Question

A gaseous substance `(AB_(3))` decomposes according to the overall equation : `AB_(3)rarr (1)/(2) A_(2)+(3)/(2) B_(2)`
The variation of the partial pressure of `AB_(3)` with time (starting with pure `AB_(3)`) is given below at `200^(@)C` :
`{:(Time//h,0,5.0,15.0,35.0),(P_(AB_(3))//mmHg,660,330,165,82.5):}`
The order of the reqaction is
A. zero
B. three
C. one
D. two

Answer 1

Correct Answer - D
From the given data, we can observe that when we start our experiment `(t=0),P_(AB)` , is `660mmHg` . At the end of `5` hours, it is reduced to half `(330mmHg)` . Therefore, half life of reaction is `5` hours. Now, when we start next with `P_(AB)` , equal to `330mmHg` , it becomes again half `(165mmHg)` but at the end of `15` hours. Thus, half life is `10hrs` .
We know
`t_(1//2)prop(1)/(a^(n-1))`
Thus `((t_(1//2))_(t))/((t_(1//2))_(2)`
It is a second order reaction.
Alternatively, note that when we start our experiment with `660mmHg` , the half life id=s `5` hours bvut when we start our experiment with `330mmHg` , the half life is `10` hours, i.e., when pressure is halved, the half life is doubled. THis implies that half life is inversely proportional to initial pressure (or concentration). This is the characteriic of a second order reaction only.