At 353 K, the vapoure pressure of pure ethylene bromide and propylene bromide are 22.93 and 16.93 K N m^(-2). Respectively and these compounds forms nearly ideal solution 3. moles of ethylen bromide and 2mole of propylene bromide are equilibrated at 353 K and at a total pressure of20.4 KN m^(-2). (a) What is the composition of the liquids phase (b) How many moles of each compound are present in the vapour phase ?

At 353 K, the vapoure pressure of pure ethylene bromide and propylene

1.	At 353 K, the vapoure pressure of pure ethylene bromide and propylene bromide are 22.93 and 16.93 K N m^(-2). Respectively and these compounds forms nearly ideal solution 3. moles of ethylen bromide and 2mole of propylene bromide are equilibrated at 353 K and at a total pressure of20.4 KN m^(-2). (a) What is the composition of the liquids phase (b) How many moles of each compound are present in the vapour phase ?
Answer» Solution :(a) Let ethylene BROMIDE `rarr to A` And propylene bromide `to B` Then from EQUATION, `P_(A)^(@)=22.93 N Km^(-2)` `P_(B)^(0) 18.93 K Nm^(-2)` `n_(A)=3` mole `n_(B)=3` mole `n_(B)=2` mole Total pressure`P_(T)=20.4 K Nm^(-2)` `:.P_(T)=P_(A)^(@)xxP_(B)^(@)X_(B)` `=P_(A)^(@)xxP_(B)^(@)(1-X_(A))=(P_(A)^(@)-P_(B)^(@))X_(A)+P_(B)^(@)` `rArr X_(A)=(P_(T)-P_(B)^(@))/(P_(A)^(@)-P_(B)^(@))=(20.4-16.93)/(22.93-16.93)=0.578` `:.X_(A)=1-0.578=0.422` (b) Let mole fraction in VAPOUR phase `=X_(A)` `X_(A)=(P_(A)^(0))/(P_(T))` `X_(A)=(22.93xx0.598)/(20.4)=0.64""...(1)` Assuming that the no. of moles of A and B that are VAPORIZED are a and b then `X_(A)=(a)/(a+b)=0.64` But composition of A in liquid phase. `X_(A)=(3-a)/((3-a)+(2-b))=0.578` `=(3-a)/(5-(a+b))=0.578 ""...(2)` Solving equation (1) and (2) `a=0.9967` mole `b=0.537` mole