Question

The following data are for the system $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ $$ \begin{array}{lcccccc} \hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}}(\text { atm }) & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{\text {B }} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\ \hline \end{array} $$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after \(30 \mathrm{~s}\) ? After \(90 \mathrm{~s}\) ?

Short answer

Answer: The system reached equilibrium at approximately 80 seconds. At 30 seconds, the forward and reverse reaction rates were equal, with both at 0.0055 atm/s. At 90 seconds, both forward and reverse reaction rates were also equal, but at 0 atm/s, indicating that the system was already at equilibrium.

Step-by-step solution

(a) Identifying the Equilibrium Point - Pressure Analysis

To determine the equilibrium point, observe the given pressure data for both A and B. Notice that at a certain time, the pressures of A and B are not changing significantly anymore. This indicates that the system has reached equilibrium. From the table, we can observe that after 80 seconds, both the pressure of A and B are not changing significantly: At 80s: \(P_A = 0.62 atm\), \(P_B = 0.76 atm\) At 100s: \(P_A = 0.62 atm\), \(P_B = 0.76 atm\) Since the pressure values remain constant between 80s and 100s, the system is in equilibrium. So, the time taken to reach equilibrium is approximately 80 seconds.

(b) Comparing forward and reverse reaction rates at 30s and 90s

To compare the forward and reverse reaction rates, we need to analyze the changes in pressures over a defined time interval. The rate of the forward reaction is given by the decrease in pressure of A, while the rate of the reverse reaction is given by the increase in pressure of B, divided by 2, as the balanced chemical equation shows a 1:2 stoichiometric ratio. First, let's calculate the average rates from 20s to 40s, which include the 30s time interval. Forward reaction rate (20s to 40s) = (\(P_A\) at 20s - \(P_A\) at 40s) / 20s = (0.83 atm - 0.72 atm) / 20s = 0.0055 atm/s Reverse reaction rate (20s to 40s) = (\(P_B\) at 40s - \(P_B\) at 20s) / (20s × 2) = (0.56 atm - 0.34 atm) / 40s = 0.0055 atm/s Now, let's calculate the average rates from 80s to 100s, which will include the 90s time interval. Forward reaction rate (80s to 100s) = (\(P_A\) at 80s - \(P_A\) at 100s) / 20s = (0.62 atm - 0.62 atm) / 20s = 0 atm/s Reverse reaction rate (80s to 100s) = (\(P_B\) at 100s - \(P_B\) at 80s) / (20s × 2) = (0.76 atm - 0.76 atm) / 40s = 0 atm/s From the results, we can conclude that: - After 30s, the forward and reverse reaction rates are equal, and both are 0.0055 atm/s. - After 90s, the forward and reverse reaction rates are equal and both are 0 atm/s, indicating that the system is already at equilibrium, and no further changes in pressure are occurring.