For the following gas phase equilibrium reaction at constant temperature, NH3(g) leftharpoons (1/2)N2(g) + (3/2)H2(g)…

Year	2026	2025	2024	Total
Questions	10	18	8	36

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Kₚ = (PN₂)1/2 (PH₂)3/2(PNH₃)

Core Logic

Let the degree of dissociation of $NH₃$ $NH_3$ be $α$ $\alpha$. Initial moles: $NH₃ = 1$ $NH_3 = 1$ Equilibrium moles: $NH₃ = 1 - α$ $NH_3 = 1 - \alpha$, $N₂ = (α)/(2)$ $N_2 = \frac{\alpha}{2}$, $H₂ = (3α)/(2)$ $H_2 = \frac{3\alpha}{2}$ Total moles = $1 - α + (α)/(2) + (3α)/(2) = 1 + α$

Step 1: Setting up Partial Pressures

Using mole fractions and total pressure $PT = \sqrt(3)$ $P_T = \sqrt{3}$: $PNH₃ = (1-α)/(1+α) PT$ $PN₂ = (α/2)/(1+α) PT$ $PH₂ = (3α/2)/(1+α) PT$

Step 2: Solving for alpha

Kₚ = ((α/2)/(1+α) PT)1/2 \cdot ((3α/2)/(1+α) PT)3/2 (1-α)/(1+α) PT

Kₚ = (α/2)1/2 (3α/2)3/2 1-α \cdot (PT²)/(PT (1+α)) \cdot (1+α)

Wait, simplify the $PT$ $P_T$ term:

Kₚ = (α/2)1/2 (3α/2)3/2 1-α \cdot ((PT)/(1+α))

Substitute $Kₚ = 9$ $K_p = 9$ and $PT = \sqrt(3)$ $P_T = \sqrt{3}$:

9 = (α/2)1/2 (3α/2)3/2 (1-α) \cdot \sqrt(3)1+α

9 = \sqrt(27) \cdot (α/2)² (1-α)(1+α) \cdot \sqrt(3)

9 = ( 9 \cdot (α²/4) )/( 1 - α² )

1 = ( α²/4 )/( 1 - α² )

1 - α² = (α²)/(4)

1 = (5α²)/(4)

α² = (4)/(5) = 0.8

α = (0.8)1/2

Step 3: Matching the given format

We are given $α = (x \times 10⁻²)-1/2$ $\alpha = (x \times 10^{-2})^{-1/2}$. Equating both forms:

(0.8)1/2 = (x \times 10⁻²)-1/2

Invert both sides:

x \times 10⁻² = (1)/(0.8)

x \times 10⁻² = 1.25

$x = 125$ $x = 125$

Pattern Recognition

When doing fractional stoichiometric power $Kₚ$ $K_p$ calculations, grouping the $α$ $\alpha$ terms early simplifies algebra massively. $33/2 \times 31/2 = 9$ $3^{3/2} \times 3^{1/2} = 9$, which perfectly cancels the $Kₚ$ $K_p$ value.

Chapter Mix

Class 11 Chemistry: Equilibrium

More Equilibrium Previous-Year Questions — Page 7

Q76 jee_main_2024_27_jan_morning Salt Hydrolysis

Given below are two statements: Statement (I): Aqueous solution of ammonium carbonate is basic. Statement (II): Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on

Kₐ

and

Kb

value of acid and the base forming it. In the light of the above statements, choose the most appropriate answer from the options given below :

A. Both Statement I and Statement II are correct
B. Statement I is correct but Statement II is incorrect
C. Both Statement I and Statement II are incorrect
D. Statement I is incorrect but Statement II is correct

Solution

Related Formula

pH of a weak acid-weak base salt system:

pH = 7 + (1)/(2)(pKₐ - pKb)

