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Equilibrium appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Degree of Dissociation and Kp.

Year 2026 2025 2024 Total
Questions 10 18 8 36
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For the following gas phase equilibrium reaction at constant temperature, NH₃(g) leftharpoons (1)/(2)N₂(g) + (3)/(2)H₂(g) If the total pressure is √(3) atm and the pressure equilibrium constant (Kₚ) is 9 atm, then the degree of dissociation is given as (x × 10⁻²)-1/2. The value of x is ____ (Nearest integer)

Numerical Answer Type:
Enter a numerical value Answer: 125 to 125 +4 marks

Solution & Explanation

Related Formula
Kₚ = (PN₂)1/2 (PH₂)3/2(PNH₃)
Core Logic

Let the degree of dissociation of NH₃ be α. Initial moles: NH₃ = 1 Equilibrium moles: NH₃ = 1 - α, N₂ = (α)/(2), H₂ = (3α)/(2) Total moles = 1 - α + (α)/(2) + (3α)/(2) = 1 + α

Step 1: Setting up Partial Pressures

Using mole fractions and total pressure PT = √(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 · ((3α/2)/(1+α) PT)3/2 (1-α)/(1+α) PT Kₚ = (α/2)1/2 (3α/2)3/2 1-α · (PT²)/(PT (1+α)) · (1+α)

Wait, simplify the PT term:

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

Substitute Kₚ = 9 and PT = √(3):

9 = (α/2)1/2 (3α/2)3/2 (1-α) · √(3)1+α 9 = √(27) · (α/2)² (1-α)(1+α) · √(3) 9 = ( 9 · (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 × 10⁻²)-1/2. Equating both forms:

(0.8)1/2 = (x × 10⁻²)-1/2

Invert both sides:

x × 10⁻² = (1)/(0.8) x × 10⁻² = 1.25

x = 125

Pattern Recognition

When doing fractional stoichiometric power Kₚ calculations, grouping the α terms early simplifies algebra massively. 33/2 × 31/2 = 9, which perfectly cancels the Kₚ value.

Chapter Mix

Class 11 Chemistry: Equilibrium

Reference Study Guides

Equilibrium

More Equilibrium Previous-Year Questions — Page 8

Q62 jee_main_2024_31_jan_morning Equilibrium Constant
For the given reaction, choose the correct expression of KC from the following :- Fe(aq)³⁺ + SCN(aq)⁻ leftharpoons (FeSCN)(aq)²⁺
  • A. KC = [FeSCN²⁺][Fe³⁺][SCN⁻]
  • B. KC = [Fe³⁺][SCN⁻][FeSCN²⁺]
  • C. KC = [FeSCN²⁺][Fe³⁺]²[SCN⁻]²
  • D. KC = [FeSCN²⁺]²[Fe³⁺][SCN⁻]

Solution

Related Formula
KC = [Products][Reactants]
Core Logic
KC = Products ion conc.Reactants ion conc. KC = [FeSCN²⁺][Fe³⁺][SCN⁻]
Chapter Mix

Class 11 Chemistry: Equilibrium

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