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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 脳 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 脳 10^{-3} s^{-1} at 64掳C and [N2O5] = 0.0240 M, we can find the rate of reaction using the rate law: Rate = k[N2O5] Rate = (4.82 脳 10^{-3} s^{-1})(0.0240 M) Rate = 1.156 脳 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 脳 0.0240 M = 0.0480 M. We can find the new rate using the rate law: New Rate = k[2 脳 N2O5] New Rate = (4.82 脳 10^{-3} s^{-1})(0.0480 M) New Rate = 2.312 脳 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 脳 0.0240 M = 0.0120 M. We can find the new rate using the rate law: New Rate = k[0.5 脳 N2O5] New Rate = (4.82 脳 10^{-3} s^{-1})(0.0120 M) New Rate = 5.78 脳 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 enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{s}^{-1}\) at \(100^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{s}^{-1}\) at \(21^{\circ} \mathrm{C}\) . (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{Cas} \mathrm{com}-\) pared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts \((\mathrm{c})\) and \((\mathrm{d}),\) what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c),\) has an activation energy of 86.8 \(\mathrm{kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) A g \(\mathrm{KOH}\) in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reac-solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving 0.335 g KOH in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) 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 \(35^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\) .

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of 0.039 \(\mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) . In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) . Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

\(\begin{array}{l}{\text { (a) What is meant by the term elementary reaction? }} \\ {\text { (b) What is the difference between a unimolecular }} \\\ {\text { and a bimolecular elementary reaction? (c) What is a }}\end{array}\) \(\begin{array}{l}{\text {reaction mechanism?}(\mathbf{d}) \text { What is meant by the term rate- }} \\ {\text { determining step? }}\end{array}\)

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{A}]^{x}\) . Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of \(8 .\)
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