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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 [N鈧侽鈧匽, where k = 4.12 x 10鈦宦 s鈦宦 at 55掳C. (b) The rate of reaction when [N鈧侽鈧匽 = 0.050 M is 2.06 x 10鈦烩伌 M/s. (c) When the concentration of N鈧侽鈧 is tripled to 0.150 M, the rate of reaction increases to 6.18 x 10鈦烩伌 M/s. (d) When the concentration of N鈧侽鈧 is reduced by 10% to 0.045 M, the rate of reaction decreases to 1.85 x 10鈦烩伌 M/s.

Step by step solution

01

(a) Write the rate law for the reaction

Since the reaction is first-order in N鈧侽鈧, the rate law for the reaction is: Rate = k [N鈧侽鈧匽 where Rate is the rate of the reaction, k is the rate constant (4.12 x 10鈦宦 s鈦宦 at 55掳C), and [N鈧侽鈧匽 is the concentration of N鈧侽鈧.
02

(b) Calculate the rate of reaction when [N鈧侽鈧匽 = 0.050 M

Using the rate law, substitute the rate constant k and the concentration of N鈧侽鈧: Rate = (4.12 x 10鈦宦 s鈦宦) (0.050 M) Rate = 2.06 x 10鈦烩伌 M/s
03

(c) What happens to the rate when the concentration of N鈧侽鈧 is tripled to 0.150 M

Since the rate law is first-order in N鈧侽鈧, 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鈦宦 s鈦宦) (0.150 M) Rate = 6.18 x 10鈦烩伌 M/s The rate of reaction increases to 6.18 x 10鈦烩伌 M/s.
04

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

When the concentration of N鈧侽鈧 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鈦宦 s鈦宦) (0.045 M) Rate = 1.85 x 10鈦烩伌 M/s The rate of reaction decreases to 1.85 x 10鈦烩伌 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

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for "R"). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

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The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. Step \(1: \quad \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (fast) Step \(2: \quad \mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)\) (slow) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is \(\mathrm{O}\) a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

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