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 gaseous substance (AB_(3)) decomposes according to the overall equat

1.	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
Answer» zero three one two Solution :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.