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

An elementary reaction between \(A\) and \(B\) is a second order reaction. Which of the followng rate equations must be correct? (a) \(r=k[A]^{2}[B]^{0}\) (b) \(r=k[A]^{3 / 2}[B]^{1 / 2}\) (c) \(r=k[A]^{0}[B]^{2}\) (d) \(r=k[A][B]\)

Short Answer

Expert verified
All options (a), (b), (c), and (d) are correct second order rate equations.

Step by step solution

01

Understanding an Elementary Second Order Reaction

An elementary reaction that is second order means that the sum of the exponents of the concentration terms in the rate equation must equal 2. This is because the order of the reaction is determined by adding the powers of the concentration terms of the reactants.
02

Evaluating Option (a)

The rate law given by option (a) is \(r=k[A]^{2}[B]^{0}\). Here, we have the concentration of A raised to the power of 2 and the concentration of B raised to the power of 0. The sum of the exponents is 2 + 0 = 2, which is a second order reaction.
03

Evaluating Option (b)

Option (b) gives the rate law as \(r=k[A]^{3/2}[B]^{1/2}\). The sum of the exponents here is \(3/2 + 1/2 = 2\), which also indicates a second order reaction.
04

Evaluating Option (c)

The rate law given by option (c) is \(r=k[A]^{0}[B]^{2}\). Similarly to option (a), the sum of the exponents is 0 + 2 = 2, showing that it is also a second order reaction.
05

Evaluating Option (d)

Option (d) offers the rate law \(r=k[A][B]\), with the concentration of A raised to the first power and the concentration of B also raised to the first power. The sum here is 1 + 1 = 2, which matches the criteria for a second order reaction.
06

Selecting the Correct Rate Equations

All options (a), (b), (c), and (d) are correct because they all have rate laws in which the sum of the exponents of the concentrations equals 2. Hence, each of these rate equations is representative of a second order 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.

Elementary Reaction
Understanding the basics of an elementary reaction is crucial when studying chemical kinetics. An elementary reaction refers to a single-step process where reactants are converted directly to products without any intermediate states. In an elementary reaction, molecules collide and react according to their stoichiometry. This means that if the stoichiometric coefficient of a reactant is 2, then two molecules of that reactant are involved in the collision.

For a second-order elementary reaction, which involves either two reactant molecules or two reactive sites, the reaction rate is directly proportional to the product of the concentrations of the reactants, each raised to the power of 1. In more complex terms, if we consider a general second-order reaction given by 2A → Products or A + B → Products, the reaction rate would be described by the rate law equation: \( r = k [A]^2 \) for the first case or \( r = k [A][B] \) for the second case, where \( k \) is the rate constant.

Second-order kinetics implies that the rate of reaction is not simply related to the concentration of one reactant but instead is influenced by the concentration of two reacting species. This consideration is essential when predicting how concentration changes over time and underscores the importance of understanding the stoichiometry in elementary reactions.
Rate Equation
The rate equation, also known as the rate law, is an expression that relates the rate of a chemical reaction to the concentration of the reactants. For an elementary reaction, the exponents in the rate equation correspond to the stoichiometric coefficients of the reactants in the balanced chemical equation.

In a second-order elementary reaction, the rate equation will take the form of \( r = k [A]^{x}[B]^{y} \), where \( x \) and \( y \) are the orders with respect to reactants \( A \) and \( B \) respectively, and their sum (\( x + y \) = order of the reaction) must equal 2. The constant \( k \) is the rate constant, a measure of the rate at which the reaction proceeds. The units of \( k \) vary with the overall order of the reaction, which in the case of a second-order reaction, are typically \( M^{-1} s^{-1} \).

It is important to note that the rate equation cannot always be deduced from the stoichiometry of the reaction; it must often be determined experimentally. However, for elementary reactions, the rate law can be predicted from the chemical equation as each step is a single molecular event.
Reaction Kinetics
Reaction kinetics is the field of chemistry that studies the rates at which chemical processes occur and the factors that affect these rates. It encompasses the analysis of how different conditions, such as concentration, temperature, and presence of catalysts, can influence the speed of chemical reactions.

In the context of reaction kinetics, understanding the order of a reaction is vital. The order indicates how the rate is affected by the concentration of reactants. For instance, in a second-order reaction, the reaction rate is proportional to the square of the concentration of one reactant, or to the product of the concentrations of two different reactants.

Moreover, reaction kinetics involves mechanisms, which detail the stepwise sequence of events that occur during a reaction. For an elementary second-order reaction, the mechanism is straightforward – two particles collide and form the products in a single step. However, many reactions proceed through complex mechanisms with several elementary steps, requiring detailed kinetic analysis to understand each phase.

Applying kinetic principles, like those outlined in the exercise, helps in designing chemical processes, controlling reaction rates for desirable outcomes, and predicting the behavior of reactions under different conditions. The aim, ultimately, is to connect the microscopic view of molecules and reactions with the macroscopic observable properties like reaction rate, thus forming a comprehensive view of dynamic chemical processes.

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

Two radioactive nuclides \(A\) and \(B\) have half-lives \(50 \mathrm{~min}\) and \(10 \mathrm{~min}\) respectively. A fresh sample contains the nuclide of \(B\) to be eight time that of \(A\). How much time should elapse so that the number of nuclides of \(A\) becomes double of \(B\) ? (a) 30 (b) 40 (c) 50 (d) 100

For the ideal gaseous reaction, the rate is generally expressed in terms of \(\frac{d P}{d t}\) instead of \(\frac{d C}{d t}\) or \(\frac{d n}{d t}\) (where \(C=\frac{n}{V}\) is concentration and \(n\) the no. of moles). What is the reaction among these three expressions if \(T\) and \(V\) are constant? (a) \(\frac{d C}{d t}=\frac{d n}{d t}=\frac{d P}{d t}\) (b) \(\frac{d C}{d t}=\frac{1}{V} \frac{d n}{d t}=\frac{1}{R T}\left(\frac{d P}{d t}\right)\) (c) \(R T \frac{d C}{d t}=\frac{d n}{d t}=\frac{d P}{d t}\) (d) None of these

The hydrolysis of sucrose was studied with the help of polarimeter and following data were collected time (min.) \(\begin{array}{lll}: 0 & 70 & \infty\end{array}\) observed rotation (degrees) \(\begin{array}{lll}: 44 & 16.5 & -11\end{array}\) when the reaction mixture will be optically inactive ? (Given: \(\ln 2=0.7, \ln 3=1.1, \ln 5=1.6\) ) (a) 16 min. (b) \(69.47 \mathrm{~min}\). (c) \(160 \mathrm{~min}\). (d) none of these

At \(300 \mathrm{~K}\) the half-life of a sample of a gaseous compound initially at \(1 . \mathrm{atm}\) is \(100 \mathrm{sec}\). When the pressure is \(0.5\) atm the half-life is \(50 \mathrm{sec}\). The order of reaction is : (a) 0 (b) 1 (c) 2 (d) 3

Identify reaction type: \({ }_{13} \mathrm{Al}^{27}+{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{13} \mathrm{Al}^{28}+{ }_{1} \mathrm{H}^{1}\) (a) \((d, p)\) (b) \((p, p)\) (c) \((p, d)\) (d) none of these

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