An enzymatic hydrolysis of fructose was allowed to proceed to equilibrium at 25 degrees C. The original concentration of fructose-1-p was .2 M, but when the system had reached equlibrium the concentration of fructose 1-p was only 6.52 * 10 -5 M. Calculate the equilibrium constant for this reaction and the standard free energy of hydrolysis of fructose 1-p.
To calculate the equilibrium constant (K) for the enzymatic hydrolysis of fructose, we can use the equilibrium concentrations of the reactants and products. In this case, the reactant is fructose-1-p (F1P), and the product is fructose (F).
The equilibrium constant (K) is defined as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients. The balanced chemical equation for the reaction is:
F1P ↔ F
Therefore, the equilibrium constant expression is:
K = [F]^1 / [F1P]^1
Given that the concentration of fructose-1-p at equilibrium is 6.52 x 10^(-5) M and assuming the concentration of fructose is 0 M initially (as it is a reactant that is produced), we can substitute the values into the equation:
K = (0 M)^1 / (6.52 x 10^(-5) M)^1
K = 0 / 6.52 x 10^(-5) M
K = 0
Therefore, the equilibrium constant for this reaction is 0.
Now, to calculate the standard free energy of hydrolysis of fructose-1-p (ΔG°), we can use the relationship between K and ΔG°, which is given by the equation:
ΔG° = -RT ln(K)
Where:
ΔG°: Standard free energy change
R: Gas constant (8.314 J/mol∙K)
T: Temperature in Kelvin (25 degrees Celsius = 298 K)
ln: Natural logarithm
Substituting the values into the equation:
ΔG° = -(8.314 J/mol∙K) x (298 K) x ln(0)
ΔG° = -(8.314 J/mol∙K) x (298 K) x (-∞)
ΔG° = ∞
Therefore, the standard free energy of hydrolysis of fructose-1-p is infinite.