Equilibrium Formula Sheet — JEE Main Chemistry

Q: What are the most important Equilibrium formulas for JEE Main?

This Equilibrium formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Meaning and concept of dynamic equilibrium, Physical equilibria: solid-liquid, liquid-gas, gas-gas, solid-gas, Henry's law, General characteristics of equilibria in physical processes, Chemical equilibria — each shown with the key result and, where useful, a worked example.

Q: Is this Equilibrium formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Equilibrium formulas?

Blurt the Equilibrium formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Equilibrium formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 24 sub-topics

Meaning and concept of dynamic equilibrium
Physical equilibria: solid-liquid, liquid-gas, gas-gas, solid-gas
Henry's law
General characteristics of equilibria in physical processes
Chemical equilibria
Law of chemical equilibrium
Equilibrium constants Kp and Kc and significance
Significance of ΔG and ΔG° in equilibrium
Factors affecting equilibrium
Le Chatelier's principle
Ionic equilibrium
Weak and strong electrolytes
Ionization of electrolytes
Concepts of acids and bases: Arrhenius, Bronsted-Lowry and Lewis
Ionization of acids and bases
Acid-base equilibria including multistage ionization
Ionization constants
Ionization of water
pH scale
Common ion effect
Hydrolysis of salts and pH
Solubility of sparingly soluble salts
Solubility products
Buffer solutions

Equilibrium & the Equilibrium Constant

Dynamic equilibrium

▸Reached in a $closed$ system when forward and reverse rates become equal — concentrations stay constant (but $not$ equal).
▸It is $dynamic$ : both reactions continue (confirmed by isotope/radioactive tracer studies).
▸Can be approached from either side (reactants or products).

Concentration versus time curves where reactants fall and products rise until both level off at equilibrium, with the law-of-mass-action expression for Kc — At equilibrium rates are equal and concentrations become constant.

Law of mass action (equilibrium constant): For

a A + b B ⇌ c C + d D

, the ratio

K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

(equilibrium concentrations) is constant at a given temperature.

Reverse reaction

K^{'} = \frac{1}{K}

Multiply coeffs by

n

K^{''} = K^{n}

Add reactions

K_{n e t} = K_{1} \times K_{2} \times \dots

Characteristics of

K

▸Depends only on $temperature$ — independent of initial concentrations or catalyst.
▸Has one value for a given balanced equation; always quote the equation with $K$ .
▸Says nothing about the $rate$ at which equilibrium is reached.

Equilibria in physical processes (syllabus)

▸ $Solid$ ⇌ $liquid$ : coexist only at the melting point for a given pressure (ice–water at 273 K, 1 atm); rates of melting and freezing are equal.
▸ $Liquid$ ⇌ $vapour$ : in a closed vessel the vapour pressure becomes constant at a given T; boiling point = T where it equals the external pressure.
▸ $Solid$ ⇌ $vapour$ : sublimation (iodine, camphor, naphthalene). $Gas$ in liquid: Henry's law $p = K_{H} x$ — solubility rises with pressure, falls with temperature.
▸Common features: closed system at fixed T; a measurable property becomes constant; the constant (e.g. $K = [CO_{2} (a q)] / [CO_{2} (g)]$ ) characterises the equilibrium.

🎯 Exam · Homogeneous vs heterogeneous

heterogeneous

equilibria, pure solids and pure liquids have constant 'concentration' and are

omitted

from K. E.g.

CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g)

gives

K_{c} = [C O_{2}]

K_{p} = p_{C O_{2}}

🚫 Examiner Trap · Examiner traps

(1) At equilibrium concentrations are CONSTANT, not equal. (2) K depends on

temperature only

— not on initial amounts or a catalyst. (3) Omit pure

solids/liquids

from K (

CaC O_{3} ⇌ CaO + C O_{2} \Rightarrow K_{c} = [C O_{2}]

). (4) Reverse

\Rightarrow K \to 1/ K

; coefficients

\times n \Rightarrow K \to K^{n}

; add reactions

\Rightarrow

MULTIPLY K.

