Chemical Equilibrium (JEE Advanced)

Question 1

For N₂O₄(g) leftharpoons 2NO₂(g) at a fixed temperature, the equilibrium mixture at a total pressure of 1 atm has a vapour density of 38.33 (relative to H₂). At the same temperature, at what total pre

Accepted Answer

A. Theoretical VD of N₂O₄ = 92/2 = 46. For A ⇌ 2B, D/d = 1 + α, so α = 46/38.33 − 1 = 0.20.
Kₚ = (4α² P)/(1−α²) = (4(0.04)(1))/0.96 = 1/6 atm.
For α = 0.5: P = (Kₚ(1−α²))/(4α²) = ((1/6)(0.75))/1 = 0.125 atm.
Key insight: Kₚ is fixed, so a larger α needs a lower pressure. 0.25 atm comes from dropping the (1−α²) factor; 0.0625 and 0.5 from using α instead of α² or forgetting the factor 4.
Exam pattern:

Question 2

A rigid evacuated vessel is first filled with NH₃ to 0.40 atm. Excess solid NH₄HS and excess solid NH₄X are then introduced at constant temperature, and both equilibria are set up: NH₄HS(s) leftharpoo

Accepted Answer

D. Let p(H₂S) = a, p(HX) = b. Then p(NH₃) = 0.40 + a + b.
a p_(NH₃) = 0.07 and b p_(NH₃) = 0.14. Adding: (a+b)(0.40 + a + b) = 0.21.
With s = a+b: s² + 0.4s − 0.21 = 0, so s = 0.30 atm; p_(NH₃) = 0.70 atm (a = 0.10, b = 0.20).
Total = 0.70 + 0.30 = 1.00 atm.
Traps: ignoring initial NH3 gives 2√0.21 = 0.92; adding 0.40 to that gives 1.32; doubling 0.70 gives 1.40.
Exam pattern: JEE Advanced – two soli

Question 3

For N₂(g) + 3H₂(g) leftharpoons 2NH₃(g) at 500 K, K_c = 0.50 L² mol⁻². A mixture has partial pressures p_(N₂) = 1.0 atm, p_(H₂) = 3.0 atm and p_(NH₃) = 0.50 atm. Take R = 0.0821 L atm mol⁻¹ K⁻¹. Which

Accepted Answer

B. Kₚ = K_c(RT)^(Δ n) with Δ n = −2: RT = 41.05, so Kₚ = 0.50/41.05² ≈ 2.97×10⁻⁴ atm⁻².
Qₚ = (0.5²)/(1.0× 3.0³) = 9.26×10⁻³ atm⁻².
Qₚ ≈ 31 Kₚ > Kₚ: net backward reaction (NH3 decomposes).
Trap (D): comparing a pressure quotient with K_c (different units/standard states) gives the wrong direction.
Exam pattern: JEE Advanced – predicting direction using Q vs K with Kp–Kc conversion

Chemical Equilibrium: questions with solutions

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