A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$

Since a plot of \(\ln [\mathrm{A}]\) versus time gives a straight line, the reaction follows first-order kinetics. A first-order reaction is characterized by having its rate dependent only on the concentration of one reactant raised to the power of 1.

The rate law for a first-order reaction is given by: $$ \text{rate} = k[\mathrm{A}] $$ where \(k\) is the rate constant and \([\mathrm{A}]\) is the concentration of the reactant A. The integrated rate law for a first-order reaction is given by: $$ \ln \frac{[\mathrm{A}]}{[\mathrm{A}]_0} = -kt $$ where \( [\mathrm{A}]_0 \) is the initial concentration and \(t\) is the time.

We are given the slope of a plot of \(\ln [\mathrm{A}]\) versus time, which is:\(-2.97\times10^{-2} \mathrm{min}^{-1}\). Since the integrated rate law expression can be rewritten as: $$ \ln [\mathrm{A}] = \ln [\mathrm{A}]_0 - kt $$ The slope of the line when plotting \(\ln [\mathrm{A}]\) against \(t\) is equal to the negative of the rate constant: $$ k = -(-2.97\times10^{-2} \mathrm{min}^{-1}) = 2.97\times10^{-2} \mathrm{min}^{-1} $$

For a first-order reaction, the half-life is given by the formula: $$ t_{1/2} = \frac{\ln 2}{k} $$ Using the rate constant calculated in the previous step, we can find the half-life: $$ t_{1/2} = \frac{\ln 2}{2.97\times10^{-2} \mathrm{min}^{-1}} = 23.4 \text{ min} $$

For this step, we need to find the time required for the concentration of A to decrease to \(2.50\times10^{-3}M\). Using the integrated rate law, we can solve for the time: $$ \ln \frac{2.50\times10^{-3}}{2.00\times10^{-2}} = -(2.97\times10^{-2} \mathrm{min}^{-1})t $$ Solve for \(t\): $$ t = \frac{\ln(2.50\times10^{-3}) - \ln(2.00\times10^{-2})}{-2.97\times10^{-2} \mathrm{min}^{-1}} = 64.8 \text{ min} $$ In conclusion: a. The rate law is given by \(\text{rate} = k[\mathrm{A}]\), the integrated rate law is given by \(\ln \frac{[\mathrm{A}]}{[\mathrm{A}]_0} = -kt\), and the rate constant is \(2.97\times10^{-2} \mathrm{min}^{-1}\). b. The half-life for the reaction is 23.4 minutes. c. It takes 64.8 minutes for the concentration of A to decrease to \(2.50\times10^{-3}M\).