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Study... Experiment...
Aditya answered this 8th Edition 11th Edition
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1,027 answers
Hi
We can consider the given deactivation of pores in the following manner
P --BP
Here P is pore and BP is the blocked pore.
Unit...
We can see in that part 1 that the unit of the decay constant is given as S^-1 . 7th Edition
View all solutions
We know that for a nth order reaction the unit of the rate constant is given by
(S^-1)*M^(1-n)
Since, the molarity term is not appearing, we can say that n=1 .
So the rate of disappearance of Pores Top Chegg tutors who can
-dp/dt = K*P
Here P is the concentration of P at any time
help right now
-dp/P = kdt
Integrating between time t=t1 , P1 and time =t2 and pressure P2 gives Omar M.
61
ln(P2/P1) = -K(t2-t1) equation 1 University of Pennsy…
We can take any two data points and then substitute the values in the above equation.
Erin S.
For simplicity I am taking rst two data points Yale University
1
P2 = 4.23*10^9 /m3
P1 =8.5*10^10 /me Lorenzo L.
4
t1 = 0 S Columbia University
t2 = 300 S
ln(0.423/8.5)=-k*300 Find me a tutor
K= 0.01 S^-1 .
Part b and C
At t=40 minutes we can use the equation 1 with the following values.
P2=unknown
P1 =8.5*10^10 /m^3
t1=0
t2 = 40*60=2400 S
ln(P2/8.5*10^10)= - 0.01*2400=24
P2=3.20/m^3
Inactivation factor is nothing but the fraction of pores inactivated.
This can be written as = 1 - fraction of pores survived
= 1-3.2/8.5*10^10
= 1 (approximately)
Survival factor =fraction of pores that survived
=3.2/8.5*10^10
=3.77*10^-11
Hope this helps
Feel free to comment in the case of any doubt.
Thank you
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