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What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)\) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

Short Answer

Expert verified
(a) The molecularity of the reaction is 2, and the rate law is \(Rate = k[\mathrm{NO}]^{2}\). (b) The molecularity of the reaction is 1, and the rate law is \(Rate = k[\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}]\). (c) The molecularity of the reaction is 1, and the rate law is \(Rate = k[\mathrm{SO}_{3}]\).

Step by step solution

01

Find the Molecularity

The molecularity is the number of reactant molecules involved in the reaction. Here, there are two molecules of NO involved in the reaction, so the molecularity is 2.
02

Write the Rate Law

The rate law for this reaction is: \[Rate = k[\mathrm{NO}]^{2}\] (b) $\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)$
03

Find the Molecularity

In this case, there is only one reactant molecule involved in the reaction, the \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g)\). Therefore, the molecularity is 1.
04

Write the Rate Law

The rate law for this reaction is: \[Rate = k[\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}]\] (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)
05

Find the Molecularity

In this case, there is only one reactant molecule involved in the reaction, the \(\mathrm{SO}_{3}(g)\). Therefore, the molecularity is 1.
06

Write the Rate Law

The rate law for this reaction is: \[Rate = k[\mathrm{SO}_{3}]\]

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Key Concepts

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

Molecularity
Molecularity refers to the number of reactant molecules that participate in an elementary reaction. It is a theoretical concept used in chemical kinetics to better understand the reaction mechanism.
Molecularity can be of three types:
  • Unimolecular reactions: Involve a single reactant molecule. For instance, the decomposition of \(\mathrm{SO}_{3} \rightarrow \mathrm{SO}_{2} + \mathrm{O} \)is a unimolecular reaction with molecularity of 1.
  • Bimolecular reactions: Involve two reactant molecules. For example, \(2 \mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O}_{2} \)is a bimolecular reaction with molecularity of 2.
  • Termolecular reactions: Involve three reactant molecules, which are rare due to the high probability requirements of three molecules simultaneously colliding.
Understanding molecularity is crucial for predicting the mechanism of a chemical reaction and helps in writing the correct rate law for elementary reactions. Note, molecularity only applies directly to elementary reactions and is not used for complex reactions that occur in more than one step.
Elementary Reactions
Elementary reactions are single-step processes that describe the exact number of molecules (atoms, ions) and their arrangement that participate in a chemical reaction.
These reactions provide the simplest depiction of what happens at the molecular level during a reaction. Each elementary step has its own molecularity, distinguishing it from more complex, multi-step reactions.
  • Elementary reactions are precise and describe a specific event, such as a collision of molecules leading to a transformation.
  • The stoichiometric coefficients of an elementary reaction equal the exponents in the rate law for that reaction.
  • They give insight into reaction mechanisms, allowing chemists to hypothesize which steps might occur as intermediate stages in a multi-step reaction pathway.
Due to their one-step nature, understanding elementary reactions is foundational for forming complex reaction mechanisms, making them essential for studying chemical kinetics.
Rate Law
The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For elementary reactions, the rate law is directly derived from the molecularity.
  • Unimolecular reactions: For reactions like \(\mathrm{H}_{2}\mathrm{C}-\mathrm{CH}_{2} \rightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3} \)and \(\mathrm{SO}_{3} \rightarrow \mathrm{SO}_{2} + \mathrm{O} \), the rate law is expressed as:\[Rate = k [A]\]where \(A\)is the concentration of the reactant.
  • Bimolecular reactions: For reactions like \(2 \mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O}_{2} \), the rate law becomes:\[Rate = k [\mathrm{NO}]^2\]indicating that the reaction rate is proportional to the square of the NO concentration.
In any rate law, \(k\)is the rate constant, which varies with temperature. Understanding the rate laws allows scientists to predict how the concentration of reactants affects reaction speed, aiding in the manipulation of conditions to achieve desired reaction rates in industrial or lab settings.

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Most popular questions from this chapter

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.

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

You perform a series of experiments for the reaction \(A \longrightarrow B+C\) 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 .

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 \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C}\) ? (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

A first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) has the rate constant \(k=3.2 \times 10^{-3} \mathrm{~s}^{-1}\). If the initial concentration of \(\mathrm{A}\) is \(2.5 \times 10^{-2} M\), what is the rate of the reaction at \(t=660 \mathrm{~s}\) ?

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