Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 13, Problem 13.142QA
Interpretation Introduction

To find:

a) Write the rate law for the given reaction and the value of the rate constant at 298oK.

b) Calculate the activation energy of the reaction from given data.

Expert Solution & Answer
Check Mark

Answer to Problem 13.142QA

Solution:

a) The rate law for the given reaction is rate= k O3C2H4 and value of rate constant is k=1.02×103M-1s-1

b) The activation energy of the reaction is22kJ/mol.

Explanation of Solution

1) Concept:

We can write the rate law for the reaction as rate=kO3x[C2H4]y, where x and y are the order of reaction with respect to O3 and C2H4 respectively.

We need to find the values of x, y to write complete rate law and then find the rate constant k of the reaction.

2) Given:

We are given the table for the values of initial concentrations and initial rates of NH2 and NO in the four experiments.

Experiment [O3]0(M) [C2H4]0(M) Initial Rate (M/s)
1 0.86×10-2 1.00×10-2 0.0877
2 0.43×10-2 1.00×10-2 0.0439
3 0.22×10-2 0.50×10-2 0.0110

3) Calculations:

a) We need to first find the order of reaction with respect to O3 and C2H4 .

Using the general equation of rate law,

rate=kO3x[C2H4]y

 We can write rate equations for all the experiments as

0.0877=k[0.86×10-2]x1.00×10-2y------------ (1) Experiment 1

0.0439=k[0.43×10-2]x1.00×10-2y------------ (2) Experiment 2

0.0110=k[0.22×10-2]x0.50×10-2y------------- (3) Experiment 3

Now by taking the ratio of equation (1) and (2), we can find value of x.

0.08770.0439=k[0.86×10-2]x1.00×10-2yk[0.43×10-2]x1.00×10-2y

2=[2]x

x=1

Now, to find value of y, we have to take ratio of equation (2) and equation (3) and put x=1.

0.04390.0110=k[0.43×10-2]11.00×10-2yk[0.22×10-2]10.50×10-2y

4=2×[2]y

2=[2]y

y=1

Thus, we can write the rate law equation as

rate=kO3C2H4

1. Using this rate law and the values of concentrations and initial rates, we can find the rate constant,

i) So from experiment 1, rate law is

0.0877=k[0.86×10-2]1.00×10-2

k=0.0877[0.86×10-2]1.00×10-2

k=1.02×103M-1s-1

ii) From experiment 2, rate law is

0.0439=k[0.43×10-2]1.00×10-2

k=0.0439[0.43×10-2]1.00×10-2

k=1.02×103M-1s-1

iii) From experiment 3, rate law is

0.0110=k[0.22×10-2]0.50×10-2

k=0.0110[0.22×10-2]0.50×10-2

k=1.00×103M-1s-1

The average value of rate constant

k=1.02×103M-1s-1

b) To find the activation energy, we have to plot the graph of lnk versus 1/T. From the graph, we can find the slope and then the activation energy usingthe formula

Ea=-slope×R

We have given table for the values of rate constants at different temperatures.

T (K) 1/T(K-1) k(M-1s-1) lnk
263 0.003802 3.28×102 5.793
273 0.003663 4.73×102 6.159
283 0.003534 6.65×102 6.4997
293 0.003413 9.13×102 6.8167

The plot of lnk versus 1/T is as follows, and the value of slope=-2632.21

Chemistry: An Atoms-Focused Approach, Chapter 13, Problem 13.142QA

The activation energy would be Ea=-slope×R

Ea=-2632.21 K×(8.314J/K.mol

Ea=21885J/mol

Ea=22kJ/mol

Conclusion:

a) The rate law for the given reaction is rate= k O3C2H4, and value of rate constant is k=1.02×103M-1s-1

b) The activation energy of the reaction is 22kJ/mol.

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Chapter 13 Solutions

Chemistry: An Atoms-Focused Approach

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