Each of the following is an elementary reaction. Write its rate law and indicate its molecularity. (a) \(\mathrm{O}+\mathrm{CF}_{2} \mathrm{Cl}_{2} \rightarrow\) \(\mathrm{ClO}+\mathrm{CF}_{2} \mathrm{Cl} ;\) (b) \(\mathrm{OH}+\mathrm{NO}_{2}+\mathrm{N}_{2} \longrightarrow \mathrm{HNO}_{3}+\mathrm{N}_{2} ;\) (c) \(\mathrm{ClO}^{-}+\) \(\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO}+\mathrm{OH}^{-}\); (d) Which of these reactions might be radical chain propagating?

Short Answer

Expert verified
Rate laws: (a) Rate = k[O][CF2Cl2] - bimolecular, (b) Rate = k[OH][NO2][N2] - termolecular, (c) Rate = k[ClO^-][H2O] - bimolecular. Reaction (a) might be radical chain propagating.

Step by step solution

01

Writing Rate Law for Reaction (a)

To write the rate law for the elementary reaction \(\mathrm{O} + \mathrm{CF}_2 \mathrm{Cl}_2 \rightarrow \mathrm{ClO} + \mathrm{CF}_2 \mathrm{Cl}\), we use the concentrations of the reactants raised to the power of their coefficients in the balanced equation. Since this is an elementary reaction, the coefficients are all one, hence the rate law is: \[\text{Rate} = k[\mathrm{O}][\mathrm{CF}_2 \mathrm{Cl}_2]\] where \(k\) is the rate constant.
02

Determining Molecularity for Reaction (a)

Molecularity is determined by the number of species involved in the reaction step. Since two reactants are involved in the reaction, it is a bimolecular reaction.
03

Writing Rate Law for Reaction (b)

For the reaction \(\mathrm{OH} + \mathrm{NO}_2 + \mathrm{N}_2 \rightarrow \mathrm{HNO}_3 + \mathrm{N}_2\), the rate law since it is an elementary step is directly proportional to the concentration of the reactants: \[\text{Rate} = k[\mathrm{OH}][\mathrm{NO}_2][\mathrm{N}_2]\]
04

Determining Molecularity for Reaction (b)

In this reaction, three reactant species are involved in the step, so it is a termolecular reaction.
05

Writing Rate Law for Reaction (c)

The rate law for the reaction \(\mathrm{ClO}^{-} + \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HClO} + \mathrm{OH}^{-}\) is derived similarly, as: \[\text{Rate} = k[\mathrm{ClO}^{-}][\mathrm{H}_2\mathrm{O}]\]
06

Determining Molecularity for Reaction (c)

Since two reactants participate in the reaction, it is characterized as a bimolecular reaction.
07

Identifying the Radical Chain Propagating Reaction

Radical chain propagating reactions typically involve radicals (atoms or molecules with an unpaired electron). In the reactions given, reaction (a) may involve radical species since \(\mathrm{O}\) is commonly found as the radical \(\mathrm{O}\cdot\) and \(\mathrm{ClO}\cdot\) could potentially be a radical as well. Therefore, reaction (a) might be a radical chain propagating reaction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molecularity
Understanding the molecularity of a reaction is crucial when studying reaction mechanisms. Molecularity refers to the number of reactant particles involved in a single step of an elementary reaction. It is always a whole number — unimolecular, bimolecular, or termolecular for one, two, or three reactant particles, respectively.

For instance, if we consider the provided exercise where reaction (a) \( \mathrm{O} + \mathrm{CF}_2 \mathrm{Cl}_2 \rightarrow \mathrm{ClO} + \mathrm{CF}_2 \mathrm{Cl} \) is described, it is bimolecular because two reactants collide to form the products. It's essential to realize that molecularity is not determined from the overall reaction but from the individual elementary steps that make up the overall process.
Elementary Reaction
  • An elementary reaction is a single-step process where reactants convert directly to products without any intermediate stages.
  • For an elementary reaction, the reaction order and molecularity are the same. This means that the exponents in the rate law of an elementary reaction match the number of molecules participating in that step.
  • In the step-by-step solution, the rate laws are derived based on the presumption that the provided reactions are elementary. For example, the rate law for reaction (c) \(\mathrm{ClO}^{-} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO} + \mathrm{OH}^{-}\) is \[\text{Rate} = k[\mathrm{ClO}^{-}][\mathrm{H}_2\mathrm{O}]\] where this direct proportionality to reactants' concentrations signifies its elementary nature.
Reaction Kinetics
Reaction kinetics involves the study of the rate at which chemical reactions occur and the factors affecting these rates. It dives deep into understanding how different conditions such as temperature, pressure, and concentration can influence the speed of a reaction.

Rate laws are mathematical statements derived from the principles of reaction kinetics. They describe the relationship between the concentration of reactants and the rate of the reaction. Taking example (b) \(\mathrm{OH} + \mathrm{NO}_2 + \mathrm{N}_2 \longrightarrow \mathrm{HNO}_3 + \mathrm{N}_2\), the termolecular nature means the reaction rate is affected by the concentration of three different reactants, which is rather rare and indicates a complex kinetic behaviour.
Radical Chain Propagation
Radical chain propagation is a sequence in a radical chain reaction where the reactive intermediates (radicals) continue to generate more reactive intermediates through a series of steps, thereby perpetuating the reaction. In the case of reaction (a) from the exercise, it's suspected to be a radical chain propagating reaction because it involves oxygen, which can form radicals capable of propagating a chain reaction. Typically, these involve the generation of a radical, its reaction with a stable molecule to form a product and another radical, which then reacts in a similar fashion. This process can lead to a reaction that continues until the radicals are terminated or consumed. Understanding this concept can explain phenomena such as the rapid spread of fire or the degradation of materials under exposure to UV radiation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) The equilibrium constant for a reaction equals the rate constant for the forward reaction divided by the rate constant for the reverse reaction. (b) In a reaction that is a series of equilibrium steps, the overall equilibrium constant is equal to the product of all the forward rate constants divided by the product of all the reverse rate constants. (c) Increasing the concentration of a product increases the rate of the reverse reaction, and so the rate of the forward reaction must then increase, too.

Express the units for rate constants when the concentrations are in moles per liter and time is in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions.

Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. $$ \begin{aligned} &\text { Step } 1 \mathrm{AC}+\mathrm{B} \longrightarrow \mathrm{AB}+\mathrm{C} \\ &\text { Step } 2 \mathrm{AC}+\mathrm{AB} \longrightarrow \mathrm{A}_{2} \mathrm{~B}+\mathrm{C} \end{aligned} $$

The half-life of a substance taking part in a third-order reaction \(\mathrm{A} \rightarrow\) products is inversely proportional to the square of the initial concentration of A. How can this half-life be used to predict the time needed for the concentration to fall to (a) onehalf; (b) one-fourth; (c) one- sixteenth of its initial value?

Derive an expression for the half-life of the reactant A that decays by a third-order reaction with rate constant \(k\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free