Question

The equilibrium concentrations were found to be [br2] = 2.2 × 10−3 m, [cl2] = 3.0 × 10−2 m, and [brcl] = 1.5 × 10−2 m. write the equilibrium expression, and calculate the equilibrium constant for this reaction at this temperature.

Answer 1

$Br_{2} + Cl_{2}---\ \textgreater \ 2BrCl K_{eq} = \frac{[BrCl]^{2}}{[Br_{2}][Cl_{2}]]} = \frac{[1.5*10^{-2}]^{2}}{[2.2*10^{-3}][3.0*10^{-2}]} =0.34*10=3.4 K_{eq}=3.4$