Intermediate

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Chemical Equilibrium — Kc, Kp & Le Chatelier's Principle

Master Chemical Equilibrium for JEE & NEET. Covers Kc, Kp, Kp=Kc(RT)^Δn, reaction quotient Q, Le Chatelier's principle, Haber process, degree of dissociation.

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Chemical equilibrium is one of the most fundamental concepts in physical chemistry — it describes the state a reversible reaction reaches when the rates of the forward and reverse reactions become equal, and concentrations stop changing. The equilibrium constant ( $K_{c}$ or $K_{p}$ ) quantifies how far a reaction proceeds, and Le Chatelier's Principle predicts how a system at equilibrium responds to disturbances. For JEE and NEET, this topic yields consistent numerical questions on $K_{c}$ , $K_{p}$ , the relationship between them, degree of dissociation, reaction quotient $Q$ , and Le Chatelier analysis. Mastering the algebraic and conceptual aspects of equilibrium is essential for both physical and inorganic chemistry.

1. Law of Mass Action and Equilibrium Constant

For a general reversible reaction: $a A + b B ⇌ c C + d D$

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

where square brackets denote molar concentrations at equilibrium.

$K_{c}$ is temperature-dependent only — not affected by pressure, volume, concentration changes, or catalysts.
Pure solids and pure liquids are excluded from $K_{c}$ expressions (their "concentrations" are constant, absorbed into $K_{c}$ ).
$K_{c}$ for the reverse reaction = $1 / K_{c}$ of the forward reaction.
If a reaction is multiplied by $n$ : new $K_{c}^{'} = (K_{c})^{n}$ .

2. $K_{p}$ and the Relationship $K_{p} = K_{c} (R T)^{Δ n}$

For reactions involving gases, $K_{p}$ is expressed in terms of partial pressures:

$K_{p} = \frac{(P_{C})^{c} (P_{D})^{d}}{(P_{A})^{a} (P_{B})^{b}}$

The relationship between $K_{p}$ and $K_{c}$ :

$K_{p} = K_{c} (R T)^{Δ n}$

where $Δ n =$ (moles of gaseous products) $-$ (moles of gaseous reactants), and $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ .

Reaction	$Δ n$	Relation
$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$	$2 - 4 = - 2$	$K_{p} = K_{c} (R T)^{- 2}$ , so $K_{p} < K_{c}$
$N_{2} O_{4} ⇌ 2 {NO}_{2}$	$2 - 1 = + 1$	$K_{p} = K_{c} (R T)^{+ 1}$ , so $K_{p} > K_{c}$
$H_{2} + I_{2} ⇌ 2 HI$	$2 - 2 = 0$	$K_{p} = K_{c}$

Worked Example

$N_{2} O_{4} ⇌ 2 {NO}_{2}$ : $K_{c} = 0.36$ at 30°C (303 K). Find $K_{p}$ .

$Δ n = + 1$ ; $K_{p} = K_{c} (R T)^{1} = 0.36 \times (0.0821 \times 303) = 0.36 \times 24.88 = 8.96$

3. Reaction Quotient $Q$ and Predicting Shift Direction

$Q$ has the same expression as $K_{c}$ , but uses concentrations at any moment (not necessarily at equilibrium).

Condition	Interpretation	Direction of shift
$Q < K_{c}$	Reaction has not yet reached equilibrium; products insufficient	Forward →
$Q = K_{c}$	System is at equilibrium	No shift
$Q > K_{c}$	System has excess products	Reverse ←

4. Le Chatelier's Principle

Statement: If a system at equilibrium is subjected to a change (stress), it will shift in the direction that partially counteracts that change.

Change (Stress)	System response	Effect on $K$
Add reactant	Shift forward (→)	No change
Remove reactant	Shift backward (←)	No change
Increase pressure (compress)	Shift to side with fewer moles of gas	No change
Decrease pressure (expand)	Shift to side with more moles of gas	No change
Increase temperature	Shift in endothermic direction	Changes $K$
Decrease temperature	Shift in exothermic direction	Changes $K$
Add catalyst	Reaches equilibrium faster; no shift	No change
Add inert gas (constant V)	No effect (partial pressures unchanged)	No change

Application — Haber Process ( $N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$ , $Δ H = - 92 {kJ mol}^{- 1}$ )

High pressure (200–500 atm): Favours forward reaction ( $Δ n = - 2$ , fewer moles on product side) → more NH₃.
Low temperature: Favours exothermic forward reaction → more NH₃. But rate is too slow at very low T.
Compromise temperature (400–500°C): Reasonable rate + acceptable yield.
Catalyst (Fe, with ${Al}_{2} O_{3}$ and $K_{2} O$ ): Increases rate without changing $K$ .

