Question

Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will cause the net reaction to proceed forward. Concentration (M)initial:change:equilibrium:[XY]0.500−x0.500−xnet→⇌[X]0.100+x0.100+x+[Y]0.100+x0.100+x The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced. Part B Based on a Kc value of 0.130 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically.

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+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 0.130 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically.

Answer 1

          XY   ⇌   X   + Y
initial 0.5      0.1   0.1
at equi (0.5-x) (0.1+x) (0.1+x)
and Kc = 0.13
Kc = [X][Y] / [XY]
0.13 = (0.1+x)*(0.1+x)/(0.5-x)
x = 0.1217
[X] = 0.1 + 0.1217 = 0.2217 = [Y]
[XY] = 0.5 - 0.1217 = 0.3783

          XY   ⇌   X   + Y
initial 0.2      0.3   0.3
at equi (0.2-x) (0.3+x) (0.3+x)
and Kc = 0.13
Kc = [X][Y] / [XY]
0.13 = (0.3+x)*(0.3+x)/(0.2-x)
x = -0.1019
[X] = 0.3 - 0.1019 = 0.1981 = [Y]
[XY] = 0.2 -(- 0.1019) = 0.3019

Which indicate reaction done in back ward direction