Consider the reaction
X₂Y( g)⇌ X₂( g)+1/2Y₂( g)
The equation representing correct relationship between the degree of dissociation (x) of X₂Y(g) with its equilibrium constant Kp is __________.
Assume x to be very very small.

Solution

Chemistry ma’am · 4 steps

Build the ICE table, write Kp, then square it to clear the half power on Y2.

1. Moles at equilibrium. Start with 1 mol of X₂Y and let a fraction x dissociate. That leaves 1-x of X₂Y and forms x of X₂ and x/2 of Y₂. The total is 1+x/2 mol.
2. Partial pressures. Each partial pressure is its mole fraction times the total pressure p. Because x is very small, 1-x ≈ 1 and 1+x/2 ≈ 1. So p_(X₂Y) ≈ p, p_(X₂) ≈ xp and p_(Y₂) ≈ x/2p.

   Kₚ = p_(X₂)(p_(Y₂))^(1/2)/p_(X₂Y) ≈ (xp)(x/2p)^(1/2)/p = (x^(3/2) p^(1/2))/√2

3. Square both sides. Squaring removes the roots: Kₚ² = (x³ p)/2, so x³ = 2Kₚ²/p.
4. So. x = ∛(2Kₚ²/p), which is option (c).

The idea to keep. When a product has a coefficient of one half, it appears under a square root in Kp. Square the whole expression before you solve for x, and keep track of the factor that the one half puts on its moles.

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