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Chemistry 11 Online
OpenStudy (dls):

In a system, A(s) <===> 2B(g)+3C(g) If the concentration of C at equilibrium is increased by a factor of 2,it will cause the equilibrium concentration of B to change to : (a)two times the original value (b)one half of its original value (c) 2√2 times the original value (d)1/2√2 times the original value

OpenStudy (dls):

@shubhamsrg @Yahoo!

OpenStudy (shubhamsrg):

write the value of eqbm const. K here..

OpenStudy (dls):

\[\LARGE [C]^{3}[B]^2\]

OpenStudy (shubhamsrg):

thats just the numerator..

OpenStudy (shubhamsrg):

ohh wait,,its solid.. yep correct..

OpenStudy (dls):

dude thats a solid :P

OpenStudy (dls):

yeah so o.O

OpenStudy (shubhamsrg):

since temp doesnt change, K will also not change.

OpenStudy (dls):

O.o

OpenStudy (dls):

o.O

OpenStudy (dls):

yes

OpenStudy (shubhamsrg):

K= C^3 . B^2 = (2C)^3 . (B')^2 solve for B'

OpenStudy (dls):

\[\LARGE 8C^{3}B^{2}=K\]

OpenStudy (dls):

\[\LARGE 8C^{3} B'^{2}=C^{3}B^{2}\]

OpenStudy (shubhamsrg):

?? any problems ??

OpenStudy (dls):

1/root 8

OpenStudy (shubhamsrg):

whats that ?

OpenStudy (dls):

B/root 8

OpenStudy (dls):

\[\LARGE B'=\frac{B}{\sqrt{8}}\]

OpenStudy (shubhamsrg):

yep..correct..

OpenStudy (dls):

so why is it 1/root 8?

OpenStudy (shubhamsrg):

o.O you just calculated that right,,why shouldnt it be 1/sqrt8 ?

OpenStudy (dls):

oh got it :P can i ask something stupid?

OpenStudy (shubhamsrg):

hmmm.. ?

OpenStudy (dls):

got it x2 :P

OpenStudy (dls):

thanks..this was a nice problem O.o uve done this type?

OpenStudy (shubhamsrg):

guess so.. :|

OpenStudy (dls):

good good o.o kitna syllabus hogaya?

OpenStudy (shubhamsrg):

hmm,,mere padhai ka hisab kitab bahut kharab hai bhai,,fb pe hi chat karein please.! :P

OpenStudy (dls):

lol :D

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