CHEM 142 Lecture Notes - Lecture 11: Reaction Coordinate, Collision Theory, 2-Step Garage

33 views7 pages

coralplatypus226

7 Jul 2016

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

To this point, we have treated all reactions as if they happen in a single step. The true story is, however, often much more complex as most reactions occur through a multi-step pathway. Consider the decomposition of ozone (o3) to molecular oxygen (o2): We normally express reactions like this as the overall reaction, and not a series of individual steps, called elementary steps. Elementary steps cannot be broken down into simpler steps they occur as they are written and describe one actual event (collision, energy change, bond breaking etc. ) that takes place in the course of a reaction. Note that elementary steps and overall reactions look the same when written down but represent very different processes and it is important to realize which one you are working with. The overall reaction omits intermediates, compounds formed in one step and consumed in a subsequent step, and catalysts.

Related Questions

Consider the reaction below:

2NO2(g) + F2(g) â¶ 2NO2F(g)

a. Based on the information provided in the table below (all measurements are at 25 ÂºC), determine the

experimental rate law and calculate the rate constant at this temperature. You can assume ideal gas

behavior for both NO2 and F2.

Initial Pressures (atm) Experiment p(NO2); p(F2) (mol L-1 s-1)

1) .038 ; .060

2) .076 ; .060

3) .076 ; .030

Initial Rate

1 0.038 0.060 1.4 ï´ 10â 5

2 0.076 0.060 2.8 ï´ 10â 5

3 0.076 0.030 1.4 ï´ 10â 5

b. Contrasting NO2 and F2, which one of these two real gases would be closer to ideal gas assumed in part a?

Draw the Lewis structures of both to support your argument.

c. Based on the rate law determined in part a, propose a reaction mechanism comprising of two bimolecular

elementary steps. Hint: F is an intermediate in the overall reaction.

d. Which elementary process in part c is rate-determining?

e. You conduct experiment 1 from the table in part a. If your initial p(NO2) is 0.038 atm and p(F2) is kept at

constant flow of 0.060 atm, how long would it take for the pressure of NO2 to drop to 0.001 atm?

f. The collision frequency and steric factor for the rate-determining step in the overall process at 25 ÂºC are

5.9x1010 and 0.16, respectively. Calculate the activation energy (in kJ/mol) for the rate determining step.

g. If the enthalpy of reaction (i.e. standard enthalpy change for the overall process) is -100 kJ/mol, sketch the

energy profile of the two-step reaction given that i) the activation energy for the fast step in part c is 5

kJ/mol and ii) equilibrium partial pressures of all species involved in the fast step are 1.0 atm

Just answer

1) The first-order rate constant for the reaction of methyl chloride (CH₃Cl) with water to produce methanol (CH₃OH) and hydrochloric acid (HCl) is 3.32 Ã10^â10 s^â1 at 25Â°C. Calculate the rate constant at 58.3Â°C if the activation energy is 116 kJ/mol.

(Enter your answer in scientific notation.)answer

2)Given the same reactant concentrations, the reaction

CO(g) + Cl₂(g) â COCl₂(g)at 259Â°C is 272 times as fast as the same reaction at 147Â°

C. Calculate the activation energy for this reaction. Assume that the frequency factor is constant.

kJ/mol answer

3)For a certain reaction, the frequency factor A is 7.5 Ã10⁹ s^â1 and the activation energy is 24.5 kJ/mol.

What is the rate constant for the reaction at 92Â°C

(Enter your answer in scientific notation.) answer

4)The rate constant of a first-order reaction is 3.25 Ã10^â4 s^â1 at 350.Â°C. If the activation energy is 107 kJ/mol,

calculate the temperature at which its rate constant is 9.85 Ã10^â4 s^â1.

5)The rate at which tree crickets chirp is 1.80 Ã10² per minute at 29Â°C but only 37.9 per minute at 6.5

Â°C. From these data, calculate the activation energy for the chirping process.

kJ/mol answer

6)Consider the following elementary step

X + 2Y âXY₂

If the initial rate of formation of XY₂ is 0.0076 M/s and the initial concentrations of X and Y are 0.38 Mand 0.64 M, what is the rate constant for the reaction?

M^â2s^{â1 answer}

⁷⁾The activation energy for the decomposition of hydrogen peroxide is 59.0 kJ/mol. When the reaction is catalyzed by the enzyme catalase, it is 6.00 kJ/mol.

2H₂O₂(aq) â2H₂O(l) + O₂(g)

Calculate the temperature that would cause the nonenzymatic catalysis to proceed as rapidly as the enzyme-catalyzed decomposition at 20.0Â°C. Assume the frequency factor, A, to be the same in both cases. Report your answer to 3 significant figures.

8)The activation energy (E_a) for the reaction

2N₂O(g) â 2N₂(g) + O₂(g) ÎH

= â164 kJ/mol

is 370.0 kJ/mol. What is the (E_a) for the reverse reaction?

kJ/mol answer

CHEM 142 Lecture Notes - Lecture 11: Reaction Coordinate, Collision Theory, 2-Step Garage

Document Summary

Get access

Related textbook solutions

Chemistry: Structure and Properties

Basic Chemistry

Principles of Chemistry Molecular Approach

Chemistry: Structure and Properties

Principles of Chemistry Molecular Approach

Chemistry: A Molecular Approach

Chemistry: A Molecular Approach

Principles of Chemistry: A Molecular Approach

Chemistry: The Central Science

Related Documents

CHEM 142 Study Guide - Midterm Guide: Intermolecular Force, Covalent Bond, Hydrogen Bond

CHEM 142 Chapter Notes - Chapter 12: Rate Equation, Chemical Kinetics, Reaction Rate

CHEM 142 Lecture Notes - Lecture 13: Rice Chart, Ibm 37Xx, Kob

Related Questions