Problem 52 - Chemistry and Chemical Reactivity

Question

Data for the following reaction are given in the table. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})$$ $$\begin{array}{clll}\text { Experiment } & \text { [NO] }(\mathrm{M}) & {\left[\mathrm{Br}_{2}\right](\mathrm{M})} &\begin{array}{l}\text { Initial Rate } \\\\(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s})\end{array} \\\\\hline 1 & 1.0 \times 10^{-2} & 2.0 \times 10^{-2} & 2.4 \times 10^{-2} \\\2 & 4.0 \times 10^{-2} & 2.0 \times 10^{-2} & 0.384 \\\3 & 1.0 \times 10^{-2} & 5.0 \times 10^{-2} & 6.0 \times 10^{-2} \\\\\hline\end{array}$$ What is the order of the reaction with respect to [NO] and \(\left[\mathrm{Br}_{2}\right],\) and what is the overall order of the reaction?

Step-by-Step Solution

Answer

Order with respect to [NO] is 2, [Br2] is 1, overall order is 3.

Step 1: Understanding the Rate Law

For a reaction of the form \(2 \text{NO}(g) + \text{Br}_2(g) \rightarrow 2 \text{NOBr}(g)\), the rate law is given by \( \text{Rate} = k [\text{NO}]^m [\text{Br}_2]^n \). Here, \(m\) and \(n\) are the orders of the reaction with respect to \([\text{NO}]\) and \([\text{Br}_2]\) respectively, and \(k\) is the rate constant. Our goal is to determine \(m\), \(n\), and the overall reaction order \(m + n\).

Step 2: Determining the Order with Respect to [NO]

To find the order \(m\) with respect to \([\text{NO}]\), compare experiments where only \([\text{NO}]\) changes. For Experiments 1 and 2, \([\text{NO}]\) increases from \(1.0 \times 10^{-2}\) to \(4.0 \times 10^{-2}\), a 4-fold increase, while \([\text{Br}_2]\) remains constant. The rate increases from \(2.4 \times 10^{-2}\) to \(0.384\), a 16-fold increase. Thus, \((4)^m = 16\), giving \(m = 2\).

Step 3: Determining the Order with Respect to [Br2]

To find the order \(n\) with respect to \([\text{Br}_2]\), compare experiments where only \([\text{Br}_2]\) changes. For Experiments 1 and 3, \([\text{Br}_2]\) increases from \(2.0 \times 10^{-2}\) to \(5.0 \times 10^{-2}\), a 2.5-fold increase, while \([\text{NO}]\) remains constant. The rate increases from \(2.4 \times 10^{-2}\) to \(6.0 \times 10^{-2}\), a 2.5-fold increase. Thus, \((2.5)^n = 2.5\), giving \(n = 1\).

Step 4: Calculating the Overall Reaction Order

The overall order of the reaction is the sum of the individual orders, \(m + n\). Thus, the overall reaction order is \(2 + 1 = 3\).

Key Concepts

Rate Law

The rate law is a crucial concept in understanding how the speed of a chemical reaction depends on the concentration of reactants. For the given reaction of nitrogen monoxide (NO) and bromine (\(\mathrm{Br}_{2}\)), the rate at which the product forms can be expressed in terms of the reactant concentrations. The mathematical representation of the rate law is:\[\text{Rate} = k [\text{NO}]^m [\text{Br}_2]^n\]Here:- \(k\) is the rate constant, unique to each reaction at a given temperature.- \([\text{NO}]\) and \([\text{Br}_2]\) are concentrations of the reactants.- \(m\) and \(n\) represent the reaction orders with respect to NO and \(\text{Br}_2\) respectively.To find \(m\) and \(n\), we analyze how changes in reactant concentrations affect the reaction rate. Understanding the rate law helps elucidate the reaction mechanism. The overall order gives insights into how sensitive the reaction rate is to concentration changes.

Reaction Kinetics

Reaction kinetics studies how fast reactions proceed, focusing on reaction speeds under varying conditions. In the context of the given reaction, the goal is to determine the influence of NO and \(\text{Br}_2\) concentrations on the reaction rate.Through kinetics, we can decipher whether a reaction will occur quickly or slowly. By analyzing Experiment 1 and Experiment 2, where the \([\text{NO}]\) increases fourfold while maintaining constant \([\text{Br}_2]\), the rate's 16-fold increase can be observed. This is solved by the equation \((4)^m = 16\), identifying \(m = 2\). Similarly, looking at Experiments 1 and 3, where \([\text{Br}_2]\) changes, we see that the changes in the concentration cause a proportional change in rate, leading us to \(n = 1\). Together, these findings depict how kinetics can shed light on reaction behavior and aid predictions. Terms like half-life, activation energy, and catalysts are also fundamental to kinetics, but this problem primarily concerns the concentration dependency.

Experimental Data Analysis

Experimental data analysis is integral for determining reaction orders, which play a pivotal role in forming the rate law. The experiments provide various initial concentrations and rates, forming a data set from which patterns are identified. This process involves comparing experiments to see how a change in one reactant alters the rate.For NO, by comparing Experiment 1 and Experiment 2, where only \([\text{NO}]\) changes, we calculate the rate change to infer \(m\). A similar technique is employed utilizing Experiment 1 and Experiment 3 to determine \(n\) for \([\text{Br}_2]\). These comparisons:- Help isolate variables, focusing analysis where only one factor changes at a time.- Provide clear insights into the reaction's behavior with varied conditions.- Allow us to conclude both individual and overall reaction orders.The careful analysis of this data underpins accurate conclusion formation and augments our understanding of underlying chemical processes.