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.
at equilibrium?
at equilibrium?
yes no yes no
at equilibrium?
at equilibrium?
yes no yes no

Explanation:

Step1: Calculate reaction quotient \(Q\) for first reaction

For the reaction \(A
ightleftharpoons B\) with \(K = 1\). Let \(n_A\) (number of \(A\) molecules) \(= 2\) and \(n_B\) (number of \(B\) molecules) \(= 6\). The reaction quotient \(Q=\frac{n_B}{n_A}\). So \(Q=\frac{6}{2}=3
eq K = 1\).

Step2: Calculate reaction quotient \(Q\) for second reaction

For the reaction \(C
ightleftharpoons D\) with \(K = 1\). Let \(n_C\) (number of \(C\) molecules) \(= 4\) and \(n_D\) (number of \(D\) molecules) \(= 6\). The reaction quotient \(Q=\frac{n_D}{n_C}\). So \(Q=\frac{6}{4}=1.5
eq K = 1\).

Step3: Calculate reaction quotient \(Q\) for third reaction

For the reaction \(E
ightleftharpoons F\) with \(K=\frac{1}{11}\). Let \(n_E\) (number of \(E\) molecules) \(= 1\) and \(n_F\) (number of \(F\) molecules) \(= 10\). The reaction quotient \(Q=\frac{n_F}{n_E}\). So \(Q = 10
eq K=\frac{1}{11}\).

Step4: Calculate reaction quotient \(Q\) for fourth reaction

For the reaction \(G
ightleftharpoons H\) with \(K = 9\). Let \(n_G\) (number of \(G\) molecules) \(= 1\) and \(n_H\) (number of \(H\) molecules) \(= 7\). The reaction quotient \(Q=\frac{n_H}{n_G}\). So \(Q = 7
eq K = 9\).

Answer:

For the first reaction: no
For the second reaction: no
For the third reaction: no
For the fourth reaction: no