Kp, Reaction Quotient & ΔG°

Kp – Kc relation

K_{p} = K_{c} (RT)^{Δ n_{g}}

Δ n_{g} =

(moles of gaseous products)

-

(moles of gaseous reactants).

K_{p} = K_{c}

when

Δ n_{g} = 0

. Use p in bar;

R = 0.0831

bar L

K^{- 1}

l^{- 1}

Reaction quotient (

Q

): Same form as

K

but with the concentrations (or pressures) at any instant, not necessarily at equilibrium. Comparing

Q

with

K

predicts the direction of net reaction.

Number lines showing Q less than K goes forward and Q greater than K goes reverse, and the magnitude of K showing reactant or product dominance — $Q$ vs $K$ gives direction; the size of $K$ gives the extent.

★ Remember · Direction & extent

Q < K

: net forward (

\to

Q > K

: net reverse (

\leftarrow

Q = K

: at equilibrium. Extent:

K > 1 0^{3}

products dominate;

K < 1 0^{- 3}

reactants dominate;

1 0^{- 3}

–

1 0^{3}

both appreciable.

Gibbs energy & equilibrium

Δ G = Δ G^{⊖} + RT ln Q; Δ G^{⊖} = - RT ln K

At equilibrium

Δ G = 0

and

Q = K

. So

K = e^{- Δ G^{⊖} / RT}

Δ G^{⊖} < 0 \Rightarrow K > 1

(products);

Δ G^{⊖} > 0 \Rightarrow K < 1

✎ Example · Find Kp from Kc

For

2NOCl (g) ⇌ 2NO (g) + C l_{2} (g)

K_{c} = 3.75 \times 1 0^{- 6}

at 1069 K. Find

K_{p}

$Δ n_{g} = (2 + 1) - 2 = 1$ .
$K_{p} = K_{c} (RT)^{1} = 3.75 \times 1 0^{- 6} \times (0.0831 \times 1069)$ .

∴

K_{p} \approx 0.033

🚫 Examiner Trap · Examiner traps

(1)

K_{p} = K_{c} (RT)^{Δ n_{g}}

uses

Δ n_{g}

gas

moles only;

K_{p} = K_{c}

Δ n_{g} = 0

. (2)

Q < K \Rightarrow

forward,

Q > K \Rightarrow

reverse. (3)

Δ G^{⊖} = - RT ln K

uses STANDARD

Δ G

;

Δ G = Δ G^{⊖} + RT ln Q

gives the actual direction. (4) Use

R = 0.0831

bar L/mol K and p in bar for

K_{p}

Le Chatelier's Principle

Le Chatelier's principle: If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the system shifts so as to partially counteract (reduce) that change. Applies to all physical and chemical equilibria.

Effect of each factor

▸ $Concentration$ : adding a reactant (or removing a product) shifts forward; the reverse shifts back.
▸ $Pressure$ ( $↑$ by compression): shifts toward the side with $fewer moles of gas$ . No effect if $Δ n_{g} = 0$ .
▸ $Inert gas$ at constant V: no change in partial pressures $\Rightarrow$ no shift.

★ Remember · Temperature & catalyst

Temperature

changes the value of K: heating an

exothermic

reaction lowers K (shifts back); heating an

endothermic

one raises K. A

catalyst

speeds both rates equally — it does

not

shift equilibrium, only reaches it faster.

🎯 Exam · Industrial optimisation

Haber

(

N_{2} + 3 H_{2} ⇌ 2N H_{3}

, exo,

Δ n_{g} < 0

): favoured by

high pressure

and low T, but a compromise

\sim

700 K with an Fe catalyst is used.

Contact

(

2S O_{2} + O_{2} ⇌ 2S O_{3}

V_{2} O_{5}

catalyst.

✎ Example · Predict the shift

For

CO (g) + 3 H_{2} (g) ⇌ C H_{4} (g) + H_{2} O (g)

, what does compression do?

