Question

1. a certain reaction has the following general form: aa→bb at a particular temperature and a₀ = 2.00×10⁻² m, concentration versus time data were collected for this reaction, and a plot of lna versus time resulted in a straight line with a slope value of -2.97×10⁻² 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 2.50×10⁻³ m?

Explanation:

Step1: Determine the reaction order

For a reaction \(aA
ightarrow bB\), if a plot of \(\ln[A]\) versus time is a straight line, the reaction is first - order.

The general rate law for a first - order reaction is \(Rate = k[A]\)

The integrated rate law for a first - order reaction is \(\ln[A]=\ln[A]_0 - kt\)

Comparing with the equation of a straight line \(y = mx + c\) (where \(y=\ln[A]\), \(x = t\), \(m=-k\) and \(c=\ln[A]_0\)), given that the slope \(m=- 2.97\times10^{-2}\text{ min}^{-1}\), so the rate constant \(k = 2.97\times10^{-2}\text{ min}^{-1}\)

Step2: Calculate the half - life

The formula for the half - life of a first - order reaction is \(t_{1/2}=\frac{\ln2}{k}\)

Substitute \(k = 2.97\times10^{-2}\text{ min}^{-1}\) into the formula:

\(t_{1/2}=\frac{\ln2}{2.97\times10^{-2}\text{ min}^{-1}}=\frac{0.693}{2.97\times10^{-2}\text{ min}^{-1}}\approx23.3\text{ min}\)

Step3: Calculate the time required for the concentration change

Using the integrated rate law \(\ln[A]=\ln[A]_0 - kt\)

We can rewrite it as \(t=\frac{\ln[A]_0-\ln[A]}{k}\)

Given \([A]_0 = 2.00\times10^{-2}M\) and \([A]=2.50\times10^{-3}M\)

\(\ln[A]_0=\ln(2.00\times10^{-2})=\ln2+\ln(10^{-2})\approx0.693 - 4.605=- 3.912\)

\(\ln[A]=\ln(2.50\times10^{-3})=\ln2.5+\ln(10^{-3})\approx0.916 - 6.908=- 5.992\)

\(t=\frac{-3.912+ 5.992}{2.97\times10^{-2}\text{ min}^{-1}}=\frac{2.08}{2.97\times10^{-2}\text{ min}^{-1}}\approx70.0\text{ min}\)

Answer:

a. Rate law: \(Rate = k[A]\), Integrated rate law: \(\ln[A]=\ln[A]_0 - kt\), Rate constant \(k = 2.97\times10^{-2}\text{ min}^{-1}\)

b. \(t_{1/2}\approx23.3\text{ min}\)

c. \(t\approx70.0\text{ min}\)