JEE Mains Chemistry · Equilibrium

Equilibrium Constant and Its Forms

Writing Kc and Kp from a balanced equation, leaving out solids, building a new K by reversing, scaling or adding equations, and converting Kp to Kc through the change in gas moles.

All Equilibrium notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, ten of them numerical, and four from 2026. Eight write K from the balanced equation, often with a solid that drops out; six build a new K by reversing, scaling or adding equations; five convert between Kp and Kc. Three rules cover the page.

Concept 1 of 3: Writing K from the equation

At equilibrium the forward and reverse rates are equal, so no amount changes any more. The amounts are constant, not equal to each other. K is products over reactants, each raised to its coefficient. A pure solid or liquid has a fixed concentration, so it is left out, and only gases and dissolved species remain.

Definition

For $a\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}$ : $K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$ .
$K_p$ has the same shape with partial pressures of the gases.
Pure solids and liquids are left out (their activity is 1).
At equilibrium both rates are equal and every concentration is constant and non-zero.
A physical equilibrium (liquid and vapour, a saturated solution) needs a closed system at a fixed temperature.

Law of mass action

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

Worked example

Solid

\mathrm{NH_4Cl}

is heated in an evacuated flask:

\mathrm{NH_4Cl(s)}\rightleftharpoons\mathrm{NH_3(g)}+\mathrm{HCl(g)}

. The total pressure at equilibrium is 0.6 atm. Find

K_p

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q50Moderate

Example 1 · Equilibrium · Equilibrium Constant and Its Forms

Solid carbon, CaO and

{CaCO}_{3}

are mixed and allowed to attain equilibrium at T K .

{CaCO}_{3}(s) \rightleftharpoons CaO(s) + {CO}_{2}(g),\quad K_{p_{1}} = 0.08\text{ atm}

C(s) + {CO}_{2}(g) \rightleftharpoons 2CO(g),\quad K_{p_{2}} = 2\text{ atm}

The partial pressure of CO is

\_\_\_\_

\times 10^{- 1}\text{ }atm

A solid left in the expression

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

K_p=p_{\mathrm{CO}}^2/p_{\mathrm{CO_2}}

. Carbon does not appear. The same holds for

\mathrm{CaCO_3}

\mathrm{CaO}

, metals and liquid water in a reaction mixture.

Equal rates, not equal amounts

A concentration-time plot reaches equilibrium when every curve goes flat. The curves do not have to meet. A plot where the reactant falls to zero shows a reaction that went to completion, not an equilibrium.

Drill 7 more on writing k from the equation

Concept 2 of 3: Reversing, scaling and adding equations

K follows the equation exactly as written. Turn the equation round and K turns upside down. Multiply every coefficient by n and every term in K is raised to the power n, so K becomes Kⁿ. Add two equations and their K values multiply, because the terms of one expression multiply the terms of the other.

Definition

Reverse the equation: $K'=\frac{1}{K}$ .
Multiply every coefficient by $n$ : $K'=K^n$ . Halving gives $\sqrt K$ ; dividing by 3 gives $K^{1/3}$ .
Add two equations: $K=K_1K_2$ .
Subtract one equation from another: $K=\frac{K_1}{K_2}$ .

Combining equilibria

K_{\text{reverse}}=\frac1K,\qquad K_{n\times}=K^n,\qquad K_{1+2}=K_1K_2

Worked example

\mathrm{A}\rightleftharpoons 2\mathrm{B}

has

K_1=16

, and

\mathrm{B}\rightleftharpoons\mathrm{C}

has

K_2=0.5

. Find K for

\mathrm{C}\rightleftharpoons\tfrac12\mathrm{A}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q30Moderate

Example 2 · Equilibrium · Equilibrium Constant and Its Forms

Consider the following reactions in which all the reactants and products are present in gaseous state

