   # Consider the following reaction: CH 3 X + Y → CH 3 Y + X At 25°C, the following two experiments were run, yielding the following data: Experiment 1: [Y] 0 = 3.0 M [CH 3 X] (mol/L) Time(h) 7.08 × 10 −3 1.0 4.52 × 10 −3 1.5 2.23 × 10 −3 2.3 4.76 × 10 −4 4.0 8.44 × l0 −5 5.7 2.75 × l0 −5 7.0 Experiment 2: [Y] 0 = 4.5 M [CH 3 X] (mol/L) Time(h) 4.50 × 10 −3 0 1.70 × 10 −3 1.0 4.19 × 10 −4 2.5 1.11 × 10 −4 4.0 2.81 × l0 −5 5.5 Experiments also were run at 85°C. The value of the rate constant at 85°C was found to be 7.88 × 10 8 (with the time in units of hours), where [CH 3 X] 0 = 1.0 × 10 −2 M and [Y] 0 = 3.0 M. a. Determine the rate law and the value of k for this reaction at 25°C. b. Determine the half-life at 85°C. c. Determine E a for the reaction. d. Given that the C8X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 11, Problem 116MP
Textbook Problem
500 views

## Consider the following reaction: CH 3 X + Y → CH 3 Y + X At 25°C, the following two experiments were run, yielding the following data:Experiment 1: [Y]0 = 3.0 M [CH3X] (mol/L) Time(h) 7.08 × 10−3 1.0 4.52 × 10−3 1.5 2.23 × 10−3 2.3 4.76 × 10−4 4.0 8.44 × l0−5 5.7 2.75 × l0−5 7.0 Experiment 2: [Y]0 = 4.5 M [CH3X] (mol/L) Time(h) 4.50 × 10−3 0 1.70 × 10−3 1.0 4.19 × 10−4 2.5 1.11 × 10−4 4.0 2.81 × l0−5 5.5 Experiments also were run at 85°C. The value of the rate constant at 85°C was found to be 7.88 × 108 (with the time in units of hours), where [CH3X]0 = 1.0 × 10−2 M and [Y]0 = 3.0 M.a. Determine the rate law and the value of k for this reaction at 25°C.b. Determine the half-life at 85°C.c. Determine Ea for the reaction.d. Given that the C8X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.

(a)

Interpretation Introduction

Interpretation: The rate law for the given reaction is to be determined by the use of the concentration verses time data. The value of rate constant at 25°C is to be solved. The half-life 85°C is to be calculated. The activation energy is to be determined. A mechanism is to be suggested to explain the results of part a and c.

Concept introduction: The rate law or rate equation is the mathematical relation between the rate of the reaction and the concentration of the reactant.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant.

To determine: The rate law for the given reaction and the value of k for the reaction at 25°C is to be determined.

### Explanation of Solution

Explanation

Given

The stated chemical equation is,

CH3X+YCH3Y+X

The initial concentration of reactant Y=3.0M

Initial value of the reactant [CH3X]0=1.0×102M

The value of reactant at time 1.5h is [CH3X]=4.2×103M

A very high initial concentration value for the reactant Y in both the experiments clearly indicates that it is not involve in the rate law. Therefore, the expression for the rate law is given by the equation,

Rate=k[CH3X]_

Where,

• [CH3X] is the concentration of the reactant CH3X .
• k is the rate constant.

The above calculated rate expression clearly indicates that the given reaction is forst order reaction

(b)

Interpretation Introduction

Interpretation: The rate law for the given reaction is to be determined by the use of the concentration verses time data. The value of rate constant at 25°C is to be solved. The half-life 85°C is to be calculated. The activation energy is to be determined. A mechanism is to be suggested to explain the results of part a and c.

Concept introduction: The rate law or rate equation is the mathematical relation between the rate of the reaction and the concentration of the reactant.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant.

To determine: The value of the half-life for the given reaction at 85°C .

(c)

Interpretation Introduction

Interpretation: The rate law for the given reaction is to be determined by the use of the concentration verses time data. The value of rate constant at 25°C is to be solved. The half-life 85°C is to be calculated. The activation energy is to be determined. A mechanism is to be suggested to explain the results of part a and c.

Concept introduction: The rate law or rate equation is the mathematical relation between the rate of the reaction and the concentration of the reactant.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant.

To determine: The activation energy Ea .

(d)

Interpretation Introduction

Interpretation: The rate law for the given reaction is to be determined by the use of the concentration verses time data. The value of rate constant at 25°C is to be solved. The half-life 85°C is to be calculated. The activation energy is to be determined. A mechanism is to be suggested to explain the results of part a and c.

Concept introduction: The rate law or rate equation is the mathematical relation between the rate of the reaction and the concentration of the reactant.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant.

To determine: The mechanism that explains the results in parts a and c.

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