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Calculating equilibrium concentrations when the net reaction proceeds in reverse


Concentration (M)

[XY] [X] [Y]

initial: 0.200 0.300 0.300

change: +x -x -x

equilibrium: 0.200+x 0.300-x 0.300-x


Based on a Kc value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Verified Answer :

scme1702

Kc = [XY] / [X][Y]
0.17 = (0.2 + x) / (0.3 - x)(0.3-x)
0.17(0.3 - x)² = 0.2 + x
0.17(0.09 - 0.6x + x²) = 0.2 + x
0.0153 - 0.102x + 0.17x² = 0.2 + x
0.17x² - 1.102x + 0.053 = 0

Solving the quadratic:
x = 0.048

Thus, equilibrium concentrations:
[XY] = 0.2 + 0.048 = 0.248
[X] = 0.3 - 0.048 = 0.252
[Y] = 0.3 - 0.048 = 0.252

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