2. Given the following data, calculate the Keq for the denaturation reaction of β-lactoglobin at 25 oC.
Δ Ho = -88 kJ/mol, Δ So = 0.3 kJ/mol
dGo = dHo - TdSo
Solve for dG.
Then dG = -RT*lnK
Solve for K.
Use 8.314 for R and 298.15 for T.
Dr. Bob222 is completely correct about how to get the answer, but I think I know why you posted the question in the first place. You are correct to be kind of confused because ΔG does not equal ΔG'°, and they are not the same thing. However, since β-lactoglobin is the reactant and product for its denaturing, ΔG= ΔG'°. Looking at the equation:
ΔG= ΔG'°+ RTlnQ
Q=products/reactants
since the products= the reactants, Q=1 and ln(1) is equal to 0, so ΔG= ΔG'°. With that being said, you can use ΔG=-RT*lnK or K'eq=10^(-ΔG'°/1.36) and solve for K'eq; both equations are the same.
1.24 * 10^31
To calculate Keq for the denaturation reaction of β-lactoglobin at 25°C, we will use the equation:
ΔG = -RT ln(Keq)
Where:
ΔG is the change in Gibbs free energy
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
First, we need to convert the given values of ΔHo and ΔSo from kJ/mol to J/mol:
ΔHo = -88 kJ/mol = -88,000 J/mol
ΔSo = 0.3 kJ/mol = 300 J/mol
Next, we convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Plugging the values into the equation:
ΔG = -RT ln(Keq)
ΔG = ΔHo - TΔSo
Now we solve for Keq:
Keq = e^(-ΔG / RT)
Substituting the values:
Keq = e^(-ΔHo / RT + ΔSo / R)
Keq = e^(-(-88,000 J/mol) / (8.314 J/(mol·K) * 298.15 K) + 300 J/mol / (8.314 J/(mol·K)))
Keq = e^(26467.4 - 36.16)
Keq = e^26391.24
Using a scientific calculator or an online tool, we can calculate the value of Keq.