A + b B ⇌ c C + d D$ , $K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$	To write equilibrium constant expressions for homogeneous reactions.
Equilibrium Constant $K_{c}$ and $K_{p}$	$K_{p} = K_{c} (R T)^{Δ n_{g}}$ where $Δ n_{g} = (c + d) - (a + b)$ for gaseous species.	When converting between concentration-based and pressure-based equilibrium constants.
Reaction Quotient $Q$	$Q_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$ (using initial concentrations)	To determine the direction a reaction will shift to reach equilibrium.
Effect of Temperature on $K$	$\ln \frac{K_{2}}{K_{1}} = - \frac{Δ H}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$ (van't Hoff equation)	To calculate $K$ at different temperatures or analyze temperature effect.
Bronsted-Lowry Acid and Base	Acid $⇌$ Conjugate base + H⁺	For proton transfer reactions in any solvent.
Ionization of Water and pH	$K_{w} = [H^{+}] [O H^{-}] = 10^{- 14}$ at 25°C, $p H = - \log [H^{+}]$ , $p O H = - \log [O H^{-}]$ , $p H + p O H = 14$	To calculate pH, pOH, or [H⁺]/[OH⁻] in aqueous solutions.
Ionization Constants $K_{a}$ and $K_{b}$	For weak acid HA: $K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]}$ ; For weak base B: $K_{b} = \frac{[B H^{+}] [O H^{-}]}{[B]}$ ; $K_{a} \times K_{b} = K_{w}$	To quantify strength of weak acids/bases and calculate pH of their solutions.
pH of Weak Acid/Base (Approximation)	$[H^{+}] = \sqrt{K_{a} C}$ (if $C / K_{a} ≳ 400$ ), $[O H^{-}] = \sqrt{K_{b} C}$ for weak base	To calculate pH of weak acid or base solutions quickly.
Henderson-Hasselbalch Equation	$p H = p K_{a} + \log \frac{[A^{-}]}{[H A]}$	To calculate pH of buffer solutions or find ratio of conjugate pair for desired pH.
Solubility Product $K_{s p}$	$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$	To determine solubility of salts or predict precipitation in ionic solutions.
Relationship Between Solubility (s) and $K_{s p}$	For $A_{x} B_{y} (s) ⇌ x A^{y +} + y B^{x -}$ , $K_{s p} = (x s)^{x} (y s)^{y} = x^{x} y^{y} s^{x + y}$	To calculate solubility from $K_{s p}$ or vice versa.
Degree of Ionization ( $α$ )	$α = \frac{number of molecules ionized}{initial number}$ ; For weak acid: $α = \sqrt{\frac{K_{a}}{C}}$	To find extent of ionization of weak electrolytes.
Hydrolysis of Salts	For salt of weak acid (A⁻) and strong base: $A^{-} + H_{2} O ⇌ H A + O H^{-}$ , $K_{h} = K_{w} / K_{a}$	To predict pH of salt solutions.
pH of Salt Solutions	Weak acid/strong base: $p H = \frac{1}{2} p K_{w} + \frac{1}{2} p K_{a} + \frac{1}{2} \log C$ ; Weak base/strong acid: $p H = \frac{1}{2} p K_{w} - \frac{1}{2} p K_{b} - \frac{1}{2} \log C$	To calculate pH of salt solutions quickly.
Titration Curves and Indicators	Equivalence point: moles acid = moles base; pH at half-equivalence = pK_a (for weak acid)	To choose appropriate indicator or analyze titration data.
Relation Between $K_{p}$ and $K_{c}$ for Gaseous Reactions	$K_{p} = K_{c} (R T)^{Δ n_{g}}$	To convert between $K_{p}$ and $K_{c}$ for gas-phase equilibria.
ICE Table Method	Example: For $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ , initial [N₂]=a, [H₂]=b, [NH₃]=0, change: -x, -3x, +2x, equilibrium: a-x, b-3x, 2x	To calculate equilibrium concentrations from initial conditions and $K$ .

