JEE Mains Chemistry · Equilibrium

Equilibrium Composition from ICE Tables

Tracking initial, change and equilibrium amounts to find K or the equilibrium mixture, in concentrations or in partial pressures, with Q deciding the direction and inert gases counted only in the total.

All Equilibrium notes JEE Mains Chemistry strategy

Why this matters

Fourteen PYQs, nine of them numerical, and two from 2026. Six track amounts through an ICE table in moles or concentration; eight work in pressures, often with an inert gas in the flask or a solid that adds no pressure. The table is the same in both.

Concept 1 of 2: ICE tables in moles and concentration

Write three rows under the equation: Initial, Change, Equilibrium. The change row follows the coefficients, so one unknown x fixes every amount. Put the equilibrium row into K and solve for x. If the mixture already holds products, compare Q with K first to see which way it moves.

Definition

Work in moles; divide by the volume to get concentrations for $K_c$ .
The change row is x times each coefficient: minus for the side that is used up, plus for the side that forms.
If $\Delta n=0$ , the volume cancels and moles can go straight into K.
$Q$ has the same form as K but uses the amounts at any moment. $Q<K$ : forward. $Q>K$ : backward. $Q=K$ : at equilibrium.
A quadratic has two roots. Keep the one that leaves every amount positive.

Reaction quotient

Q=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\ \text{(any moment)},\qquad Q<K\Rightarrow\text{forward}

Worked example

2 mol each of A and B are placed in a 5 L flask. For

\mathrm{A+B\rightleftharpoons C+D}

K_c=9

. Find the equilibrium concentration of C.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q32Moderate

Example 1 · Equilibrium · Equilibrium Composition from ICE Tables

One mole each of He and

A(g)

are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.

A(g) \rightleftharpoons B(g)

K_{C}

for this reaction at 400 K is 4.0. The partial pressures (in atm ) of He and

B(g)

are respectively (at equilibrium). (Assume

He,A(g)

and

B(g)

behave as ideal gases) (Given :

R = 0.082\text{ }L\text{ }atm{\text{ }K}^{- 1}{\text{ }mol}^{- 1}

)

The change row follows the coefficients

\mathrm{H_2+I_2\rightleftharpoons 2HI}

, forming 3 mol of HI uses 1.5 mol each of

\mathrm{H_2}

and

\mathrm{I_2}

, not 3 mol. Write x in front of each coefficient before filling the row.

Check the direction before you write the table

If all four species of

\mathrm{A+B\rightleftharpoons C+D}

start at 1 M and

K=100

, then

Q=1<K

, so C and D grow. A table that lets them shrink gives a root that leaves a negative amount.

Drill 5 more on ice tables in moles and concentration

Concept 2 of 2: ICE tables in partial pressures

At a fixed volume and temperature, pressure is proportional to moles. So an ICE table can be written directly in atmospheres, and the total pressure tells you x. When only the total pressure is given, the gas law turns it into total moles. An inert gas adds to the total moles and pressure but never appears in K.

Definition

Partial pressure: $p_i=\frac{n_iRT}{V}=x_iP$ , where $x_i$ is the mole fraction and P the total pressure.
Total moles at equilibrium: $n=\frac{PV}{RT}$ .
At fixed V and T, write the ICE table in pressures and use the total to find x.
An inert gas (He, Ar, $\mathrm{N_2}$ when it does not react) counts in the total, not in K.
A solid adds nothing to the pressure: in $\mathrm{C(s)+CO_2\rightleftharpoons 2CO}$ , the total is $p_0+x$ .

Partial pressure

p_i=\frac{n_iRT}{V}=x_i\,P

Worked example

A sealed flask holds

\mathrm{N_2O_4}

at 0.6 atm. It reaches

\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)}

at constant volume and temperature, and the total pressure becomes 0.9 atm. Find

K_p

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q37Moderate

Example 2 · Equilibrium · Equilibrium Composition from ICE Tables

A vessel at 1000 K contains

CO_{2}

with a pressure of 0.5 atm . Some of

CO_{2}

is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then

K_{P}

is :

The inert gas is in the total, not in K

Subtract the inert gas's moles before you find x, and leave it out of K. Using the total pressure as if it came only from the reacting gases gives the wrong x.

Carbon adds nothing to the pressure

\mathrm{C(s)+CO_2\rightleftharpoons 2CO}

, x of

\mathrm{CO_2}

gives 2x of CO, so the total rises by x, not by 2x.

Drill 7 more on ice tables in partial pressures

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 (2)

ICE tables in moles and concentration
Reaction quotient
$Q=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\ \text{(any moment)},\qquad Q<K\Rightarrow\text{forward}$
ICE tables in partial pressures
Partial pressure
$p_i=\frac{n_iRT}{V}=x_i\,P$

Watch out for (4)

The change row follows the coefficients
→ ICE tables in moles and concentration
Check the direction before you write the table
→ ICE tables in moles and concentration
The inert gas is in the total, not in K
→ ICE tables in partial pressures
Carbon adds nothing to the pressure
→ ICE tables in partial pressures

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 14 questions