Q.87 Chemistry · Section B ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 87 Solution

Topic: Equilibrium · Official NTA Booklet Code T1

Question

Consider the following reaction in a sealed vessel at equilibrium with concentrations of N₂ = 3.0×10⁻³ M, O₂ = 4.2×10⁻³ M and NO = 2.8×10⁻³ M. 2NO(g) ⇌ N₂(g) + O₂(g). If 0.1 mol L⁻¹ of NO(g) is taken in a closed vessel, what will be degree of dissociation (α) of NO(g) at equilibrium?

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Verified Against NTA Final Answer Key

Option (2) : 0.717

📚 Equilibrium Medium

📋 Given Parameters & Problem Setup

Equilibrium reaction 2 NO(g) ⇌ N2(g) + O2(g)

Given equilibrium concentrations [N2] = 3.0 × 10^-3 M, [O2] = 4.2 × 10^-3 M, [NO] = 2.8 × 10^-3 M

New initial condition 0.1 mol/L of NO(g) taken in a closed vessel

⚡

The 30-Second Exam Hall Shortcut: ⚡ Square Root Equilibrium Shortcut

⏱️ Save ~60 seconds

K_c = [N2][O2]/[NO]² = (3e-3 × 4.2e-3)/(2.8e-3)² = 12.6/7.84 = 1.607. For 2NO ⇌ N2 + O2, √K_c = α / [2(1 - α)]. √1.607 ≈ 1.268 -> α ≈ 0.717!

1 Compute K_c from first equilibrium: K_c = (3.0 × 10⁻³ × 4.2 × 10⁻³) / (2.8 × 10⁻³)² = 12.6 / 7.84 = 1.6071.

2 For dissociation of NO: 2 NO ⇌ N2 + O2. At equilibrium: [NO] = C(1 - α), [N2] = Cα/2, [O2] = Cα/2.

3 K_c = (Cα/2)² / [C(1 - α)]² = α² / [4(1 - α)²]. Note that concentration C cancels completely because Δn_g = 0!

4 Take square root: √K_c = α / [2(1 - α)].

5 Substitute: √1.6071 ≈ 1.2677 -> 2.5354(1 - α) = α -> 2.5354 = 3.5354 α.

6 α = 2.5354 / 3.5354 ≈ 0.717.

7 Option (2) in under 45 seconds!

🛡️

Why You Can 100% Trust This Shortcut: NCERT Chemistry · Equilibrium (Homogeneous chemical equilibria in gas phase and calculation of degree of dissociation α).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determination of Equilibrium Constant Kc

For the gas-phase reaction 2 NO(g) ⇌ N2(g) + O2(g), the equilibrium constant expression is:

K_c = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} = \frac{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^2} = \frac{12.6 \times 10^{-6}}{7.84 \times 10^{-6}} = 1.6071

Step 2

ICE Table for Initial NO Concentration of 0.1 M

Let initial concentration of NO be C = 0.1 mol/L and degree of dissociation be α: - Initial: [NO] = C, [N2] = 0, [O2] = 0 - Change: [NO] = -Cα, [N2] = +Cα/2, [O2] = +Cα/2 - Equilibrium: [NO] = C(1 - α), [N2] = Cα/2, [O2] = Cα/2

[\text{NO}] = C(1 - \alpha), \quad [\text{N}_2] = \frac{C\alpha}{2}, \quad [\text{O}_2] = \frac{C\alpha}{2}

Step 3

Setting up Equilibrium Expression and Algebraic Solution

Substituting equilibrium expressions into Kc:

K_c = \frac{\left(\frac{C\alpha}{2}\right)\left(\frac{C\alpha}{2}\right)}{[C(1 - \alpha)]^2} = \frac{\alpha^2}{4(1 - \alpha)^2}

Step 4

Solving for α

Taking square roots on both sides:

\sqrt{K_c} = \frac{\alpha}{2(1 - \alpha)} \implies \sqrt{1.6071} = 1.2677 = \frac{\alpha}{2(1 - \alpha)}

Step 5

Final Arithmetic Simplification

Expanding and solving the linear equation in α:

2 \times 1.2677 \times (1 - \alpha) = \alpha \implies 2.5354 - 2.5354\alpha = \alpha \implies 3.5354\alpha = 2.5354 \implies \alpha = \frac{2.5354}{3.5354} \approx 0.717

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Result of omitting the stoichiometric factor of 2 in the denominator (writing Kc = α² / (1-α)²).

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting that 2 NO produces 1/2 N2 and 1/2 O2.

2 (2)

✓ 100% Correct

Verdict & Scientific Reason:

Exact solution to the quadratic equilibrium expression.

💡

Why this is airtight: None.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Off by two orders of magnitude due to improper square root extraction.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming α ≪ 1 when Kc > 1.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Decimal place calculation error.

⚠️

Student Fallacy (Why candidates pick this trap): Dividing by initial concentration C = 0.1 erroneously.

⚠️

NTA Trap Alert: Approximating (1 - α) ≈ 1 When Kc > 1

Because Kc = 1.607 is of order unity, α is very large (~72%). You CANNOT make the approximation 1 - α ≈ 1! You must solve the exact square-root equation.

🛡️ How to avoid the -1 mark: Only approximate 1 - α ≈ 1 when Kc is extremely small (e.g. < 10⁻⁵).

📖

Core NEET Syllabus Standard

Chemistry Core Physical Chemistry · Equilibrium Standard Benchmark

Core Scientific Principle Tested:

"For a symmetric reaction 2A ⇌ B + C, Kc = α² / [4(1-α)²], and the degree of dissociation α is independent of total pressure and initial concentration."

Syllabus Focus: Homogeneous Equilibrium: Evaluation of Equilibrium Constants

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Two moles to two moles: C disappears, root of 1.6 yields zero-point-seven-one-seven in clear cheers!

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NEET 2024 Chemistry Question 87 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Square Root Equilibrium Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Determination of Equilibrium Constant Kc

ICE Table for Initial NO Concentration of 0.1 M

Setting up Equilibrium Expression and Algebraic Solution

Solving for α

Final Arithmetic Simplification

360° Option Elimination Engine Key Feature

NTA Trap Alert: Approximating (1 - α) ≈ 1 When Kc > 1

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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