Question

e two substances in each sample can interconvert. that is, each kind of molecule can turn into the other. the equilibrium e equilibrium is shown below the sketch decide whether each solution is at equilibrium. at equilibrium? yes no at equilibrium? yes no at equilibrium? yes no at equilibrium? yes no

Explanation:

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

For the reaction \(A
ightleftharpoons B\) (\(K = 1\)), assume each molecule represents a concentration unit. Let's say the number of \(A\) - type molecules (three - ball chains) is \(n_A=5\) and \(B\) - type molecules (three - ball clusters) is \(n_B = 5\). The reaction quotient \(Q=\frac{[B]}{[A]}\). Since \(Q=\frac{5}{5}=1\) and \(K = 1\), so \(Q = K\).

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

For the reaction \(C
ightleftharpoons D\) (\(K=\frac{1}{9}\)), assume the number of \(C\) - type molecules (three - ball clusters) is \(n_C = 9\) and \(D\) - type molecules (two - ball purple) is \(n_D=1\). The reaction quotient \(Q=\frac{[D]}{[C]}\). So \(Q=\frac{1}{9}\) and \(K=\frac{1}{9}\), so \(Q = K\).

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

For the reaction \(E
ightleftharpoons F\) (\(K=\frac{3}{7}\)), assume the number of \(E\) - type molecules (red clusters) is \(n_E = 6\) and \(F\) - type molecules (blue clusters) is \(n_F=2\). The reaction quotient \(Q=\frac{[F]}{[E]}\). So \(Q=\frac{2}{6}=\frac{1}{3}
eq\frac{3}{7}\), so \(Q
eq K\).

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

For the reaction \(G
ightleftharpoons H\) (\(K=\frac{1}{3}\)), assume the number of \(G\) - type molecules (dark - grey clusters) is \(n_G = 8\) and \(H\) - type molecules (light - grey clusters) is \(n_H=2\). The reaction quotient \(Q=\frac{[H]}{[G]}\). So \(Q=\frac{2}{8}=\frac{1}{4}
eq\frac{1}{3}\), so \(Q
eq K\).

Answer:

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