Consider the initial-rate data at a certain temperature in the table for the reaction described by 2 NO,(g) +0,(2)...

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Consider the initial-rate data at a certain temperature in the table for the reaction described by 2 NO,(g) +0,(2) NO.(8) + O

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Answer 1

Answer #1

see experiment 1 and 2:

[NO2] becomes 1.10/0.650 = 1.7 times

[O3] is constant

rate becomes 3.96/2.34 = 1.7 times

so, order of NO2 is 1

see experiment 2 and 3:

[NO2] becomes 1.76/1.1 = 1.6

[O3] 1.4/0.8 = 1.75

rate becomes 11.09/3.96 = 2.80

Since 1.16*1.75 = 2.80

Order of NO2 and O3 are both 1

so, order of O3 is 1

Rate law is:

rate = k*[NO2]*[O3]

Put values from 1st row of table in rate law

rate = k*[NO2]*[O3]

2.34*10^4 = k*0.65*0.8

k = 4.50*10^4 M-1.s-1

Answer: 4.50*10^4 M-1.s-1

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Answer 2

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Consider the initial-rate data at a certain temperature in the table for the reaction described by 2 NO,(g) +0,(2)...

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