Q. 24

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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)


2019-11-29.png


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)


2019-11-29.png


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


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