The vapour pressure of two miscible liquids (A) and (B) are `300mm` 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 minute, `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 adn ideal behavoiour for the final solution.

Answer

`k = 1.004 xx 10^(-4) min^(-1)`

Explanation

Initial moles of `A = 10`
Let the number of moles of `A` when polymerization is arrested be `n`.
Moles of `B = 12`
Moles of solute added `= 0.525`
Total moles `= (n+12 + 0.525)`
`= (n+12.525)`
`chi_(A) = (n)/((n+12.525))`
`chi_(B) = (12)/((n+12.525))`
`:. P = P_(A).^(@)chi_(A) + P_(B).^(@)chi_(B)`
or `400 = 300 xx (n)/((n+12.525)) + 500 xx(12)/((n+12.525))`
Solving, we get
`n = 9.9`
For the first order polymerization,
`k = (2.303)/(t)log.(A_(0))/(A)` or `k = (2.303)/(t)log.(a)/(a-x)`
`= (2.303)/(100) log.(10)/(9.9)`
Solving, we get
`k ~~ 1.004 xx 10^(-4) min^(-1)`