
phalloidin
Registered: 07/03/04
Posts: 865

Biochemistry Help
#4609948  09/01/05 09:17 PM (11 years, 9 months ago) 


hey, anyone have any ideas how to approach this problem

phalloidin
Registered: 07/03/04
Posts: 865

Re: Biochemistry Help [Re: phalloidin]
#4610181  09/01/05 10:31 PM (11 years, 9 months ago) 


if anyone is actually working this, assume standard temperature is 37C for part A. My answer for A is 16.6 kJ/mol so the Keq=626.9 which gives an answer of 0.00015 for part C. Part B is what really bugs me though. I can't see how to determine the Keq at 32C.

DieCommie
Registered: 12/11/03
Posts: 28,165

Re: Biochemistry Help [Re: phalloidin]
#4610797  09/02/05 12:49 AM (11 years, 9 months ago) 


Try physicsforums.com
They have a forums for many scientific disciplines, which are suited for homework questions.

phalloidin
Registered: 07/03/04
Posts: 865

Re: Biochemistry Help [Re: DieCommie]
#4610830  09/02/05 12:57 AM (11 years, 9 months ago) 


cool thanks

LiquidSmoke
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Re: Biochemistry Help [Re: phalloidin]
#4622029  09/05/05 11:58 AM (11 years, 9 months ago) 


HAHAHAHA oh maan, entry level biochemistry is fun.
Okay for a.) It's quite simple.
You have the standard free energy for G6P hydrolysis and ATP hydrolysis.
It's basically asking you to calculate the change in total free energy for the combined reactions.
But since the hydrolysis of the G6P is in reverse, just change the 13.9 kJ/mol into just positive 13.9 kJ/mol.
THen you simply just add it with the free energy of ATP hydrolysis, so you get
13.9  30.5 = 16.6 kJ/mol total for the reaction.
b.) now that you have the total free energy, you can calculate the equilibrium constant simply by using t he free energy formula:
Delta G = R*T*ln(K)
Where Delta G is the number you just calculated, R is the gas constant (1.987 cal/mol) and T is the temprature in Kelvins (Celsius + 273) then just solve for K using reverse natural log.
c.) Now that you have the equilibrium constant, just incorporate K as it's definition in a coupled reaction.
When A + B > C + D
K = [C][D]/[A][B'] with [B'] being 11 fold of [D]. Just simply calculate out the ratios of [C] to [A], and that will give you the ratio of G6P to G6.
It's pretty simple stuff.
 "Shmokin' weed, Shmokin' wizz, doin' coke, drinkin' beers. Drinkin' beers beers beers, rollin' fatties, smokin' blunts. Who smokes tha blunts? We smoke the blunts"  Jay and Silent Bob strike Back

phalloidin
Registered: 07/03/04
Posts: 865


yeah, I got part A and then derived Keq at 37C and got part C. My problem was that for part B the temperature is at 32C whereas part A the temp was 37C. Could you still use the delta G that you get from part A? Wouldn't it change with the temperature?

LiquidSmoke
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Re: Biochemistry Help [Re: phalloidin]
#4623599  09/05/05 08:29 PM (11 years, 9 months ago) 


nah the REGULAR delta G is soley based on the machanisms of the reaction, i.e. the relative stabilities of the beginning and ending compounds.
The STANDARD FREE energy (Delta G^0') the delta G with that crazy lookin sign at the top of it, is what varies with temperature.
So once yo usolve for the free energey change (regular delta G), you can use it as a constant, independent of temperature.
Notice that the regular equation can be calculated simply from the changes in enthalpy. You were probably confused because it's also involved in the other equation consisting of the standard free energy and the (negative RTlnK).
 "Shmokin' weed, Shmokin' wizz, doin' coke, drinkin' beers. Drinkin' beers beers beers, rollin' fatties, smokin' blunts. Who smokes tha blunts? We smoke the blunts"  Jay and Silent Bob strike Back

LiquidSmoke
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Re: Biochemistry Help [Re: phalloidin]
#4624038  09/05/05 10:38 PM (11 years, 9 months ago) 


oh and as a clarification.
The original Delta G becomes temperature independent, despite the formula:
Delta G = Delta H  T Delta S
Because there is no change in entropy, there is no increase or decrease in the number of particles from the reactions to products (you start with 2, you end with 2).
So Delta S is 0, which means T Delta S is also 0, which is why the enthalpy change is the only contributor to the gibbs free energy.
The value you got for KeQ before solving for b is not the right value.
You had to first be able to find the Keq from problem b before you could correctly calculate the answer for question C.
 "Shmokin' weed, Shmokin' wizz, doin' coke, drinkin' beers. Drinkin' beers beers beers, rollin' fatties, smokin' blunts. Who smokes tha blunts? We smoke the blunts"  Jay and Silent Bob strike Back

phalloidin
Registered: 07/03/04
Posts: 865


cool man, thanks for the help

