/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Given the following initial-rate... [FREE SOLUTION] | 91Ó°ÊÓ

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Given the following initial-rate data at \(300 \mathrm{~K}\) for the reaction described by the chemical equation $$ 2 \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{O}_{2}(g) $$ \begin{tabular}{ccc} \hline & & Initial rate of formation \\ {\(\left[\mathrm{NO}_{2}\right]_{0} / \mathbf{M}\)} & {\(\left[\mathbf{O}_{3}\right]_{\mathbf{0}} / \mathbf{M}\)} & of \(\mathbf{O}_{2}(g) / \mathbf{M} \cdot \mathrm{s}^{-1}\) \\ \hline \(0.65\) & \(0.80\) & \(2.61 \times 10^{4}\) \\ \(1.10\) & \(0.80\) & \(4.40 \times 10^{4}\) \\ \(1.70\) & \(1.55\) & \(1.32 \times 10^{5}\) \\ \hline \end{tabular} determine the reaction rate law and the value of the rate constant.

Short Answer

Expert verified
Rate law: \( \text{Rate} = 5.02 \times 10^4 [\text{NO}_2][\text{O}_3] \) with rate constant \( k = 5.02 \times 10^4 \text{ M}^{-1}\text{s}^{-1} \).

Step by step solution

01

Write the General Rate Law Expression

The general rate law expression for the reaction can be written as: \[\text{Rate} = k [\text{NO}_2]^m [\text{O}_3]^n\] where \( k \) is the rate constant, and \( m \) and \( n \) are the reaction orders with respect to \( \text{NO}_2 \) and \( \text{O}_3 \), respectively. Our goal is to determine \( m \), \( n \), and \( k \).
02

Determine the Order of Reaction with Respect to NOâ‚‚

Compare experiments 1 and 2, where the initial concentration of \( \text{O}_3 \) is constant (0.80 M), but \([\text{NO}_2]_0\) is varied. From experiment 1 to 2:- \([\text{NO}_2]_0\) goes from 0.65 to 1.10 M - Rate goes from \(2.61 \times 10^4 \) to \(4.40 \times 10^4 \) \( \text{M}\cdot\text{s}^{-1} \) Using the rate law: \[\frac{4.40 \times 10^4}{2.61 \times 10^4} = \left(\frac{1.10}{0.65}\right)^m\] \[1.686 = 1.692^m\] Thus, \( m \approx 1 \). Hence, the reaction is first order with respect to \( \text{NO}_2 \).
03

Determine the Order of Reaction with Respect to O₃

Compare experiments 2 and 3, where the initial concentration of \( \text{NO}_2 \) is constant (1.10 M), but \([\text{O}_3]_0\) is varied. From experiment 2 to 3: - \([\text{O}_3]_0\) goes from 0.80 to 1.55 M - Rate goes from \(4.40 \times 10^4 \) to \(1.32 \times 10^5 \) \( \text{M}\cdot\text{s}^{-1} \)Using the rate law: \[\frac{1.32 \times 10^5}{4.40 \times 10^4} = \left(\frac{1.55}{0.80}\right)^n\] \[3 = \left(1.9375\right)^n\] Thus, \( n \approx 1 \). Therefore, the reaction is first order with respect to \( \text{O}_3 \).
04

Write the Overall Rate Law and Calculate the Rate Constant

Based on the determined orders, the rate law can be written as: \[\text{Rate} = k [\text{NO}_2]^1 [\text{O}_3]^1\] \[\text{Rate} = k [\text{NO}_2][\text{O}_3]\] Insert values from experiment 1 to find \( k \):- \( \text{Rate} = 2.61 \times 10^4 \), \([\text{NO}_2]_0 = 0.65\), and \([\text{O}_3]_0 = 0.80\)\[2.61 \times 10^4 = k (0.65)(0.80)\] \[k = \frac{2.61 \times 10^4}{0.520} = 5.02 \times 10^4 \text{ M}^{-1}\text{s}^{-1}\] Thus, the rate constant \(k\) is \(5.02 \times 10^4\, \text{M}^{-1}\text{s}^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Constant
A rate constant, denoted as \( k \), is a crucial part of the reaction rate law that gauges the speed of a reaction at a given temperature. For the reaction: \[ 2 \text{NO}_{2}(g) + \text{O}_{3}(g) \rightarrow \text{N}_{2}\text{O}_{5}(g) + \text{O}_{2}(g) \] the rate law is determined as \( \text{Rate} = k [\text{NO}_2]^1 [\text{O}_3]^1 \). Here, the rate constant \( k \) quantifies the inherent speed without the influence of reactant concentrations. - **Temperature Dependency**: The rate constant is dependent on temperature. It typically increases with rising temperatures. This means at higher temperatures, reactions tend to be faster due to increased molecular interactions. - **Units**: The units of the rate constant are determined by the overall order of the reaction. For our example, a second-order reaction yields units of \( \text{M}^{-1}\text{s}^{-1} \). In the provided exercise, using experiment 1, the rate constant is calculated as \( \mathbf{5.02 \times 10^4 \, \text{M}^{-1}\text{s}^{-1}} \). This value helps predict how fast \( \text{O}_{2} \) forms under similar conditions.
Reaction Order
Reaction order with respect to a reactant indicates how the concentration of that reactant influences the rate of reaction. For our reaction here, both \( \text{NO}_2 \) and \( \text{O}_3 \) have an order of 1, making this a second-order overall reaction.- **Determining Reaction Order**: Reaction order is found experimentally. By holding one concentration constant and varying the other, changes in rate reveal the order. For \( \text{NO}_2 \), comparing experiments where only \([\text{NO}_2]\) changes showed the rate doubling, confirming first order. Similar steps confirmed the first order for \( \text{O}_3 \).- **General Impact**: If a reaction is first order with respect to \( \text{NO}_2 \), the rate will directly change in line with \( \left[\text{NO}_2\right] \). This means doubling \( \left[\text{NO}_2\right] \) also doubles the rate.Understanding reaction order is pivotal in planning chemical reactions, as it predicts how adjustments in concentrations alter reaction speed.
Chemical Kinetics
Chemical kinetics studies the rate at which chemical processes occur and the factors influencing them. It provides insights into reaction mechanisms and the steps leading to the formation of products.- **Rate Laws**: These expressions tie the rate of reaction to concentrations of reactants. Specifically, for our two-component reaction, the rate law: \( \text{Rate} = k [\text{NO}_2][\text{O}_3] \) gives a concise view of reactant influence.- **Mechanism Exploration**: By examining kinetics, chemists can theorize about the steps a reaction might take. Kinetic data provide clues about whether a reaction proceeds via a single step or multiple intermediary ones.Chemical kinetics is essential for designing industrial processes and predicting how different conditions like temperature and concentration will impact the chemical reaction's speed.

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