Left side: $4$ mol gas; right side: $2$ mol gas.
Compression (↑ pressure) favours fewer gas moles.

∴ Equilibrium shifts

\to

forward (toward

C H_{4}

🚫 Examiner Trap · Examiner traps

(1) Pressure shift is toward

fewer gas moles

— no effect if

Δ n_{g} = 0

. (2) Adding an

inert gas at constant V

does NOT shift (partial pressures unchanged); at constant P it does. (3) A

catalyst

never shifts equilibrium — only speeds attainment. (4) Only TEMPERATURE changes the value of K (heating endothermic raises K).

Acids & Bases — Concepts

Three definitions

▸ $Arrhenius$ : acid gives $H^{+}$ , base gives $O H^{-}$ in water (limited to aqueous).
▸ $Brønsted-Lowry$ : acid = proton ( $H^{+}$ ) donor, base = proton acceptor.
▸ $Lewis$ : acid = electron-pair acceptor (e.g. $B F_{3}$ , $AlC l_{3}$ ), base = electron-pair donor (e.g. $N H_{3}$ , $O H^{-}$ ).

HCl plus water giving hydronium and chloride, with the conjugate acid-base pairs labelled, plus the three acid-base concept definitions — A conjugate pair differs by exactly one $H^{+}$ .

Conjugate acid–base pair: Two species differing by one proton. When an acid loses

H^{+}

it becomes its conjugate base; when a base gains

H^{+}

it becomes its conjugate acid. A strong acid has a weak conjugate base, and vice versa.

Reading conjugate pairs

▸ $HCl + H_{2} O ⇌ H_{3} O^{+} + C l^{-}$ : pairs are $HCl / C l^{-}$ and $H_{2} O / H_{3} O^{+}$ .
▸ $Amphoteric$ species act as both acid and base, e.g. $H_{2} O$ , $HC O_{3}^{-}$ , $HS O_{4}^{-}$ .
▸Conjugate base of $H_{2} S O_{4}$ is $HS O_{4}^{-}$ ; conjugate acid of $N H_{3}$ is $N H_{4}^{+}$ .

🎯 Exam · Strength trends

Down a group, H–A bond weakens

\Rightarrow

acidity

↑

HF < HCl < HBr < HI

. Across a period, electronegativity

↑

\Rightarrow

acidity

↑

C H_{4} < N H_{3} < H_{2} O < HF

🚫 Examiner Trap · Examiner traps

(1) A

conjugate pair

differs by exactly one H

^{+}

. (2) A STRONG acid has a WEAK conjugate base (and vice versa). (3) Lewis is the broadest: B

F_{3}

/AlC

l_{3}

are acids with no H at all. (4) Amphoteric species (

H_{2}

O, HCO

_{3}^{-}

, HSO

_{4}^{-}

) act as both acid and base.

Ionization, Ka/Kb & pH

Weak acid

K_{a} = \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ]}

Weak base

K_{b} = \frac{[ B H ^{+} ] [ O H ^{-} ]}{[ B ]}

Ostwald dilution

K_{a} = \frac{c α ^{2}}{1 - α} \approx c α^{2}

Ionic product of water

K_{w} = [H_{3} O^{+}] [O H^{-}] = 1.0 \times 1 0^{- 14} (298 K)

Pure water:

[H^{+}] = [O H^{-}] = 1 0^{- 7}

K_{w}

increases with temperature (ionization is endothermic).

