Question

Be sure to answer all parts. The equilibrium constant Kc for the reaction below is 0.00599 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentrations are [Br2] = 0.0344 M and [Br] = 0.0508 M, calculate the concentrations of these species at equilibrium. [Br2] = M [Br] = M

Answer 1

Br₂ <---------> 2Br

I 0.0344M 0.0508M

C -x +2x

E   (0.0344 -x)    (0.0508 + 2x)

Equilibrium constant (Kc) = 0.00599 = [Br]² / [Br₂]

0.00599 = [ 0.0508 + 2x]² / [0.0344 -x]

0.00599 [0.0344 -x]= [ 0.00258 + 4x² + 0.4064x]

0.0002 - 0.00599x = 0.00258 + 4x² + 0.4064x

4x² + 0.41239x + 0.00238 = 0 ======1)

Equation 1) is a quadratic equation its solution gives value f x as:

x = - 0.00613

Hence [Br₂] = 0.0344 - (0.00613) = 0.04053M

[Br] = [0.0508 + (2 X0.00613)] = 0.0385M