JEE Advance - Chemistry (2001 - No. 1)

The vapour pressure of the two miscible liquids (A) and (B) are 300 and 500 mm of Hg respectively. In a flask 10 moles of (A) is mixed with 12 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 minutes, 0.525 mole of a solute is dissolved which arrests the polymerization completely. The final vapour pressure of the solution is 400 mm of Hg. Estimate the rate of constant of the polymerization reaction. Assume negligible volume change on mixing and polymerization and ideal behaviour for the final solution.

1. 004 x 10^-2 min^-1

1. 004 x 10^-3 min^-1

1. 004 x 10^-4 min^-1

1. 04 x 10^-4 min^-1

1. 4 x 10^-5 min^-1

Explanation

Given, initial moles of A = 10; initial moles of B = 12. Let, the number of moles of A = n, when polymerization is arrested. Moles of solute added = 0.525.

$$\therefore$$ Total number of moles = (n + 12 + 0.525) = (n + 12.525)

$$\therefore$$ Mole fraction of $$A({x_A}) = {n \over {n + 12.525}}$$ and

Mole fraction of $$B({x_B}) = {{12} \over {n + 12.525}}$$

The total vapour pressure of the solution = $$P_A^0{X_A} + P_B^0{X_B}$$

or, $$400 = 300 \times {n \over {n + 12.525}} + 500 \times {{12} \over {n + 12.525}}$$ or, n = 9.9

For a first order reaction, $$k = {{2.303} \over t}\log {{{{[A]}_0}} \over {[A]}}$$

or, $$k = {{2.303} \over {100}}\log {a \over {a - x}} = {{2.303} \over {100}}\log {{10} \over {9.9}}$$ [$$\because$$ a $$-$$ x = n = 9.9]

or, $$k = 1.004 \times {10^{ - 4}}$$ min^$$-$$1

JEE Advance - Chemistry (2001 - No. 1)

Explanation

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