Q. 24

# <span lang="EN-US

Answer :

(a) Given, reaction is of 1st order,

Time, t = 40 mins

Let, [R0] be the initial concentration.

[R] at 75% completion = [R0] - [R0] = = For 1st order reaction, K = K = = = 0.0346 min-1

Now, for a 1st order reaction, K = t1/2 = = 20.02 mins 20 mins

(b)

(i) From graph, it is clear that t1/2 does not depend on initial concentration i.e. [R0]. We can say it is a 1st order reaction because t1/2 for a 1st order reaction depends only on value of K.

t1/2 = (ii) From graph, it is clear that value of t1/2 varies linearly with value of initial concentration i.e. [R0]. We can say it is a zero order reaction from the following formula-

t1/2 = i.e. t1/2 [R0]

OR

2 NO + O22 NO2

(a) let the rate law be

rate = k[NO]x[O2]y

From dividing rate of experiment 1 and 3, 0.25 = (0.5)y

y = 2

From dividing rate of experiment 2 and 4, 0.25 = (0.25)x

x = 1

The order of reaction with respect to NO and O2 is 1 and 2 respectively.

(b) Rate law, rate = k[NO]1[O2]2

And overall order of reaction = 1+2 = 3

(c) k = Substituting values in above equation from experiment 1,

k = k = 0.06 mol-1/2 l1/2 s-1

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

Note: In the follChemistry - Exemplar

Which of the follChemistry - Exemplar