Based on a K c value of 0.190 and the given data table, what are the equilibrium

Question

Based on a Kc value of 0.190 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will cause the net reaction to proceed forward net? Concentration (M) [XY initial: change: equilibrium: + Y 0.100 +x 0.100+r 0.500 0.100 0.500-r 0.100+ The change in concentration, , is negative for the reactants because they are consumed and positive for the products because they are produced.

Explanation / Answer

Kc = [X][Y]/[XY]
0.190 = (0.100+x)*(0.100+x) / (0.500-x)
0.190*(0.500-x) = (0.100+x)*(0.100+x)
0.0950 - 0.190*x = 0.0100 + x^2 + 0.200*x
x^2 + 0.390*x - -0.085 = 0

This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 0.39
c = -8.5*10^-2

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 0.4921

roots are :
x = 0.156 and x = -0.5457

since x can't be negative, at it will make final [X] negative
x = 0.156

At equilibrium,
[XY] = 0.500-x = 0.500 - 0.156 = 0.344 M
[X] = 0.100 + x = 0.100 + 0.156 = 0.256 M
[Y] = 0.100 + x = 0.100 + 0.156 = 0.256 M

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