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In the reaction 2N(2)O(5) rarr 4NO(2) + ...

In the reaction `2N_(2)O_(5) rarr 4NO_(2) + O_(2)`, initial pressure is `500 atm` and rate constant `K` is `3.38 xx 10^(-5) sec^(-1)`. After `10` minutes the final pressure of `N_(2)O_(5)` is

A

490 atm

B

350 atm

C

480 atm

D

420 atm

Text Solution

Verified by Experts

The correct Answer is:
B
Since `-(1)/(2) (d[N_(2)O_(5)])/(dt) =k [N_(2)O_(5)] rArr (-d[N_(2)O_(5)])/(dt) =2k[N_(2)O_(5)]=k' [N_(2)O_(5)]`
`therefore " "t_(1//2) =(0.693)/(k') =(0.693)/(2xxk) =(0.693)/(2xx3.38xx10^(-2)), " "t_(1//2) =(0.693)/(6.76xx10^(-2)) ~~10` m in
So after 10 min pressure will drop to half of its initial pressure.
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