Law of Chemical Equilibrium And Equilibrium Constant

Q: What is the difference between physical and chemical equilibrium?

Physical equilibrium involves a change of state without any change in chemical identity, such as ice melting to water or water evaporating into vapour. Chemical equilibrium involves a reversible chemical reaction where reactant molecules and product molecules interconvert. Both types share the feature that rates of the forward and reverse processes become equal.

Q: Why is chemical equilibrium called dynamic?

At equilibrium the forward and reverse reactions do not stop; they continue at the molecular level. Reactant molecules keep forming products and product molecules keep re-forming reactants, but at equal rates. The net observable change is zero but activity at the microscopic level continues, which is why it is called dynamic rather than static.

Q: When are Kp and Kc equal?

Kp equals Kc when the change in the total number of moles of gases is zero, meaning the total moles of gaseous reactants equals the total moles of gaseous products. Examples include H2 + I2 to 2HI and N2 + O2 to 2NO. In these cases the factor RT raised to delta n becomes 1.

Q: Does a catalyst change the value of the equilibrium constant?

No. A catalyst lowers the activation energy of both the forward and reverse reactions equally, so it speeds up the approach to equilibrium without changing the position of equilibrium. The value of K depends only on temperature; catalyst presence, initial concentrations, and pressure changes do not affect it.

Q: What does a very large value of K signify?

A very large equilibrium constant (typically K greater than 10 to the power 3) means the reaction is strongly product-favoured. The forward reaction goes nearly to completion and the equilibrium mixture contains mostly products with negligible amounts of reactants. A large K does not tell us how fast equilibrium is reached; that is decided by the rate constants and activation energy.

Q: Why are pure solids and pure liquids excluded from the equilibrium constant expression?

The active mass of a pure solid or pure liquid is proportional to its density, which remains constant throughout the reaction. Since these constants can be absorbed into the equilibrium constant itself, their effective activities are conventionally taken as unity and they are omitted from the K expression. For example, Kc for CaCO3(s) giving CaO(s) plus CO2(g) is simply the concentration of CO2.

Q: How does the equilibrium constant change if the equation is reversed or multiplied?

Reversing the equation replaces K with 1/K. Multiplying the entire equation by a factor n raises K to the power n. Adding two equations gives an overall K equal to the product K1 times K2. These rules follow directly from the mathematical form of the K expression and are widely used to relate equilibrium constants of related reactions.

Q: Can the equilibrium constant have units?

In practice Kc carries units of mol per litre raised to delta n and Kp has units of atm raised to delta n. When delta n is zero K is dimensionless. Rigorously, K is defined using activities (dimensionless quantities) so the true thermodynamic K is always unitless. Both conventions are used in exams; follow the numerical convention consistent with the problem.

ChemistryEquilibriumFor JEE aspirants

The law of chemical equilibrium states that for a reversible reaction $a A + b B ⇌ c C + d D$ at a given temperature, the ratio $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ is a constant, called the equilibrium constant. This constant tells us the extent to which the reaction proceeds. A large $K_{c}$ means products dominate; a small $K_{c}$ means reactants dominate. The equilibrium constant depends only on temperature - it does not change with initial concentration, pressure, volume, or the presence of a catalyst.

Key Formulas - Quick Reference

Equilibrium constant (concentration): $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$
Equilibrium constant (partial pressure): $K_{p} = \frac{P _{C}^{c} P _{D}^{d}}{P _{A}^{a} P _{B}^{b}}$
Relation between $K_{p}$ and $K_{c}$ : $K_{p} = K_{c} (R T)^{Δ n_{g}}$ , where $Δ n_{g} = (c + d) - (a + b)$ counts only gaseous species
Reversed reaction: $K_{reverse} = \frac{1}{K _{forward}}$
Reaction multiplied by $n$ : $K^{'} = (K)^{n}$ (where $n$ can be a fraction)
Two reactions added: $K_{net} = K_{1} \times K_{2}$
Degree of dissociation from vapour density: $α = \frac{D - d}{( n - 1 ) d}$ , where $D$ = initial VD, $d$ = VD at equilibrium, $n$ = moles of products per mole of reactant

1. What is Chemical Equilibrium?

Consider a reversible reaction carried out in a closed vessel:

$A + B ⇌ C + D$

At $t = 0$ , only reactants are present. As $A$ and $B$ react, $C$ and $D$ start forming. The moment products appear, the reverse reaction also begins. Initially the forward rate is high (because $[A]$ and $[B]$ are high) and the reverse rate is zero. As time passes, the forward rate drops and the reverse rate rises. Eventually the two rates become equal - this is the state of chemical equilibrium.

