Question

tiny samples of aqueous solutions are sketched below, as if under a microscope so powerful that individual molecules could be seen. (the water molecules are not shown.)
the two substances in each sample can interconvert. that is, each kind of molecule can turn into the other. the equilibrium constant ( k ) for each interconversion equilibrium is shown below the sketch.
decide whether each solution is at equilibrium.

Explanation:

Step1: Calculate the reaction quotient \(Q\) for each sample

• For the first sample:
• Let's assume the left - hand molecule is \(A\) and the right - hand molecule is \(B\). There are \(n(A)=9\) and \(n(B) = 1\). The reaction is \(A

ightleftharpoons B\), and \(Q=\frac{[B]}{[A]}\) (since in the absence of volume information for a homogeneous aqueous solution, we can use the number of molecules as a relative measure of concentration). So \(Q=\frac{1}{9}\approx0.11\). Given \(K = 5\), since \(Q
eq K\), the solution is not at equilibrium.

• For the second sample:
• Let the left - hand molecule be \(C\) and the right - hand molecule be \(D\). There are \(n(C)=6\) and \(n(D)=4\). The reaction is \(C

ightleftharpoons D\), and \(Q=\frac{[D]}{[C]}=\frac{4}{6}=\frac{2}{3}\). Given \(K=\frac{2}{3}\), since \(Q = K\), the solution is at equilibrium.

• For the third sample:
• Let the left - hand molecule be \(E\) and the right - hand molecule be \(F\). There are \(n(E)=1\) and \(n(F)=8\). The reaction is \(E

ightleftharpoons F\), and \(Q=\frac{[F]}{[E]} = 8\). Given \(K = 9\), since \(Q
eq K\), the solution is not at equilibrium.

• For the fourth sample:
• Let the left - hand molecule be \(G\) and the right - hand molecule be \(H\). There are \(n(G)=4\) and \(n(H)=6\). The reaction is \(G

ightleftharpoons H\), and \(Q=\frac{[H]}{[G]}=\frac{6}{4}=\frac{3}{2}\). Given \(K=\frac{3}{2}\), since \(Q = K\), the solution is at equilibrium.

Answer:

First sample: no; Second sample: yes; Third sample: no; Fourth sample: yes.