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Outcome 2: Assignment
1. Calculate the equilibrium constant for the reaction at pH 7.0 and 25°C
(ΔG°’ = - 9.7 kJ.mol-1
).
Glycerol-3-phosphate + H2O Glycerol + Pi
2. The final step in the pathway for the synthesis of glucose from lactate (gluconeogenesis) is:
Glucose-6-phosphate + H2O Glucose + Pi
When glucose-6-phosphate is incubated with the proper enzyme and the reaction runs until
equilibrium has been reached, the final concentrations are found to be:
glucose-6-phosphate (0.035 mM), glucose (100 mM), and Pi (100 mM).
Calculate ΔG°’ at 25C and pH 7.
3. The Go' for the reaction Citrate Isocitrate is +6.64 kJmol-1
. The Go' for the reaction
Isocitrate α-Ketoglutarate is -267 kJmol-1
.
Calculate the Go' for the conversion of citrate to α-ketoglutarate?
Help with number 3 please
To calculate the equilibrium constant (K) for a reaction, you need to use the equation:
ΔG°’ = -RTln(K)
Where:
ΔG°’ is the change in Gibbs free energy under standard conditions
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
K is the equilibrium constant
For Outcome 2, question 1:
Given ΔG°’ = -9.7 kJ/mol, we need to convert it to J/mol:
ΔG°’ = -9.7 kJ/mol × 1000 J/1 kJ = -9700 J/mol
Now, we can solve for K:
-9700 J/mol = -8.314 J/mol·K × 298 K × ln(K)
Simplifying the equation:
ln(K) = (-9700 J/mol) / (-8.314 J/mol·K × 298 K)
ln(K) = 3.896
Take the exponential of both sides to solve for K:
K = e^(3.896)
K ≈ 49.145
Therefore, the equilibrium constant for the reaction at pH 7.0 and 25°C is approximately 49.145.
For Outcome 2, question 2:
To calculate ΔG°’ at 25°C and pH 7, we can use the equation:
ΔG°’ = ΔG° + RTln(Q)
Where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.
Given the final concentrations:
glucose-6-phosphate = 0.035 mM
glucose = 100 mM
Pi = 100 mM
The reaction can be written as:
glucose-6-phosphate + H2O ↔ glucose + Pi
The reaction quotient (Q) is given by:
Q = ([glucose] × [Pi]) / ([glucose-6-phosphate] × [H2O])
Plugging in the values:
Q = (100 mM × 100 mM) / (0.035 mM × 1 M)
Q ≈ 285714.2857
Now, we can calculate ΔG°’:
ΔG°’ = -RTln(K)
ΔG°’ = ΔG° + RTln(Q)
Let's calculate ΔG° first:
ΔG° = -9.7 kJ/mol × 1000 J/1 kJ = -9700 J/mol
Now, using the equation:
ΔG°’ = ΔG° + RTln(Q)
ΔG°’ = -9700 J/mol + (8.314 J/mol·K × 298 K × ln(285714.2857))
Calculating the value:
ΔG°’ ≈ -7400 J/mol
Therefore, ΔG°’ at 25°C and pH 7 is approximately -7400 J/mol.
For Outcome 2, question 3:
To calculate the ΔG°' for the conversion of citrate to α-ketoglutarate, we can use the equation:
ΔG°' = ΔG°'c - ΔG°'i
Where ΔG°' is the standard Gibbs free energy change, and the subscripts "c" and "i" represent the products and reactants, respectively.
Given values:
ΔG°' for Citrate → Isocitrate = +6.64 kJ/mol = +6,640 J/mol
ΔG°' for Isocitrate → α-Ketoglutarate = -267 kJ/mol = -267,000 J/mol
Substituting the values:
ΔG°' = +6,640 J/mol - (-267,000 J/mol)
ΔG°' ≈ 273,640 J/mol
Therefore, the ΔG°' for the conversion of citrate to α-ketoglutarate is approximately 273,640 J/mol.