Chemical equilibrium is the state of a reversible reaction in which the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of reactants and products remain constant with time.

Figure 1: Approach to chemical equilibrium. The forward rate

r_{f}

decreases while the reverse rate

r_{b}

increases until they meet at

r_{f} = r_{b}

, defining the equilibrium state.

Figure 2: Concentrations of reactants fall and products rise until both level off at their equilibrium values, indicating no further net change.

2. Reversible vs Irreversible Reactions

Feature	Reversible reaction	Irreversible reaction
Direction	Proceeds in both forward and reverse directions	Proceeds only in the forward direction
Notation	Double arrow ( $⇌$ )	Single arrow ( $\to$ )
Extent	Reaches equilibrium; never completes fully	Reactants almost completely convert to products
Condition	Closed vessel	Often in open vessel, or products escape
Example	$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$	$A g N O_{3} (a q) + N a C l (a q) \to A g C l (s) + N a N O_{3} (a q)$

A reaction that is reversible in a closed vessel may become effectively irreversible if carried out in an open one. For example, $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ reaches equilibrium in a sealed container, but in an open lime kiln the $C O_{2}$ escapes, driving the reaction to completion.

3. Types of Equilibria

Homogeneous equilibrium

All reactants and products are in the same phase. Examples:

$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ - all gaseous
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ - all gaseous (Haber's process)
$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$ - all liquids

Heterogeneous equilibrium

Reactants and products exist in more than one phase. Examples:

$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ - solid and gas
$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ - solid and gas
$3 F e (s) + 4 H_{2} O (g) ⇌ F e_{3} O_{4} (s) + 4 H_{2} (g)$ - solid and gas

Important convention: In writing the equilibrium constant expression, the concentrations (or partial pressures) of pure solids and pure liquids are taken as unity. This is because their densities, and hence their "active masses"

=

moles per litre

\propto

density, remain effectively constant during the reaction. So for

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

, we write

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

and

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

4. Characteristics of Chemical Equilibrium

Dynamic in nature: Both forward and reverse reactions continue at the molecular level; only the net change is zero.
Rate equality: Rate of forward reaction equals rate of reverse reaction.
Constant observable properties: Concentrations, pressure, colour, density, and refractive index remain constant with time.
Attainable from either side: The same equilibrium state is reached whether we start from pure reactants or pure products.
Catalyst effect: A catalyst speeds up the approach to equilibrium but does not shift the position of equilibrium.
Free energy change: At equilibrium, $Δ G = 0$ .
Requires a closed system: If any product escapes, equilibrium cannot be maintained.

Dynamic vs static equilibrium: A metal rod with a hot end and a cold end has a steady temperature profile - but heat flows only in one direction (hot to cold). This is a steady state, not a dynamic equilibrium. In chemical equilibrium, matter and energy flow equally in both directions.

5. Law of Mass Action

Proposed by Guldberg and Waage (1864): the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its stoichiometric coefficient.

For a general reaction $a A + b B \to c C + d D$ :

$Rate \propto [A]^{a} [B]^{b} \Rightarrow Rate = k [A]^{a} [B]^{b}$

where $k$ is the rate constant at that temperature.

Active mass is the molar concentration of a reacting species:

Active mass = \frac{number of moles}{volume in litres}

. For pure solids and pure liquids, active mass is taken as unity.

