Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
The molecularity of the elementary reaction \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) is 2. The rate law for this reaction is \(Rate = k[\mathrm{NO}]^{2}\).

Step by step solution

01

Determine the molecularity of the elementary reaction

We can determine the molecularity of the elementary reaction by looking at the reactants. The given reaction is: \[2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\] There are 2 molecules of NO(g) required for the reaction, we can see it from the coefficient "2" in front of NO(g). Thus, the molecularity of the reaction is 2.
02

Write the rate law for the elementary reaction

To write the rate law for the elementary reaction, we can use the formula: \[Rate = k[\mathrm{Reactant}]^{m}\] Where Rate is the rate of the reaction, k is the rate constant, [Reactant] is the concentration of the reactant, and m is the order of the reaction with respect to the reactant. Since this is an elementary reaction with molecularity 2, the order of the reaction with respect to NO(g) is the same as its stoichiometric coefficient, which is 2. Therefore, the rate law for this reaction is: \[Rate = k[\mathrm{NO}]^{2}\]

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
When discussing molecularity, we refer to the minimum number of molecules, atoms, or ions required to collide for an elementary reaction to occur. In simple terms, it counts how many reactant particles are involved in a single step of a reaction.
For example, if two molecules of a reactant are needed, like in the case of the reaction provided in the exercise \[2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\], the molecularity is two.
  • Unimolecular: Involves one molecule. For instance, the isomerization of cyclopropane to propene is a unimolecular reaction.
  • Bimolecular: Involves two molecules. Most reactions, like the one given (where 2 NO molecules react), are bimolecular.
  • Termolecular: Involves three molecules. This is quite rare, as it's unlikely for three molecules to collide simultaneously.
Molecularity is always a whole number and cannot be zero or fractional since it must represent actual physical particles colliding in a single step. Unlike reaction order, it is derived directly from the reaction equation of an elementary step.
Elementary Reaction
Elementary reactions are fundamental processes where reactants are directly converted to products in a single step. These are the simplest types of reactions and serve as the building blocks of more complex reactions. The provided example \[2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\] is an elementary reaction.
  • Direct Process: The reaction occurs in a single event. Unlike complex reactions, there are no intermediates or multiple steps involved.
  • Rate Law Derivation: The rate law for an elementary reaction can be written directly using the stoichiometry of the reaction. This makes predicting the rate law straightforward.
Understanding elementary reactions helps in the study of complex mechanisms by providing insight into how molecules interact and lead to the formation of products. Each step in a mechanism, when broken down to its simplest form, is described as an elementary reaction.
Reaction Order
Reaction order indicates how the rate of a reaction depends on the concentration of its reactants. For elementary reactions, reaction order can be determined from the stoichiometry provided in the balanced reaction equation.
In the reaction \[2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\], the reaction order with respect to NO is 2 because the coefficient of NO in the balanced equation is 2.
  • Order of Reaction: It's the sum of the powers of the concentration terms in the rate law equation. For the example given, the rate law is \[Rate = k[\mathrm{NO}]^{2}\] and thus it is a second-order reaction overall.
  • Determination from Mechanisms: In complex reactions, the overall order may not align with the stoichiometric coefficients, as it depends on the rate-determining step, which is often an elementary reaction like those discussed here.
The concept of reaction order is critical in the field of chemical kinetics, as it helps chemists understand and predict how changes in concentration impact the speed of a reaction. It also aids in the design of industrial processes where controlling reaction rates is crucial.

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

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

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.

For the generic reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) that is zero order in \(\mathrm{A}\), what would you graph in order to obtain the rate constant?

Americium-241 is used in smoke detectors. It has a first order rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine- \(125,\) which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the halflives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a 1.00 -mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00 -mg sample of each isotope remains after 4 days?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{hr}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

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