Class 12th Chemistry 2017 Set3 Delhi Board Paper Solution
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 )
For first order reaction:
k = 2.303 / t Log a0 /a
where a0 = 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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