Question

Given a K value of 0.43 for the following aqueous equilibrium, suppose sample Z is placed into water such that it’s original concentration is 0.033 M. Assume there was zero initial concentration of either A(aq) or B(aq). Once equilibrium has occurred, what will be the equilibrium concentration of Z?
2A(aq) + B(aq) <> 2Z (aq)

Answer 1

Answer:

[Z] = 0.00248M

Explanation:

Based in the reaction:

2A(aq) + B(aq) ⇄ 2Z (aq)

K of the reaction is defined as:

K = [Z]² / [A]²[B] = 0.43

If you add, in the first, just 0.033M of  Z, concentrations in equilibrium are:

[Z] = 0.033M - 2X

[A] = 2X

[B] = X

Replacing in K equation:

0.43 = [0.033M - 2X]² / [2X]² [X]

0.43 = [0.033M - 2X]² / [2X]² [X]

0.43 = 4X² -0.132X + 0.001089 / 4X³

1.72X³ - 4X² + 0.132X - 0.001089 = 0

Solving for X:

X = 0.01526M

Replacing, concentration in equilibrium of Z is:

[Z] = 0.033M - 2*0.01526M = 0.00248M

Answer 2

Answer:

Less than 0.033 M

Explanation:

Hello,

In this case, given the equilibrium:

$2A(aq) + B(aq) \rightleftharpoons 2Z (aq)$

Thus, the law of mass action is:

$K=\frac{[Z]^2}{[A]^2[B]}$

Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:

$\frac{1}{K} =\frac{[A]^2[B]}{[Z]^2}=2.33$

Know, by introducing the change  due to the reaction extent, we can write:

$2.33=\frac{(2x)^2*x}{(0.033M-2x)^2}$

Which has the following solution:

$x_1=2.29M\\x_2=0.0181M\\x_3= 0.0153M$

But the correct solution is $x=0.0153M$  since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:

$[Z]_{eq}=0.033M-2*0.0153M=0.0024M$

Which is of course less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).

Regards.