Q28 of 30 Page 1

(a) A reaction is second order in A and first order in B.

(i) Write the differential rate equation.


(ii) How is the rate affected on increasing the concentration of A three times?


(iii) How is the rate affected when the concentrations of both A and B are doubled?


(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1,2 for this reaction.


(Given log 1.428 =0.1548)


OR


(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.


(b) Rate constant 'k' of a reaction varies with temperature 'T' according to the equation:



Where, Ea is the activation energy. When a graph is plotted for log k Vs. a straight line with a slope of –4250 K is obtained. Calculate 'Ea' for the reaction.


(R = 8.314 JK–1 mol-1).



(a)


(i) The reaction is of second order in A and first order in B. So, the differential rate equation is-


Rate = = k[A][B]2


k is the rate constant.


(ii) When the concentration of A is increased three times,


Rate = = k[3A]2[B]2


=K[9A][B]2


=9K[A][B]2


So, rate increases 9 times.


(iii) On increasing the concentration of A and B as 2A and 2B. Rate = = K[2A][2B]2


= 8K[A][B]2


So, the rate of reaction becomes 8 times of the initial rate.


(b) We know, for first order kinetics,


k =


where, k = rate constant.


t = time taken.


a =initial concentration.


a-x = concentration left after t amount of time.


So, it takes 40 minutes for 30% decomposition and 70% of the initial concentration is remaining.


k = = 0.00891 min-1.


For calculating half life in a first order reaction we know,


t1/2 = = = 77.78 minutes.


So, half life is 77.78 minutes.


OR


(a) For a first order reaction, the time required for 99% completion is-


t1 =


= [As 1% is remaining]


=


= 2×


Similarly for a first order reaction, the time required for 90% completion is-


t2 =


= [As 10% is remaining]


=


= 1×


Hence, t1 = 2t2. The time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.


(b)


The equation is like . If the graph is plotted between y () and x (), the slope can be calculated.



So, slope is = = -4250K


As R = 8.314 JK–1 mol-1


∴ EA = (-4250×2.303×8.314)


= -81375.35 kJ mol-1.


So EA is -81375.35 kJ mol-1.


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