A certain reaction has the form $$ \mathrm{aA} \longrightarrow $$ At a particular temperature, concentration versus time data were collected. A plot of 1\(/[\mathrm{A}]\) versus time (in seconds) gave a straight line with a slope of \(6.90 \times 10^{-2} .\) What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If \([\mathrm{A}]_{0}\) for this reaction is \(0.100 M,\) what is the first half-life (in seconds)? If the original concentration (at \(t=0 )\) is \(0.100 M,\) what is the second half-life (in seconds)?

Short Answer

Expert verified
The differential rate law for this reaction is \(-\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^2\), the integrated rate law is \(\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A}]_0} = kt\), the rate constant \(k = 6.90 \times 10^{-2}\) s\(^{-1}\), the first half-life is approximately \(14.49\) seconds, and the second half-life is approximately \(28.98\) seconds.

Step by step solution

01

Identify the order of the reaction

Since the plot of \(1/[\mathrm{A}]\) versus time gave a straight line, we can identify the order of the reaction. A straight line in the given plot indicates that this reaction is of second order.
02

Write the differential rate law for a second-order reaction

Since we know that it's a second-order reaction, we can write the differential rate law as follows: \(-\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^2\) where \(k\) is the rate constant and \([\mathrm{A}]\) is the concentration of reactant A at any time t.
03

Write the integrated rate law for a second-order reaction

For a second-order reaction, we can integrate the differential rate law and write the integrated rate law as follows: \(\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A}]_0} = kt\) Here, \([\mathrm{A}]_0\) is the initial concentration of the reactant A, and \(t\) is the time.
04

Determine the rate constant, k, for the reaction

According to the given information, the slope of the plot of \(1/[\mathrm{A}]\) versus time is \(6.90 \times 10^{-2}\) s\(^{-1}\). For a second-order reaction, the slope of this plot is equal to the rate constant, k. Therefore, the value of k is: \(k = 6.90 \times 10^{-2}\) s\(^{-1}\)
05

Calculate the first half-life

For a second-order reaction, the half-life (\(t_{1/2}\)) can be calculated using the following formula: \(t_{1/2} = \frac{1}{k[\mathrm{A}]_0}\) Given that the initial concentration at \(t=0\) is \([\mathrm{A}]_0 = 0.100\) M, we can calculate the first half-life: \(t_{1/2} = \frac{1}{(6.90 \times 10^{-2} \ \mathrm{s}^{-1})(0.100 \ \mathrm{M})} = 14.49\) s Therefore, the first half-life is approximately \(14.49\) seconds.
06

Calculate the second half-life

Since half-life in a second-order reaction depends on the initial concentration, the second half-life will be different from the first half-life. The second half-life starts when the concentration reaches half of the initial concentration. Therefore, for the second half-life calculation, \([\mathrm{A}]_0 = 0.050\) M. Using the formula for a second-order reaction, we can calculate the second half-life: \(t_{1/2} = \frac{1}{k[\mathrm{A}]_0} = \frac{1}{(6.90 \times 10^{-2} \ \mathrm{s}^{-1})(0.050 \ \mathrm{M})} = 28.98\) s The second half-life is approximately \(28.98\) seconds. In conclusion, the differential rate law for this reaction is \(-\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^2\), the integrated rate law is \(\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A}]_0} = kt\), the rate constant \(k = 6.90 \times 10^{-2}\) s\(^{-1}\), the first half-life is approximately \(14.49\) seconds, and the second half-life is approximately \(28.98\) seconds.

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