In equilibrium the initial concentration of hydrogen and bromine is 0.50mol/L what is the equilibrium concentration of hydrogen bromide ( H2 + Br2 <> 2HBr ) kc= 50.0

The concentration of HBr at the equilibrium is   0.780 M

Explanation:

Step 1: Data given

The initial concentration of H2 = 0.50 mol/L

The initial concentration of Br2 = 0.50 mol/L

Kc = 50.0

Step 2: The balanced equation

H2 + Br2 ⇆ 2HBr

Step 3: The concentration at equilibrium

[H2] = 0.50 - X M

[Br2] = 0.50 - X M

[HBr] = 2X

Step 4: Calculate concentration of HBr

Kc = [HBr]² [H2][Br2]

50.0 = 4X² / (0.50 - X)(0.50 -X)

X = 0.390

[H2] = 0.50 - 0.390 = 0.110 M

[Br2] = 0.50 - 0.390 = 0.110 M

[HBr] = 2*0.390 = 0.780 M

i'm assuming that it's c.) red bank..