Question

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

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

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.

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.

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.

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\(^{-1}\)

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.

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.

Key concepts

Differential Rate Law

The differential rate law is a fundamental concept in chemical kinetics used to describe how the rate of a reaction relates to the concentration of reactants. For a second-order reaction, such as the one examined in the original exercise, this law is expressed as: \[ -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^2 \] This equation tells us that the rate at which the reactant \([\mathrm{A}]\) decreases is directly proportional to the square of its concentration. Here, \(k\) is the rate constant, and \([\mathrm{A}]\) is the concentration of the reactant. - The negative sign indicates the decrease in concentration over time.- This form of rate law implies the reaction rate increases with increasing concentration as both are raised to the power of 2. This is distinct from first-order reactions, where the rate depends linearly on the concentration, and zero-order reactions, where the rate is independent of concentration. Understanding the differential rate law helps in predicting how quickly a reaction will proceed under given conditions.

Integrated Rate Law

The integrated rate law provides a relationship that allows us to determine the concentration of a reactant at any given time during the reaction. For a second-order reaction, the integrated rate law is formulated as follows: \[ \frac{1}{[\mathrm{A}]} - \frac{1}{[\mathrm{A}]_0} = kt \] - \([\mathrm{A}]_0\) represents the initial concentration of the reactant.- \(t\) represents the time that has passed. This equation can be transformed to graphically represent the reaction's progression—a plot of \(1/[\mathrm{A}]\) versus time forms a straight line, indicating a second-order process. - The slope of this line is equal to the rate constant, \(k\).- The straight line supports the confirmation of the reaction's order.This law is incredibly useful because it allows researchers to predict a reaction's future progress from early data and to determine reaction order from experimental findings.

Rate Constant

The rate constant \(k\) is a critical factor in the study of chemical kinetics. It acts as a proportionality factor in the rate laws. For the specific second-order reaction discussed, \(k\) is also the slope of the line from the plot of \(1/[\mathrm{A}]\) versus time. In the given case, the rate constant is \( 6.90 \times 10^{-2} \ \mathrm{s}^{-1} \). - This numerical value provides insight into how quickly the reaction proceeds.- A larger \(k\) means a faster reaction rate, while a smaller \(k\) implies a slower rate.It's essential for the rate constant to be experimentally determined under standard conditions such as temperature, since it can vary if experimental conditions are altered. Understanding the rate constant helps chemists control reaction environments effectively.

Half-Life

The concept of half-life in second-order reactions is notably different from that in first-order reactions. In second-order kinetics, the half-life depends on the initial concentration of the reactant, \([\mathrm{A}]_0\). The formula used is: \[ t_{1/2} = \frac{1}{k[\mathrm{A}]_0} \] The unique dependence on initial concentration leads to an interesting effect:- The first half-life is \(14.49\) seconds for \([\mathrm{A}]_0 = 0.100 \ \mathrm{M}\).- The second half-life, when the concentration reduces to \(0.050 \ \mathrm{M}\), is \(28.98\) seconds. This increasing half-life signifies that as the reaction progresses, it takes longer for each successive half of the remaining reactant to be consumed, which contrasts with first-order reactions where the half-life remains constant. - Such behavior highlights the critical role initial concentrations play in controlling reaction timelines. Understanding half-life variations helps chemists to plan time-dependent reactions and scales effectively.