5. Degree of Dissociation

For ${PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ : If $α$ = degree of dissociation, starting with 1 mol ${PCl}_{5}$ :

	PCl₅	PCl₃	Cl₂	Total
Initial	1	0	0	1
Equilibrium	$1 - α$	$α$	$α$	$1 + α$

$K_{p} = \frac{α^{2} P}{1 - α^{2}} \approx α^{2} P (for small α)$

Effect of pressure: Increasing pressure → decreases $α$ (equilibrium shifts left to reduce moles of gas). Increasing temperature → increases $α$ (dissociation is endothermic).

6. Significance of $K_{c}$ Values

$K_{c}$ value	Interpretation
$K_{c} ≫ 1$ (e.g., $> 10^{3}$ )	Reaction strongly favours products; essentially complete
$K_{c} \approx 1$	Significant amounts of both reactants and products at equilibrium
$K_{c} ≪ 1$ (e.g., $< 10^{- 3}$ )	Reaction strongly favours reactants; negligible products formed

$K_{p} = K_{c} (R T)^{Δ n}$ — Key Cases at a Glance

$Δ n$	$K_{p}$ vs $K_{c}$	Example
$Δ n > 0$	$K_{p} > K_{c}$	$N_{2} O_{4} ⇌ 2 {NO}_{2}$
$Δ n = 0$	$K_{p} = K_{c}$	$H_{2} + I_{2} ⇌ 2 HI$
$Δ n < 0$	$K_{p} < K_{c}$	$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$

JEE / NEET Power Tips — Chemical Equilibrium

Only temperature changes $K$ . Adding reactants, removing products, changing pressure, adding catalyst — none of these change $K_{c}$ or $K_{p}$ . Only changing temperature changes the value of $K$ . This is the single most tested conceptual point.
Adding an inert gas at constant volume has NO effect. Partial pressures of reacting species don't change → $Q$ unchanged → no shift. But adding inert gas at constant pressure decreases partial pressures of all species → shift toward more moles of gas.
$K_{c}$ expressions exclude pure solids and liquids. For ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$ : $K_{c} = [{CO}_{2}]$ only. A common error is including ${CaCO}_{3}$ or $CaO$ in the expression.
When a reaction is reversed, $K^{'} = 1 / K$ . When multiplied by $n$ , $K^{'} = K^{n}$ . Always adjust $K$ when manipulating reactions before adding them (Hess's law for $K$ ).
Degree of dissociation $α$ and $K$ are related — increasing $T$ always increases $K$ for endothermic reactions and decreases it for exothermic reactions. This is because $\ln K = - Δ H / R T + const$ (van't Hoff equation).

Practice Questions

Q1 (JEE Main / NEET): For $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ , $K_{c} = 0.36$ at 30°C. Find $K_{p}$ at the same temperature. ( $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ )

Explanation:

First, find the change in the number of moles of gas ( $Δ n$ ) and convert the temperature to Kelvin:
$Δ n = 2 - 1 = + 1$
$T = 30 + 273 = 303 K$

Use the relationship between $K_{p}$ and $K_{c}$ :
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{p} = 0.36 \times (0.0821 \times 303)^{1}$
$K_{p} = 0.36 \times 24.88 =$ 8.96

Q2 (JEE Main): At equilibrium, the concentrations are: $[H_{2}] = 0.5 M$ , $[I_{2}] = 0.5 M$ , $[HI] = 2 M$ . Calculate $K_{c}$ for $H_{2} + I_{2} ⇌ 2 HI$ . If $[HI] = 3 M$ is suddenly imposed, which way will the reaction shift?

Explanation:

First, calculate the equilibrium constant ( $K_{c}$ ):
$K_{c} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} = \frac{(2)^{2}}{(0.5) (0.5)} = \frac{4}{0.25} =$ 16

After suddenly imposing $[HI] = 3 M$ , calculate the reaction quotient ( $Q$ ):
$Q = \frac{(3)^{2}}{(0.5) (0.5)} = \frac{9}{0.25} = 36$

Since $Q > K_{c}$ ( $36 > 16$ ), the reaction shifts backward (←) to consume the excess HI and re-establish equilibrium.

