dGo = dHo - TdSo
Solve for dG.
Then dG = -RT*lnK
Solve for K.
Use 8.314 for R and 298.15 for T.
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
3 answers
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.
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
1 answer