Question

1. for a first order reaction, a plot of which variables will result in a straight line with a slope equal to -k?

a. loga vs 1/time
b. lna vs time
c. a vs time
d. a² vs time
e. 1/a vs time
indicate multiple choice selections in the table below

Explanation:

To determine which plot gives a straight line with slope \(-k\) for a first - order reaction, we start with the integrated rate law for a first - order reaction. The integrated rate law for a first - order reaction is \(\ln[A]=\ln[A]_0 - kt\), where \([A]\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration of the reactant, and \(k\) is the rate constant.

Step 1: Analyze the form of the equation

The equation \(\ln[A]=\ln[A]_0 - kt\) is in the form of a linear equation \(y = mx + c\), where \(y=\ln[A]\), \(x = t\), \(m=-k\) (the slope), and \(c = \ln[A]_0\) (the y - intercept).

Step 2: Analyze other options

• For option a: The integrated rate law for a first - order reaction is not of the form that would give a linear plot between \(\log[A]\) and \(1/\text{time}\). The correct form for \(\log\) (base 10) would be \(\log[A]=\log[A]_0-\frac{kt}{2.303}\), and it is not related to \(1/\text{time}\) in a linear way for a first - order reaction.
• For option c: For a first - order reaction, the concentration \([A]\) does not have a linear relationship with time. The concentration of the reactant in a first - order reaction decreases exponentially with time (\([A]=[A]_0e^{-kt}\)).
• For option d: \([A]^2\) vs time is not a linear relationship for a first - order reaction. From \([A]=[A]_0e^{-kt}\), \([A]^2 = [A]_0^2e^{-2kt}\), which is not a linear function of time.
• For option e: The plot of \(1/[A]\) vs time is the linear plot for a second - order reaction (the integrated rate law for a second - order reaction is \(\frac{1}{[A]}=\frac{1}{[A]_0}+kt\), with slope \(k\)).

Answer:

b. \(\ln[A]\) vs time