Water decomposes at 2300 K
H₂O(g)→ H₂(g)+1/2O₂(g)
The percent of water decomposing at 2300 K and 1 bar is ___________ (Nearest integer).
Equilibrium constant for the reaction is 2× 10⁻³ at 2300 K.

Solution

Chemistry ma’am · 4 steps

Let α split, write Kₚ from mole fractions, and the small-α shortcut does the rest.

1. Set up moles. Start with 1 mol of H₂O and let α decompose. At equilibrium you have 1-α of H₂O, α of H₂ and α/2 of O₂. The total is 1+α/2.
2. Write Kp. At P = 1 bar, each partial pressure equals its mole fraction. Since Kₚ is tiny, α is small, so take 1-α ≈ 1 and 1+α/2 ≈ 1.

   Kₚ = (p_(H₂) (p_(O₂))^(1/2))/p_(H₂O) ≈ α(α/2)^(1/2) = α^(3/2)/√2

3. Solve for alpha. α^(3/2) = √2 × 2 × 10⁻³ = 2.83 × 10⁻³. Since (0.02)^(3/2) = 0.02 × 0.1414 = 2.83 × 10⁻³, this gives α = 0.02.
4. So. α = 0.02, so 2% of the water decomposes at 2300 K and 1 bar.

The common slip: 6. Leaving out the square root on the O₂ term gives α²/2 = 2 × 10⁻³, so α ≈ 0.063, or about 6, not 2.

The idea to keep. When K is very small, the degree of dissociation is small too. Drop α next to 1 and the equation becomes a simple power of α. Keep the stoichiometric power on every pressure term.

Answer from the NTA final answer key. Solution by PrepWiser.