Question

At 1,500 K, the equilibrium constant for the reaction below is Kc=0.0819. Br2(g) --> 2 Br(g) What is the percent dissociation of Br2 at 1,500 K, if the initial concentration of Br2 is 2.25 mol/L? Give your answer accurate to two significant figures. Do not include the % sign as part of your answer. Hint: If the initial concentration of Br2 is Co and the concentration of Br2 decreases by x (because of dissociation), then the percent dissociation of Br2 is equal to 100(x / Co).

Answer 1

Br2 (g)   -----------------> 2 Br (g)

2.25                                  0

2.25 - x                             x

Kc = [Br]^2 / [Br2]

0.0819 = x^2 / 2.25 - x

0.1843 - 0.0819 x = x^2

x = 0.390

% dissociation =( x / Co ) x 100

                       = 0.390 / 2.25 ) x 100

% dissociation = 17.3 %