(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant A. The rate equation can be written as: \(r = k [A]\) where r is the rate of reaction, k is the rate constant, and [A] is the concentration of A. To find a linear relationship for this reaction, we will use the integrated rate law for a first-order reaction: \(\ln[A] = -kt + \ln[A]_0\) where [A] is the concentration of A at time t, and [A]₀ is the initial concentration of A. This equation resembles the formula for a straight line (y = mx + b), where: - y = \(\ln[A]\) - m = -k - x = t - b = \(\ln[A]_0\) Thus, plotting ln[A] versus time (t) will yield a straight line for a first-order reaction.

Now that we've determined that the linear relationship for a first-order reaction involves plotting ln[A] versus time, we can use the slope of the straight line from the graph to calculate the rate constant (k). As previously stated, the equation of the straight line for the first-order reaction is: \(\ln[A] = -kt + \ln[A]_0\) From this equation, we can see that the slope of the straight line (-k) is the rate constant multiplied by -1. To calculate the rate constant k, we take the negative of the slope of the straight line from the graph, which is the relationship between ln[A] and time: k = -slope So, by plotting ln[A] versus time for a first-order reaction, we can obtain the rate constant k by taking the negative of the slope of the resulting straight line.