Open the Equilibrium formula page

What JEE asks

Chemical Equilibrium. JEE often tests the concept of dynamic equilibrium with graphs of concentration vs. time, asking to identify equilibrium point.
Law of Mass Action. JEE asks to write $K_{c}$ or $K_{p}$ expressions, often with solids/liquids omitted.
Equilibrium Constant $K_{c}$ and $K_{p}$ . JEE often gives $K_{c}$ and asks for $K_{p}$ at a given temperature, or vice versa.
Reaction Quotient $Q$ . JEE gives initial concentrations and $K$ , asks to find direction or equilibrium concentrations.
Le Chatelier's Principle. JEE asks to predict shift direction for given changes, e.g., increasing pressure favors side with fewer gas moles.
Effect of Temperature on $K$ . JEE gives $K$ at one temperature and $Δ H$ , asks for $K$ at another temperature.
Acid and Base (Arrhenius). JEE may ask to identify Arrhenius acids/bases from given compounds.
Bronsted-Lowry Acid and Base. JEE asks to identify conjugate pairs or determine relative strengths.
Lewis Acid and Base. JEE asks to classify species as Lewis acid/base, e.g., BF₃, NH₃.
Ionization of Water and pH. JEE often gives $K_{w}$ at a non-standard temperature and asks for pH of neutral solution.
Ionization Constants $K_{a}$ and $K_{b}$ . JEE gives $K_{a}$ and concentration, asks for pH of weak acid solution using approximation or quadratic.
pH of Weak Acid/Base (Approximation). JEE often expects approximation but may require quadratic for borderline cases.
Buffer Solution. JEE asks to identify buffer pairs or calculate pH of buffer after addition of acid/base.
Henderson-Hasselbalch Equation. JEE gives buffer composition and asks pH, or asks for ratio to achieve a given pH.
Solubility Product $K_{s p}$ . JEE gives $K_{s p}$ and asks for solubility in pure water or in presence of common ion.
Relationship Between Solubility (s) and $K_{s p}$ . JEE asks to find solubility of salts like Ag₂CrO₄ from $K_{s p}$ .
Common Ion Effect. JEE asks to calculate solubility of AgCl in 0.1 M NaCl solution using $K_{s p}$ .
Degree of Ionization ( $α$ ). JEE gives $K_{a}$ and concentration, asks for $α$ or pH.
Hydrolysis of Salts. JEE asks to find pH of solutions like NaHCO₃, NH₄Cl, or CH₃COONa.
pH of Salt Solutions. JEE gives salt concentration and $K_{a} / K_{b}$ , asks for pH.
Titration Curves and Indicators. JEE asks to identify indicator for a given titration or calculate pH at various points.
Relation Between $K_{p}$ and $K_{c}$ for Gaseous Reactions. JEE gives $K_{c}$ and temperature, asks for $K_{p}$ or vice versa.
ICE Table Method. JEE often requires ICE table for equilibrium problems, sometimes needing quadratic solution.

Common traps

Equilibrium is dynamic, not static; reactions continue at equal rates. Chemical Equilibrium
Active mass is molar concentration (mol/L), not total moles. Law of Mass Action
Use R = 0.0821 L atm mol⁻¹ K⁻¹ when pressure in atm. Equilibrium Constant $K_{c}$ and $K_{p}$
Do not confuse $Q$ with $K$ ; $Q$ changes as reaction proceeds. Reaction Quotient $Q$
Only applies to systems at equilibrium, not to reactions in progress. Le Chatelier's Principle
Temperature change always changes $K$ , unlike concentration/pressure. Effect of Temperature on $K$
Limited to aqueous solutions only. Acid and Base (Arrhenius)
Water can act as both acid and base (amphoteric). Bronsted-Lowry Acid and Base
Do not classify Lewis acids by H⁺ donation: the Lewis definition is about accepting an electron pair, so it is broader than the Brønsted one (BF₃ and Al³⁺ have no H⁺ to donate yet are Lewis acids). Lewis Acid and Base
$K_{w}$ is temperature dependent; at 100°C, $K_{w} \approx 5.5 \times 10^{- 13}$ (neutral pH ≈ 6.1; $10^{- 12}$ is only a rounded value). Ionization of Water and pH
Strong acids have very large $K_{a}$ (complete ionization). Ionization Constants $K_{a}$ and $K_{b}$
Always check the 5% rule: if $[H^{+}] / C < 0.05$ , approximation is valid. pH of Weak Acid/Base (Approximation)
Buffer capacity is highest when [acid] ≈ [conjugate base]. Buffer Solution
Use concentrations at equilibrium, not initial (but often initial used if no reaction). Henderson-Hasselbalch Equation

Topics in Equilibrium

Law of Mass Action
Le Chatelier's Principle
Ionic Equilibrium
pH and Buffers
Solubility Product
Acids and Bases

Previous-year questions on Equilibrium

JEEGenie has 489 previous-year questions on Equilibrium. A few worked examples follow.

