Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0 .200+x0.200+x←net⇌[X]0.300−x0.300−x +[Y]0.300−x0.300−x The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed.

This is an incomplete question, here is a complete question.

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse.

The chemical reaction is:

Concentration(M)        [XY]            [X]            [Y]

(M)initial:                     0.200        0.300      0.300

change:                         +x               -x              -x

equilibrium:             0.200+x      0.300-x     0.300-x

The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a Kc value of 0.140 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answer : The Equilibrium concentrations of XY, X, and Y is, 0.104 M, 0.204 M and 0.204 M respectively.

Explanation :

The chemical reaction is:

initial:                      0.200      0.300   0.300

change:                    +x             -x           -x

equilibrium:      (0.200+x)  (0.300-x)   (0.300-x)

The equilibrium constant expression will be:

Now put all the given values in this expression, we get:

By solving the term 'x', we get:

x = 0.0963 and x = 0.644

We are neglecting the value of x = 0.644 because the equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.0963

Equilibrium concentrations of XY = 0.200+x = 0.200+0.0963 = 0.104 M

Equilibrium concentrations of X = 0.300-x = 0.300-0.0963 = 0.204 M

Equilibrium concentrations of X = 0.300-x = 0.300-0.0963 = 0.204 M

the awnser is group 12

it is a solid

explanation: