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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

`250` atm

C

`480` atm

D

`420` atm

Text Solution

Verified by Experts

The correct Answer is:
A
`p_(0)=500` atm
`k=(2.303)/(t)log_(10).(p_(0))/(p_(t))`
`3.38xx10^(-5)=(2.303)/(10xx60)log.(500)/(p_(t))`
or `0.00880=log.(500)/(p_(t))implies(500)/(1.02)=490` atm .
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