calculation problem

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kabuto
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calculation problem

Post by kabuto » Sat Apr 15, 2006 1:01 pm

The following reaction occur in glycolysis.

Fructose-6-phosphate --> glucose-6-phosphate K’ = 1.97

What is the ΔG for the reaction at 25 C.
If the concentration of fructose-6-phosphate is 1.5 M and the concentration of glucose-6-phosphate is 0.5 M, what is ΔG?
(R=8.315 J/mol K)
Explain ΔG0 and ΔG are different?

:lol: :lol: :lol: :lol: :lol: :lol: :lol:
can anyone show me how to solve this??

raju
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Post by raju » Sat Apr 15, 2006 2:07 pm

dude i think this is ur home work any how i can just give u a hint ant the name of book to reffer
hints:
1 . search for one standard equation that relates G and G0
2 , its famous name is gibbs equation for entrophy
books:
leninger
voit& voit
more than sufisient
or any bioenergitics or biochemistry or even thermodynamics book wil do all the values r been given subtitute to get the answer :D
the organisms that survive r not the one which r strongest nor the one which r briliant THEY R THE ONE WHICH RESPOND TO CHANGES IN NATURE FIRST

sdekivit
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Re: calculation problem

Post by sdekivit » Sat Apr 15, 2006 4:43 pm

kabuto wrote:The following reaction occur in glycolysis.

Fructose-6-phosphate --> glucose-6-phosphate K’ = 1.97

What is the ΔG for the reaction at 25 C.
If the concentration of fructose-6-phosphate is 1.5 M and the concentration of glucose-6-phosphate is 0.5 M, what is ΔG?
(R=8.315 J/mol K)
Explain ΔG0 and ΔG are different?

:lol: :lol: :lol: :lol: :lol: :lol: :lol:
can anyone show me how to solve this??


delta G = delta G(0) + RT ln [glucose-6-phosphate]/[fructose-6-phosphate]

delta G(0) is the standard Gibbs free energy under standard conditions and is therefore different delta G which isn't under standard conditions.

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kabuto
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Post by kabuto » Sun Apr 16, 2006 1:07 pm

thanks raju, sdekivit
i think i get it now

delta G = delta G(0) + RT ln [glucose-6-phosphate]/[fructose-6-phosphate]
deltaG= 1.97 + (8.315x25) ln(0.5/1.5)

another question.
what are delta G and delta G(0) use for?

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