Water decomposes at 2300 K
The percent of water decomposing at 2300 K and 1 bar is __________ (Nearest integer).
Equilibrium constant for the reaction is at 2300 K.
The percent of water decomposing at 2300 K and 1 bar is __________ (Nearest integer).
Equilibrium constant for the reaction is at 2300 K.
Show answer and solution
Answer: 02
Solution
Let degree of dissociation = α. Starting with 1 mol H₂O, at equilibrium: moles of H₂O = 1−α, H₂ = α, O₂ = α/2. Total moles = 1 + α/2.
Mole fractions: x(H₂O) = (1−α)/(1+α/2), x(H₂) = α/(1+α/2), x(O₂) = (α/2)/(1+α/2).
At total pressure P = 1 bar, partial pressures equal mole fractions (since P = 1 bar).
Kp = [α/(1+α/2)] × [α/(2(1+α/2))]^(1/2) / [(1−α)/(1+α/2)]
= α × (α/2)^(1/2) / [(1−α)(1+α/2)^(1/2)]
Since α << 1: (1−α) ≈ 1, (1+α/2)^(1/2) ≈ 1, so:
Kp ≈ α^(3/2) / √2 = 2×10⁻³
α^(3/2) = 2√2 × 10⁻³ = 2.828 × 10⁻³
α = (2.828×10⁻³)^(2/3)
ln(2.828×10⁻³) = ln(2.828) + ln(10⁻³) = 1.04 − 6.908 = −
Mole fractions: x(H₂O) = (1−α)/(1+α/2), x(H₂) = α/(1+α/2), x(O₂) = (α/2)/(1+α/2).
At total pressure P = 1 bar, partial pressures equal mole fractions (since P = 1 bar).
Kp = [α/(1+α/2)] × [α/(2(1+α/2))]^(1/2) / [(1−α)/(1+α/2)]
= α × (α/2)^(1/2) / [(1−α)(1+α/2)^(1/2)]
Since α << 1: (1−α) ≈ 1, (1+α/2)^(1/2) ≈ 1, so:
Kp ≈ α^(3/2) / √2 = 2×10⁻³
α^(3/2) = 2√2 × 10⁻³ = 2.828 × 10⁻³
α = (2.828×10⁻³)^(2/3)
ln(2.828×10⁻³) = ln(2.828) + ln(10⁻³) = 1.04 − 6.908 = −
[JEE Main 2023 · 29 Jan · Shift 1]