Question

The decomposition of `N_(2)O_(5)` at an initial pressure of `380 mm` and `50^(@)C` is `50%` complete in `56` minutes and `71%` complete in `100` minutes. What is the order of reaction? How much of `N_(2)O_(5)` will decomposes in `100` minutes at `50^(@)C` but at an initial pressure of `500 mm`?

Answer 1

The first order rate equation is,
`K=2.303/t"log"a/((a-x))`
We have for `50%` conmpletion of decomposition in `56` min.
`K=2.303/56"log"100/50`
`=2.303/56xx0.3010`
`=0.0124 min^(-1)`
For `71%` complrtion of decomposition in `100` min, we have
`K=2.303/100"log"100/29`
`=2.303/100xx0.5376`
`=0.0124 min^(-1)`
Aconstant value of `K` indicates the order of reaction to be one.
Also for `I` order `t_(1//n) prop (a)^(0)`. Thus `71%` of the reaction will be completed at initial pressure of `500 mm`.