Answer:
Less than 0.033 M:
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Explanation:
Hello,
In this case, the described equilibrium is:
[tex]2A+B\rightarrow 2Z[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]
Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]
Which has the following solution:
[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]
But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
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.
