The Go’ value for the hydrolysis of glucose-6-phosphate is -13.8 kJ/mol. Assuming a concentration of 4 mM for this molecule, what would the phosphate concentration be at equilibrium (G = 0). Remember to convert kJ to J.

Answer 1

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.

Answer 2

that are all the information given

Answer 3

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!!!!!!!

Answer 4

In the manipulation of the equation, replace

10^(-∆G0'/RT)=Reactants/Products

with

10^(-∆G0'/RT)=Products/Reactants