Respuesta :
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:
[tex]XY(g)\rightleftharpoons X(g)+Y(g)[/tex]
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:
[tex]XY(g)\rightleftharpoons X(g)+Y(g)[/tex]
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:
[tex]K_c=\farc{[X][Y]}{[XY]}[/tex]
Now put all the given values in this expression, we get:
[tex]0.140=\frac{(0.300-x)\times (0.300-x)}{(0.200+x)}[/tex]
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