Solved Example 1

Four 10 L vessels contain: (A) 16 g

C H_{4}

, (B) 18 g

H_{2} O

, (C) 35.5 g

C l_{2}

, (D) 44 g

C O_{2}

. Find the active mass (molar concentration) in each.

Solution:

Active mass $= \frac{mass in grams}{molar mass \times volume in L}$ .

(A) $[C H_{4}] = \frac{16}{16 \times 10} = 0.1 M$

(B) $[H_{2} O] = \frac{18}{18 \times 10} = 0.1 M$

(D) $[C O_{2}] = \frac{44}{44 \times 10} = 0.1 M$

6. Law of Chemical Equilibrium - Deriving $K_{c}$

For the reversible reaction $a A + b B ⇌ c C + d D$ , applying the law of mass action to both directions:

Forward rate: $r_{f} = k_{f} [A]^{a} [B]^{b}$

Reverse rate: $r_{b} = k_{b} [C]^{c} [D]^{d}$

At equilibrium, $r_{f} = r_{b}$ , so:

$k_{f} [A]^{a} [B]^{b} = k_{b} [C]^{c} [D]^{d}$

Rearranging:

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

This ratio of rate constants is called the equilibrium constant in terms of concentrations, $K_{c}$ . It is a constant at a given temperature.

Solved Example 2

A

and

B

are mixed at temperature

T

such that the initial concentration of

A

is twice that of

B

. After equilibrium is reached in

A + B ⇌ C + D

, the concentration of

C

is three times the equilibrium concentration of

B

. Calculate

K_{c}

Solution:

Let $[B]_{0} = a$ , so $[A]_{0} = 2 a$ . Let $x$ mol/L of $A$ (and of $B$ ) react:

$A B C D initial 2 aa 00 eqbm 2 a - x a - xxx$

Given $[C]_{e q} = 3 [B]_{e q}$ : $x = 3 (a - x) \Rightarrow x = \frac{3 a}{4}$

$K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{x \cdot x}{( 2 a - x ) ( a - x )} = \frac{( 3 a /4 ) ^{2}}{( 2 a - 3 a /4 ) ( a - 3 a /4 )} = \frac{9 a ^{2} /16}{( 5 a /4 ) ( a /4 )} = \frac{9}{5} = 1.8$

7. Equilibrium Constant in Terms of Partial Pressures ( $K_{p}$ )

For a gaseous reaction, the equilibrium constant can be written using partial pressures instead of concentrations:

$K_{p} = \frac{P _{C}^{c} P _{D}^{d}}{P _{A}^{a} P _{B}^{b}}$

Partial pressures at equilibrium are related to concentrations by the ideal gas law: $P = (\frac{n}{V}) R T = C \cdot R T$ .

8. Relation Between $K_{p}$ and $K_{c}$

Substituting $P_{i} = [i] \cdot R T$ into the $K_{p}$ expression for $a A (g) + b B (g) ⇌ c C (g) + d D (g)$ :

$K_{p} = \frac{([ C ] R T ) ^{c} ([ D ] R T ) ^{d}}{([ A ] R T ) ^{a} ([ B ] R T ) ^{b}} = K_{c} \cdot (R T)^{(c + d) - (a + b)}$

$K_{p} = K_{c} (R T)^{Δ n_{g}}$

where $Δ n_{g} = (moles of gaseous products) - (moles of gaseous reactants)$ .

Sign of $Δ n_{g}$	Relation	Example
$Δ n_{g} = 0$	$K_{p} = K_{c}$	$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
$Δ n_{g} > 0$	$K_{p} > K_{c}$	$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
$Δ n_{g} < 0$	$K_{p} < K_{c}$	$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

Solved Example 3

27 ° C

K_{p}

for

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

0.1

atm. Calculate

K_{c}

Solution:

Only $C O_{2}$ is a gas; pure solids are omitted from $Δ n_{g}$ . So $Δ n_{g} = 1$ .

