Gaseous decomposition of A follows 1st order kinetics. Pure A(g) is taken in a sealed flask where decomposition occurs as
$A (g) ⟶ 2 B (g) + C (g)$
After 10 sec., a leak was developed in the flask. On analysis of the effused gaseous mixture (Obeying Graham's law) coming out initially, moles of B(g) were found to be double of A. What is rate constant in
$\sec^{- 1}$ .
Given : Molecular weight of A = 16; Molecular weight of B = 4 ; Molecular weight of C = 8.
$[\ln 3 = 1.1; \ln 2 = 0.7]$

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answer is 4.

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Detailed Solution

$\underset{a - x}{A} ⟶ \underset{2 x}{2 B} + \underset{x}{C}$

$\begin{array}{l} 2 x = a - x \\ \Rightarrow x = \frac{a}{3} \end{array}$

$\begin{array}{l} 10 \times k = \ln (\frac{a}{2 a}) \\ k = \frac{(1.1 - 0.7)}{10} = 0.04 \sec^{- 1} \end{array}$

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