JEE Mains Chemistry · Equilibrium

Le Chatelier's Principle

Predicting which way an equilibrium moves when concentration, pressure, temperature, an inert gas or a catalyst changes, and computing the new equilibrium after a species is added.

All Equilibrium notes JEE Mains Chemistry strategy

Why this matters

Eleven PYQs, eight of them multiple choice, and three from 2026. Seven ask which way an equilibrium moves when pressure, temperature, an inert gas or a catalyst changes; four add a species and ask for the new amounts.

Concept 1 of 2: Which way the equilibrium shifts

A system at equilibrium that is disturbed moves in the direction that partly undoes the change. Add a reactant and some of it is used up; squeeze the gas and it moves to fewer gas moles; heat it and it moves in the direction that absorbs heat. Only a change of temperature changes K itself.

Definition

Concentration and pressure change the position, never K.
Temperature changes K: an exothermic forward reaction has a smaller K when hot.
A catalyst speeds both directions equally. It changes neither K nor the equilibrium mixture; equilibrium is only reached sooner.
Pure solids and liquids are not in K, so adding more of one does not disturb the equilibrium.
An inert gas matters only if it changes the partial pressures, which happens at constant pressure, not at constant volume.

Change	Which way it shifts	Effect on K
Add a reactant gas or solute	Forward, using up some of it	None
Add a pure solid or liquid	No shift	None Adding $\mathrm{Fe_2O_3}$ to the blast-furnace equilibrium changes nothing.
Raise the pressure by compressing	Toward fewer gas moles	None
Compress when $\Delta n=0$ , as in $\mathrm{H_2+I_2\rightleftharpoons 2HI}$	No shift	None
Add an inert gas at constant volume	No shift	None
Add an inert gas at constant pressure	Toward more gas moles	None The volume grows, so every partial pressure falls, like a pressure drop.
Heat an exothermic forward reaction	Backward	K falls
Heat an endothermic forward reaction	Forward	K rises
Add a catalyst	No shift; equilibrium comes sooner	None
Raise the pressure on ice and water at 0 °C	Toward water, which takes less volume	Melting point falls

Only temperature changes K. Every other change moves the mixture while K stays fixed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q106Moderate

Example 1 · Equilibrium · Le Chatelier's Principle

Consider the equilibrium

CO(\text{ }g) + 3H_{2}(\text{ }g) \rightleftharpoons CH_{4}(\text{ }g) +H_{2}O(\text{ }g)

If the pressure applied over the system increases by two fold at constant temperature then (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below :

Pressure moves the mixture, not K

Compressing

\mathrm{CO+3H_2\rightleftharpoons CH_4+H_2O}

raises every concentration and shifts it forward, but K is unchanged. "K increases because products increase" is the offered wrong statement.

Read how the inert gas is added

At constant volume the partial pressures do not change, so nothing moves. At constant pressure the volume grows and the reaction shifts toward more gas moles. If the question does not say, the standard answer assumes constant volume.

Drill 6 more on which way the equilibrium shifts

Concept 2 of 2: Finding the new equilibrium

Le Chatelier tells you the direction; K tells you how far. K does not change when a species is added at the same temperature. So find K from the old mixture, add the new amount, and solve an ICE table that starts from the disturbed mixture.

Definition

Step 1: K from the old equilibrium amounts.
Step 2: add the species and compute Q. $Q>K$ : backward; $Q<K$ : forward.
Step 3: ICE table from the disturbed mixture, solved with the same K.
If $\Delta n=0$ , moles can be used in place of concentrations.
The new mixture only partly undoes the change: an added species ends above its old value.

Same K before and after

K=\frac{[\mathrm{B}]_{\text{old}}}{[\mathrm{A}]_{\text{old}}}=\frac{[\mathrm{B}]_{\text{new}}}{[\mathrm{A}]_{\text{new}}}\quad(\mathrm{A\rightleftharpoons B},\ \text{same }T)

Worked example

In a 1 L flask,

\mathrm{A(g)\rightleftharpoons B(g)}

is at equilibrium with

[\mathrm{A}]=0.4

M and

[\mathrm{B}]=0.8

M. Then 0.2 mol of B is added. Find the new concentrations.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q29Moderate

Example 2 · Equilibrium · Le Chatelier's Principle

Observe the following equilibrium in a 1 L flask.

A(g) \rightleftharpoons B(g)

T(K)

, the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A are added into the flask and heated to

T(K)

to establish the equilibrium again. The new equilibrium concentrations (in

M

) of

A

and

B

are respectively.

Start from the disturbed mixture

The Initial row of the new table holds the amounts just after the addition, not the original starting amounts and not the old equilibrium without the addition.

The change is only partly undone

An added species ends above its old equilibrium value. In the example, B was 0.8 M, rose to 1.0 M on adding, and settles at 0.933 M. An answer that returns B to 0.8 M or below is wrong.

Drill 3 more on finding the new equilibrium

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Finding the new equilibrium
Same K before and after
$K=\frac{[\mathrm{B}]_{\text{old}}}{[\mathrm{A}]_{\text{old}}}=\frac{[\mathrm{B}]_{\text{new}}}{[\mathrm{A}]_{\text{new}}}\quad(\mathrm{A\rightleftharpoons B},\ \text{same }T)$

Reference tables (1)

Which way the equilibrium shifts10 rows

Change	Which way it shifts	Effect on K
Add a reactant gas or solute	Forward, using up some of it	None
Add a pure solid or liquid	No shift	None Adding $\mathrm{Fe_2O_3}$ to the blast-furnace equilibrium changes nothing.
Raise the pressure by compressing	Toward fewer gas moles	None
Compress when $\Delta n=0$ , as in $\mathrm{H_2+I_2\rightleftharpoons 2HI}$	No shift	None
Add an inert gas at constant volume	No shift	None
Add an inert gas at constant pressure	Toward more gas moles	None The volume grows, so every partial pressure falls, like a pressure drop.
Heat an exothermic forward reaction	Backward	K falls
Heat an endothermic forward reaction	Forward	K rises
Add a catalyst	No shift; equilibrium comes sooner	None
Raise the pressure on ice and water at 0 °C	Toward water, which takes less volume	Melting point falls

Only temperature changes K. Every other change moves the mixture while K stays fixed.

Watch out for (4)

Pressure moves the mixture, not K
→ Which way the equilibrium shifts
Read how the inert gas is added
→ Which way the equilibrium shifts
Start from the disturbed mixture
→ Finding the new equilibrium
The change is only partly undone
→ Finding the new equilibrium

Test yourself on Equilibrium

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Equilibrium chapter test Drill all 11 questions