Question

A certain reaction has the following general form: $$\mathrm{aA} \longrightarrow \mathrm{bB}$$ At a particular temperature and \([\mathrm{A}]_{0}=2.80 \times 10^{-3} \mathrm{M},\)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 A to decrease to \(7.00 \times 10^{-4} \mathrm{M} ?\)

Short answer

The reaction is a first-order reaction with a rate law of \(rate = k[\mathrm{A}]\). The rate constant, \(k\), is \(3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). The half-life of the reaction is 19.25 seconds, and the time required for the concentration of A to decrease to \(7.00 \times 10^{-4} \mathrm{M}\) is approximately 27 seconds.

Step-by-step solution

Identify the reaction order and derive the rate law

Since a plot of \(1 /[\mathrm{A}]\) versus time results in a straight line, this is a 1st order reaction. The rate law for the 1st order reaction can be given as: $$rate = k[\mathrm{A}]^n$$ where n = 1 (since it's a 1st order reaction), k = rate constant, [𝐴] = concentration of A. Then the rate law can be simplified as: $$rate = k[\mathrm{A}]$$

Derive the integrated rate law and calculate the rate constant

For a 1st order reaction, the integrated rate law is given by: $$\ln[\mathrm{A}] = -kt + \ln[\mathrm{A}]_0$$ We also know that the plot of \(1 /[\mathrm{A}]\) versus time resulted in a straight line with a slope of \(+3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). Since the reaction is of 1st order, the slope of a plot of \(1 /[\mathrm{A}]\) versus time is equal to the rate constant k. Therefore, we can obtain the rate constant as: $$k = 3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$$

Calculate the half-life for this reaction

For a 1st order reaction, the half-life can be calculated using the following formula: $$t_{1/2} = \frac{0.693}{k}$$ Using the rate constant we found in step 2, we can calculate the half-life: $$t_{1/2} = \frac{0.693}{3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}} = 19.25 \, \mathrm{s}$$ The half-life of this reaction is 19.25 seconds.

Determine the time required for the concentration of A to decrease to the given value

The initial concentration of A is given as \([\mathrm{A}]_{0}=2.80 \times 10^{-3} \mathrm{M}\). We need to find the time required for the concentration of A to decrease to \(7.00 \times 10^{-4} \mathrm{M}\) using the 1st order integrated rate law: $$\ln[\mathrm{A}] = -kt + \ln[\mathrm{A}]_0$$ Plugging in the values: $$\ln(7.00 \times 10^{-4} \mathrm{M}) = -k \cdot t + \ln(2.80 \times 10^{-3} \mathrm{M})$$ Now, we will solve for t: $$t = \frac{\ln(7.00 \times 10^{-4}) - \ln(2.80 \times 10^{-3})}{-k}$$ $$t = \frac{\ln(7.00 \times 10^{-4}) - \ln(2.80 \times 10^{-3})}{-3.60 \times 10^{-2} \mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}} \approx 27 \, \mathrm{s}$$ The time required for the concentration of A to decrease to \(7.00 \times 10^{-4} \mathrm{M}\) is approximately 27 seconds.

Key concepts

Rate Law

Understanding the rate law is crucial for studying the speed of chemical reactions. It describes how the rate of a reaction is related to the concentration of its reactants. In its simplest form, the rate law can be expressed as \( rate = k[\mathrm{A}]^n \), where \( k \) represents the rate constant, \( [\mathrm{A}] \) is the concentration of reactant A, and \( n \) is the reaction order with respect to A. This relationship allows chemists to predict how changes in concentration will affect the reaction rate. For instance, if \( n \) equals 1, the rate is directly proportional to the concentration of \( A \), implying that doubling \( A \) will double the reaction rate. This formula is not only fundamental in kinetics but also essential for controlling reactions in industrial processes.

In the given exercise, the appearance of a straight line when plotting \( 1 / [\mathrm{A}] \) versus time indicates a first-order reaction (where \( n = 1 \) ). Therefore, we simplify the rate law to \( rate = k[\mathrm{A}] \), highlighting the direct proportionality between the rate and the concentration of A.

Integrated Rate Law

While the rate law gives us the instant look at how the reaction progresses, the integrated rate law provides us with a bigger picture by correlating concentrations of reactants over time. For a first-order reaction, we use the formula \( \ln[\mathrm{A}] = -kt + \ln[\mathrm{A}]_0 \), where \( \ln[\mathrm{A}]_0 \) is the natural logarithm of the initial concentration of A, \( k \) is the rate constant, and \( t \) is the time elapsed. By integrating the rate law over time, this equation allows us to calculate how much reactant remains after a certain period or how long it will take for the reaction to reach a specific stage.

In the exercise, with the given slope from the plot, we identify the rate constant, which then assists in formulating the integrated rate law for that specific reaction. We're able to establish the decline in concentration of A over time, essential for predicting the course of the reaction.

Reaction Order

Reaction order is a term that reveals the dependency of the rate on the concentration of each reactant. It's usually discovered by experimentation and is incredibly useful as it determines the form of the rate law. In other words, it tells us by what factor the rate will increase or decrease when the concentration of a reactant is modified. The reaction orders can be zero, first, second, or even fractional, which are indicative of different kinetic behaviors.

For instance, a zero-order reaction rate isn't affected by the concentration of the reactant, while a first-order reaction, like the one described in our textbook problem, means that the rate is directly proportional to the concentration of a single reactant. Knowing the reaction order helps chemists control the reaction rate and design chemical processes effectively.

Half-Life of a Reaction

The half-life of a reaction, represented as \( t_{1/2} \), is the time required for the concentration of a reactant to reach half of its initial value. It is an intrinsic characteristic of a reaction, depicting how quickly reactants are consumed. For first-order reactions, the half-life is constant and does not depend on the concentration of the reactant, which is a unique attribute among reaction orders. The formula for calculating the half-life of a first-order reaction is \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant.

In our exercise, when we use the determined rate constant in this formula, we find the half-life for the reaction, giving us insight into the reaction's kinetics. This constant half-life characteristic is particularly useful for pharmacokinetics and radioisotope dating, where it assists in determining how long a substance will remain active or detectable.

Rate Constant

The rate constant \( k \) is a coefficient in the rate law that provides the speed factor for a reaction at a given temperature. This numerical value accounts for the specific conditions under which the reaction occurs. It is influenced by the nature of the reactants, the presence of a catalyst, and environmental conditions like temperature and pressure. The units of the rate constant vary depending on the reaction order and are essential for ensuring the rate law equation is dimensionally consistent.

In practical terms, a larger rate constant signifies a quicker reaction. For the problem in question, we extracted this rate constant from the slope of the plot of \( 1 / [\mathrm{A}] \) against time. With the value of the rate constant in hand, we can easily calculate other critical aspects of a reaction, such as its half-life and the time needed for a certain amount of reactant to be consumed.