Kinetics of a Polymerization RXN

Chemistry Level 4

The vapour pressure of two miscible liquids $(A)$ and $(B)$ are $300$ and $500$ $\text{mm of Hg}$ respectively. In a flask $10$ $\text{moles}$ of $(A)$ is mixed with $12$ $\text{moles}$ of $(B)$ . However, as soon as $(B)$ is added, $(A)$ starts polymerizing into a completely insoluble solid. The polymerization follows first-order kinetics. After $100$ $\text{minutes}$ , $0.525$ $\text{mole}$ of a solute is dissolved which arrests the polymerization completely. The final vapour pressure of the solution is $400$ $\text{mm of Hg}$ . Estimate the rate constant $k$ ( in $min^{-1}$ ) of the polymerization reaction. Assume negligible volume change on mixing and polymerization and ideal behavior for the final solution.

If you get the rate constant $\large{k = 10^{-\alpha}} min^{-1}$ give your answer as $\alpha$ .

The answer is 4.

1 solution

Ayon Ghosh
Dec 27, 2017

Let after $100$ $\text{min}$ we have $x$ $\text{moles}$ of A.Then since reaction follows first order kinetics : $k = \dfrac{2.303}{t} log \dfrac{10}{x}$ .Also at the end of $t=100$ $\text{minutes}$ the vapour pressure of the solution is $p_{sol}^{0} = p_{A}^{0}x_{A} + p_{B}^{0}x_{B} = 300(\dfrac{x}{x+12}) + 500 (\dfrac{12}{x+12})...(i)$ .

After adding $0.525$ $\text{moles}$ of the solute the vapour pressure gets lowered to $400$ $\text{mm of Hg}$ .Hence now we can apply Raoults Law to it.

$\dfrac{p_{sol}^{0} - 400}{p_{sol}^{0}} = \dfrac{n}{n+N} = \dfrac{0.525}{0.525 + x + 12}...(ii)$

Solve eqs $(i)$ and $(ii)$ we get $x = 9.90$ $\text{moles}$ .

Simply put this into the starting equation for $k$ (the rate constant) to get : $\large{\boxed{k \approx 10^{-4} \text{min^{-1}}}}$ .

It will be better if you don't ask past jee questions.

Harry Jones - 3 years, 4 months ago

@Harry Jones True I also found later that its in JEE 2001 but actually I found it in RC Mukherjee and liked the problem.

Ayon Ghosh - 3 years, 4 months ago

Kinetics of a Polymerization RXN

The answer is 4.

1 solution

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