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The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(55^{\circ} \mathrm{C}\) the rate constant is \(4.12 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.050 \mathrm{M} ?(\mathbf{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is tripled to \(0.150 \mathrm{M}\) ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is reduced by \(10 \%\) to \(0.045 \mathrm{M}\) ?

Short Answer

Expert verified
(a) The rate law for the reaction is: Rate = k [N₂O₅], where k = 4.12 x 10⁻³ s⁻¹ at 55°C. (b) The rate of reaction when [N₂O₅] = 0.050 M is 2.06 x 10⁻⁴ M/s. (c) When the concentration of N₂O₅ is tripled to 0.150 M, the rate of reaction increases to 6.18 x 10⁻⁴ M/s. (d) When the concentration of N₂O₅ is reduced by 10% to 0.045 M, the rate of reaction decreases to 1.85 x 10⁻⁴ M/s.

Step by step solution

01

(a) Write the rate law for the reaction

Since the reaction is first-order in N₂O₅, the rate law for the reaction is: Rate = k [N₂O₅] where Rate is the rate of the reaction, k is the rate constant (4.12 x 10⁻³ s⁻¹ at 55°C), and [N₂O₅] is the concentration of N₂O₅.
02

(b) Calculate the rate of reaction when [N₂O₅] = 0.050 M

Using the rate law, substitute the rate constant k and the concentration of N₂O₅: Rate = (4.12 x 10⁻³ s⁻¹) (0.050 M) Rate = 2.06 x 10⁻⁴ M/s
03

(c) What happens to the rate when the concentration of N₂O₅ is tripled to 0.150 M

Since the rate law is first-order in N₂O₅, if the concentration is tripled, the rate of reaction will also triple. Use the rate law to calculate the new rate: Rate = (4.12 x 10⁻³ s⁻¹) (0.150 M) Rate = 6.18 x 10⁻⁴ M/s The rate of reaction increases to 6.18 x 10⁻⁴ M/s.
04

(d) What happens to the rate when the concentration of N₂O₅ is reduced by 10% to 0.045 M

When the concentration of N₂O₅ is reduced by 10% to 0.045 M, the rate of reaction will be 90% of the rate at 0.050 M concentration. Calculate the new rate using the rate law: Rate = (4.12 x 10⁻³ s⁻¹) (0.045 M) Rate = 1.85 x 10⁻⁴ M/s The rate of reaction decreases to 1.85 x 10⁻⁴ M/s when the concentration is reduced by 10% to 0.045 M.

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Key Concepts

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

First Order Reaction
A first-order reaction refers to a chemical reaction where the rate is directly proportional to the concentration of a single reactant. In the context of the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), the rate is proportional to the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) itself.
This means that if we increase the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\), the rate of reaction will also increase by the same factor. Conversely, when the concentration decreases, the reaction rate decreases proportionally.
Here, the rate law can be expressed as:
  • \(\text{Rate} = k[\mathrm{N}_{2} \mathrm{O}_{5}]\), where \(k\) is the rate constant.
Understanding this relationship helps in predicting how changes in reactant concentration affect the reaction speed, which is crucial in both laboratory and industrial settings.
Kinetics
Kinetics is the branch of chemistry that deals with the rates at which chemical reactions occur. It helps us understand how different factors influence the speed of reactions. In our example of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposition, kinetics plays a role in determining how quickly the reaction proceeds at a given temperature and concentration of the reactants.
Kinetic studies provide insights into the reaction mechanism and help in optimizing conditions to achieve desired reaction speeds. The analysis of such data can reveal the order of the reaction and the influences of various parameters like temperature, pressure, and concentration.
Rate Constant
The rate constant, denoted as \(k\), is a critical factor in the rate law of chemical reactions. It gives us a measure of how quickly a reaction takes place, independent of reactant concentrations.
For the first-order decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), \(k\) is given as \(4.12 \times 10^{-3} \mathrm{~s}^{-1}\) at \(55^{\circ} \mathrm{C}\). This constant is unique for any given reaction and depends on factors like temperature. A higher \(k\) value implies a faster reaction.
It's important to note that the units of the rate constant change with the order of reaction; in a first-order reaction, it is \(\mathrm{s}^{-1}\). Understanding the rate constant helps in determining how adjustments in conditions might accelerate or slow down a reaction.
Chemical Decomposition
Chemical decomposition is a process where a single compound breaks down into two or more simpler compounds or elements. In this reaction, \(2\mathrm{~N}_{2} \mathrm{O}_{5}\) decomposes into \(4 \mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\).
This decomposition reaction, like most, is affected by factors such as temperature and concentration, both of which can influence the speed and extent of the reaction.
Such processes are crucial in various applications, including waste management and the chemical industry, where specific compounds are broken down and repurposed in more useful forms. Recognizing the rules governing these reactions, like rate laws, is essential for controlling and optimizing these processes.

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Most popular questions from this chapter

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism.(c) Identify anyintermediatesin the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(\) alc \()\) has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.1 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) Predict the rate constant for the reaction at \(30^{\circ} \mathrm{C}\). (b) A solution of KOH in ethanol is made up by dissolving \(0.500 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(500 \mathrm{~mL}\) of solution. Similarly, \(1.500 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(500 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(30^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? ((d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(40^{\circ} \mathrm{C} .\)

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l} \text { Step } 1: \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X} \\\ \text { Step } 2: \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D} \end{array} $$ \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}\), \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c),\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

As shown in Figure 14.23 , the first step in the heterogeneous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed explanation for the "sticking" of ethylene to a metal surface is the interaction of the electrons in the \(\mathrm{C}-\mathrm{C} \pi\) bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene?

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