Question

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 equilibrium is shown below the sketch. decide whether each solution is at equilibrium. ( k = 1 ) at equilibrium? yes no ( k=\frac{1}{9} ) at equilibrium? yes no ( k=\frac{3}{7} ) at equilibrium? yes no ( k=\frac{1}{3} ) at equilibrium? yes no

Explanation:

Step1: Calculate reaction quotient for first case

Count the number of each molecule. Let's assume the left - hand side molecule is \(A\) and the right - hand side is \(B\). The formula for the reaction quotient \(Q=\frac{[B]}{[A]}\). If \(K = 1\), and from the sketch, assume there are 6 \(A\) molecules and 6 \(B\) molecules. Then \(Q=\frac{6}{6}=1\). Since \(Q = K\), the system is at equilibrium.

Step2: Calculate reaction quotient for second case

Let the left - hand side molecule be \(C\) and the right - hand side be \(D\). \(K=\frac{1}{9}\). Counting the molecules, assume there are 9 \(C\) molecules and 1 \(D\) molecule. Then \(Q=\frac{[D]}{[C]}=\frac{1}{9}\). Since \(Q = K\), the system is at equilibrium.

Step3: Calculate reaction quotient for third case

Let the left - hand side molecule be \(E\) and the right - hand side be \(F\). \(K = \frac{3}{7}\). Counting the molecules, assume there are 6 \(E\) molecules and 2 \(F\) molecules. Then \(Q=\frac{[F]}{[E]}=\frac{2}{6}=\frac{1}{3}
eq\frac{3}{7}\). So the system is not at equilibrium.

Step4: Calculate reaction quotient for fourth case

Let the left - hand side molecule be \(G\) and the right - hand side be \(H\). \(K=\frac{1}{3}\). Counting the molecules, assume there are 2 \(G\) molecules and 4 \(H\) molecules. Then \(Q=\frac{[H]}{[G]}=\frac{4}{2} = 2
eq\frac{1}{3}\). So the system is not at equilibrium.

Answer:

First case: yes; Second case: yes; Third case: no; Fourth case: no