Chemical Kinetics and Nuclear Chemistry

Concept:For a reversible first-order reaction $R \rightleftharpoons P$ at constant temperature, the equilibrium concentrations are determined by the ratio of rate constants, and the concentrations change exponentially until equilibrium is reached.Explanation:Given $k_b = 4k_f$, the equilibrium constant is $K_{eq} = \frac{k_f}{k_b} = \frac{1}{4}$.Let $x$ be the amount of $R$ converted to $P$ at equilibrium.Then $[P] = x$ and $[R] = [R]_0 - x$.From $K_{eq} = \frac{[P]}{[R]} = \frac{x}{[R]_0 - x} = \frac{1}{4}$, we get $4x = [R]_0 - x$, so $x = \frac{[R]_0}{5}$.At equilibrium: $\frac{[P]}{[R]_0} = \frac{1}{5} = 0.2$ and $\frac{[R]}{[R]_0} = \frac{4}{5} = 0.8$.At $t=0$: $\frac{[R]}{[R]_0}=1$, $\frac{[P]}{[R]_0}=0$.As reaction proceeds, $[R]/[R]_0$ decreases from 1 to 0.8, and $[P]/[R]_0$ increases from 0 to 0.2, both approaching their equilibrium values exponentially.Answer:The correct graphical representation is the one where $\frac{[R]}{[R]_0}$ starts at 1 and asymptotically approaches 0.8, while $\frac{[P]}{[R]_0}$ starts at 0 and asymptotically approaches 0.2.