1. Tardigrade
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  3. Chemistry
  4. Given gaseous decomposition of A follows first order kinetics. Pure A(g) is taken in a sealed flask where decomposition occurs as, A(. g .) arrow 2B(. g .)+C(. g .) a leak was developed in the flask, after 10 sec the leaking gaseous mixture obeys Graham's law. On analysis of the effused gaseous mixture coming out initially, moles of B(g) were found to be double of A. Calculate rate constant in s e c- 1 . Given that Molecular weight of A=16 Molecular weight of B=4 Molecular weight of C=8 [. ln 3=1.1; ln 2=0.7]. Write your answer by multiplying rate constant with 100.

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Q. Given gaseous decomposition of A follows first order kinetics. Pure $A(g)$ is taken in a sealed flask where decomposition occurs as,
$A_{\left(\right. g \left.\right)} \rightarrow 2B_{\left(\right. g \left.\right)}+C_{\left(\right. g \left.\right)}$
a leak was developed in the flask, after $10$ sec the leaking gaseous mixture obeys Graham's law. On analysis of the effused gaseous mixture coming out initially, moles of $B(g)$ were found to be double of A. Calculate rate constant in $s e c^{- 1}$ .
Given that Molecular weight of $A=16$
Molecular weight of $ B=4$
Molecular weight of $C=8 $
$ \left[\right.\ln 3=1.1;\ln 2=0.7\left]\right.$
Write your answer by multiplying rate constant with $100$.

NTA AbhyasNTA Abhyas 2022

Solution:

$r_{A}:r_{B}:r_{C}$
$=\frac{P_{A}}{\sqrt{M_{A}}}:\frac{P_{B}}{\sqrt{M_{B}}}:\frac{P_{C}}{\sqrt{M_{C}}}=\frac{P_{0} - P}{4}:\frac{2 P}{2}:\frac{P}{2 \sqrt{2}}$
Also $\frac{P_{0} - P}{4 P}=\frac{1}{2}$
$P_{0}=3Pk=\frac{2 . 303 log \left(3/2\right)}{10}=0.0405\,\sec^{- 1}$
$0.0405\times 100=4$