Chemical Kinetics and Nuclear Chemistry

Concept:For a first-order reaction $\mathrm{R} \rightarrow \mathrm{P}$ at constant temperature, the concentration changes and the rate constant have characteristic graphical representations.Explanation:Option A: Product concentration at time $t$ is $[\mathrm{P}]_t = [\mathrm{R}]_0 (1 - e^{-kt})$.The graph of $[\mathrm{P}]$ vs $t$ starts at zero and rises exponentially to approach $[\mathrm{R}]_0$ asymptotically.Option B: The rate law is $-\frac{d[\mathrm{R}]}{dt} = k[\mathrm{R}]$.Integrating gives $\ln[\mathrm{R}] = \ln[\mathrm{R}]_0 - kt$, so $\ln[\mathrm{R}]$ vs $t$ is a straight line with slope $-k$.Option C: Rate of formation of product: $\frac{d[\mathrm{P}]}{dt} = k[\mathrm{R}] = k[\mathrm{R}]_0 e^{-kt}$.Thus $\frac{d[\mathrm{P}]}{dt}$ vs $t$ shows an exponential decay from $k[\mathrm{R}]_0$ to zero.Option D: The rate constant $k$ depends only on temperature.At a fixed temperature, $k$ is constant, so a graph of $k$ vs $t$ is a horizontal straight line.Answer:Options A, B, C, and D are all correct.