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:

Rather than jumping straight to the equilibrium point, try setting up how [R] changes using both the forward and backward reactions together, then use the fact that [R] plus [P] stays constant to eliminate [P] from that equation. Solving it tells you not just where the curves settle but also whether they should rise or fall smoothly without any overshoot.

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: \[ \boxed{\text{(C) Figure C}} \]

Answer 2

Concept:

• Rather than relying only on the equilibrium condition, solving the actual rate equation confirms not just the equilibrium values but also the shape of the curves, which is exactly what is needed to pick the correct graph among several similar-looking options.
• Since $[R]+[P]=[R]_0$ holds at every instant (not just at equilibrium), $[P]$ can be eliminated to turn the two coupled rate equations into a single linear differential equation in $[R]$ alone.

Step 1: Write the rate law for $[R]$ and eliminate $[P]$
$\dfrac{d[R]}{dt} = -k_f[R] + k_b[P]$. Using $[P] = [R]_0 - [R]$ at all times:
$\dfrac{d[R]}{dt} = -k_f[R] + k_b([R]_0-[R]) = k_b[R]_0 - (k_f+k_b)[R]$

Step 2: Solve this linear differential equation with an integrating factor
$\dfrac{d[R]}{dt} + (k_f+k_b)[R] = k_b[R]_0$, integrating factor $= e^{(k_f+k_b)t}$.
Integrating from $0$ to $t$ with $[R](0)=[R]_0$:
$[R](t) = \dfrac{k_b[R]_0}{k_f+k_b} + [R]_0\dfrac{k_f}{k_f+k_b}\,e^{-(k_f+k_b)t}$

Step 3: Read off the shape of the curve
The second term is a single positive exponential that dies away smoothly with time, with no overshoot or oscillation. So $[R]/[R]_0$ must fall smoothly and monotonically from $1$ down to a plateau at $\dfrac{k_b}{k_f+k_b}$, and since $[P]=[R]_0-[R]$, the ratio $[P]/[R]_0$ must rise smoothly and monotonically from $0$ up to a matching plateau.

Step 4: Substitute $k_b = 4k_f$
$\dfrac{[R]_{eq}}{[R]_0} = \dfrac{k_b}{k_f+k_b} = \dfrac{4k_f}{k_f+4k_f} = \dfrac{4}{5} = 0.8$
$\dfrac{[P]_{eq}}{[R]_0} = 1 - 0.8 = 0.2$

Step 5: Match to the graphs
The only figure showing $[R]/[R]_0$ starting at $1$ and smoothly decreasing to $0.8$, while $[P]/[R]_0$ starts at $0$ and smoothly rises to $0.2$, with no overshoot, is Figure C.

Final Answer: Figure C