Year	Questions
2026	28
2025	30
2024	18
2023	30
2022	29
2021	37
2020	24
2019	47
2018	20
2017	7
2016	8
2015	6
2014	7
2013	13
2012	6
2011	5
2010	9
2009	3
2008	6
2007	3
2006	8
2005	7
2004	11
2003	7
2002	9
2001	2
2000	4
1999	5
1998	5
1997	4
1996	5
1995	1
1994	4
1993	5
1992	2
1991	3
1990	4
1989	4
1988	4
1987	4
1986	4
1985	9
1984	16
1983	3
1982	9
1981	11
1979	3

Question 1 medium

The solubility product constants of

A g_{2} C r O_{4}

and

A g B r

are

32 x

and

4 y

respectively at 298 K. The value of

(\frac{molarity of A g_{2} C r O_{4}}{molarity of A g B r})

can be expressed as :

$\frac{2 \sqrt[3]{x}}{y}$
$2 \sqrt{\frac{x}{y}}$
$\sqrt{\frac{x}{y}}$
$\frac{\sqrt[3]{x}}{\sqrt{y}}$

Answer: D

Explanation. For

A g_{2} C r O_{4}

A g_{2} C r O_{4} (s) ⇌ 2 A g^{+} + C r O_{4}^{2 -}

Let molar solubility be

s_{1}

. Then

[A g^{+}] = 2 s_{1}

[C r O_{4}^{2 -}] = s_{1}

K_{s p} = (2 s_{1})^{2} (s_{1}) = 4 s_{1}^{3}

Given

K_{s p} = 32 x

4 s_{1}^{3} = 32 x ⟹ s_{1}^{3} = 8 x ⟹ s_{1} = 2 x^{1 / 3}

For

A g B r

A g B r (s) ⇌ A g^{+} + B r^{-}

Let molar solubility be

s_{2}

. Then

[A g^{+}] = [B r^{-}] = s_{2}

K_{s p} = s_{2}^{2}

Given

K_{s p} = 4 y

s_{2}^{2} = 4 y ⟹ s_{2} = 2 y^{1 / 2}

Now the ratio:

\frac{molarity of A g_{2} C r O_{4}}{molarity of A g B r} = \frac{s_{1}}{s_{2}} = \frac{2 x^{1 / 3}}{2 y^{1 / 2}} = \frac{x^{1 / 3}}{y^{1 / 2}}

\frac{\sqrt[3]{x}}{\sqrt{y}}

**Answer: D**

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 easy

An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would

(

Given

\log 2 = 0.30)

increase to 1.3
reduce to 0.5
increase to 2
remain same

Answer: A

Explanation. Initial pH = 1.0 means

[H^{+}] = 10^{- 1} M

. Diluting with equal volume halves the concentration:

[H^{+}]_{new} = \frac{10^{- 1}}{2} = 5 \times 10^{- 2} M

. New pH =

- \log (5 \times 10^{- 2}) = - \log 5 + 2

. Since

\log 5 = \log (10 / 2) = 1 - \log 2 = 1 - 0.30 = 0.70

, pH =

2 - 0.70 = 1.30

. Thus pH increases to 1.3. The correct answer is option A.

From NTA/IIT papers: JEE Main 2025 · 7 Apr Shift 1

Question 3 easy

The equilibrium constant for the reaction

{S O}_{3} (g) ⇌ {S O}_{2} (g) + \frac{1}{2} O_{2} (g)

K_{c} = 4.9 \times 10^{- 2}

. The value of

K_{c}

for the reaction given below is

2 {S O}_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)

is :

49
416
41.6
4.9

Answer: B

Explanation. For the reaction

{S O}_{3} ⇌ {S O}_{2} + \frac{1}{2} O_{2}

K_{c} = 4.9 \times 10^{- 2}

. Reverse it:

{S O}_{2} + \frac{1}{2} O_{2} ⇌ {S O}_{3}

K_{c}^{'} = \frac{1}{4.9 \times 10^{- 2}} = \frac{1}{0.049} \approx 20.408

. Multiply by 2:

2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}

K_{c}^{''} = (K_{c}^{'})^{2} = (20.408)^{2} \approx 416.5

. Thus,

K_{c} \approx 416

. The answer is option B.

