1. Given the following equation:
Glycerol-3-phosphate glycerol + Pi Go’ = -9.7 kJ/mol
At equilibrium the concentrations of both glycerol and inorganic phosphate are 1 mM. Under these conditions, calculate the final concentration of glycerol-3-phosphate. Remember to convert kJ to J.
2. Using the data for the reaction in question 1, calculate the G value (Go’ = -13.8 kJ/mol).
I'm not sure if you typed in the reaction correctly. If the reaction is as followed:
Glycerol-3-phosphate----> Glycerol + Pi
∆G0'= -9.7kj
Then this what you I believe you should do:
The removal of the phosphate is the only way that I can see the reaction being energetically favorable, and at equilibrium, ∆G =0.
R=0.0083145 kJ/mol
T=273.15+25=298.15K
∆G=0
∆Go'= -9.7kj/mol
Pi=1mM
Glycerol= 1mM
Q=reactants/products
Glycerol-3-phosphate=G3P=?
Q=reactants/products=[1mM][1mM]/G3P
Plug in your values and solve for Glycerol 3-phosphate
∆G=∆G0'+RTlnQ
0=∆G0'+RTlnQ
-∆G0'=RTlnQ
-∆G0'/RT=lnQ
10^(-∆G0'/RT)=Reactants/Products
Reactants/10^(-∆G0'/RT)=Products
[1mM][1mM]/10^(-∆G0'/RT)=G3P
[1mM][1mM]/(9.7kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=G3P
For the second part, use the value that you obtained for G3P, and the values that are given to you for Glycerol, ∆G0' (-13.8 kJ/mol), and Pi and plug into
the equation below and solve.
∆G=∆G0'+RTlnQ
Last part of the equation for the first step should be,
[1mM][1mM]/10^[(9.7kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=G3P
I have a bad typo in my original post that may lead to confusion: replace the words reactants with products and vis versa. Everything else is okay except for that.
Meaning, Q=products/reactants, not the other way around. for 10^(-∆G0'/RT)=Reactants/Products part of the equation manipulation.
To solve these questions, we need to use the relationship between the standard free energy change (∆G°), the equilibrium constant (K), and the concentrations of the reactants and products. The equation relating these quantities is as follows:
∆G° = -RT ln(K)
Where:
- ∆G° is the standard free energy change in joules (J).
- R is the gas constant, which has a value of 8.314 J/(mol·K).
- T is the temperature in Kelvin (K).
- ln is the natural logarithm.
Let's address each question separately:
1. To calculate the final concentration of glycerol-3-phosphate, we can use the following equation:
∆G° = -RT ln(K)
First, let's convert the given value of ∆G° from kilojoules per mole (kJ/mol) to joules per mole (J/mol):
-9.7 kJ/mol × 1000 J/kJ = -9700 J/mol
The equilibrium constant (K) is related to the concentrations of the reactants and products at equilibrium. So, to calculate the concentration of glycerol-3-phosphate, we need to determine the value of K.
In this case, since the concentrations of glycerol and inorganic phosphate are both 1 mM, we can assume their concentrations at equilibrium are 1 mM as well. So, K would be equal to:
K = [glycerol-3-phosphate] / ([glycerol] × [Pi])
With the given values of [glycerol] and [Pi] = 1 mM, we can substitute them into the equation:
K = [glycerol-3-phosphate] / (1 mM × 1 mM)
K = [glycerol-3-phosphate] / 1 mM^2
Now, we can rearrange the equation to solve for [glycerol-3-phosphate]:
[glycerol-3-phosphate] = K × (1 mM^2)
Plug in the value of K and solve for [glycerol-3-phosphate].
2. To calculate the G value using the given ∆G° value, we can rearrange the equation as follows:
∆G° = -RT ln(K)
Rearranging the equation, we have:
ln(K) = -∆G° / (RT)
Substitute the values of ∆G°, R, and T into the equation:
ln(K) = - (-13.8 kJ/mol × 1000 J/kJ) / (8.314 J/(mol·K) × T)
Now, we need to determine the value of ln(K) by calculating the natural logarithm of K:
K = e^(ln(K))
Plug in the value of ln(K) and solve for K.
Note: In both cases, ensure that all units are consistent throughout the calculations.