Question 2

For the reversible reaction $R \rightleftharpoons P$, at equilibrium, the rate of the forward reaction equals the rate of the backward reaction:

$k_f [R]_{eq} = k_b [P]_{eq}$

Given $k_b = 4k_f$, we have:

$k_f [R]_{eq} = 4k_f [P]_{eq} \Rightarrow [R]_{eq} = 4[P]_{eq}$

From the principle of mass conservation, the total concentration remains constant:

$[R]_{eq} + [P]_{eq} = [R]_0 + [P]_0$

Since $[P]_0 = 0$, $[R]_{eq} + [P]_{eq} = [R]_0$.

Substituting $[R]_{eq} = 4[P]_{eq}$:

$4[P]_{eq} + [P]_{eq} = [R]_0 \Rightarrow 5[P]_{eq} = [R]_0 \Rightarrow \frac{[P]_{eq}}{[R]_0} = \frac{1}{5} = 0.2$

And $\frac{[R]_{eq}}{[R]_0} = \frac{4}{5} = 0.8$.

In a first-order reversible reaction, the concentration follows an exponential approach to equilibrium. Graph (C) correctly shows the concentration ratio $\frac{[R]}{[R]_0}$ starting at $1$ and decreasing to $0.8$, while $\frac{[P]}{[R]_0}$ starts at $0$ and increases to $0.2$ exponentially.