Consider that calcium metal reacts with oxygen gas in the air to form calcium oxide. suppose we react 6.84 mol calcium with 4.00
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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