Question

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

Concentration (M)initial:change:equilibrium:[XY](0.500) (−x) (0.500−x) net→⇌[X](0.100) (+x) (0.100+x) +[Y](0.100) (+x) (0.100+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.

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

Concentration (M)initial:change:equilibrium:[XY](0.200) (+x) (0.200+x)←net⇌[X](0.300) (−x) (0.300−x)+[Y](0.300) (−x) (0.300−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

I'm gonna do part B and then, you try to do part C in a similar way:

For part B, we know the following reaction:

r: XY <---------> X + Y Kc = 0.130

i: 0.5 0.1 0.1

e: 0.5-x (0.1+x)²

0.130 = (0.1+x)² / 0.5-x

0.130(0.5-x) = 0.01 + 0.2x + x²

0.065 - 0.13x = 0.01+0.2x+x²

x²+0.33x-0.055 = 0

Using the quadratic formula:

x = -0.33 (0.33² + 4*0.055)^1/2 / 2

x = -0.33 0.57 / 2

x₁ = 0.12 M; x₂ = -0.45 M

We'll take the positive value of 0.12 M

[XY] = 0.5-0.12 = 0.38 M

[X] = [Y] = 0.1+0.12 = 0.22 M

To do part C, just use the same procedure, but with the reverse reaction:

X + Y <-------> XY

Follow the same procedure as I did in part B and you should get the answer.

Hope this helps