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.

Question

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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

rA:rB:rC =(PA/√MA):(PB/√MB):(PC/√MC)=(P0 - P/4):(2 P/2):(P/2 √2) Also (P0 - P/4 P)=(1/2) P0=3Pk=(2 . 303 log (3/2)/10)=0.0405 sec- 1 0.0405× 100=4

rA​:rB​:rC​ =MA​​PA​​:MB​​PB​​:MC​​PC​​=4P0​−P​:22P​:22​P​ Also 4PP0​−P​=21​ P0​=3Pk=102.303log(3/2)​=0.0405sec−1 0.0405×100=4

$r_{A} : r_{B} : r_{C}$
$= \frac{P _{A}}{M _{A}} : \frac{P _{B}}{M _{B}} : \frac{P _{C}}{M _{C}} = \frac{P _{0} - P}{4} : \frac{2 P}{2} : \frac{P}{2 2}$
Also $\frac{P _{0} - P}{4 P} = \frac{1}{2}$
$P_{0} = 3 P k = \frac{2.303 l o g ( 3/2 )}{10} = 0.0405 sec^{- 1}$
$0.0405 \times 100 = 4$