Question

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:

• A. Figure A
• B. Figure B
• C. Figure C
• D. Figure D

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.

Answer 1

The Correct Option is C

Concept: For a reversible first-order reaction, \[ R \rightleftharpoons P \] the forward reaction rate is proportional to the concentration of \(R\), while the backward reaction rate is proportional to the concentration of \(P\). Thus, \[ \text{Forward rate} = k_f [R] \] and \[ \text{Backward rate} = k_b [P] \] At equilibrium, the forward and backward rates become equal. Hence, \[ k_f [R]_{eq} = k_b [P]_{eq} \] This relation is extremely important because it directly connects the equilibrium concentrations with the rate constants. The equilibrium constant for the reaction is: \[ K = \frac{[P]_{eq}}{[R]_{eq}} = \frac{k_f}{k_b} \] The problem gives: \[ k_b = 4k_f \] Therefore, \[ K = \frac{k_f}{4k_f} = \frac{1}{4} \] Thus, \[ \frac{[P]_{eq}}{[R]_{eq}} = \frac{1}{4} \] This tells us that at equilibrium, the concentration of \(R\) must be four times the concentration of \(P\).

Step 1: Understanding the initial condition. Initially, \[ [P]_0 = 0 \] and \[ [R]_0 = [R]_0 \] So at time \(t=0\), \[ \frac{[R]}{[R]_0} = 1 \] and \[ \frac{[P]}{[R]_0} = 0 \] This means the graph for \(R\) must start from \(1\), while the graph for \(P\) must start from \(0\).

Step 2: Finding equilibrium concentrations. Since the total concentration remains conserved, \[ [R] + [P] = [R]_0 \] At equilibrium, \[ [R]_{eq} + [P]_{eq} = [R]_0 \] Also, \[ \frac{[P]_{eq}}{[R]_{eq}} = \frac{1}{4} \] Let \[ [P]_{eq} = x \] Then, \[ [R]_{eq} = 4x \] Using conservation of concentration, \[ 4x + x = [R]_0 \] \[ 5x = [R]_0 \] \[ x = \frac{[R]_0}{5} \] Therefore, \[ [P]_{eq} = \frac{[R]_0}{5} \] and \[ [R]_{eq} = \frac{4[R]_0}{5} \] Dividing both by \( [R]_0 \), \[ \frac{[P]_{eq}}{[R]_0} = \frac{1}{5} = 0.2 \] and \[ \frac{[R]_{eq}}{[R]_0} = \frac{4}{5} = 0.8 \] Thus, at equilibrium: \[ [R]/[R]_0 \rightarrow 0.8 \] and \[ [P]/[R]_0 \rightarrow 0.2 \]

Step 3: Matching these values with the graphical options. Now we carefully analyze the required behavior of the graphs:
• \( [R]/[R]_0 \) must start from \(1\) and decrease to \(0.8\)
• \( [P]/[R]_0 \) must start from \(0\) and increase to \(0.2\)
• Both curves must gradually approach constant equilibrium values Let us examine the options:
• Figure A shows both curves approaching \(0.5\), which is incorrect.
• Figure B shows \( [P]/[R]_0 \) approaching nearly \(0.8\), which contradicts the equilibrium ratio.
• Figure C correctly shows: \[ [R]/[R]_0 \rightarrow 0.8 \] and \[ [P]/[R]_0 \rightarrow 0.2 \] Hence it satisfies all conditions.
• Figure D shows continuous change without reaching the proper equilibrium values. Therefore, the correct graphical representation is Figure C. Final Answer: \[ \boxed{\text{(C) Figure C}} \]