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
Question
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Chapter 13, Problem 13.135QA
Interpretation Introduction

To find:

a) State whether the rate law is consistent with the observed order of the reaction for NO and O3 if the given reaction were to occur in a single step.

b) Find the value activation energy of the reaction.

c) Find the rate of the reaction at 25oC when NO=3×10-6 M and O3=5×10-9 M

d) Find the values of the rate constant at 10oC and 35oC.

Expert Solution & Answer
Check Mark

Answer to Problem 13.135QA

Solution:

a) Yes, if the given reaction occur in a single step the rate law would be consistent with the observed order of the reaction for NO and O3

b) The value activation energy of the reaction is 62.5kJ/mol

c) The rate of the reaction at 25oC when NO=3×10-6 M and O3=5×10-9 M is 1.2×10-12 M/s

d) The value of the rate constant at 10oC is 21 M-1s-1 and at 35oC it is

1.8×10-2M-1s-1.

Explanation of Solution

1) Concept:

From the given reaction, we will check, how the rate of reaction depends on concentration of NO(g) and O3 (g). We could use the given data to calculate activation energy of the reaction. We could use the rate law and data to determine the rate of reaction.

We have to use the formula for calculating activation energy.

lnk1k2=EaR1T2-1T1

2) Given:

We have given the reaction

NO(g)+O3 (g) NO2 (g)+O2 (g)

This reaction is first order with respect to NO and O3.

Temperature (oC) Temperature (oK) Rate constant (M-1s-1)
25 298 80
75 348 3000

At 25oC, NO=3×10-6 M and O3=5×10-9 M

3) Formula:

lnk1k2=EaR1T2-1T1

Where,

k1 and k2= rate constant

Ea= Activation energy

R= Gas constant = 8.314J/moloC

T1 and T2= Temperature

4) Calculations:

a) The given reaction is,

NO(g)+O3 (g) NO2 (g)+O2 (g)

Here one molecule of  NO(g) collides with one molecule of O3 (g). The molecules must collide to exchange atoms. Therefore the rate of collision depends on the number of collisions per unit time. Assume a closed system in which there is one molecule of NO(g) and more number of O3 (g) molecules. Here one molecule of NO(g) collides with different O3 (g) molecules and we get the product. If the concentration of NO(g) molecules is doubled (that means two molecules), then those NO(g) molecules will collide with different O3 (g) molecules. Therefore doubling the concentration of NO(g) doubles the rate of collision. The rate of collision also doubles, if the concentration of O3 (g) is doubled. Thus the rate law for the reaction is,

Rate=k NO[O3]

Therefore, if the given reaction occurs in a single step, the rate law would be consistent with the observed order of the reaction for NO and O3.

b) To calculate activation energy, we have formula,

lnk1k2=EaR1T2-1T1

So, by putting the values,

ln(80 M-1s-1)(3000 M-1s-1)=Ea8.314J/moloC1348oK-1298oK

-3.6243=Ea×(-5.799×10-5)J/mol

Ea=62498J/mol

Ea=62.5kJ/mol

Therefore, the value of activation energy of the reaction is 62.5kJ/mol.

c) This reaction is first order with respect to NO and O3. Therefore the rate law of the reaction is

rate=kNO[O3]

At 25oC, NO=3×10-6 M and O3=5×10-9 M

rate=(80 M-1s-1)×(3×10-6 M)×(5×10-9 M)

rate=1.2×10-12M/s

Therefore, at 25oC, when NO=3×10-6 M and O3=5×10-9 M, the rate of reaction is 1.2×10-12M/s.

d) The values of the rate constant at 10oC and 35oC.

e)

i. Rate constant at 10oC:

We are using the formula of activation energy. By putting value of rate constant at 25oC asK1 we can find the value of K2 which is the rate constant at 10oC=283oK.

lnk1k2=EaR1T2-1T1

ln(80 M-1s-1)k2=(62498J/mol)(8.314J/moloC)1283oK-1298oK

ln(80 M-1s-1)k2=(1.3370)

(80 M-1s-1)k2=3.8079

k2=21 M-1s-1

ii. Rate constant at 35oC:

We are using the formula of activation energy. By putting value of rate constant at 25oC asK1 we can find the value of K2 which is the rate constant at 35oC=308oK.

lnk1k2=EaR1T2-1T1

ln(80 M-1s-1)k2=(62498J/mol)(8.314J/moloC)1308oK-1298oK

ln(80 M-1s-1)k2=(-0.8190)

(80 M-1s-1)k2=0.4480

k2=1.8×102M-1s-1

At 10oC and 35oC, the value of rate constant is 21 M-1s-1 and 1.8×102M-1s-1 respectively.

Conclusion:

a) Yes, if the given reaction occurs in a single step the rate law would be consistent with the observed order of the reaction for NO and O3

b) The value activation energy of the reaction is 62.5kJ/mol

c) The rate of the reaction at 25oC when NO=3×10-6 M and O3=5×10-9 M is 1.2×10-12 M/s

d) The value of the rate constant at 10oC is 21 M-1s-1 and at 35oC it is 1.8×10-2M-1s-1.

e)

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Chapter 13 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

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