When glucose-6-phosphate and fructose-6-phosphate are mixed with the enzyme phosphohexose isomerase under standard conditions, the final equilibrium concentration of glucose-6-phosphate is 1.33 M and the final concentration of fructose-6-phosphate is 0.66 M. What is the free energy change for the conversion of glucose-6-phosphate
→ fructose-6-phosphate?
The answer is deltaG=+ 1.7 kJ/mol
dG = -RTlnK
K = 0.66/1.33
R = 8.314
T = 298
dG = 1.7 kJ
To calculate the free energy change (ΔG) for the conversion of glucose-6-phosphate to fructose-6-phosphate, we can use the formula:
ΔG = -RT ln(Keq)
Where:
ΔG is the free energy change
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
ln is the natural logarithm
Keq is the equilibrium constant
Now, let's break down the steps to calculate the ΔG:
Step 1: Convert the given temperature to Kelvin.
The standard conditions for biochemical reactions are usually 25°C or 298 K.
Step 2: Calculate the equilibrium constant (Keq).
The equilibrium constant is the ratio of the concentrations of products to reactants at equilibrium. In this case, it can be written as:
Keq = [fructose-6-phosphate] / [glucose-6-phosphate]
Using the given final concentrations:
Keq = 0.66 M / 1.33 M
Step 3: Substitute the values into the ΔG formula.
ΔG = -RT ln(Keq)
Substituting the values:
ΔG = - (8.314 J/(mol·K)) * 298 K * ln (0.66 M / 1.33 M)
Step 4: Convert the units.
The given answer is in kilojoules per mole (kJ/mol), so we need to convert from Joules to kilojoules.
ΔG = - (8.314 J/(mol·K)) * 298 K * ln (0.66 M / 1.33 M) / 1000 J/kJ
After completing the calculation, the answer comes out to be approximately +1.7 kJ/mol.