Q3 (NEET): For the reaction $2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$ , $Δ H < 0$ . What happens to $K_{c}$ when the temperature is increased?

A) Increases
B) Decreases
C) Remains unchanged
D) First increases then decreases

Answer: B) Decreases.

Explanation: The reaction is exothermic ( $Δ H < 0$ ). According to Le Chatelier's Principle, increasing the temperature shifts the equilibrium in the endothermic (backward) direction. This reduces the concentration of the product ( ${SO}_{3}$ ) and increases the reactants ( ${SO}_{2}$ and $O_{2}$ ), which mathematically causes $K_{c}$ to decrease. Temperature is the only factor that can change the actual value of the equilibrium constant.

Q4 (JEE Main): ${PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ . At equilibrium, the degree of dissociation is $0.3$ at total pressure $P$ . If the pressure is doubled, how does $α$ change?

Explanation:

The equilibrium constant for this reaction in terms of the degree of dissociation ( $α$ ) and total pressure ( $P$ ) is given by:
$K_{p} = \frac{α^{2} P}{1 - α^{2}}$

Since temperature is unchanged, $K_{p}$ remains perfectly constant. If the total pressure is doubled ( $2 P$ ), the new degree of dissociation ( $α^{'}$ ) must adjust to satisfy the equation:
$K_{p} = \frac{α^{' 2} (2 P)}{1 - α^{' 2}}$

For the expression to equal the same $K_{p}$ value while $P$ is multiplied by 2, the fraction $\frac{α^{' 2}}{1 - α^{' 2}}$ must be halved. This means $α^{'}$ must be smaller than $α$ . Therefore, doubling the pressure decreases the degree of dissociation. (This aligns with Le Chatelier's Principle: increasing pressure shifts the reaction to the left, toward the side with fewer moles of gas).

Q5 (Board / NEET MCQ): Which of the following does NOT affect the equilibrium constant $K_{c}$ ?

A) Temperature
B) Adding a catalyst
C) Changing concentration of reactant
D) Both B and C

Answer: D) Both B and C.

Explanation: For a given specific reaction, only temperature changes the value of the equilibrium constant ( $K_{c}$ ). Adding a catalyst speeds up both the forward and reverse reaction rates equally, allowing the system to reach equilibrium faster, but it does not alter $K_{c}$ . Changing the concentration of a reactant changes the reaction quotient ( $Q$ ) and causes a temporary shift, but it does not change the fundamental value of $K_{c}$ .

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Est. Read Time26 minutes

CategoryEquilibrium

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Chemical Equilibrium — Kc, Kp & Le Chatelier's Principle

1. Law of Mass Action and Equilibrium Constant

2. $K_{p}$ and the Relationship $K_{p} = K_{c} (R T)^{Δ n}$

Worked Example

3. Reaction Quotient $Q$ and Predicting Shift Direction

4. Le Chatelier's Principle

Application — Haber Process ( $N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$ , $Δ H = - 92 {kJ mol}^{- 1}$ )

5. Degree of Dissociation

6. Significance of $K_{c}$ Values

$K_{p} = K_{c} (R T)^{Δ n}$ — Key Cases at a Glance

Practice Questions

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Practice Questions

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Chemical Equilibrium — Kc, Kp & Le Chatelier's Principle

1. Law of Mass Action and Equilibrium Constant

2. Kp and the Relationship Kp=Kc(RT)Δn

Worked Example

3. Reaction Quotient Q and Predicting Shift Direction

4. Le Chatelier's Principle

Application — Haber Process (N2+3H2⇌2NH3, ΔH=−92 kJ mol−1)

5. Degree of Dissociation

6. Significance of Kc Values

Kp=Kc(RT)Δn — Key Cases at a Glance

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2. $K_{p}$ and the Relationship $K_{p} = K_{c} (R T)^{Δ n}$

3. Reaction Quotient $Q$ and Predicting Shift Direction

Application — Haber Process ( $N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$ , $Δ H = - 92 {kJ mol}^{- 1}$ )

6. Significance of $K_{c}$ Values

$K_{p} = K_{c} (R T)^{Δ n}$ — Key Cases at a Glance