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 0.210 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically. [XY], [X], [Y] =

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 0.210 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

[XY], [X], [Y] =