The decompoistion of `N_(2)O_(5)` according to the equation `2N_(2)O_(5)(g) rarr 4NO_(2)(g)+O_(2)(g)`
is a first order reaction. After `30 min`, form the start of the decompoistion in a closed vessel, the total pressure developed is found t be `284.5 mm Hg`. On complete decompoistion, the total pressure is `584.5 mm Hg`. Calculate the rate constant of the reaction.

Explanation

`5.2 xx10^(-3)" m in"^(-1)`
For the reaction : `2N_(2)O_(5) rarr 4NO_(2) +O_(2)`
If `p_(0)` is the initial pressure, the total pressure after completion of reaction would be `(5)/(2) p_(o)`
`rArr 584.5=(5)/(2) p_(o) rArr p_(o) =233.8` m m
Let the pressure of `N_(2)O_(5)` decreases by 'p' amount after 30 min. Therefore,
`2N_(2)O_(5) rarr 4NO_(2)+O_(2)`
At 30 min : `p_(o)- p" "2p" "(p)/(2)`
Total pressure `=p_(o) +(3)/(2)p=284.5 =p=(2)/(3) (284.5 -233.8)=33.8`
How, kt = In `(p_(o))/(p_(o)-p) rArr k=(1)/(30)" In "(233.8)/(233.8-33.8)" m in"^(-1) =5.2xx10^(-3)" m in"^(-1)`