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 reaction and the standard free energy of hydrolysis of fructose 1-p (ΔG°), we can use the following equation:
ΔG° = -RTln(K)
where:
- ΔG° is the standard free energy change of the reaction,
- R is the gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin,
- K is the equilibrium constant.
First, let's find the equilibrium concentration of fructose-1-p using the given concentration:
[fructose-1-p]eq = 6.52 × 10^(-5) M
Now, let's find the concentration of fructose-1-p at the start of the reaction (initial concentration):
[fructose-1-p]initial = 0.2 M
Next, we can write the expression for the equilibrium constant (K) as the ratio between the equilibrium concentration of products over the equilibrium concentration of reactants:
K = [fructose-1-p]eq / [fructose-1-p]initial
K = (6.52 × 10^(-5)) / (0.2)
Now, since we know the temperature is 25 degrees Celsius, we need to convert it to Kelvin:
T = 25 + 273.15 = 298.15 K
Finally, we can substitute the values into the equation to calculate the standard free energy of hydrolysis:
ΔG° = -RT ln(K)
ΔG° = -(8.314 J/(mol·K)) * (298.15 K) * ln(K)
Calculate the ln(K) value using the natural logarithm function:
ln(K) = ln((6.52 × 10^(-5)) / (0.2))
Now, substitute the ln(K) value into the equation for ΔG°:
ΔG° = -(8.314 J/(mol·K)) * (298.15 K) * ln((6.52 × 10^(-5)) / (0.2))
Calculate the final value of ΔG° to get the answer.
Note: Make sure to perform all calculations using appropriate units and pay attention to significant figures for the final answer.