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

Explain why the coefficients of an elementary step equal the reaction orders of its rate law but those of an overall reaction do not.

Short Answer

Expert verified

When two molecules collide, an elementary step is created in a single step. The rate of an elementary step is proportional to the product of the concentrations of the reactants with the same exponents (order) as the stoichiometric coefficients in the equation for this step. The complete reaction may have a number of such stages with varying speeds. On the other hand, the speed is computed using the slowest step. As a result, the reaction orders of its rate law do not necessarily match the stoichiometric coefficients of the overall equation.

Step by step solution

01

Step 1: What is the reaction of an elementary?

A chemical reaction in which one or more chemical species react directly to create products in a single reaction step and with a single transition state is known as an elementary reaction.

02

Why do the coefficients of an elementary step match the reaction orders of its rate law

The collision of two molecules creates an elementary step in a single step. The product of the concentrations of the reactants with the same exponents (order) as the stoichiometric coefficients in the equation for this step is proportional to the rate of an elementary step. A number of such phases with varying speeds may be included in the total reaction. However, the pace is calculated using the slowest step. As a result, the reaction orders of its rate law do not always match the overall equation's stoichiometric coefficients.

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!

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

Question:Sulfonation of benzene has the following mechanism:

(1)2H2SO4โ†’H3O++HSO4-+SO3[fast](2)SO3+C6H6โ†’H(C6H5+)SO3-[slow](3)H(C6H5+)SO3-+HSO4-โ†’C6H5SO3-+H2SO4[fast](4)C6H5SO3-+H3O+โ†’C6H5SO3H+H2O[fast]

(a) Write an overall equation for the reaction.

(b) Write the overall rate law for the initial rate of the reaction

A gas reacts with a solid presence in large chunks. Then the reaction is run again with the solid pulverized. How does the increase in the surface area of the solid affect the rate of its reaction with the gas? Explain.

Carbon disulfide, a poisonous, flammable liquid, is an excellent solvent for phosphorus, sulfur, and some other nonmetals. A kinetic study of its gaseous decomposition reveals these data:

Experiment

Initial[CS2] mol/L

Initial rate (mol/Ls-1 )

1

0.100

2.7ร—10-7

2

0.080

2.2ร—10-7

3

0.055

1.5ร—10-7

4

0.044

1.2ร—10-7

(a) Write the rate law for the decomposition ofCS2 .

(b) Calculate the average value of the rate constant

Reaction rate is expressed in terms of changes in concentration of reactants and products. Write a balanced equation for

Rate=-โˆ†[CH4]โˆ†t=12โˆ†[O2]โˆ†t=12โˆ†[H2O]โˆ†t=โˆ†[CO2]โˆ†t

While developing a catalytic process to make ethylene glycol from synthesis gas CO+H2 , a chemical engineer finds the rate is fourth-order in gas pressure. The uncertainty in the pressure reading is 5%. When the catalyst is modified, the rate increases by 10%. If you were the company patent attorney, would you file for a patent on this catalyst modification? Explain.

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