Question

Consider mixture B, which will cause the net reaction to proceed forward.

Concentration (M) XY X Y

initial:    .500    .100 .100

change: -x +x +x

equilibrium:    .500-x .100+x    .100+x

The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced. Part B Based on a Kc value of 0.220 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively? Consider mixture C, which will cause the net reaction to proceed in reverse.

Concentration XY    = X Y

initial:    .200 .300    .300

change: +x    -x -x

equilibrium:[    .200 +x    .300+x    .300+x

The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Part C Based on a Kc value of 0.220 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answer 1

Concentration (M)       XY                   X                         Y

initial:                         .500                .100                   .100

change:                         -x                   +x                       +x

equilibrium:              .500-x                 .100+x              .100+x

Kc = [X][Y]/[XY]

0.220 = (0.1+x)(0.1+x)/(0.5-x)

x = 0.169

At equlibrium

[XY] = 0.5 - 0.169 = 0.331

[X] = 0.1 + x = 0.1 + 0.169 = 0.269

[Y] = 0.1 + x = 0.1 + 0.169 = 0.269

Concentration       XY          =             X                       Y

initial:                    .200                     .300                  .300

change:                   +x                        -x                       -x

equilibrium:[            .200 +x            .300-x                .300-x

Kc = [X][Y]/[XY]

0.220 = (0.3-x)(0.3-x)/(0.2+x)

x = 0.0606

At equlibrium

[XY] = 0.2 + x = 0.2 + 0.0606 = 0.2606

[X] = 0.3 - X = 0.3 - 0.0606 = 0.2394

[Y] = 0.3 - x = 0.3 - 0.0606 = 0.2394