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Chapter 14: Problem 29

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 \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{s}^{-1}\) (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 M ?(\mathbf{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to 0.0480\(M ?(\mathbf{d})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to 0.0120 \(\mathrm{M} ?\)

Short Answer

Expert verified
(a) Rate = k[N2O5] (b) Rate = 1.156 × 10^{-4} M s^{-1} (c) Doubling the concentration of N2O5 doubles the rate of the reaction. (d) Halving the concentration of N2O5 halves the rate of the reaction.

Step by step solution

01

(a) Write the rate law for the reaction

Since the reaction is first order in N2O5, the rate law can be written as: Rate = k[N2O5], where k is the rate constant and [N2O5] is the concentration of N2O5.
02

(b) Find the rate of reaction when [N2O5] = 0.0240 M

Given the rate constant k = 4.82 × 10^{-3} s^{-1} at 64°C and [N2O5] = 0.0240 M, we can find the rate of reaction using the rate law: Rate = k[N2O5] Rate = (4.82 × 10^{-3} s^{-1})(0.0240 M) Rate = 1.156 × 10^{-4} M s^{-1}
03

(c) Determine the effect on the rate when the concentration of N2O5 is doubled

When the concentration of N2O5 is doubled, [N2O5] = 2 × 0.0240 M = 0.0480 M. We can find the new rate using the rate law: New Rate = k[2 × N2O5] New Rate = (4.82 × 10^{-3} s^{-1})(0.0480 M) New Rate = 2.312 × 10^{-4} M s^{-1} This is twice the initial rate, so doubling the concentration of N2O5 doubles the rate of the reaction.
04

(d) Determine the effect on the rate when the concentration of N2O5 is halved

When the concentration of N2O5 is halved, [N2O5] = 0.5 × 0.0240 M = 0.0120 M. We can find the new rate using the rate law: New Rate = k[0.5 × N2O5] New Rate = (4.82 × 10^{-3} s^{-1})(0.0120 M) New Rate = 5.78 × 10^{-5} M s^{-1} This is half the initial rate, so halving the concentration of N2O5 also halves the rate of the reaction.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the study of how quickly chemical reactions occur and the factors that affect these rates. While the completion of a reaction might seem instant in some cases, understanding the rate at which reactants turn into products gives us a deeper insight into the mechanism of the reaction.

Several factors influence the speed of a reaction, such as temperature, concentration of reactants, surface area, and the presence of catalysts. Through studying kinetics, chemists can also derive rate laws, which are mathematical expressions that describe the relationship between the rate of a reaction and the concentration of reactants. In the given exercise, knowing the rate law allows us to predict how changes in the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will affect the rate of the decomposition reaction.
Reaction Rate
The reaction rate measures the speed at which a chemical reaction proceeds. Specifically, it refers to the change in concentration of a reactant or product per unit time. In our textbook example, the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) can be quantified by the rate at which its concentration decreases over time.

Mathematically, the rate is often expressed as \(\Delta[Product]/\Delta t\) or \(\Delta[Reactant]/\Delta t\), where \(\Delta\) signifies a change over time. Rate plays a crucial role in everything from industrial chemical synthesis to biological processes, making it a vital concept in chemical kinetics. The step-by-step solution for the exercise illustrates the direct proportionality between \(\mathrm{N}_{2} \mathrm{O}_{5}\)'s concentration and the reaction rate, consistent with a first order reaction.
First Order Reaction
A first order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of one of the reactants. In the equation \(\text{Rate} = k[\text{Reactant}]\), \(k\) represents the rate constant, and \(\text{Reactant}\) denotes the concentration of the reactant.

In the context of the reaction involving \(\mathrm{N}_{2} \mathrm{O}_{5}\), doubling the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) doubles the rate, while halving the concentration halves the rate. This direct relationship is characteristic of first order kinetics. In real-world applications, understanding the order of a reaction is essential for controlling reaction conditions in various industries, such as pharmaceuticals or environmental engineering, and is pivotal in reaction modeling and simulation.

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Most popular questions from this chapter

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) . Catalysts that convert these species to \(\mathrm{N}_{2}\) are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of NO, NO \(_{2}\) and \(\mathrm{N}_{2} .(\mathbf{b})\) Using a resource such as Table \(8.3,\) look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2},\) describing what wavelengths of light need to be monitored as a function of time.

The addition of NO accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$\begin{array}{c}{\mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g)} \\ {2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)}\end{array}$$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(N_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

A flask is charged with 0.100 mol of A and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(A(g) \longrightarrow \mathrm{B}(g) .\) The following data are collected:(a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 s interval in units of mol/s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A?

Consider the following reaction: $$\mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q)$$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-} .\) When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M},\) the reaction rate at 298 \(\mathrm{K}\) is 0.0432 \(\mathrm{M} / \mathrm{s}\) . (a) What is the value of the rate constant? (\mathbf{b} )What are the units of the rate constant? (c) What would happen to the rate if the concentration of OH \(^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

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: $$ \begin{array}{l}{\text { Step } 1 : \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}} \\ {\text { Step } 2 : \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}}\end{array}$$ \(X\) 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. (ii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).
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