Posted by Lena on Monday, February 11, 2013 at 11:44am.
The hydrolysis of glucose-6-phosphate produces phosphate and glucose, so I will have to assume that glucose and phosphate are at equal concentrations for the reaction in order to do this. If you were given the information for glucose, let me or someone else know.
that are all the information given
Okay then this what you I believe you should do:
Glucose-6-phosphate----> Glucose + Pi
∆G0'= -13.8kj
Since the reaction is at equilibrium, ∆G =0.
R=0.0083145 kJ/mol
T=273.15+25=298.15K
∆G=0
∆Go'= -13.8kj/mol
Pi=x
Glucose= x
Glucose-6-phosphate=4mM
Q=products/reactants
Q=products/reactants=[x][x]/4mM
Plug in your values and solve for x
∆G=∆G0'+RTlnQ
0=∆G0'+RTlnQ
-∆G0'=RTlnQ
-∆G0'/RT=lnQ
10^(-∆G0'/RT)=Reactants/Products
Reactants*[10^(-∆G0'/RT]=Products
(4mM)*10^(-∆G0'/RT)=x^2
4mM*[10^(13.8kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=x^2
Take the square root of your answer to solve for one x.
x=Pi concentration=glucose concentration
*****The concentration of glucose and Pi will be the same.
Good luck!!!!!!!
In the manipulation of the equation, replace
10^(-∆G0'/RT)=Reactants/Products
with
10^(-∆G0'/RT)=Products/Reactants
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