Q15 of 37 Page 1

Following data are obtained for the reaction:

N2O5→ 2NO2 + 1/2 O2
















t/s



0



300



600



[N2O5]/mol L–1



1.6 × 10–2



0.8 × 10–2



0.4 × 10–2



A. Show that it follows first order reaction.


B. Calculate the half-life.


(Given log 2 = 0.3010 log 4 = 0.6021)

A. Lets use hit and trial method .


We know that , for a first order reaction



k = rate constant


t = time


Ro = initial concentration


R = concentration at particular time t


So, by hit and trial, lets substitute values in the table


t = 300, Ro = 1.6 × 10-2, R = 0.8 × 10-2



similarly, by taking t as 600, Ro as 1.6 × 10-2 and R as 0.4 × 10-2


we get k = 2.31 10-3 s-1


hence, its first order reaction as the rate constant remains same.


B. We know that half life of a first order reaction is


t1/2 = ,


k = rate constant


substituting the above-obtained k, we get half-life = 300s


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