Question

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) , what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part line? (b) What is the slope of the straight line from part (a)? (c) Does the half-life of a second-order reaction increase, decrease, or remain the same as the reaction proceeds?

Short answer

(a) For a generic second-order reaction A → B, the quantity that, when graphed versus time, will yield a straight line is \(\frac{1}{[A]}\). (b) The slope of the straight line from part (a) is the rate constant \(k\). (c) The half-life of a second-order reaction increases as the reaction proceeds.

Step-by-step solution

Derive the integrated rate law for a second-order reaction

For a second-order reaction A → B, the rate law can be written as: \[r = k[A]^2\] To find the integrated rate law, first, rewrite the rate law in terms of concentration and differential: \[\frac{d[A]}{dt} = -k[A]^2\] Now, separate variables and integrate both sides: \[\int \frac{d[A]}{[A]^2} = -k \int dt\] \[\frac{-1}{[A]} = -kt + C\]

Find the quantity that gives a straight line

To find the quantity that, when plotted against time, will give a straight line, we want a linear relationship between that quantity and time. In the integrated rate law, we have: \[\frac{1}{[A]} = kt + \frac{1}{[A]_0}\] The relationship between \(\frac{1}{[A]}\) and \(t\) is linear. Therefore: (a) The graph of \(\frac{1}{[A]}\) versus time (\(t\)) will yield a straight line for a second-order reaction.

Determine the slope of the straight line

From the above linear equation of the form \(y = mx + b\), where \(y\) is \(\frac{1}{[A]}\), and \(x\) is \(t\), and the slope (m) is: (b) The slope of the straight line from part (a) is \(k\), the rate constant of the reaction.

Analyze the half-life for second-order reactions

The equation for the half-life (\(t_{1/2}\)) for a second-order reaction is: \[ t_{1/2} = \frac{1}{k[A]_0} \] As the reaction progresses, the concentration of A, [A]₀, decreases. Since the half-life is inversely proportional to the initial concentration, as [A]₀ decreases, the half-life will increase. Therefore: (c) The half-life of a second-order reaction increases as the reaction proceeds.

Key concepts

Integrated Rate Law

In understanding second-order reactions, the integrated rate law plays a pivotal role. For a reaction like \( \text{A} \rightarrow \text{B} \), the rate at which the concentration of \([A]\) changes is squared. This is reflected in the rate law expression:

• The rate law: \( r = k[A]^2 \)

To derive the integrated rate law, we start by expressing the change in concentration over time:

• \( \frac{d[A]}{dt} = -k[A]^2 \)

By integrating this expression, we arrive at:

• \( \frac{1}{[A]} = kt + \frac{1}{[A]_0} \)

This provides us a formula showing the relationship between the concentration of \( [A] \), time \( t \), and the rate constant \( k \).

Notice how \( \frac{1}{[A]} \) changes linearly with time. By plotting \( \frac{1}{[A]} \) versus time, we get a straight line, illustrating the direct relationship for second-order reactions. This nature of the graph helps in determining reaction characteristics, such as rate constant and progress.

Rate Constant

The rate constant, denoted as \( k \), is an essential parameter in second-order reactions. It not only indicates the speed of the reaction but also appears as the slope in the linearized form of the integrated rate law:

• \( y = mx + b \rightarrow \frac{1}{[A]} = kt + \frac{1}{[A]_0} \)
• Slope \( (m) = k \)

Here, \( m \), the slope of the line on the graph of \( \frac{1}{[A]} \) versus \( t \), represents the rate constant \( k \).

The rate constant can be determined from experimental data by measuring how concentrations of \( A \) change with time and then plotting these on a graph. Understanding \( k \) is key to quantitatively predicting how fast a reaction will proceed under given conditions. It allows chemists to compare reaction rates and determine the efficiency of a process.

In enzyme kinetics and industrial applications, \( k \) becomes a crucial parameter in adjusting reactions to desired speeds.

Half-Life

The concept of half-life \( (t_{1/2}) \) in second-order reactions is different from that in first-order reactions. For a second-order reaction, the half-life equation is:

• \( t_{1/2} = \frac{1}{k[A]_0} \)

This formula illustrates that the half-life is inversely proportional to the initial concentration of \([A]_0\) and directly proportional to the inverse of the rate constant \( k \).

As the reaction progresses and \([A]_0\) decreases, the half-life \( t_{1/2} \) increases. This means that each successive half-life is longer than the previous one, which is opposite to what occurs in first-order reactions.

Understanding this behaviour is essential when predicting the time required for a reaction to reach a certain extent, particularly in processes such as pharmaceutical kinetics, where timing of drug release and decay is crucial.