Calculating equilibrium concentrations when the net reaction proceeds forward Co

Question

Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixbure B, whidh will cause the net reaction to proceed forward net Concentration (M) Initial: change: equilibrium: KY) 0.500 -I 0.500 0.100 0.100 +3 0.100+z 0.100+z T, is negative for the reactants because they are consumed and positive for the products becaune they are produced PartB Based on a K, value of 0.200 and the given data table, what are the equilibrium concentrations of XY. X. and Y, respectively? Express the molar concentrations numerically View Available Hin Submit

Explanation / Answer

Initially,
Qc = [X][Y]/[XY]
= 0.100*0.100/0.500
= 0.02

Since Qc is less than Kc, equilibrium will move in forward direction and x must be positive

Kc = [X][Y]/[XY]
0.200 = (0.100+x)*(0.100+x) / (0.500-x)
0.100 - 0.200x = (0.100+x)^2
0.100 - 0.200x = 0.010 + x^2 + 0.200x
x^2 + 0.400x - 0.090 = 0

This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 0.4
c = -9*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 = 0.52

roots are :
x = 0.1606 and x = -0.5606

since x can't be negative, the possible value of x is
x = 0.161

At equilibrium:
[XY] = 0.500-x = 0.500 - 0.161 = 0.339 M
[X] = 0.100-x = 0.100 + 0.161 = 0.261 M
[Y] = 0.100-x = 0.100 + 0.161 = 0.261 M

Answer: 0.339,0.261,0.261 M

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