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:
Less than 0.033 M
Explanation:
Hello,
In this case, given the equilibrium:
[tex]2A(aq) + B(aq) \rightleftharpoons 2Z (aq)[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}[/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}=2.33[/tex]
Know, by introducing the change due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033M-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=0.0153M[/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=0.0024M[/tex]
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.
