The vapour pressure of two miscible liquids A and B are 300 and 500 mmHg respectively. In a flask 10 moles of A are mixed with 12 moles 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 400mmHg. Estimate the rate constant of the polymerisation reaction. Assume negligible volume change on mixing and polymerisation, and ideal behaviour for the final solution.

Explanation

10 "mole"
`nA(l)overset(B(l)(12"mole"))underset("polymerisation")(to)An(s)`
After 100 min : x moles (say)
Rate law : Rate `=-(1)/(n) (d[A])/(dt)=k.[A]`
or `-(1)/(n)(dx)/(dt)=k.(x)`
On integration : `nk.=k=(2.303)/(t)log.(a)/(x)`............(1)
[In this problem n is not given. As n i.e., no. of molecules of a polumerising to give one molecule is a constant for this reaction and so the multiplication of n with the rate constant k. gives another constant k, which may also be called rate constant, though not according to its definition]
Let us now calculate x to get the value of k from Equation (1) .
When `0.525` mole of a solute is added , polymerisation stops and x moles of A remain. Just before the addition of the solute
moles of solvent `A=x`
moles of solvent `B=12`
`:.` vapour pressure of the mixture of solvents `=x_(A)p_(A)^(0)+x_(B)p_(B)^(0)`
`p^(0)=(x)/(x+12)xx300+(12)/(x+12)xx500`.............(2)
Vapour pressure of the solution `p=400`............(3)
(after the addition of `0.525` mole of solute)
Applying Raoult.s law
`(p^(0)-p)/(p^(0))=` mole fraction of the solute
`=(0.525)/(0.525+(x+12))` ...........(4)
Solving equations (2), (3) and (4) we get
`x=2.84`
Substituting `x=2.84`, `t=100m` and `a=10` in Equation (1)
`k=(2.303)/(100)log.(10)/(2.84)=1.2xx10^(-2)min^(-1)`