Question

The equilibrium constant Kc for the reaction below is 0.00622 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentrations are [Br2] = 0.0672 M and [Br] = 0.0563 M, calculate the concentrations of these species at equilibrium.

Answer 1

First we should decide which way the reaction proceeds

Qc = [Br]² / [Br₂] = 0.0563² / 0.0672 = 0.0471

Since Qc > Kc, the reaction will go backwards

Draw ICE table

	Br2(g)	2Br(g)
Initial	0.0672	0.0563
Change	+x (since backwards)	-2X
Equilibrium	0.0672 + x	0.0563 - 2x

Kc = [Br]² / [Br₂]

0.00622 =   [0.0563 - 2x]² / [0.0672 + x ]

x=0.01672

At equilibrium

Concentration of Br2 = 0.0672 + 0.01672 = 0.08392 M

Concentration of Br = 0.0563 - 2 * 0.01672 = 0.02286 M