$K_{p} = K_{c} (R T)^{Δ n_{g}} \Rightarrow K_{c} = \frac{K _{p}}{R T} = \frac{0.1}{0.0821 \times 300} \approx 4.06 \times 1 0^{- 3} M$

Solved Example 4

Given

K_{c} = 50

700 K

for

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

, calculate

K_{c}

for

H I (g) ⇌ \frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g)

Solution:

The new reaction is the reverse of the original, then divided (multiplied) by $\frac{1}{2}$ .

Reversing: $K^{'} = \frac{1}{K _{c}} = \frac{1}{50}$ . Multiplying by $\frac{1}{2}$ : $K^{''} = (K^{'})^{1/2} = \frac{1}{50} = \frac{1}{50} \approx 0.141$ .

9. Characteristics of the Equilibrium Constant

Constant at a given temperature. $K_{c}$ (or $K_{p}$ ) has a fixed value for a particular reaction at a fixed temperature.
Independent of initial concentrations. Starting with any amounts of reactants and products, the same $K_{c}$ is obtained at equilibrium.
Independent of catalyst. A catalyst does not change $K_{c}$ - it only shortens the time to reach equilibrium.
Independent of pressure and volume changes (as long as temperature is unchanged).
Depends on the stoichiometry. If the equation is written differently, $K_{c}$ changes accordingly.
Reversal rule: If the equation is reversed, $K_{c}^{'} = 1/ K_{c}$ .
Multiplication rule: If the equation is multiplied by $n$ , $K_{c}^{'} = (K_{c})^{n}$ .
Addition rule: If two equations are added, the resultant $K_{c} = K_{c_{1}} \times K_{c_{2}}$ .
Depends only on temperature. Any change in temperature changes $K_{c}$ , following the van't Hoff equation.

Solved Example 5

Given equilibrium constants for the three reactions at temperature

T

:
(i)

A ⇌ B, K_{1} = 2

(ii)

B ⇌ C, K_{2} = 5

(iii)

C ⇌ D, K_{3} = 3

Find

K

for

A ⇌ D

Solution:

Adding all three: $A ⇌ D$ . By the addition rule, $K = K_{1} \times K_{2} \times K_{3} = 2 \times 5 \times 3 = 30$ .

10. Significance of the Magnitude of $K$

Very large $K$ ( $> 1 0^{3}$ ): Forward reaction goes nearly to completion. Products dominate at equilibrium. Concentration of reactants can often be neglected.
Very small $K$ ( $< 1 0^{- 3}$ ): Forward reaction hardly proceeds. Reactants dominate at equilibrium. Concentration of products can often be neglected.
Intermediate $K$ ( $1 0^{- 3} < K < 1 0^{3}$ ): Both reactants and products are present in appreciable amounts at equilibrium.

K

tells us where equilibrium lies, not how fast it is reached. A reaction with a huge

K

can still be extremely slow (e.g.,

2 H_{2} + O_{2} ⇌ 2 H_{2} O

has enormous

K

but does not proceed at room temperature without a spark).

11. Units of the Equilibrium Constant

Depending on $Δ n$ , $K_{c}$ carries units of $(mol/L)^{Δ n}$ and $K_{p}$ carries units of $(atm)^{Δ n_{g}}$ .

Reaction	$Δ n$	Unit of $K_{c}$	Unit of $K_{p}$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$	$- 2$	$L^{2} m o l^{- 2}$	$a t m^{- 2}$
$H_{2} + I_{2} ⇌ 2 H I$	$0$	Dimensionless	Dimensionless
$P C l_{5} ⇌ P C l_{3} + C l_{2}$	$+ 1$	$m o l L^{- 1}$	$a t m$
$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$	$+ 1$	$m o l L^{- 1}$	$a t m$

Strictly speaking, the true (thermodynamic) equilibrium constant is expressed in terms of activities, which are dimensionless (each concentration or pressure divided by a standard reference value:

1 M

for solutions,

1 a t m

for gases). In that framework,

K

is always unitless. For JEE and NEET calculations, we use the practical units shown above unless otherwise stated.

