Question

Catalytic decomposition of nitrous oxide by gold at `900^(@)C` at an initial pressure of `200 mm` was `50%` in `53` minute and `73%` in `100` minute.
(a) What is the order of reaction?
(b) Calculate velocity constant.
(c ) How much will it decompose in `100` minute at the same temperature but at an initial pressure of `600 mm`?

Answer 1

We know,
`K=2.303/t"log"_(10) a/((a-x)) …(1)`
Case `I: a prop 200 mm`,
`x prop 200 xx50/100` or `x prop 100 mm`
`t_(1//2)=53 "minute"`
`:. K_(1)=2.303/53"log"_(10) 200/(200-100)`
`=1.307xx10^(-2) "minute"^(-1)`
Case `II: a prop 200 mm`,
`x prop 200 xx73/100` or `x prop 146 mm`
`t_(73%)=100 "minute"`
`:. f_(2)=2.303/100"log"_(10) 200/(200-146)`
`=1.309xx10^(-2) minute^(-1)`
(a) The eq. (1) gives same value of `K` and thus, reaction obeys `I` order.
(b) `K=(K_(1)+K_(2))/(2)=((1.307+1.309)xx10^(-2))/(2)`
`=1.308xx19^(-2) "minute"^(-1)`
(c ) `t_(1//n) prop (a)^(0)` for first order reaction and thus, reaction will proceed to `73%` completion in `100` minute if initial pressure is `600 mm`.