InterviewSolution
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InterviewSolution
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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. |
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Answer» Correct Answer - `1.0 xx 10^(-4)` `A+B rarr A_(n)+B` `P_(M)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)` Let a mole of A are left due to polymerization after 100 min. `P_(M) = 300 ((a)/(12+a))+500((12)/(12+a))` ….(i) `k =(2.303)/(100)"log"(10)/(a)` ……(ii) after 100 minute solute is added & final vapour pressure is 400 mm Hg i.e. `P_(s) = 400` `(P_(M)-400)/(400) = (0.525)/((a+12))` ....(iii) from equation (i) and (iii) `a = 9.9` putting this in eq. (ii) k `= (2.303)/(100)"log"(10)/(9.9) = 1.0xx10^(-4)` |
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