Calculate the equilibrium concentration of B2(g) in a vessel that initially contains 1.00 M AB(g). 2AB(g) <--> A2(g) + B2(g) kc=4.0a?

Equilibrium constant.....

2AB(g) <==> A2(g) + B2(g) ................ Kc = 4.0
1.00M ............0 .........0 ............. Initial
-2x ................ +x ...... +x ........... Change
1.00-2x ........... x ........ x ........... Equilibrium

Kc = [A2][B2] / [AB]²
4.0 = x² / (1.00-2x)²
2.0 = x / (1.00-2x)
x = 0.40

[B2] = x = 0.40M

_______________ 2AB(g) __ ⇌ __ A₂(g) __ + __ B₂(g) ____ Kc = 4.0
Initial: _________ 1.00 M _______ 0 M ______ 0 M
Change: _________ -2y M _______ +y M _____ +y M
Equilibrium: __ (1.00 - 2y) M _____ y M ______ y M

At equilibrium: Kc = [A₂] [B₂] / [AB]²
4.0 = y² / (1.00 - 2y)²
4 (1 - 2y)² = y²
4 (1 - 4y + 4y²) = y²
15y² - 16y + 4 = 0
(5y - 2) (3y - 2) = 0
y = 0.4 or y = 0.667 (rejected for 0.100 - 2y must be positive)
Hence, y = 0.4

[B₂] at equilibrium = y M = 0.400 M