A_((g))⇌ B_((g))+C/2(g) The correct relationship between K_P,α and equilibrium pressure P is

Solution

Chemistry ma’am · 3 steps

Build the mole table from 1 mol of A, turn moles into partial pressures, and the 2's cancel out neatly.

1. Mole table. Start with 1 mol of A and let a fraction α dissociate. Each mole of A gives 1 mol of B and 1/2 mol of C, so the total at equilibrium is 1 - α + α + α/2 = (2+α)/2.

   Species	Moles at equilibrium	Mole fraction	Partial pressure
   A	1-α	2(1-α)/(2+α)	2(1-α)P/(2+α)
   B	α	2α/(2+α)	(2α P)/(2+α)
   C	α/2	α/(2+α)	(α P)/(2+α)

2. Write Kp. The coefficient of C is 1/2, so p_C appears under a square root. The factor 2P/(2+α) in p_B cancels the same factor in p_A, leaving α/(1-α).

   K_P = (p_B p_C^(1/2))/p_A = α/(1-α)((α P)/(2+α))^(1/2) = α^(3/2)P^(1/2)/(2+α)^(1/2)(1-α)

3. So. The relationship matches option (b).

The idea to keep. For any dissociation, write moles from 1 mol, find the total, then partial pressure = mole fraction × P. Fractional coefficients become fractional powers in K_P.

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