12. Degree of Dissociation ( $α$ )

The degree of dissociation,

α

, is the fraction of one mole of the reactant that has dissociated at equilibrium:

α = \frac{moles dissociated}{moles taken initially}

. Percentage dissociation

= α \times 100

For a generic dissociation $A_{n} (g) ⇌ n A (g)$ starting with $a$ moles:

$A_{n} A initial a 0 eqbm a (1 - α) n a α$

Total moles at equilibrium $= a (1 - α) + n a α = a [1 + (n - 1) α]$ .

Solved Example 6

One mole of

N_{2}

is mixed with three moles of

H_{2}

in a

4 L

container. If

25%

N_{2}

is converted to

N H_{3}

227 ° C

, calculate

K_{c}

and

K_{p}

. Use

R = 0.0821 L a t m K^{- 1} m o l^{- 1}

Solution:

Reaction: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ . Moles of $N_{2}$ reacted $= 0.25$ ; hence $H_{2}$ reacted $= 0.75$ , $N H_{3}$ formed $= 0.50$ .

Equilibrium moles: $N_{2} = 0.75, H_{2} = 2.25, N H_{3} = 0.50$ .

Equilibrium concentrations (in a $4 L$ vessel):

$[N_{2}] = 0.1875 M, [H_{2}] = 0.5625 M, [N H_{3}] = 0.125 M$

$K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} = \frac{( 0.125 ) ^{2}}{( 0.1875 ) ( 0.5625 ) ^{3}} \approx 0.468 L^{2} m o l^{- 2}$

$Δ n_{g} = 2 - 4 = - 2$ , $T = 500 K$ . $(R T)^{- 2} = (0.0821 \times 500)^{- 2} = (41.05)^{- 2} \approx 5.93 \times 1 0^{- 4}$ .

$K_{p} = K_{c} (R T)^{- 2} \approx 0.468 \times 5.93 \times 1 0^{- 4} \approx 2.78 \times 1 0^{- 4} a t m^{- 2}$

13. Vapour Density and Degree of Dissociation

For gaseous dissociation reactions where $Δ n \neq = 0$ , vapour density measurements provide a direct way to find $α$ . Since total moles change on dissociation, so does the average molecular mass - and vapour density is directly proportional to molecular mass.

For $A_{n} (g) ⇌ n A (g)$ starting with $a$ moles at fixed $T, V$ : total moles $= a [1 + (n - 1) α]$ . Since vapour density is inversely proportional to number of moles (at constant $T$ and $V$ ):

$\frac{D}{d} = 1 + (n - 1) α \Rightarrow α = \frac{D - d}{( n - 1 ) d}$

where $D$ = initial vapour density (undissociated) and $d$ = vapour density at equilibrium.

Equivalent forms using molecular masses: $\frac{M _{theoretical}}{M _{observed}} = 1 + (n - 1) α$ .

This method is not applicable when

Δ n = 0

(e.g.,

H_{2} + I_{2} ⇌ 2 H I

) because the total number of moles does not change, so vapour density is unaffected.

Solved Example 7

When

P C l_{5}

is heated, it dissociates:

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

. The vapour density of the mixture at

200 ° C

70.2

. Find the degree of dissociation.

Solution:

Molar mass of $P C l_{5} = 31 + 5 (35.5) = 208.5$ . Initial vapour density $D = 208.5/2 = 104.25$ .

Here $n = 2$ (one mole of $P C l_{5}$ gives two moles of products).

$α = \frac{D - d}{( n - 1 ) d} = \frac{104.25 - 70.2}{( 2 - 1 ) ( 70.2 )} = \frac{34.05}{70.2} \approx 0.485$

So $P C l_{5}$ is about $48.5%$ dissociated at $200 ° C$ .

14. Simultaneous Equilibria JEE Advanced

Sometimes two or more equilibria share a common species in the same container. The equilibrium concentration of the shared species is the total from all equilibria, and this shared value appears in each $K$ expression.

