The vapour pressure to two miscible liquids (A) and (B) are 300 and 500 mm of Hg respectively. In a flask 10 mole of (A) is mixed with 12 mole of (B). However, as soon as (B) is added, (A) starts polymerising into a completely insoluble solid. The polymerisation follows first-order kinetics. After 100 minutes, `0.525` mole of a solute is dissolved which arrests the polymerisation completely. The final vapour pressure of the solution is 400 mm of Hg. Estimate the rate constant of the polymerisation reaction. Assume negligible volume change on mixing and polymerisation and ideal behaviour for the final solution.

Explanation

`1.005xx10^(-4)" m in"^(-1)`
Let the number of moles of A left after 100 m in x
Total number of moles after 100 m x 12 0.525
`P_("m ix")=P_(A) +P_(B) =((x)/(12+x) xx300)+((12)/(12+x)xx500) =(6000+300 x)/(12+x)`
According to Raoult's law
`(((6000+300 x)/(12+x))-400)/((6000+300x)/(12+x))=(0.525)/(x+12+0.525)`
On solving, we get x 9.9
Now according to first-order kinetics, `k=(2.303)/(100)" log "(10)/(9.9) =1.005 xx10^(-4)" m in"^(-1)`