Consider mixture B, which will cause the net reaction to proceed forward. Concen
ID: 591464 • Letter: C
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?
Explanation / Answer
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
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