From NTA/IIT papers: JEE Main 2024 · 4 Apr Shift 2

Question 4 hard

On decreasing the

p H

from

7

2

, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from

10^{- 4} m o l L^{- 1}

10^{- 3} m o l L^{- 1}

. The

p K_{a}

of HX is

Answer: B

Explanation. For the sparingly soluble salt

M X

, the dissolution equilibrium is

M X (s) ⇌ M^{+} + X^{-}

with solubility product

K_{s p} = [M^{+}] [X^{-}]

Since

X^{-}

is the conjugate base of the weak acid

H X

, it reacts with

H^{+}

X^{-} + H^{+} ⇌ H X

At lower pH, more

H^{+}

is available, so more

X^{-}

is converted into

H X

. This decreases free

[X^{-}]

, shifting the dissolution equilibrium forward and increasing solubility. Let the total dissolved salt concentration be

S

. Then

[M^{+}] = S

and the total anion concentration is

S = [X^{-}] + [H X]

Using the acid dissociation constant,

K_{a} = \frac{[H^{+}] [X^{-}]}{[H X]}

we get

S = [X^{-}] (1 + \frac{[H^{+}]}{K_{a}})

Also,

K_{s p} = [M^{+}] [X^{-}] = S [X^{-}]

[X^{-}] = \frac{K_{s p}}{S}

Thus,

S = \frac{K_{s p}}{S} (1 + \frac{[H^{+}]}{K_{a}})

S^{2} = K_{s p} (1 + \frac{[H^{+}]}{K_{a}})

At pH

7

[H^{+}] = 10^{- 7}

, and

S_{1} = 10^{- 4}

(10^{- 4})^{2} = K_{s p} (1 + \frac{10^{- 7}}{K_{a}})

At pH

2

[H^{+}] = 10^{- 2}

, and

S_{2} = 10^{- 3}

(10^{- 3})^{2} = K_{s p} (1 + \frac{10^{- 2}}{K_{a}})

Dividing the two equations:

\frac{10^{- 6}}{10^{- 8}} = \frac{1 + \frac{10^{- 2}}{K_{a}}}{1 + \frac{10^{- 7}}{K_{a}}}

100 = \frac{1 + \frac{10^{- 2}}{K_{a}}}{1 + \frac{10^{- 7}}{K_{a}}}

Since

10^{- 7} / K_{a} ≪ 1

, approximate the denominator as

1

. Then

100 \approx 1 + \frac{10^{- 2}}{K_{a}}

\frac{10^{- 2}}{K_{a}} \approx 99

K_{a} \approx \frac{10^{- 2}}{99} \approx 10^{- 4}

Therefore,

p K_{a} = - \log K_{a} \approx 4

Answer: (B)

From NTA/IIT papers: JEE Advanced 2023

Frequently asked questions

How much of JEE Main comes from Equilibrium?: JEEGenie estimates about 5% of JEE Main marks come from Equilibrium, based on recent papers. The share varies from year to year.
Which formulas matter most in Equilibrium?: The Equilibrium sheet on this page has 17 formula entries, starting with Law of Mass Action, Equilibrium Constant K_c and K_p, Reaction Quotient Q, Effect of Temperature on K, Bronsted-Lowry Acid and Base. Each one comes with when to use it and the trap to avoid.
What topics does Equilibrium cover for JEE?: Equilibrium covers Law of Mass Action, Le Chatelier's Principle, Ionic Equilibrium, pH and Buffers, Solubility Product, Acids and Bases.
How should I revise Equilibrium before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Equilibrium questions on JEEGenie.
How many previous-year questions are there on Equilibrium?: JEEGenie has 489 previous-year questions on Equilibrium, and this page shows a few with worked explanations.
Is practice on Equilibrium free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.