Core Logic

Ammonium carbonate, $(NH₄)₂CO₃$ $(\text{NH}_4)_2\text{CO}_3$, is formed from a weak acid ( $H₂CO₃$ $\text{H}_2\text{CO}_3$, $Kₐ \approx 4.3 \times 10⁻⁷$ $K_a \approx 4.3 \times 10^{-7}$) and weak base ( $NH₄OH$ $\text{NH}_4\text{OH}$, $Kb \approx 1.8 \times 10⁻⁵$ $K_b \approx 1.8 \times 10^{-5}$). Since $Kb > Kₐ$ $K_b > K_a$, the aqueous medium accumulates an excess of hydroxyl particles over hydronium, forming a basic system ( $pH > 7$ $\text{pH} > 7$). Both statements are structurally accurate descriptions.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q84 jee_main_2024_29_jan_morning Kp and Kc Relationship

For the reaction

N₂O₄(g) leftharpoons 2NO₂(g)

Kₚ = 0.492

atm at

300K

Kc

for the reaction at same temperature is

______ \times 10⁻²

(Given: R = 0.082 L atm mol⁻¹ K⁻¹)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Kₚ = Kc \cdot (RT)Δ ng

Core Logic

For the given gaseous equilibrium reaction:

N₂O₄(g) leftharpoons 2NO₂(g)

First, find the change in the number of moles of gas ( $Δ ng$ $\Delta n_g$):

Δ ng = nₚ - nᵣ = 2 - 1 = 1

Step 1: Calculation

Substitute the given values into the $Kₚ - Kc$ $K_p - K_c$ relationship: $Kₚ = 0.492$ $K_p = 0.492$ $R = 0.082$ $R = 0.082$ $T = 300 K$ $T = 300\text{ K}$

0.492 = Kc \cdot (0.082 \times 300)¹

Kc = (0.492)/(0.082 \times 300)

Kc = (0.492)/(24.6)

$Kc = 0.02$ $K_c = 0.02$

Converting to the requested format ( $x \times 10⁻²$ $x \times 10^{-2}$):

Kc = 2 \times 10⁻²

So, the value is 2.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q85 jee_main_2024_30_january_evening Buffer Solutions

The pH of an aqueous solution containing 1M benzoic acid

(pKₐ = 4.20)

and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ______ mL.

Numerical Answer. Answer: 100 to 100

Solution

Related Formula

Henderson-Hasselbalch Equation for Acidic Buffers:

pH = pKₐ + ( [Salt][Acid] )

Core Logic

Let the volume of 1M Benzoic acid be $Vₐ$ $V_a$ mL and the volume of 1M Sodium benzoate be $Vₛ$ $V_s$ mL. Total volume = $Vₛ + Vₐ = 300 mL$ $V_s + V_a = 300\,\text{mL}$.

Millimoles of acid = $1 \times Vₐ = Vₐ$ $1 \times V_a = V_a$ Millimoles of salt = $1 \times Vₛ = Vₛ$ $1 \times V_s = V_s$

Applying Henderson's Equation:

4.5 = 4.2 + ((Vₛ)/(Vₐ))

Step 1: Calculate Volume Ratio

((Vₛ)/(Vₐ)) = 4.5 - 4.2 = 0.3

Since $2 \approx 0.3$ $\log 2 \approx 0.3$, we have:

(Vₛ)/(Vₐ) = 2

$Vₛ = 2 Vₐ$ $V_s = 2 V_a$

Step 2: Substitute and Solve

We know $Vₛ + Vₐ = 300$ $V_s + V_a = 300$ Substituting $Vₛ = 2 Vₐ$ $V_s = 2 V_a$:

2 Vₐ + Vₐ = 300

$3 Vₐ = 300$ $3 V_a = 300$

Vₐ = 100 mL

Chapter Mix

Class 11 Chemistry: Equilibrium

Q82 jee_main_2024_30_jan_morning Solubility Product

The pH at which

Mg(OH)₂

[Kₛₚ=1\times 10⁻¹¹]

begins to precipitate from a solution containing

0.10 M Mg²⁺

ions is

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Kₛₚ = [Mg²⁺][OH^-]²

pOH = - [OH^-]

$pH + pOH = 14$ $pH + pOH = 14$

Core Logic

Precipitation begins just when the ionic product equals the solubility product ( $Qₛₚ = Kₛₚ$ $Q_{sp} = K_{sp}$).

