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Decomposition. of non-volatile solute 'A' into another non-volatile solute B and C when dissolved in water follows first order kinetics as :
$A ⟶ 2 B + C$
When one mole of A is dissolved in 180 gm of water and left for decomposition, the vapour pressure of solution was found to be 20 mm Hg after 12 hrs. Determine the vapour pressure of the solution (in mm of Hg) after 24 hrs. Assume constant temperature of 25° C, throughout. The vapour pressure of pure water at 25° C is 24 mm Hg. [Fill your answer by multiplying it with 100]

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answer is 1920.

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Detailed Solution

In the reaction we have,

$A ⟶ 2 B + C$

$\begin{array}{cccc} t = 0 & 1 & - & - \\ t = t & 1 - x & 2 x & x \end{array}$

$\begin{array}{l} n_{T} = 1 + 2 x \\ P_{S} = x_{solvent} \times P_{solvent}^{0} \\ 20 = (\frac{10}{11 + 2 x}) \times 24 \end{array}$

$\begin{array}{l} \Rightarrow x = \frac{1}{2} \\ \Rightarrow t_{1 / 2} = 12 hrs . \\ After 24 hrs., n_{A} \to \frac{1}{4} \\ \Rightarrow x = \frac{3}{4} \\ \Rightarrow P_{8} = (\frac{10}{11 + \frac{3}{2}}) \times 24 = 19.2 mm \end{array}$

Ans. $\to 1920$

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