Example: solid $N H_{4} H S$ decomposes and, in the presence of a catalyst, part of the $H I$ product also dissociates:

$N H_{4} H S (s) ⇌ N H_{3} (g) + H I (g); K_{p_{1}}$

$2 H I (g) ⇌ H_{2} (g) + I_{2} (g); K_{p_{2}}$

At equilibrium, if we let the initial pressure of $H I$ from step 1 be $P$ and $x$ atm of it further dissociates in step 2, then $P_{H I, eq} = P - x$ , $P_{H_{2}} = x /2$ , $P_{I_{2}} = x /2$ , and $P_{N H_{3}} = P$ (unchanged, since step 2 does not affect $N H_{3}$ ).

Solved Example 8

Solid

N H_{4} I

dissociates as

N H_{4} I (s) ⇌ N H_{3} (g) + H I (g)

400 K

with

K_{p_{1}} = 16 a t m^{2}

. In the presence of a catalyst, part of the

H I

dissociates:

2 H I ⇌ H_{2} + I_{2}

. If

P_{H_{2}}

at equilibrium is

1 a t m

, calculate

K_{p_{2}}

for the second equilibrium.

Solution:

Let $P$ atm of $N H_{3}$ (and initially $H I$ ) be produced from step 1, and $x$ atm of $H I$ dissociate in step 2. Then $P_{H_{2}} = P_{I_{2}} = x /2$ . Given $x /2 = 1$ , so $x = 2$ atm.

At equilibrium: $P_{N H_{3}} = P$ , $P_{H I} = P - 2$ .

From step 1: $K_{p_{1}} = P_{N H_{3}} \cdot P_{H I} = P (P - 2) = 16 \Rightarrow P^{2} - 2 P - 16 = 0$

$P = \frac{2 + 4 + 64}{2} = \frac{2 + 68}{2} \approx 5.12$ atm.

Therefore $P_{H I} \approx 3.12$ atm and:

$K_{p_{2}} = \frac{P _{H_{2}} \cdot P _{I_{2}}}{P _{H I}^{2}} = \frac{1 \times 1}{( 3.12 ) ^{2}} \approx 0.103$

Common Mistakes to Avoid

Watch out

Including pure solids or pure liquids in the $K$ expression. For $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ , write $K_{c} = [C O_{2}]$ , not $\frac{[ C a O ] [ C O _{2} ]}{[ C a C O _{3} ]}$ .
Confusing the rate constant $k$ with the equilibrium constant $K$ . $K$ is the ratio $k_{f} / k_{b}$ ; $k$ alone tells you nothing about equilibrium.
Assuming a catalyst changes $K$ . A catalyst speeds up both forward and reverse reactions equally - $K$ is unchanged.
Using the wrong $Δ n$ in $K_{p} = K_{c} (R T)^{Δ n_{g}}$ . Count only gaseous species. Solids and liquids do not contribute to $Δ n_{g}$ .
Applying the vapour density formula when $Δ n = 0$ . For $H_{2} + I_{2} ⇌ 2 H I$ , the total moles do not change; vapour density stays the same regardless of $α$ .
Forgetting to raise $K$ to the correct power when the equation is multiplied. Doubling the equation squares $K$ ; halving it takes the square root.
Assuming a large $K$ means a fast reaction. $K$ tells you how far equilibrium lies, not how fast it is reached - those are separate questions.
Writing partial pressures in Pa instead of atm. The standard state for gases is $1 a t m$ , so $K_{p}$ values in JEE problems assume atm. Using SI (Pa) gives a different numerical value.

Frequently Asked Questions

What is the difference between physical and chemical equilibrium?

Physical equilibrium involves a change of state without any change in chemical identity - for example, ice melting to water, or water evaporating into vapour. Chemical equilibrium involves a reversible chemical reaction, where reactant molecules and product molecules interconvert. Both types share the feature that rates of the forward and reverse processes become equal.

Why is chemical equilibrium called dynamic?

At equilibrium, the forward and reverse reactions do not stop - they continue at the molecular level. Reactant molecules keep forming products, and product molecules keep re-forming reactants, but at equal rates. The net observable change is zero, but activity at the microscopic level continues, which is why it is called "dynamic" rather than "static".

When are $K_{p}$ and $K_{c}$ equal?

