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