At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction:?

Initial concentration of HBr, [HBr]ₒ = (2.00 mol) / (4.00 L) = 0.500 mol/L

____________ 2HBr(g) ____ ⇌ ____ H₂(g) ____ + ____ Br₂(aq) ___ Kc
Initial: ______ 0.500 M __________ 0 M ____________ 0 M
Change: _____ -2y M ___________ +y M ___________ +y M
At eqm: __ (0.500 - 2y) M ________ y M ____________ y M

At equilibrium, [Br₂] = y M = 0.0955 M
Hence, y = 0.0955

Kc
= [H₂] [Br₂] / [HBr]²
= y × y / (0.500 - 2y)²
= 0.0955² / (0.500 - 2×0.0955)²
= 0.0955

.....2HBr(g) = H2(g) + Br2(g)
I ...0.5.............0 ..........0
C..-0.191.......0.0955 0.0955 .
E....0.309 .....0.0955 0.0955 .
Kc = .0.0955 * 0.0955 / .0.309^2 =0.0955