Step 1: Calculating required [OH-]

[Mg²⁺][OH^-]² = 10⁻¹¹

Given $[Mg²⁺] = 0.10 M$ $[Mg^{2+}] = 0.10 \text{ M}$

0.10 \times [OH^-]² = 10⁻¹¹

[OH^-]² = 10⁻¹⁰

[OH^-] = 10⁻⁵ M

Step 2: Finding pH

pOH = - (10⁻⁵) = 5

$pH = 14 - pOH$ $pH = 14 - pOH$ $pH = 14 - 5 = 9$ $pH = 14 - 5 = 9$

Chapter Mix

Class 11 Chemistry: Equilibrium

Q71 jee_main_2024_31_jan_evening Equilibrium Constants (Kp and Kc)

A(g) leftharpoons B(g) + (C)/(2)(g)

The correct relationship between

KP

α

and equilibrium pressure

P

A. $KP = α(1)/(2)P(1)/(2)(2 + α)(1)/(2)$
B. $KP = α(3)/(2)P(1)/(2)(2 + α)(1)/(2)(1 - α)$
C. $KP = α(1)/(2)P(3)/(2)(2 + α)(3)/(2)$
D. $KP = α(1)/(2)P(1)/(2)(2 + α)(3)/(2)$

Solution

Related Formula

KP = PB \cdot (PC)(1)/(2)PA

where $Pᵢ$ $P_i$ is the partial pressure of component $i$ $i$.

Step 1: Setting up the ICE Table

For the reaction $A(g) leftharpoons B(g) + (1)/(2) C(g)$

Let initial moles of $A = 1$ $A = 1$. At equilibrium: Moles of $A = 1 - α$ $A = 1 - \alpha$ Moles of $B = α$ $B = \alpha$ Moles of $C = (α)/(2)$ $C = \frac{\alpha}{2}$

Total moles at equilibrium = $(1 - α) + α + (α)/(2) = 1 + (α)/(2) = (2 + α)/(2)$

Step 2: Calculating Partial Pressures

Using mole fraction $\times$ $\times$ Total Pressure ( $P$ $P$): $PA = (1 - α)/(1 + (α)/(2)) \cdot P$ $PB = (α)/(1 + (α)/(2)) \cdot P$ $PC = ((α)/(2))/(1 + (α)/(2)) \cdot P$

Step 3: Calculating Kp

KP = PB \cdot (PC)(1)/(2)PA

KP = ( (α)/(1 + α/2) P ) \cdot ( (α/2)/(1 + α/2) P )1/2(1 - α)/(1 + α/2) P

KP = α \cdot (α/2)1/2 \cdot P3/2(1 + α/2)3/2 \cdot (1 + α/2)/((1 - α) P)

KP = α3/2 \cdot P1/2\sqrt(2) \cdot (1 + α/2)1/2 \cdot (1 - α)

Since $1 + α/2 = (2+α)/(2)$ $1 + \alpha/2 = \frac{2+\alpha}{2}$, the $\sqrt(2)$ $\sqrt{2}$ in denominator cancels out perfectly leaving:

KP = α(3)/(2) P(1)/(2)(2 + α)(1)/(2)(1 - α)

Chapter Mix

Class 11 Chemistry: Equilibrium

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up Partial Pressures

Step 2: Solving for alpha

Step 3: Matching the given format

Pattern Recognition

Chapter Mix

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Volume Ratio

Step 2: Substitute and Solve

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating required [OH-]

Step 2: Finding pH

Chapter Mix

Solution

Related Formula

Step 1: Setting up the ICE Table

Step 2: Calculating Partial Pressures

Step 3: Calculating Kp

Chapter Mix

More Equilibrium Questions — jee_main_2026_23_january_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back