28 Replies
hey couple of qs, ways of thinking:
1) would the half-life change at all?
2) rate of formation oif P increases ie. now for every mole of R, 3 moles of P are being formed. But the rate at which R reduces, that would be the same, right?
What do the rest of you think about this?
1) no, it wouldn't change, will still be 20days
2) -dR/dT = kR, will still be the same
the only thing changing, the final moles of P, and rate of formation of P
i think the half life will change
as the slope of concentration of P would be steeper
half life is determined by the reactant
so even if the reaction wll be R->2P
it will not change
accha i get it
you mean to say rate constant
as T1/2 = ln2/K
not really, but wait ill try explaining
i was going with the deriavation, but i dont relly know how to explain itt without it
Sir need help! :)
Go for it
oh yea, from the product side, the rate constant changes
i didn't want to use deriavation to keep it simple
@ree ^
so t1/2 will be same?
yes
let me verify
alr
t1/2 is not 20 days
ou, well ill think about it, rn i dont have anythiung on my mind :/
@!Nimay¡ any inputs on this? Office ja raha hoon and will prolly look at stuff properly only after 9th, isiliye aapko tag kar raha hoon 🙂
@Pistulya agar aapke koi inputs ho then most welcome too!
Ji Sir
It's asking for the same graph
the ans is B
P -> 3P
c 0
c-x 3x
at t=20days -> c-x = 3x (concentration is same as per graph)
x=c/4
using first order formula ln(Cnot/C)=kt
ln(C not / 3Cnot/4) = k 20
k = ln(4/3) / 20
now t1/2 = ln2/k
put k here
t1/2 = 20 ln2 / ln(4/3)
which is 48.()
text solution khichadi, if any doubt in above, im here
Max points for effort dude. Thanks soham
correct answer
thanks man
😄
lol was gonna try but I guess someone already solved it
noice
still try, basic stuff of kinetics, will need no time
+fsolved
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