Question

Consider mixture C, which will cause the net reaction to proceed in reverse.

Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+x←net⇌[X]0.300−x0.300−x+[Y]0.300−x0.300−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 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answer 1

Given that          XY    <------------>    X       +    Y                      Kc = 0.17

at equilibrium 0.2 +x                        0.3 -x      0.3-x

             Kc = (0.3 -x) ( 0.3 -x ) / (0.2 +x)

            0.17 = (0.3 -x) ( 0.3 -x ) / (0.2 +x)

            0.034 + 0.17 x = x² - 0.6x + 0.09

              x²- 0.77 x + 0.056 = 0

   On solving,

                           x = 0.0813 M

Therefore, equilibrium concentrations are

[XY] = 0.2 + x = 0.2 + 0.0813 = 0.2813 M

[X] = 0.3-x = 0.3 - 0.0813 = 0.2187 M

  [Y] = 0.3-x = 0.3 - 0.0813 = 0.2187 M