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.
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
To determine the phosphate concentration at equilibrium, we need to use the equation relating the standard free energy change (ΔG°), the equilibrium constant (Keq), and the reaction quotient (Q):
ΔG° = -RT ln(Keq)
where R is the gas constant (8.314 J/mol*K) and T is the temperature in Kelvin.
First, let's convert the given ΔG° value from kJ/mol to J/mol:
ΔG° = -13.8 kJ/mol × 1000 J/kJ = -13800 J/mol
Next, we need to rearrange the equation to solve for Keq:
ΔG° = -RT ln(Keq)
Solving for Keq:
Keq = exp(-ΔG° / RT)
Now, we can calculate Keq. Assuming room temperature (25°C or 298 K):
Keq = exp(-(-13800 J/mol) / (8.314 J/mol*K × 298 K))
Keq = exp(55.44)
By definition, at equilibrium, the concentration of products divided by the concentration of reactants (raised to the power of their stoichiometric coefficients) equals the equilibrium constant (Keq). In this case, the hydrolysis reaction of glucose-6-phosphate forms one phosphate ion. Therefore, at equilibrium, the concentration of phosphate ([P]) is equal to Keq multiplied by the initial concentration of glucose-6-phosphate ([G-6-P]).
Given that the initial concentration of glucose-6-phosphate [G-6-P] = 4 mM,
[P] = Keq × [G-6-P]
[P] = exp(55.44) × 4 mM
Calculating this value will give you the phosphate concentration at equilibrium.