Problem 96 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

Step-by-Step Solution

Answer

In conclusion, both Reaction (1) and Reaction (2) follow a 2nd order rate law based on their half-life trends. For Reaction (1), the half-life getting longer as the reaction proceeds indicates a 2nd order rate law, while for Reaction (2), the half-life getting shorter as the reaction proceeds also represents a 2nd order rate law, but with a decreasing concentration of another reactant, leading to an overall faster rate as the reaction progresses.

Step 1: Overview of Rate Laws and Half-Lives

Rate laws can be described as: rate = k [A]^n [B]^m, where k is the rate constant and n and m are the reaction orders for reactants A and B, respectively. Half-life is the time it takes for the concentration of a reactant to decrease to half its initial value. Now, let's study the half-life behaviors of 0th, 1st, and 2nd order reactions.

Step 2: 0th Order Reactions

For a 0th order reaction, the rate law is: rate = k [A]^0 = k. The half-life can be calculated using the equation: t(1/2) = [A]0 / (2k), where [A]0 is the initial concentration of reactant A. Since the half-life depends on the initial concentration and is independent of the reaction progress, this does not match any of the given reaction characteristics.

Step 3: 1st Order Reactions

For a 1st order reaction, the rate law is: rate = k [A]^1 = k[A]. The half-life can be calculated using the equation: t(1/2) = ln(2) / k. The half-life is constant and does not change as the reaction proceeds, so this does not match any of the given reaction characteristics either.

Step 4: 2nd Order Reactions

For a 2nd order reaction, the rate law is: rate = k [A]^2. The half-life can be calculated using the equation: t(1/2) = 1 / (k[A]0), where [A]0 is the initial concentration of reactant A. Here, the half-life is inversely proportional to the initial concentration of the reactant. Thus, the half-life increases as the concentration decreases, and vice versa.

Step 5: Conclusion

By comparing the half-life behaviors of 0th, 1st, and 2nd order reactions, we can conclude that: Reaction (1): The half-life getting longer as the reaction proceeds indicates that this reaction follows a 2nd order rate law. Reaction (2): The half-life getting shorter as the reaction proceeds also indicates a 2nd order rate law, but with a decreasing concentration of the other reactant in addition to reactant A, leading to an overall faster rate as the reaction progresses. Therefore, both reactions are 2nd order based on their half-life trends.

Key Concepts

Half-Life

In the realm of chemical kinetics, the half-life concept defines the time required for the concentration of a reactant to reduce to half of its initial amount. It's a crucial measure for understanding the speed and mechanism of chemical reactions.
In simpler reactions like the first-order reactions, the half-life remains constant, meaning it does not change as the reaction progresses. For this type of reaction, we use the equation: \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. Because it's independent of the initial concentration, you get the same result no matter how far the reaction has progressed.

• Zero Order Reaction: Here, the half-life is directly reliant on the starting concentration: \( t_{1/2} = \frac{[A]_0}{2k} \). As the reaction proceeds, the half-life decreases since it depends on \( [A]_0 \).
• Second Order Reaction: For these reactions, the initial concentration is inversely related to the half-life: \( t_{1/2} = \frac{1}{k[A]_0} \). This implies that as the concentration decreases, the half-life increases.

Hence, by observing how half-life changes as a reaction proceeds, you can infer a lot regarding the reaction's order.

Rate Laws

To evaluate how quickly a reaction occurs, the rate law is the tool at your disposal. It describes the relationship between the concentration of reactants and the rate of the reaction, generally portrayed as: \( ext{rate} = k[A]^n[B]^m \). The constants \( n \) and \( m \) found in this equation are known as the reaction orders, while \( k \) is the rate constant.

• Zero Order: Here, the rate law is independent of the concentration: \( ext{rate} = k \).
• First Order: The rate varies linearly with the concentration: \( ext{rate} = k[A] \).
• Second Order: This rate law changes with the square of the concentration: \( ext{rate} = k[A]^2 \).

These laws allow us to calculate the reaction rate at any given time in relation to reactant concentrations. By comparing these laws with half-life trends, one can determine the order of a reaction, as observed in the original exercise. For example, if a half-life increases as the reaction proceeds, it often hints at a second-order reaction.

Reaction Orders

The concept of reaction orders uncovers how a reaction's rate is influenced by the concentration of its reactants. The order of a reaction, indicated by the exponents such as \( n \) and \( m \) in the rate law equation \( ext{rate} = k[A]^n[B]^m \), provides depth understanding of the reaction kinetics.

• Zero Order: The concentration of reactants has no effect on the rate, and the reaction proceeds at a constant rate.
• First Order: The reaction rate is proportional to the concentration of a single reactant. It implies each molecule of reactant plays an equivalent part in the reaction rate.
• Second Order: Either the square of one reactant's concentration influences the reaction rate, or the product of two reactant concentrations does.

Understanding reaction orders not only helps in estimating how a reaction progresses but also in determining the appropriate rate law. It was evident in the exercise that a changing half-life reflects the reaction order, consequently leading to an accurate conclusion about the reaction type, such as second order.