Question

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

Short answer

a. The reaction is second-order with rate law: \(\textrm{Rate} = k[\textrm{A}]^2\), integrated rate law: \(\frac{1}{[\mathrm{A}]} = kt + \frac{1}{[\mathrm{A}]_{0}}\), and rate constant: \(k = 3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). b. The half-life of the reaction is: \(t_{1/2} = \frac{1}{(3.60 \times 10^{-2} \mathrm{L/mol \cdot s})(2.80 \times 10^{-3} \mathrm{M})}\) c. To find the time required for the concentration of A to decrease to \(7.00 \times 10^{-4} M\), use the equation: \(\frac{1}{7.00 \times 10^{-4} \mathrm{M}} = (3.60 \times 10^{-2} \mathrm{L/mol} \cdot \mathrm{s})\times t + \frac{1}{2.80 \times 10^{-3} \mathrm{M}}\). Solve for t.

Step-by-step solution

Determine the order of the reaction

Since the problem states that the plot of 1\(/[\mathrm{A}]\) versus time resulted in a straight line, this indicates that the reaction is of the second order. The general form of the second-order rate law is: $$\textrm{Rate} = k[\textrm{A}]^2$$

Find the integrated rate equation for the second order reaction

For a second-order reaction, the integrated rate equation is: $$\frac{1}{[\mathrm{A}]} = kt + \frac{1}{[\mathrm{A}]_{0}}$$

Calculate the rate constant

The problem statement provides the slope of the line of the graph of 1\(/[\mathrm{A}]\) versus time as \(+3.60 \times 10^{-2}\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). This slope corresponds to the rate constant k: $$k = 3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$$

Calculate the half-life for the second order reaction

The half-life for a second-order reaction is given by the equation: $$t_{1/2} = \frac{1}{k[\mathrm{A}]_{0}}$$ Using the initial concentration \([\mathrm{A}]_{0}=2.80 \times 10^{-3} M\) and the calculated value of k, the half-life can be determined: $$t_{1/2} = \frac{1}{(3.60 \times 10^{-2} \mathrm{L/mol \cdot s})(2.80 \times 10^{-3} \mathrm{M})}$$ Compute \(t_{1/2}\) to obtain the value for half-life.

Calculate the time required for the concentration of A to decrease to a certain value

We are given the final concentration of A, \([\mathrm{A}] = 7.00 \times 10^{-4} M\). We can use the integrated rate equation for the second-order reaction to calculate the time required: $$\frac{1}{7.00 \times 10^{-4} \mathrm{M}} = (3.60 \times 10^{-2} \mathrm{L/mol} \cdot \mathrm{s})\times t + \frac{1}{2.80 \times 10^{-3} \mathrm{M}}$$ Solve for time, t, to find how long it takes for the concentration of A to reach \(7.00 \times 10^{-4} M\).

Key concepts

Rate Law

Understanding the rate law is crucial in determining the speed of a chemical reaction. For a second-order reaction like the one in our problem, the rate law can be represented as:

• Rate = k[ A]^2

Here, **k** is the rate constant, and [A] is the concentration of reactant A. This law suggests that the rate of reaction is directly proportional to the square of the concentration of A. When the concentration of A changes, the reaction rate varies significantly, demonstrating the impact of concentration on reaction speed.
By examining the dimensions of the rate constant provided, L/mol·s, we can confirm the second-order nature as it aligns with the dimensions expected for a second-order reaction. The graph of 1/[A] vs. time yielding a straight line further verifies this, as it indicates that the relationship is second-order rather than zero- or first-order.
To connect the dots, always observe the units used in the rate law, which guide you in understanding the reaction order and its kinetics.

Integrated Rate Law

For reactions that do not occur in a single, simple step, integrated rate laws become essential. These equations help us correlate concentration with time, which is particularly useful for reactions monitored over periods.
For a second-order reaction like our example, the integrated rate law is given by:

• \( \frac{1}{[\text{A}]} = kt + \frac{1}{[\text{A}]_{0}} \)

\([\text{A}]_{0}\) represents the initial concentration, while [A] indicates concentration at any given time t. This formula enables us to determine the concentration of a reactant left after a certain time has elapsed.
By examining the straight line from a plot of 1/[A] versus time, we see the constant, linear relationship between the variables - characteristic of second-order kinetics. Knowing the slope of this line provides the rate constant (k), essential for calculating how reactant concentrations change unexpectedly over time.

Half-Life

The half-life of a reaction is a helpful measure describing the time required for half of a reactant to be consumed. For second-order reactions, the half-life is distinct from first-order reactions where it remains constant.
In our context, the half-life for a second-order reaction is determined by:

• \( t_{1/2} = \frac{1}{k[A]_0} \)

Here, the half-life depends on both the initial concentration \([A]_0\) and the rate constant k. This dependency means that as the concentration of A decreases, so does the half-life, resulting in a progressively shorter time for each subsequent reaction half-life.
When calculating half-life, remember that this phenomenon reveals how robustly the initial reactant concentration impacts the speed and dynamics of the reaction over time. For our problem, knowing the initial concentration and the rate constant is necessary to derive an accurate half-life measurement.

Reaction Kinetics

Reaction kinetics play a key role in understanding how fast reactions occur and the factors influencing this speed. Kinetics explore the time aspect of reactions—how quickly products form and reactants deplete over time.
The study is vital for fields ranging from pharmaceuticals to environmental science. In this exercise, kinetics are reflected in the second-order characteristics demonstrated by the rate and integrated rate laws, which help predict concentrations at various time points.
Key components affecting kinetics include:

• Temperature: Often increases reaction speed as molecules possess more kinetic energy to collide
• Catalysts: Substances that change the speed of a reaction without being consumed
• Concentration: As seen in the rate laws, higher concentrations often increase reaction speed

Understanding these parameters provides insight into controlling reaction conditions to optimize desired outcomes. This means that for industrial processes, ensuring optimal concentrations and temperatures can lead to increased reaction efficiency and yield.