The vapour pressure of 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` minute, `0.525` mole of a solute is dissolved whivh 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.

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)`