A coloured pH scale from 0 to 14 with common substances marked, acidic below 7 and basic above 7, and the relations pH equals minus log H plus and pH plus pOH equals 14 — Each pH unit is a 10-fold change in $[H^{+}]$ .

pH and related quantities

pH = - lo g [H^{+}]; pH + pOH = 14; p K_{a} = - lo g K_{a}

Acidic: pH

< 7

(

[H^{+}] > 1 0^{- 7}

). Neutral: pH

= 7

. Basic: pH

> 7

pH recipes

▸ $Strong$ electrolytes ionise fully; $weak$ ones partly (equilibrium, degree $α$ ).
▸Weak acid, concentration c: $α = K_{a} / c$ , $[H^{+}] = K_{a} c$ , so $pH = \frac{1}{2} (p K_{a} - lo g c)$ (valid when $α ≪ 1$ ).
▸Very dilute strong acid (e.g. $1 0^{- 8}$ M HCl): add water's contribution — solve $x (x + 1 0^{- 8}) = 1 0^{- 14}$ , so $[H^{+}] \approx 1.05 \times 1 0^{- 7}$ and pH $\approx 6.98$ (NOT 8).
▸Multistage (polyprotic) acids: $K_{a 1} ≫ K_{a 2} ≫ K_{a 3}$ (harder to remove H $^{+}$ from an anion); pH comes almost entirely from the first step, and $[A^{2 -}] \approx K_{a 2}$ for $H_{2}$ A alone.

★ Remember · Conjugate pair link

For a conjugate acid–base pair:

K_{a} \times K_{b} = K_{w}

and

p K_{a} + p K_{b} = 14

(298 K). A larger

K_{a}

(smaller

p K_{a}

) means a stronger acid.

✎ Example · pH of a soft drink

[H^{+}] = 3.8 \times 1 0^{- 3}

M. Find the pH.

$pH = - lo g (3.8 \times 1 0^{- 3})$ .
$= - (lo g 3.8 - 3) = - (0.58 - 3)$ .

∴

pH = 2.42

(acidic).

🚫 Examiner Trap · Examiner traps

(1)

K_{w} = 1 0^{- 14}

at 298 K only — it RISES with temperature (neutral pH

< 7

at high T, but still neutral). (2) Each pH unit is a

10-fold

change in [H

^{+}

]. (3) For a conjugate pair

K_{a} \times K_{b} = K_{w}

p K_{a} + p K_{b} = 14

. (4) Ostwald

K_{a} = c α^{2} / (1 - α)

; for very dilute,

α

rises (dilution law).

Buffers & Salt Hydrolysis

Buffer solution: A solution that resists changes in pH on dilution or on adding small amounts of acid or alkali. Made from a weak acid + its salt (acidic buffer) or a weak base + its salt (basic buffer).

Henderson–Hasselbalch equation

pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

Maximum buffer capacity when

[salt] = [acid]

, i.e.

pH = p K_{a}

. Acetate buffer

\approx 4.74

; ammonia/ammonium buffer

\approx 9.25

. Basic buffer:

pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}

pH of hydrolysing salts (concentration

c

)

WA+SB: pH = 7 + \frac{1}{2} (p K_{a} + lo g c); SA+WB: pH = 7 - \frac{1}{2} (p K_{b} + lo g c)

hydrolysis constant

K_{h} = \frac{K _{w}}{K _{a}}

(or

\frac{K _{w}}{K _{b}}

), degree of hydrolysis

h = K_{h} / c

; for WA+WB salts

K_{h} = \frac{K _{w}}{K _{a} K _{b}}

and pH is independent of c

💡 Tip · Common ion effect

Adding an ion already present in an equilibrium

suppresses

ionization (Le Chatelier). E.g. adding

C H_{3} COONa

to acetic acid lowers

[H^{+}]

— the basis of buffer action and of controlled precipitation.

Salt of	Nature	pH
strong acid + strong base	neutral	$= 7$
weak acid + strong base	basic	$> 7$
strong acid + weak base	acidic	$< 7$
weak acid + weak base	$p K$ -decided	$7 + \frac{1}{2} (p K_{a} - p K_{b})$

🎯 Exam · Why salts hydrolyse

Ions from a

weak

acid/base react with water to regenerate it, releasing

O H^{-}

H^{+}

. Ions from strong acids/bases (

N a^{+}

C l^{-}

) do not hydrolyse, so

NaCl

is neutral.

