Question

given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43

Answer 1

Answer:

Less than 0.033 M:

$[Z]_{eq}=2.4x10^{-3}M$

Explanation:

Hello,

In this case, the described equilibrium is:

$2A+B\rightarrow 2Z$

Thus, the law of mass action is:

$K=\frac{[Z]^2}{[A]^2[B]}=0.43$

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

$\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33$

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

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

Which has the following solution:

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

But the correct solution is $x_3=0.0152M$ since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:

$[Z]_{eq}=0.033M-2(0.0153M)$

$[Z]_{eq}=2.4x10^{-3}M$

Which is clearly 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.