$K_{p} = K_{c}$ when $Δ n_{g} = 0$ , meaning the total number of moles of gaseous reactants equals the total moles of gaseous products. Examples include $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ and $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ . In these cases the factor $(R T)^{Δ n_{g}}$ becomes $(R T)^{0} = 1$ .

Does a catalyst change the value of the equilibrium constant?

No. A catalyst lowers the activation energy of both the forward and reverse reactions equally, so it speeds up the approach to equilibrium without changing the position of equilibrium. The value of $K$ depends only on temperature; catalyst presence, initial concentrations, and pressure changes do not affect it.

What does a very large value of $K$ signify?

A very large equilibrium constant (typically $K > 1 0^{3}$ ) means the reaction is strongly product-favoured - the forward reaction goes nearly to completion, and the equilibrium mixture contains mostly products with negligible amounts of reactants. Note that a large $K$ does not tell us anything about how fast equilibrium is reached; that is decided by the rate constants and activation energy.

Why are pure solids and pure liquids excluded from the equilibrium constant expression?

The "active mass" of a pure solid or pure liquid is proportional to its density, which remains constant throughout the reaction (as long as some of the pure substance is present). Since these constants can be absorbed into the equilibrium constant itself, their effective activities are conventionally taken as unity, and they are omitted from the $K$ expression. For example, $K_{c}$ for $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ is simply $[C O_{2}]$ .

How does the equilibrium constant change if the equation is reversed or multiplied?

Reversing the equation replaces $K$ with $1/ K$ . Multiplying the entire equation by a factor $n$ raises $K$ to the power $n$ : $K^{'} = K^{n}$ . Adding two equations gives an overall $K = K_{1} \times K_{2}$ . These rules follow directly from the mathematical form of the $K$ expression and are widely used to relate equilibrium constants of related reactions.

Can the equilibrium constant have units?

In practice, $K_{c}$ carries units of $(mol/L)^{Δ n}$ and $K_{p}$ has units of $(atm)^{Δ n_{g}}$ . When $Δ n = 0$ , $K$ is dimensionless. Rigorously (thermodynamically), $K$ is defined using activities - dimensionless quantities equal to concentration or pressure divided by a standard reference - so the true thermodynamic $K$ is always unitless. Both conventions are used in exams; follow the numerical convention consistent with the problem.

Previous year questions on Law of Chemical Equilibrium And Equilibrium Constant

28 questions from past papers, each with a step-by-step solution.

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1. What is Chemical Equilibrium?

2. Reversible vs Irreversible Reactions

3. Types of Equilibria

Homogeneous equilibrium

Heterogeneous equilibrium

4. Characteristics of Chemical Equilibrium

5. Law of Mass Action

6. Law of Chemical Equilibrium - Deriving Kc​

7. Equilibrium Constant in Terms of Partial Pressures (Kp​)

8. Relation Between Kp​ and Kc​

9. Characteristics of the Equilibrium Constant

10. Significance of the Magnitude of K

11. Units of the Equilibrium Constant

12. Degree of Dissociation (α)

13. Vapour Density and Degree of Dissociation

14. Simultaneous Equilibria JEE Advanced

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Frequently Asked Questions

What is the difference between physical and chemical equilibrium?

Why is chemical equilibrium called dynamic?

When are Kp​ and Kc​ equal?

Does a catalyst change the value of the equilibrium constant?

What does a very large value of K signify?

Why are pure solids and pure liquids excluded from the equilibrium constant expression?

How does the equilibrium constant change if the equation is reversed or multiplied?

Can the equilibrium constant have units?

Previous year questions on Law of Chemical Equilibrium And Equilibrium Constant

Ready to master Equilibrium?

6. Law of Chemical Equilibrium - Deriving $K_{c}$

7. Equilibrium Constant in Terms of Partial Pressures ( $K_{p}$ )

8. Relation Between $K_{p}$ and $K_{c}$

10. Significance of the Magnitude of $K$

12. Degree of Dissociation ( $α$ )

When are $K_{p}$ and $K_{c}$ equal?

What does a very large value of $K$ signify?