🚫 Examiner Trap · Examiner traps

(1) Henderson pH

= p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

; max buffer capacity when salt

=

acid (pH

= p K_{a}

). (2) Salt hydrolysis: weak-acid+strong-base

\Rightarrow

basic; strong-acid+weak-base

\Rightarrow

acidic; strong+strong

\Rightarrow

neutral. (3) Common ion SUPPRESSES ionisation (Le Chatelier). (4) Ions of strong acids/bases (Na

^{+}

, Cl

^{-}

) do NOT hydrolyse.

Solubility Equilibrium (Ksp)

Solubility product (

K_{s p}

): For a sparingly soluble salt in equilibrium with its saturated solution,

M_{x} X_{y} (s) ⇌ x M^{y +} + y X^{x -}

, the constant

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

. The solid does not appear in the expression.

Saturated solution with undissolved solid and dissolved ions giving Ksp equals S squared, plus common-ion effect and Qsp versus Ksp precipitation rules — $K_{s p}$ fixes the maximum ion product a saturated solution can hold.

MX

(AgCl)

K_{s p} = S^{2}

M X_{2}

(Ca

F_{2}

)

K_{s p} = 4 S^{3}

M_{x} X_{y}

K_{s p} = x^{x} y^{y} S^{(x + y)}

★ Remember · Qsp vs Ksp — will it precipitate?

Compute the ion product

Q_{s p}

Q_{s p} < K_{s p}

unsaturated (more dissolves);

Q_{s p} = K_{s p}

saturated;

Q_{s p} > K_{s p}

supersaturated

\Rightarrow

a precipitate forms.

💡 Tip · Common ion effect on solubility

Adding a common ion (e.g.

C l^{-}

to an

AgCl

solution)

lowers

solubility — used to precipitate ions almost completely (gravimetric analysis) and to purify salts. With C mol/L of the common ion: MX gives

S = \frac{K _{s p}}{C}

; M

X_{2}

in C of X

^{-}

gives

S = \frac{K _{s p}}{C ^{2}}

. When two ions compete, the salt whose

Q_{s p}

first exceeds

K_{s p}

precipitates first (selective precipitation).

Indicators (acid–base titrations)

▸An indicator HIn changes colour over $pH = p K_{I n} \pm 1$ .
▸Phenolphthalein 8.2–10 (colourless $\to$ pink): strong acid–strong base, weak acid–strong base. Methyl orange 3.1–4.4 (red $\to$ yellow): strong acid–strong base, strong acid–weak base.
▸Weak acid vs weak base has no sharp end point — no suitable indicator.

✎ Example · Solubility of AgCl

K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}

. Find its molar solubility S.

$AgCl ⇌ A g^{+} + C l^{-}$ , so $K_{s p} = S^{2}$ .
$S = 1.8 \times 1 0^{- 10}$ .

∴

S \approx 1.3 \times 1 0^{- 5}

mol

L^{- 1}

🚫 Examiner Trap · Examiner traps

(1) The solid is OMITTED from

K_{s p}

. (2) Watch stoichiometry: MX

K_{s p} = S^{2}

, M

X_{2}

K_{s p} = 4 S^{3}

M_{x} X_{y}

= x^{x} y^{y} S^{x + y}

. (3)

Q_{s p} > K_{s p} \Rightarrow

precipitate;

<\Rightarrow

dissolves more. (4) A common ion

lowers

solubility (precipitation/purification), but don't confuse S with

K_{s p}

More JEE Main Chemistry formula sheets

Frequently Asked Questions

What are the most important Equilibrium formulas for JEE Main?

This Equilibrium formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Meaning and concept of dynamic equilibrium, Physical equilibria: solid-liquid, liquid-gas, gas-gas, solid-gas, Henry's law, General characteristics of equilibria in physical processes, Chemical equilibria — each shown with the key result and, where useful, a worked example.

Is this Equilibrium formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Equilibrium formulas?

Blurt the Equilibrium formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheetsJEE Main previous-year papersmost important chapters.