Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
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Chapter 13, Problem 13.138QA
Interpretation Introduction

To determine:

a) Calculate the rate constant for the given reaction.

b) Determine the rate of the reaction at 23oC and PCH3CO3NO2=10.5 torr.

c) Draw a graph showing PPAN as a function of time from 0 to 200 h.

Expert Solution & Answer
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Answer to Problem 13.138QA

Solution:

d) The rate constant for the given second order reactionis k=9.52×10-4torr-1h-1.

a) The rate of the reaction at 23oC and PCH3CO3NO2=10.5 torr is 0.10 torr/h.

b) A graph showing PPAN as a function of time from 0 to 200 h is given below.

Chemistry: An Atoms-Focused Approach (Second Edition), Chapter 13, Problem 13.138QA , additional homework tip  1

Explanation of Solution

1) Concept:

The given process is second order in peroxyacetyl nitrate, so write rate law for the reaction. From given data and formula for half-life period for second order reaction, rate constant for the reaction can be calculated.

2) Given:

The given reaction is second order in peroxyacetyl nitrate.

t1/2=100 h

PCH3CO3NO2=10.5 torr

3) Formulae:

t12=1k[X]0

t12= half-life period

k = rate constant

[X]0= initial pressure

1[X]t=kt+ 1[X]0

[X]t= pressure after time t.

4) Calculations:

a) The half-life of the second order reaction, that is,100 h and initial pressure of CH3CO3NO2 is 10.5 torr.

t12=1k[PCH3CO3NO2]

k=1t12[PCH3CO3NO2]

k=1100 h(10.5 torr)

Rate constant

k=9.52×10-4torr-1h-1

b) We can write the general rate law equation for the given second order reaction as

rate=k[PCH3CO3NO2]2

rate=(9.52×10-4torr-1h-1)(10.5 torr)2

rate=0.10 torr/h.

c) To draw the graph, we have to first find the values of [PCH3CO3NO2]t at different time intervals from 0  to 200 h. For that, we can use the integrated rate law equation of second order reaction.

1[PCH3CO3NO2]t=kt+1[PCH3CO3NO2]0

So at 0 h[PCH3CO3NO2]t=10.5 torr

1. At 50 h

1[PCH3CO3NO2]50=(9.52×10-4torr-1h-1)(50 h)+110.5 torr

1[PCH3CO3NO2]50=0.1428

[PCH3CO3NO2]50=7.000 torr

2. At 100 h

1[PCH3CO3NO2]100=(9.52×10-4torr-1h-1)(100 h)+110.5 torr

1[PCH3CO3NO2]100=0.1904

[PCH3CO3NO2]100=5.2510 torr

3. At 150 h

1[PCH3CO3NO2]150=(9.52×10-4torr-1h-1)(150 h)+110.5 torr

1[PCH3CO3NO2]150=0.2380

[PCH3CO3NO2]150=4.2010 torr

4. At 200 h

1[PCH3CO3NO2]200=(9.52×10-4torr-1h-1)(200 h)+110.5 torr

1[PCH3CO3NO2]200=0.3273

[PCH3CO3NO2]200=3.500 torr

Now we have to plot the graph of [PCH3CO3NO2]t versus time

Time (h) [PCH3CO3NO2]t(torr)
0 10.5
50 7.000
100 5.2510
150 4.2010
200 3.500

Chemistry: An Atoms-Focused Approach (Second Edition), Chapter 13, Problem 13.138QA , additional homework tip  2

We have plotted time on X-axis and [PCH3CO3NO2]t(torr) on Y-axis.

Conclusion:

a) The rate constant for the given second order reaction is,k=9.52×10-4torr-1h-1.

b) The rate of the reaction at 23oC and PCH3CO3NO2=10.5 torr is,0.10 torr/h.

c) A graph showing PPAN as a function of time from 0 to 200 h is plotted.

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Chapter 13 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

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