For a first-order reaction R P at a given temperature, k is the...

For a first-order reaction $\text{R} \rightarrow \text{P}$, the rate law is given by:

$-\dfrac{\text{d}[\text{R}]}{\text{d}t} = \dfrac{\text{d}[\text{P}]}{\text{d}t} = k[\text{R}]$

Integrating this, the concentration of the reactant at time $t$ is $[\text{R}] = [\text{R}]_0 e^{-kt}$.

The concentration of the product at time $t$ is $[\text{P}] = [\text{R}]_0 - [\text{R}] = [\text{R}]_0 (1 - e^{-kt})$.

The plot of $[\text{P}]$ versus $t$ follows the equation $[\text{P}] = [\text{R}]_0 (1 - e^{-kt})$. This represents a curve that starts from the origin and is concave down, approaching a constant value $[\text{R}]_0$. The graph in (A) is concave up, making it incorrect.

The plot of $\dfrac{\text{d}[\text{R}]}{\text{d}t}$ versus $[\text{R}]$ follows the equation $\dfrac{\text{d}[\text{R}]}{\text{d}t} = -k[\text{R}]$. This is a straight line passing through the origin with a negative slope ($-k$). The graph in (B) has a positive slope, making it incorrect.

The plot of $\dfrac{\text{d}[\text{P}]}{\text{d}t}$ versus $t$ follows the equation $\dfrac{\text{d}[\text{P}]}{\text{d}t} = k[\text{R}]_0 e^{-kt}$. This represents an exponentially decreasing curve with respect to time. The graph in (C) matches this behavior, making it correct.

The rate constant $k$ depends only on the temperature and is independent of time. Therefore, the plot of $k$ versus $t$ is a horizontal straight line. The graph in (D) matches this, making it correct.

Answer: ;