2xy \rightleftharpoons x_{2}+y_{2}{\text{ }K}_{1}= 2.5 \times10^{5}

xy +\frac{1}{2}z_{2}\rightleftharpoons xyz\ K_{2}= 5 \times10^{- 3}

The value of

K_{3}

for the equilibrium

\frac{1}{2}x_{2}+\frac{1}{2}y_{2}+\frac{1}{2}z_{2}\rightleftharpoons xyz

is :

Added equations multiply their K

For

\mathrm{X\rightleftharpoons Y}

\mathrm{Y\rightleftharpoons Z}

and

\mathrm{Z\rightleftharpoons W}

with

K=1,2,4

, the K for

\mathrm{X\rightleftharpoons W}

1\times2\times4=8

, not the sum 7.

Scaling is a power, not a factor

Dividing every coefficient by 3 turns K into

K^{1/3}

, not

K/3

. For

K=2.7\times10^{-5}

that is

3\times10^{-2}

. Cubing it, or taking a square root, are the offered wrong answers.

Drill 5 more on reversing, scaling and adding equations

Concept 3 of 3: Kp and Kc through the change in gas moles

A partial pressure is a concentration times RT, because p = (n/V)RT. So each gas term in Kp carries one factor of RT. The factors cancel between top and bottom except for the extra gas moles, which is why only Δn, the gas moles of products minus the gas moles of reactants, appears.

Definition

$K_p=K_c(RT)^{\Delta n}$ , where $\Delta n$ = gas moles of products − gas moles of reactants.
Count gases only. Solids and liquids add nothing to $\Delta n$ .
With $K_p$ in atm, use $R=0.0821\ \mathrm{L\,atm\,K^{-1}mol^{-1}}$ .
$\Delta n=0$ : $K_p=K_c$ , and neither R nor T matters.
A fractional $\Delta n$ gives a root: $\Delta n=-\tfrac12$ makes $K_p/K_c=1/\sqrt{RT}$ .

Kp and Kc

K_p=K_c\,(RT)^{\Delta n}

Worked example

For

\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}

K_c=0.04

at 500 K. Find

K_p

. (

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

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q45Moderate

Example 3 · Equilibrium · Equilibrium Constant and Its Forms

Consider the general reaction given below at 400 K

xA(g) \rightleftharpoons yB(g)

The values of

K_{p}

and

K_{c}

are studied under the same condition of temperature but variation in x and y . (I)

K_{p}= 85.87

and

K_{c}= 2.586

appropriate units (II)

K_{p}= 0.862

and

K_{c}= 28.62

appropriate units The value of

x

and

y

in (I) and (II) respectively are:

The sign of Δn

For

\mathrm{CO+\tfrac12O_2\rightleftharpoons CO_2}

\Delta n=1-\tfrac32=-\tfrac12

, so

K_p/K_c=1/\sqrt{RT}

. Writing reactants minus products gives

\sqrt{RT}

, which is always offered.

Reading Δn from a ratio

At 400 K,

RT\approx32.8

. If

K_p/K_c\approx33

, then

\Delta n=+1

; if it is about

1/33

\Delta n=-1

. Work out the ratio first, then match coefficients.

Drill 4 more on kp and kc through the change in gas moles

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 (3)

Writing K from the equation
Law of mass action
$K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$
Reversing, scaling and adding equations
Combining equilibria
$K_{\text{reverse}}=\frac1K,\qquad K_{n\times}=K^n,\qquad K_{1+2}=K_1K_2$
Kp and Kc through the change in gas moles
Kp and Kc
$K_p=K_c\,(RT)^{\Delta n}$

Watch out for (6)

A solid left in the expression
→ Writing K from the equation
Equal rates, not equal amounts
→ Writing K from the equation
Added equations multiply their K
→ Reversing, scaling and adding equations
Scaling is a power, not a factor
→ Reversing, scaling and adding equations
The sign of Δn
→ Kp and Kc through the change in gas moles
Reading Δn from a ratio
→ Kp and Kc through the change in gas moles

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.

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