Problem 8 - Chemistry The Central Science In Si Units

Question

Which of the following linear plots do you expect for a reaction \(\mathrm{A} \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4\(]\)

Step-by-Step Solution

Answer

Zero-order: [A] vs. t; First-order: ln[A] vs. t; Second-order: 1/[A] vs. t.

Step 1: Understanding Zero-Order Kinetics

For a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of reactants. The integrated rate law for a zero-order reaction is given by: \[ [A] = [A]_0 - kt \]where \([A]\) is the concentration of \(A\) at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. The plot of \([A]\) versus time \(t\) should be a straight line with a negative slope of \(-k\).

Step 2: Understanding First-Order Kinetics

For a first-order reaction, the rate of reaction depends on the concentration of the reactant \(A\). The integrated rate law is:\[ \ln [A] = \ln [A]_0 - kt \]A plot of \(\ln [A]\) versus time \(t\) will yield a straight line. The slope of the line is \(-k\), indicating a linear relationship on a semi-logarithmic scale.

Step 3: Understanding Second-Order Kinetics

For a second-order reaction, the rate depends on the square of the concentration of \(A\). The integrated rate law is:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]In this case, a plot of \(\frac{1}{[A]}\) versus time \(t\) should be a straight line. The slope of this line is equal to \(k\).

Key Concepts

Zero-Order Kinetics

Zero-order kinetics describe a scenario where the rate of reaction does not change with the concentration of the reactant. This means that the reaction proceeds at a constant rate until the reactant is depleted. The plot that represents a zero-order reaction is a straight line when the concentration of the reactant \([A]\) is plotted against time \(t\). The formula for this relationship is given by the integrated rate law:

• \([A] = [A]_0 - kt\)

In this equation, \([A]_0\) is the initial concentration, and \(k\) is the rate constant with a negative slope of -k. This linear plot reflects a direct decrease in concentration over time. Thus, the higher the value of \(k\), the steeper the decline in concentration, signaling a faster reaction rate.

First-Order Kinetics

First-order kinetics represent reactions whose rate is directly proportional to the concentration of one reactant. These are quite common in chemical reactions. For first-order kinetics, a plot of the natural logarithm of the reactant concentration \(\ln[A]\) against time is used to gain a linear relationship. The integrated rate law for a first-order reaction is:

• \(\ln [A] = \ln [A]_0 - kt\)

This equation shows that the natural logarithm of concentration decreases linearly over time. The negative slope \(-k\) indicates the rate at which the reaction occurs. This logarithmic relationship ensures that even with decreasing concentrations, the reaction continues to proceed at a reproducible rate.

Second-Order Kinetics

In second-order kinetics, the reaction rate is proportional to the square of the concentration of the reactant, or to the product of the concentrations of two reactants. These reactions tend to proceed more slowly compared to first-order reactions because the change in concentration must be more significant to affect the rate. The integrated rate law for second-order kinetics is expressed as:

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

For a second-order reaction, plotting \(\frac{1}{[A]}\) against time will yield a straight line. Here, the slope is \(k\), which signifies that a greater change in concentration results in greater alterations in reaction rate. This type of kinetic behavior is typical in reactions where two molecules of the same or different substances must collide for the reaction to progress.