Class 12th Chemistry 2017 Set3 Delhi Board Paper Solution

Question 18

Following data are obtained for the reaction:

N2O5 → 2NO2 + 1/2 O2

t/s 0 300 600
[N2O5] / mol L-1 1.6 x 10-2 0.8 x 10-2 0.4 x 10-2

(a) Show that it follows first order reaction.

(b) Calculate the half-life.

(Given: log 2 = 0.3010 ,  log 4 = 0.6021 )

Answer

For first order reaction:

k = 2.303 / t Log  a0 /a 

where a= initial concentration

and a = remaining concentration

At t = 300 seconds

K1 = 2.303/300 log 1.6 x 10-2 / 0.8 x 10-2

 K1 = 2.303/300 log 2

K1 = 2.303/300 x 0.3010 

K1 = 0.00231 s-1

 

Now at t = 600 seconds

K2 = 2.303/600 log 1.6 x 10-2 / 0.4 x 10-2

 K2 = 2.303/600 log 4

K2 = 2.303/600 x  0.6021

K2 = 0.00231 s-1

Since the value of K is contant in both the situation so it is a first order reaction.

K = 0.00231 s-1

t1/2  = 0.693 / 0.00231 = 300 seconds. 

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