Consider mixture B, which will cause the net reaction to proceed forward. The ch

Question

Consider mixture B, which will cause the net reaction to proceed forward. The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced. 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. The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a K_c 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.

Explanation / Answer

For the given reaction and the ICE chart,

Part B) Kc = 0.130

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

feeding the equilibrium values,

0.130 = (0.1 + x)(0.1 + x)/(0.5 - x)

0.130 = (0.01 + 0.2x + x^2)/(0.5 - x)

0.065 - 0.130x = 0.01 + 0.2x + x^2

x^2 + 0.330x - 0.055 = 0

x = 0.122 M

So equilibrium concentrations for,

[XY] = 0.5 - 0.122 = 0.38 M

[X] = [Y] = 0.1 + 0.122 = 0.222 M

So,

[XY],[X],[Y] = 0.38, 0.222, 0.222

Part C) Kc = 0.130

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

feeding the equilibrium values,

0.130 = (0.3 - x)(0.3 - x)(0.2 + x)

0.130 = (0.09 - 0.6x + x^2)/(0.2 + x)

0.026 + 0.130x = x^2 - 0.6x + 0.09

x^2 - 0.73x + 0.064 = 0

x = 0.10 M

So equilibrium concentrations for,

[XY] = 0.2 + 0.10 = 0.30 M

[X] = [Y] = 0.3 - 0.10 = 0.20 M

So,

[XY],[X],[Y] = 0.30, 0.20, 0.20

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