For the reaction H2+I2----2HI Kc is 49.?

Initial concentrations:
[H₂]ₒ = [I₂]ₒ = (1 mol) / (2 L) = 0.5 mol/L

___________ H₂ ____ + ____ I₂ __⇌ ____ 2HI ____ Kc = 49
Initial: _____ 0.5 _________ 0.5 ________ 0 ____ (mol/L)
Change: ____ -y __________ -y _______ +2y ____ (mol/L)
At eqm: __ (0.5 - y) _____ (0.5 - y) ______ 2y ____ (mol/L)

Kc = [HI]² / ([H₂] [I₂])
(2y)² / (0.5 - y)² = 49
2y / (0.5 - y) = √49
2y / (0.5 - y) = 7
2y = 3.5 - 7y
9y = 3.5
2y = 3.5 × (2/9)
2y = 0.78

[HI] at equilibrium = 0.78 mol/L