Question

o determine equilibrium concentrations from in Calculating equilibrium concentrations when the net reaction proceeds forward conditions The reversible reaction has a reaction quotient Qc defined as Consider mixture B, which will cause the net reaction to proceed forward net→ Concentration (M) [XY] (X 0.100 Y) 0.100 Concentration (M) XY initial 0.500 Qc [XY] equilibrium 0.500- 0.100 + π 0.100+a Because the reaction is reversible, both the forward and reverse reactions will occur simultaneously. The reaction will eventually reach equilibrium, at which point the concentrations do not change, and Qc is equal to a constant known as Ko 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 Figure 1 of 1 Based on a K value of 0.220 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically Hints Reaction forms products Reaction forms reactants Equilbrium Submit My Answers Give Up Calculating equilibrium concentrations when the net reaction proceeds in reverse

B) Based on a Kc value of 0.220 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

C) Based on a Kc value of 0.220 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Express the molar concentrations numerically.

Answer 1

The reaction given is XY = X + Y

Equilibrium constant Kc = [x][y]/[xy]

Given that, [x] = [y] =0.1+x and [xy] = 0.5-x and Kc = 0.220

0.220 = (0.10+x)^2/(0.50-x)

Solving for x using quadratic equation gives x = 0.169 M

Since in the reaction 1mole of XY gives 1mole of X and 1mole of Y

the equilibrium concentrations of XY, X, and Y, are

[XY] = 0.500-0.169 = 0.331 M

[X] = 0.1+0.169 = 0.269 M

[Y] = 0.1+0.169 = 0.269 M