(a) A first order reaction is 75% completed in 40 minutes. Calculate its t1/2.

(b) Predict the order of the reaction in the given plots:


2019-11-29.png


Where, [R]o is the initial concentration of reactant. (Given: log 2 = 0.3010, log 4 = 0.6021)


OR


The following data were obtained for the reaction:


2 NO + O22 NO2



(a) Find the order of reaction with respect to NO and O2.


(b) Write the rate law and overall order of reaction.


(c) Calculate the rate constant (k).


(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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