For a reversible reaction \( R \rightleftharpoons P \), at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants \( k_f \) and \( k_b \), respectively. At time zero, the concentration of \( R \) is \( [R]_0 \) and the concentration of \( P \) is zero. At any given time, \( [R] \) and \( [P] \) are the concentrations of \( R \) and \( P \), respectively. If \( k_b = 4k_f \), the correct graphical representation of the reaction is:
Show Hint
For a reversible first-order reaction,
\[
R \rightleftharpoons P
\]
remember the important relation:
\[
\frac{[P]_{eq}}{[R]_{eq}} = \frac{k_f}{k_b}
\]
After finding the equilibrium ratio, always apply conservation of total concentration:
\[
[R] + [P] = \text{constant}
\]
This makes equilibrium concentration calculations extremely fast in graphical and numerical problems.
Step 1: Understanding the Concept:
In a reversible first-order reaction \( R \xrightleftharpoons[k_b]{k_f} P \), the system reaches a dynamic equilibrium. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction:
\[ k_f[R]_{\text{eq}} = k_b[P]_{\text{eq}} \]
The equilibrium constant \(K_c\) is defined as:
\[ K_c = \frac{[P]_{\text{eq}}}{[R]_{\text{eq}}} = \frac{k_f}{k_b} \] Step 3: Detailed Explanation:
Given \(k_b = 4k_f\).
Then, \(K_c = \frac{k_f}{k_b} = \frac{k_f}{4k_f} = \frac{1}{4} = 0.25\).
From mass balance, the sum of concentrations is constant:
\[ [R] + [P] = [R]_0 + [P]_0 = [R]_0 \]
At equilibrium:
\[ [P]_{\text{eq}} = 0.25 [R]_{\text{eq}} \]
Substitute this into the mass balance:
\[ [R]_{\text{eq}} + 0.25 [R]_{\text{eq}} = [R]_0 \]
\[ 1.25 [R]_{\text{eq}} = [R]_0 \implies [R]_{\text{eq}} = \frac{[R]_0}{1.25} = 0.8 [R]_0 \]
Then:
\[ [P]_{\text{eq}} = [R]_0 - [R]_{\text{eq}} = [R]_0 - 0.8 [R]_0 = 0.2 [R]_0 \]
So, the concentration ratio \([R]/[R]_0\) should approach 0.8, and \([P]/[R]_0\) should approach 0.2.
Looking at the graphs:
- In Graph (A), the dashed line (R) approaches 0.8 and the solid line (P) approaches 0.2. This matches our calculated equilibrium concentrations. Step 4: Final Answer:
Graph (A) correctly depicts the concentration of R decaying to 80% of its initial value and P rising to 20%, consistent with \(k_b = 4k_f\).
