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

For the consecutive unimolecular-type first-order reaction: \(\mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{R} \stackrel{k_{2}}{\longrightarrow} \mathrm{S}\), the concentration of component ' \(\mathrm{R}\) ', \(C_{\mathrm{R}}\), at any time, ' \(t\) ' is given by \(C_{\mathrm{R}}=C_{\mathrm{AO}} \cdot K_{1}\left[\frac{e^{-k_{1} t}}{\left(k_{2}-k_{1}\right)}+\frac{e^{-k_{2} t}}{\left(k_{1}-k_{2}\right)}\right]\) If \(C_{\mathrm{A}}=C_{\mathrm{AO}}, C_{\mathrm{R}}=C_{\mathrm{s}}=0\) at \(t=0\), the time at which the maximum concentration of 'R' occurs is (a) \(t_{\max }=\frac{k_{2}-k_{1}}{\ln \left(k_{2} / k_{1}\right)}\) (b) \(t_{\max }=\frac{\ln \left(k_{2} / k_{1}\right)}{k_{2}-k_{1}}\) (c) \(t_{\max }=\frac{e^{k_{2} / k_{1}}}{k_{2}-k_{1}}\) (d) \(t_{\max }=\frac{e^{k_{2}-k_{1}}}{k_{2}-k_{1}}\)

Short Answer

Expert verified
The maximum concentration of 'R' occurs at the time given by option (a): \( t_{\max} = \frac{\ln\left(k_{2} / k_{1}\right)}{k_{2} - k_{1}} \).

Step by step solution

01

Understand the given function for concentration of R

The concentration of R as a function of time, represented by \( C_{\mathrm{R}} \), is given by the formula \( C_{\mathrm{R}} = C_{\mathrm{AO}} \cdot K_{1}\left[\frac{e^{-k_{1} t}}{\left(k_{2}-k_{1}\right)} + \frac{e^{-k_{2} t}}{\left(k_{1}-k_{2}\right)}\right] \). To find the maximum concentration of R, we need to analyze this function and find the time \( t \) at which the first derivative of \( C_{\mathrm{R}} \) with respect to \( t \) is zero.
02

Find the first derivative of \( C_{\mathrm{R}} \) with respect to time \( t \)

The first derivative of \( C_{\mathrm{R}} \) with respect to \( t \) is calculated by differentiating the given equation for \( C_{\mathrm{R}} \). The derivative will help us locate the time at which \( C_{\mathrm{R}} \) reaches its maximum value. Use the properties of the exponential function and the chain rule to differentiate each term separately.
03

Set the first derivative equal to zero to find critical points

To find when the concentration of R is at a maximum, set the first derivative from step 2 equal to zero. This will provide us with an equation that we can solve for \( t \).
04

Solve for \( t \)

The critical points obtained from setting the first derivative to zero give potential times at which \( C_{\mathrm{R}} \) could be maximized. Solve the equation for \( t \) to find the time \( t_{\max} \) at which the concentration of R is at its maximum.
05

Check if the found critical point gives a maximum

To confirm that the critical point found indeed corresponds to a maximum, we can use either the second derivative test or simply check that the first derivative changes sign around the critical point. If the first derivative goes from positive to negative, then the critical point is a maximum.

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.

Chemical Kinetics
Chemical kinetics is the branch of chemistry that focuses on the rates of chemical reactions and the factors that affect them. Understanding chemical kinetics is essential for predicting how quickly a chemical reaction will occur, which is crucial in various applications such as the synthesis of new materials, pharmaceuticals, and in environmental modeling.

The rate of a reaction is often expressed as a change in concentration of a reactant or product per unit time. Factors influencing reaction rates include temperature, pressure, concentration, and the presence of catalysts. In our specific case of a unimolecular first-order reaction, the reaction rate is proportional to the concentration of a single reactant oligomer.
Reaction Rate
The reaction rate is a measure of how quickly the concentration of a reactant or product in a chemical reaction changes over time. It is an essential aspect of chemical kinetics because it directly affects how long a reaction takes to reach completion or to reach a state of dynamic equilibrium.

In the context of a unimolecular first-order reaction, the reaction rate is directly proportional to the concentration of the reactant that is transforming into a product. Also, because it is a first-order reaction, its rate law can be expressed as rate = k[Reactant], where 'k' is the rate constant, and '[Reactant]' refers to the concentration of the starting chemical species.
Concentration-Time Relationship
In chemical kinetics, the concentration-time relationship describes how the concentration of reactants and products changes over time during a reaction. This is especially pertinent in a unimolecular first-order reaction where the rate of reaction depends on the concentration of the single reactant.

To express this relationship mathematically, we use rate equations that are derived from the laws of mass action. For example, in the given exercise, the concentration of the intermediate product 'R' is described by a complex expression involving exponential terms. This equation illustrates how 'R's concentration evolves as the reaction progresses and why at a certain point in time, labeled as 't_max', the concentration of 'R' reaches its maximum value due to the interplay between the rates of its production and consumption.

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

For the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) \(\rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) under certain conditions of temperature and partial pressure of the reactants, the rate of formation of \(\mathrm{NH}_{3}\) is \(10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\). The rate of consumption of \(\mathrm{H}_{2}\) under same condition is (a) \(1.5 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\) (b) \(1.76 \times 10^{-4} \mathrm{~kg} \mathrm{~h}^{-1}\) (c) \(6.67 \times 10^{-4} \mathrm{~kg} \mathrm{~h}^{-1}\) (d) \(3 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\)

For the reaction: \(2 \mathrm{HI} \rightarrow \mathrm{H}_{2}+\mathrm{I}_{2}\), the expression, \(-\frac{1}{2} \frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{d} t}\) represents (a) the rate of formation of \(\mathrm{HI}\) (b) the rate of disappearance of \(\mathrm{HI}\) (c) the instantaneous rate of the reaction (d) the average rate of the reaction

For irreversible elementary reactions in parallel: \(\mathrm{A} \stackrel{K_{1}}{\longrightarrow} \mathrm{R}\) and \(\mathrm{A} \stackrel{K_{2}}{\longrightarrow} \mathrm{S}\), the rate of disappearance of reactant ' \(\mathrm{A}\) ' is (a) \(\left(k_{1}-k_{2}\right) C_{\mathrm{A}}\) (b) \(\left(k_{1}+k_{2}\right) C_{\mathrm{A}}\) (c) \(1 / 2\left(k_{1}+k_{2}\right) C_{\mathrm{A}}\) (d) \(k_{1} C_{\mathrm{A}}\)

The incorrect statement is (a) Rate law is an experimental fact whereas law of mass action is a theoretical proposal. (b) Rate law is always different from the expression of law of mass action. (c) Rate law is more informative than law of mass action for the development of mechanism. (d) Order of a reaction is equal to the sum of powers of concentration terms in the rate law.

Azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}\), decomposes according to the following equation: \(\mathrm{CH}_{3}-\mathrm{N} \equiv \mathrm{N}-\mathrm{CH}_{3}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g})\) The initial concentration of azomethane was \(1.50 \times 10^{-2}\) M. After 10 min, the concentration decreased to \(1.29 \times 10^{-2} \mathrm{M}\). The average rate of reaction during this time interval is (a) \(3.5 \times 10^{-6} \mathrm{Ms}^{-1}\) (b) \(2.1 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3.5 \times 10^{-6} \mathrm{M} \mathrm{h}^{-1}\) (d) \(2.1 \times 10^{-3} \mathrm{Mmin}^{-1}\)

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