Physical Biochemistry
Kwan Hee Lee, Ph.D.
Handong Global University
Week 7
ΔG° = -RT ln K
ΔG = ΔG° + RT ln Q
For HOAc, the activity coefficients γ+ and γ- are dependent on the concentrations of the solution and approach 1 as the solution
becomes very dilute. K= (cH+)(cOAc-)/(cHOAc) x γ+γ-/γHOAc = Kc γ+γ-/γHOAc =Kc γ±2/γHOAc
Therefore, for very dilute solutions, K and Kc are equal, and the experimental
determination of one gives the other.
Equilibrium constant and the standard Gibbs Free Energies of the reactants and products
For more concentrated solutions, we need to know the activity coefficients to calculate K
cfrom K, or vice versa.
Again we need to use mean ionic activity coefficients.
HOAc(aq ) H
+(aq) + OAc
-(aq)
ΔG° = μ°
H++ μ°
OAc-- μ°
HOAcEquilibrium constant and the standard Gibbs Free Energies of the reactants and products
The standard states of all species are the
extrapolated 1-M hypothetical state-that is, 1- M concentration in water but with the
properties of a very dilute solution.
For the Gibbs free-energy at any arbitrary concentration, we use general equation.
HOAc(10-4M,aq) → H+(10-4M, aq) + OAc- (10-4 M, aq)
ΔG=ΔG° +RT lnQ=ΔG°+RT ln(10-4)(10-
4)/(10-4) = ΔG° -22,820 J/mol
Equilibrium constant and the standard Gibbs Free Energies of the reactants and products
The calculation above tells us that dilution by a factor of 104 decreases the Gibbs free energy by 22,820 J/mol at 25°C, making the dissociation reaction that much more favorable.
From the Le Chatelier’s principle, we expect that the lower the co0ncentration, the more acetic acid dissociates.
The decision among the possible standard states and equilibrium constants is simply to choose
the most convenient for the system being studied.
Equilibrium constant and the standard Gibbs Free Energies of the reactants and products
Questions in Life Science
How to make buffer solutions of any pH
How much product is obtained from a reaction
How much metal ions is bound in a protein complex.
We assume that we know the equilibrium constants for all of the equilibria involved, and we assume for the time being that all solutions are ideal.
Calculation of Equilibrium
Concentration: Ideal solutions
Need to have same number of equations with the unknown variables.
Can consider the following equations
Conservation of mass
Conservation of charge
Equilibrium constants
ApproximationCalculation of Equilibrium
Concentration: Ideal solutions
cA mol of acetic acid and cS mol of sodium
acetate to water to form 1 L of aqueous buffer solution.
Mass balance: [Na+] = cS
[HOAc] + [OAc-] = cA+cS
Charge balance: [Na+] + [H+] = [OAc-] + [OH- ]
KHOAc = [H+][OAc-]/[HOAc] = [H+]cS/cA
pH=pKA + log cS/cA: Henderson Hasselbalch eqn.
Henderson-Hasselbalch equation
What amount of solid sodium acetate is needed to prepare a buffer at pH 5.00 from 1L of 0.10 M acetic acid?
Answer: pH=pK
HOAc+ log c
S/c
A5.00 = 4.75 + log c
S/0.10
Log c
S= -0.75 c
S= 0.18 mol/L
Sample test
Can we get equilibrium constant at different temperature rather than standard state?
Need to look at the temperature dependence of equilibrium constant.
dΔG/dT = - ΔS (P=constant)
dΔG°/dT = - ΔS° at standard state
ΔG°= -RT lnK
-RT ln K - RT d ln K/dT = - ΔS°
-RT2 d ln K/dT = -TΔS° + RT ln K
Temperature Dependence of the
equilibrium constant
-RT2 d ln K/dT = -TΔS° - ΔG°
= -TΔS° - (ΔH° -T ΔS° ) = - ΔH°
Can rewrite: d ln K/ d(1/T) = - ΔH° /R : van’t hoff equation
If a reaction is exothermic under standard conditions, then ln K and hence K itself
increases with 1/T.
As the temperature decreases and 1/T increases, ln K and therefore K becomes larger: an exothermic reaction is favored when the temperature is lowered.
When ΔH° is approximately a constant over the temperature range of interest,
∫ d ln K = - ΔH° /R ∫ d 1/T
ln K
2/K
1= - ΔH° /R (1/T
2– 1/T
1) (ΔH° = constant)
