Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Q41 E (page 706)

The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L/mol/s. If the initial concentration is 0.500 mol/L, what is the value of t1/2?

Short Answer

Expert verified

Half-life periods may be defined as the time period at which half the concentration of reactant converts into the product. The half-life period of the second-order decomposition of the A is 10.33×10-2seconds which has a rate constant 2.42L/mole/s.

Step by step solution

01

Reaction Rate

The reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural, which occur in the surrounding environment, whereas it can be artificially done in the laboratory to form a desired required product.

The reaction rate can be defined as the reaction speed to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take less than a millisecond to produce products, or it can take years to produce the desired product.

The half-life period can be defined as the time period at which half the concentration of the reactants gets converted into a product.

02

Explanation

The half-life period of the second-order decomposition of the A is:

\({\bf{Half - life period = }}\frac{{\left( {\bf{A}} \right){\bf{^\circ }}}}{{{\bf{2k}}}}\)

Where \(\left( {\bf{A}} \right){\bf{^\circ }}\) = initial concentration of reactant

K = Rate constant

Initial concentration of reactant,\(\left( {\bf{A}} \right){\bf{^\circ }}\) =0.500 mole/L

Rate constant for the decomposition of \({{\bf{O}}_{\bf{3}}}\)= 2.42L/mol/s

.\(\begin{align}Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.500mole/L}}{{2 \times \left( {2.42L/mole/s} \right)}}\\Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.500mole/L}}{{4.84L/mole/s}}\\Half - life{\bf{ }}period{\bf{ }} &= 10.33 \times {\bf{ }}{10^{ - 2}}\sec ond\end{align}\).

Half-life period = 10.33×10-2 second.

Therefore, the half-life period of the second-order decomposition of the A is 10.33×10-2 seconds which has a rate constant is 2.42 L/mol/s.

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.

Start your free trial

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

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

Ingeneral, can we predict the effect of doubling the concentration of A on the rate of the overall reaction A + B⟶C? Can we predict the effect if the reaction is known to be an elementary reaction?

Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of \({\bf{3 \times 1}}{{\bf{0}}^{\bf{4}}}\)g/mole that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalysed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (\({\bf{0}}{\bf{.15 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.

(Penicillin) (M)

Rate (mole/L/min)

\({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\) \(\)

\({\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\)

\({\bf{3}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)

\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\)

\({\bf{4}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)

\({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\)

Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as \({\bf{CC}}{{\bf{l}}_{\bf{2}}}{{\bf{F}}_{\bf{2}}}\), catalyse the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:

\(\begin{aligned}{}{{\bf{O}}_{\bf{3}}}\overset{sunlight}{\rightarrow}{}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\\{{\bf{O}}_{\bf{3}}}{\rm{ }} + {\rm{ }}{\bf{Cl}}\to {{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{ClO}}\\{\bf{ClO}}{\rm{ }} + {\rm{ }}{\bf{O}}\to {\bf{Cl}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\end{aligned}\)

(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:

\({\bf{2}}{{\bf{O}}_{\bf{3}}}\mathop {}\limits^{}\to {\bf{3}}{{\bf{O}}_{\bf{2}}}\)

(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism: Is NO a catalyst for the decomposition? Explain your answer.

\(\begin{aligned}{}{{\bf{O}}_{\bf{3}}}\overset{sunlight}{\rightarrow}{\rm{ }}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\\{{\bf{O}}_{\bf{3}}}{\rm{ }} + {\rm{ }}{\bf{NO}}\rightarrow {\bf{N}}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\\{\bf{N}}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\rightarrow {\bf{NO}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\end{aligned}\)

In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction \({\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{F}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2Cl}}{{\bf{F}}_{\bf{3}}}{\bf{(g)}}\). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of \({\bf{C}}{{\bf{l}}_{\bf{2}}}\) and \({{\bf{F}}_{\bf{2}}}\) and the formation of \({\bf{Cl}}{{\bf{F}}_{\bf{3}}}\).

Thefollowingdatahave been determined for the reaction: \({{\bf{I}}^{\bf{ - }}}{\bf{ + OC}}{{\bf{l}}^{\bf{ - }}} \to {\bf{I}}{{\bf{O}}^{\bf{ - }}} + {\bf{C}}{{\bf{l}}^{\bf{ - }}}\)

1

2

3

\({{\bf{(}}{{\bf{I}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.10

0.20

0.30

\({{\bf{(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.050

0.050

0.010

Rate(mol/l/s)

\({\bf{3}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{6}}.{\bf{2*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{1}}.{\bf{83*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

Determine the rate equation and the rate constant for this reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation
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