Calculating equilibrium concentrations when the net reaction proceeds in reverse

Question

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+xnet[X]0.300x0.300x+[Y]0.300x0.300x 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 0.160 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically.

Explanation / Answer

XY --------> X + Y

Kc = [X] [Y] / [XY]

0.16 = [0.3-x] [ 0.3-x] / [0.2+x]

0.032 + 0.16x = 0.09 + x^2 - 0.6x

x^2 - 0.76x + 0.058 = 0

x = 0.086

COncentration of XY = 0.2+x = 0.286

Concentration of X = 0.3-x = 0.214

Concentration of Y = 0.3-x = 0.214

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