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
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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..
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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
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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 ??
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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..
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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
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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