This is from my biochem I class.. A reaction A + B --> C + D has deltaG^o' = 8 KJ/mole In the cell however, under steady state conditions we find [A] = 2 mM, [B] = 2 mM, [C] = 10 microM & [D] = 40 microM Find delta G' in KJ/mole
what does deltaG^o' mean?
\[DeltaG'^{o}\] is the biochemical standarad free-energy change Where \[DeltaG ^{o}=DeltaG'^{o}+RT \ln (Keq)\]
ok now i get it i didnt get that G^0 but ok, if its steady state do you mean in equilibrium cause in equilibrium delta G = 0
id say you calculate Keq = [C][D]/[A][B] and then delta G if you know the temperature
deltaG^'0 is the gibbs free energy or reaction potencial at cell conditions, PH=7 T=36-37 C and G^0 its at STP so it would be delta G^0=8+8.314(gas constant)*25+273.15(Kelvin)*(A*B/C*D)
So I'm working on problem 8
just make sure the molarity units are consistent
oh...I think I was not considering cell conditions
;D
i got 30,8 kJ/mol
ok so I guess it's just my math, but I'm getting a very different answer than the multiple choices
dont forget that 8 kJ/mol is 8000 J/mol so you add to that value you get from the rest of the equation
I have that the answer is suppose to be -14.8, choice c
seems legit, negative gibbs energy means that the reaction happends by it self without need of applying energy
let me try again: i get that Keq = 1*10^-4 -> lnKeq = 9,21 T = 298 K and R = 8,314 so i dont see how its negative and only if its negative you can get -14,8! now only way i see is that you make assumption that its negative cause reaction is moved towards products...
i really dont have a clue how it can be negative sorry...
I think I might have gotten it. Using wolfram I get that the [A][B]/[C][D] portion = 1e-4 , and going by that you get -14.8 KJ. However, I still don't think I could get this by hand and a simple calculator
I didnt calculate it, if ur saying that the answer is -14.8 then it must be. but i agree with Kryten i dont see how its gonna be negative
I think my professor is just trying to make sure that we know the formula |dw:1336620456918:dw|
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