Question

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.80 \times 10^{-3} M\), concentration versus time data were collected for this reaction, and a plot of \(1 /[\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(+3.60 \times 10^{-2} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) 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 \(\mathrm{A}\) to decrease to \(7.00 \times 10^{-4} M ?\)

Short answer

The reaction is second order with a rate law of \(\text{Rate} = k[\mathrm{A}]^2\), an integrated rate law of \(\frac{1}{[\mathrm{A}]_{t}} - \frac{1}{[\mathrm{A}]_{0}} = kt\), and a rate constant \(k = 3.60 \times 10^{-2} \frac{L}{mol \cdot s}\). The half-life is approximately 97.2 seconds, and it takes about 203.3 seconds for the concentration of A to decrease to \(7.00 \times 10^{-4} M\).

Step-by-step solution

Determine the order of reaction

To determine the order of reaction, we examine the 1/[A] versus time plot. A straight line with a positive slope result from this plot implies it is a second-order reaction. Therefore, we can express the integrated rate law and rate law for a second-order reaction.

Express integrated rate law and rate law for a second-order reaction

For a second-order reaction: Integrated rate law: \[ \frac{1}{[\mathrm{A}]_{t}} - \frac{1}{[\mathrm{A}]_{0}} = kt \] Rate law: \[ \text{Rate} = k[\mathrm{A}]^2 \]

Find the rate constant (k)

Given the straight-line slope from the plot, we can set it equal to the rate constant, k. \[ k = 3.60 \times 10^{-2} \frac{L}{mol \cdot s} \]

Calculate half-life

For a second-order reaction, the half-life equation is: \[t_{1/2} = \frac{1}{k[\mathrm{A}]_{0}}\] Substitute the initial concentration of A and the rate constant into the equation: \[t_{1/2} = \frac{1}{(3.60 \times 10^{-2} \frac{L}{mol \cdot s})(2.80 \times 10^{-3} M)}\] Calculate t1/2: \[t_{1/2} \approx 97.2 \, s\] The half-life of this reaction is about 97.2 seconds.

Calculate the time for [A] to decrease to the desired concentration

We are asked to find the time required for the concentration of A to decrease to 7.00 x 10^{-4} M. We can use the integrated rate law to find this: \[\frac{1}{[\mathrm{A}]_{t}} - \frac{1}{[\mathrm{A}]_{0}} = kt \] Plug in the known values and solve for t: \[\frac{1}{7.00 \times 10^{-4}M} - \frac{1}{2.80 \times 10^{-3}M} = (3.60 \times 10^{-2} \frac{L}{mol \cdot s})t\] \[t \approx 203.3 \, s\] Thus, it takes around 203.3 seconds for the concentration of A to decrease to 7.00 x 10^{-4} M.

Key concepts

Rate Law

The rate law is a mathematical equation that relates the rate of a chemical reaction to the concentration of its reactants. For many reactions, the rate is proportional to the concentration of one or more reactants, raised to a power. In the case of a second-order reaction, the rate law expression takes the form:
\[\begin{equation}Rate = k[A]^2ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline\end{equation}\]where

• k is the rate constant,
• [A] represents the concentration of reactant A, and
• The exponent '2' indicates that the reaction is second-order with respect to A.

Understanding the rate law helps us predict how changes in concentration affect the speed of the reaction.

Integrated Rate Law

The integrated rate law provides a relationship between the concentration of reactants and time. For a second-order reaction, this relationship is illustrated by plotting the reciprocal of the concentration of reactant A against time, which yields a straight line if the reaction follows second-order kinetics. The equation for the integrated rate law for a second-order reaction is
\[\begin{equation}\frac{1}{[A]_t} - \frac{1}{[A]_0} = ktewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline\end{equation}\]where

• t is time,
• k is the rate constant, and
• [A]_t and [A]_0 are the concentrations of A at time t and initially, respectively.

By using this law, one can calculate the concentration of reactant A at any point in time.

Rate Constant

The rate constant, denoted by the symbol k, is a proportionality constant in the rate law equation. It is specific to each chemical reaction and varies with temperate and the presence of catalysts. For the second-order reaction in our example, the rate constant was determined from the slope of the plot of
\[\begin{equation}1/[A]\end{equation}\]against time, with the value of
\[\begin{equation}3.60 \times 10^{-2} \frac{L}{mol \cdot s}.\end{equation}\]The magnitude of k gives an indication of how quickly the reaction proceeds; a larger value of k means a faster reaction. Understanding the rate constant is crucial when comparing the rates of different reactions or predicting how the rate will change under varied conditions.

Reaction Half-life

The term half-life, represented as
\[\begin{equation}t_{1/2},\end{equation}\]refers to the time required for half of the reactant to be consumed in a chemical reaction. For second-order reactions, half-life depends on both the rate constant k and the initial concentration of the reactant [A]_0. The half-life equation for a second-order reaction is
\[\begin{equation}t_{1/2} = \frac{1}{k[A]_0}.\end{equation}\]From our example, the half-life is calculated to be approximately 97.2 seconds using the initial concentration and the rate constant. This concept is especially useful in comparing the lifetimes of reactive species under certain conditions and in applications such as pharmacokinetics and radioactivity.

Concentration-Time Relationship

The concentration-time relationship in chemical kinetics describes how the concentration of reactants changes over the course of a reaction. By understanding this relationship, one can predict the amount of reactant or product present at any given time. In the context of a second-order reaction, the integrated rate law shows that the reciprocal of the concentration of A changes linearly with time. This relationship was used in our exercise to calculate the time it takes for the concentration of A to decrease to
\[\begin{equation}7.00 \times 10^{-4} \end{equation}\]from an initial concentration of
\[\begin{equation}2.80 \times 10^{-3} M.\end{equation}\]The time was found to be approximately 203.3 seconds. Such calculations are fundamental in fields like chemical manufacturing, where precise reaction timing can be critical.