At constant temperature and volume, X decomposes as 2" X "(g) to 3" Y "(g)+2" Z "(g). P_(x) is the partial pressure of X. {:("Observation No.",,,"Time (in minutes)",,,P_(x)("in mm of Hg")),(1,,,0,,,800),(2,,,100,,,400),(3,,,200,,,200):} (i)What is the order of reaction with respect with respect of X ? (ii) Find the time for 75% completion of the reaction. (iii) Find the total pressure when pressure of X is 700 mm of Hg.

At constant temperature and volume, X decomposes as 2" X "(g) to 3" Y

1.	At constant temperature and volume, X decomposes as 2" X "(g) to 3" Y "(g)+2" Z "(g). P_(x) is the partial pressure of X. {:("Observation No.",,,"Time (in minutes)",,,P_(x)("in mm of Hg")),(1,,,0,,,800),(2,,,100,,,400),(3,,,200,,,200):} (i)What is the order of reaction with respect with respect of X ? (ii) Find the time for 75% completion of the reaction. (iii) Find the total pressure when pressure of X is 700 mm of Hg.
Answer» Solution :(i) As pressure of X is changing with time, it cannot be a zero order REACTION. LET us now check it for 1st order. `{:("At",,,t=100" min,",,,k=(2.303)/(100)log""(P_(0))/(P_(t))=(2.303)/(100)log""(800)/(400)=6.932xx10^(-3)"min"^(-1)),("At",,,t=200" min,",,,k=(2.303)/(200)log""(800)/(200)=(2.303)/(800)log4=6.932xx10^(-3)"min"^(-1)):}` As k comes out to be constant, hence it is a reaction of 1st order. (II) `t_(75%)=(2.303)/(k)log""(100)/(100-75)=(2.303)/(6.932xx10^(-3)"min"^(-1))log4=200" min".` (iii) `2X(g) to 3" Y"(g)+2" Z"(g)` `{:("Initial Pressure",,,800" mm",,,0,,,0),("Pressure after time t",,,800-2" p",,,3" p",,,2" p"):}` When pressure of X is 700 mm, `800-2" p"=700" or "p=50" mm"` Total pressure `=(800-2" p")+3" p"+2" p"=800+3" p"=800+3xx50=950" mm"`