Problem 39 - Chemistry and Chemical Reactivity

Question

For the reaction \(2 \mathrm{C}_{2} \mathrm{F}_{4} \longrightarrow \mathrm{C}_{4} \mathrm{F}_{8},\) a graph of \(1 /\left[\mathrm{C}_{2} \mathrm{F}_{4}\right]\) versus time gives a straight line with a slope of \(+0.04 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate law for this reaction?

Step-by-Step Solution

Answer

Rate law: \( \text{rate} = 0.04 [\text{C}_2\text{F}_4]^2 \).

Step 1: Identify Reaction Order

A straight-line graph of \(1/[\text{C}_2\text{F}_4]\) versus time indicates a second-order reaction. For a second-order reaction, the integrated rate law has the form \(1/[A]_t = kt + 1/[A]_0\), which matches the given graph.

Step 2: Determine the Rate Constant

From the slope of the graph, we know that it represents the rate constant \(k\). Therefore, the rate constant \(k\) is \(0.04 \text{ L/mol} \cdot \text{s}\).

Step 3: Write the Rate Law

For a second-order reaction \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\), the rate law is written as: \( \text{rate} = k [\text{C}_2\text{F}_4]^2 \). Using the determined \(k = 0.04 \text{ L/mol} \cdot \text{s}\), the rate law is \( \text{rate} = 0.04 [\text{C}_2\text{F}_4]^2 \).

Key Concepts

Rate Constant

The rate constant, often symbolized as \( k \), is a crucial component in the study of chemical kinetics. It is a proportionality factor in the rate law that helps predict the speed of a reaction at any given concentration of reactants. In our exercise, the reaction \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\) illustrates this concept well. Here, the slope of the line from the graph of \(1/[\text{C}_2\text{F}_4]\) versus time gives us the rate constant \( k \).To calculate \( k \) for a second-order reaction, you simply identify the slope of the line. In this case, we found that \( k \) equals \(0.04 \text{ L/mol} \cdot \text{s}\). This value tells us how rapidly the reaction proceeds under the given conditions. The units of \( \text{L/mol} \cdot \text{s} \) are essential in confirming that the rate constant is appropriate for a second-order reaction.

Integrated Rate Law

The integrated rate law offers a mathematical relationship between the concentration of reactants and time. For a second-order reaction like \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\), the integrated rate law is expressed as:\[\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\]where:

• \([A]_t\) is the concentration of the reactant at time \( t \).
• \([A]_0\) is the initial concentration of the reactant.
• \( k \) is the rate constant.

This form of the rate law is particularly useful because it allows us to determine the concentration of a reactant at any given time, assuming we know the initial concentration and the rate constant. In the context of our exercise, plotting \(1/[\text{C}_2\text{F}_4]\) versus time resulted in a straight line, confirming the reaction's second-order nature and enabling us to derive \( k \) from the graph's slope.

Reaction Order

Understanding the reaction order is vital in grasping how different factors influence the reaction rate. The order of a reaction refers to the power to which the concentration of a reactant is raised in the rate law expression. In the case of our reaction \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\), the reaction order is determined to be second-order.For second-order reactions, the rate of reaction is directly proportional to the square of the concentration of one reactant, as shown in the rate law:\[\text{rate} = k [\text{C}_2\text{F}_4]^2\]This rate law tells us that if you double the concentration of \(\text{C}_2\text{F}_4\), the rate of reaction increases by a factor of four. Recognizing a second-order reaction from the graph of \(1/[\text{C}_2\text{F}_4]\) versus time is key, as the linear relationship directly indicates the second-order nature of the reaction.