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Chapter 15: Problem 9

The reaction between ozone and nitrogen dioxide at \(231 \mathrm{K}\) is first order in both \(\left[\mathrm{NO}_{2}\right]\) and \(\left[\mathrm{O}_{3}\right].\) $$2 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ (a) Write the rate equation for the reaction. (b) If the concentration of \(\mathrm{NO}_{2}\) is tripled (and \(\left[\mathrm{O}_{3}\right]\) is not changed), what is the change in the reaction rate? (c) What is the effect on reaction rate if the concentration of \(\left.\mathbf{O}_{3} \text { is halved (with no change in }\left[\mathrm{NO}_{2}\right]\right) ?\)

Short Answer

Expert verified
(a) Rate = k [NO2] [O3]; (b) Rate triples; (c) Rate is halved.

Step by step solution

01

Understanding the Rate Equation

Since the reaction is first order in both \([\mathrm{NO}_{2}]\) and \([\mathrm{O}_{3}]\), the rate equation can be expressed as: \[ \text{Rate} = k [\mathrm{NO}_2] [\mathrm{O}_3] \]where \( k \) is the rate constant.
02

Analyzing the Effect of Tripling [NO2]

If \([\mathrm{NO}_{2}]\) is tripled, replace \([\mathrm{NO}_2]\) with \(3[\mathrm{NO}_2]\) in the rate equation:\[ \text{New Rate} = k \cdot 3[\mathrm{NO}_2] \cdot [\mathrm{O}_3] = 3 \times (k [\mathrm{NO}_2] [\mathrm{O}_3]) \]The rate of reaction will triple when \([\mathrm{NO}_2]\) is tripled, given \([\mathrm{O}_3]\) stays constant.
03

Evaluating the Effect of Halving [O3]

If \([\mathrm{O}_3]\) is halved, replace \([\mathrm{O}_3]\) with \(0.5[\mathrm{O}_3]\) in the rate equation:\[ \text{New Rate} = k [\mathrm{NO}_2] (0.5[\mathrm{O}_3]) = 0.5 \times (k [\mathrm{NO}_2] [\mathrm{O}_3]) \]The rate of reaction will be halved when \([\mathrm{O}_3]\) is halved, as \([\mathrm{NO}_2]\) remains unchanged.

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Key Concepts

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

First Order Reactions
In chemistry, a first-order reaction is one where the rate of reaction is directly proportional to the concentration of one of the reactants. This means as the concentration of the reactant changes, the rate at which the reaction proceeds changes in a directly proportional manner.
For example, if the concentration of a reactant is doubled in a first-order reaction, the rate of reaction also doubles. In the context of the reaction between nitrogen dioxide and ozone, both reactants are first-order.
This is a relatively straightforward type of reaction to predict and calculate. You can simply multiply the change in concentration to find the change in reaction rate, which makes understanding these reactions simpler for beginners.
Rate Equation
The rate equation expresses how the concentration of reactants affects the reaction rate. It is usually written as: \[\text{Rate} = k [ ext{reactant}_1]^m [ ext{reactant}_2]^n \]where:
  • \( k \) is the rate constant, a value that is specific to the reaction and conditions like temperature.
  • \([\text{reactant}_1]\) and \([\text{reactant}_2]\) represent the molarity (concentration) of the reactants.
  • \( m \) and \( n \) are the orders of the reaction with respect to each reactant.

In the exercise example, for the mixture involving nitrogen dioxide \([ ext{NO}_2]\) and ozone \([ ext{O}_3]\), the reaction is first order with respect to each reactant. The rate equation simplifies to:
\[\text{Rate} = k [ ext{NO}_2][ ext{O}_3] \]This helps you determine how changes in concentration impact the rate of reaction.
Effect of Concentration on Rate
The concentration of reactants has a direct effect on the rate of a reaction, particularly for first-order reactions. In these reactions, changing the concentration has a predictable impact on the reaction rate.
For instance, if the concentration of nitrogen dioxide \([ ext{NO}_2]\) is tripled, the reaction rate will also triple. This is straightforward because the reaction is first-order with respect to \([ ext{NO}_2]\).
Similarly, if the concentration of ozone \([ ext{O}_3]\) is halved, the reaction rate will reduce to half its original rate. Understanding these relationships can be crucial in laboratory settings and industrial chemical processes. It allows chemists to control reactions more efficiently by adjusting reactant concentrations.

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Most popular questions from this chapter

The rate equation for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (giving \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) ) is Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) The value of \(k\) is \(6.7 \times 10^{-5} \mathrm{s}^{-1}\) for the reaction at a particular temperature. (a) Calculate the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (b) How long does it take for the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration to drop to one tenth of its original value?

In the synthesis of ammonia, if \(-\Delta\left[\mathrm{H}_{2}\right] / \Delta t=\) \(4.5 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{min},\) what is \(\Delta\left[\mathrm{NH}_{3}\right] / \Delta t ?\) $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$

The ozone in the earth's ozone layer decomposes according to the equation $$2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g})$$ The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a slow step: Step 1: Fast, reversible \(\quad \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}(\mathrm{g})\) Step 2: Slow \(\mathrm{O}_{3}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) Show that the mechanism agrees with this experimental rate law: $$\text { Rate }=-(1 / 2) \Delta\left[\mathrm{O}_{3}\right] / \Delta t=k\left[\mathrm{O}_{3}\right]^{2} /\left[\mathrm{O}_{2}\right]$$

Chlorine atoms contribute to the destruction of the earth's ozone layer by the following sequence of reactions: $$\begin{array}{l}\mathrm{Cl}+\mathrm{O}_{3} \rightarrow \mathrm{ClO}+\mathrm{O}_{2} \\ \mathrm{ClO}+\mathrm{O} \rightarrow \mathrm{Cl}+\mathrm{O}_{2}\end{array}$$ where the O atoms in the second step come from the decomposition of ozone by sunlight: $$\mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ What is the net equation on summing these three equations? Why does this lead to ozone loss in the stratosphere? What is the role played by Cl in this sequence of reactions? What name is given to species such as ClO?

To determine the concentration dependence of the rate of the reaction $$\mathrm{H}_{2} \mathrm{PO}_{3}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ you might measure \(\left[\mathrm{OH}^{-}\right]\) as a function of time using a pH meter. (To do so, you would set up conditions under which \(\left[\mathrm{H}_{2} \mathrm{PO}_{3}^{-}\right]\) remains constant by using a large excess of this reactant.) How would you prove a second-order rate dependence for \(\left[\mathrm{OH}^{-}\right] ?\)
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