Physical Biochemistry

Physical Biochemistry

Kwan Hee Lee, Ph.D.

Handong Global University

Week 4

 S₂-S₁ = ∫dq_rev/T = ∫CdT/T

 At constant P, ΔS = ∫C_PdT/T = C_PlnT₂/T₁

 At conmstant V, ΔS = ∫C_VdT/T = C_VlnT₂/T₁

 Because C_P and C_V are always positive, both of above equations show that

raising temperature will always increase the entropy.

Temperature dependence of

Entropy

 ΔS⁰(T₂) = ΔS⁰(T₁) + ∫ ΔC_P dT/T, ΔC_P = C_P(products) – C_P(reactants)

Temperature dependence of the

entropy change for a chemical reaction

 There is a entropy change for a phase transition at equilibrium temperature.

 q_P = q_rev = ΔH_tr

 ΔS_tr = ΔH_tr/T_tr

 S⁰(25℃, 1 atm) = S⁰(0K) + ∫C_P(s)dT/T + ΔH_m/T_m + ∫C_P(l)dT/T, T_m is the melting point at 1 atm.

Entropy change for a phase

transition

 Ignore the direct effect of pressure on entropy for liquids and solids.

 At constant temperature, dS = dq_rev/T = (dE-dw_rev)/T

 For an ideal gas, pressure change from P₁ to P₂, dE is zero because E is independent of P at constant temperature. dS = -

dw_rev/T = PdV/T

Pressure dependence of Entropy

 D(PV) = PdV + VdP = 0 (constant temperature)

and PdV = -VdP (constant temperature)

 Therefore dS = PdV/T = -VdP/T = -nRdP/P

 ΔS = -nR∫(P₁ to P₂)

 ΔS=S(P₂)-S(P₁) = -nRlnP₂/P₁

If either P₁ or P₂ is zero, the entropy of n mol of gas becomes infinite.

 As the pressure decreases the entropy will increase.

Pressure dependence of Entropy

 Mixing two gases

 nA mol of A and nB mol of B at the same pressure and temperature.

 After the mixing, there should not be any pressure change because they do not react each other and they behave like ideal gases.

 ΔS_A = -n_ARln(X_AP/P) = -n_ARlnX_A

 ΔS_B= -n_BRln(X_BP/P) = -n_BRlnX_B

 Δs_mix = ΔS_A + ΔS_B = n_ARlnX_A + n_BRlnX_B

 If the two gases are identical, there is no entropy change upon mixing.

Pressure dependence of Entropy

 The sign of the entropy change tells whether the reaction can occur. Very few reactions occur at constant energy and volume and thus do not affect the surroundings.

 The D form and the L form have identical energies, volumes, and entropies, but the mixture has a larger entropy and nearly the same energy and volume.

 The conversion reaction is spontaneous.

Spontaneous chemical reactions

 G = H-TS

 Focus on the system that is free to exchange heat with the surroundings to remain at room temperature, and it can expand or contract in volume to remain at atmospheric pressure.

 If ΔG is negative, the process is spontaneous.

 If ΔG is positive, the process is not spontaneous.

 If ΔG is zero, the process is at equilibrium.

Gibbs Free Energy

 dG = dH-TdS – SdT

=dE + PdV + VdP – TdS –SdT (H=E+PV)

=dq_rev + dw_rev + PdV + VdP –TdS –SdT

=TdS + dw_rev + PdV + VdP –TdS – SdT

= dw_rev + PdV + VdP –SdT

 dw_rev = -PdV + dw*_rev

 dG = VdP – SdT + dw*_rev

dw*_rev: all other forms of work

 dG = dw*_rev (at constant T and P, dT=dP=0)

 ΔG = w*_rev

ΔG and a system’s capacity to do

nonexpansion work

 ΔG < 0, spontaneous at constant T and P

 ΔG > 0, nonspontaneous at constant T and P

 ΔG = 0, system is at equilibrium at constant T and P

 ΔG = ΔH - T ΔS, so need to consider both the entropy and enthalpy.

Spontaneous reactions at constant

T and P

 If ΔG of a reaction is known at 25˚C, how do we obtain its value at a different

temperature?

 If the temperature is not far off from the standard state, we assume there is not much change in enthalpy and entropy of the standard state.

 What about at physiological temperature, 37˚C?

Temperature dependence of Gibbs

Free Energy

 ΔG(T) - ΔG (25˚C) = -(T-298) ΔS(25˚C).

 dG=VdP-SdT

 At constant pressure, VdP=0, so dG/dT = - S

 ΔG=G(T₂-T₁) = -∫SdT

 For general reactions, dΔG/dT = - ΔS

 ΔG(T₂)- ΔG (T₁) = - ∫ ΔSdT (from T₁ to t₂)

 ΔG(T₂)- ΔG (T₁) = - ΔS (T₂-T₁)

Temperature dependence of Gibbs

Free Energy

 ΔG(T₂)/T₂ - ΔG(T₁)/T₁ = ∫ ΔH(T)/T₂ dT:

Gibbs-Helmoltz equation

 ΔG(T₂)/T₂ - ΔG(T₁)/T₁ = ΔH[1/ T₂ - 1/ T₁]

Temperature dependence of Gibbs

Free Energy

 ΔG(T₂)/T₂- ΔG (T₁)/T₁ = -∫(ΔH(T)/T² )dT

 d(G/T)/dT = G[d(1/T)/dT] + (1/T))dG/dT)

= -G/T² + (1/T)(dG/dT)

 d(G/T)/dT = -G/T²-S/T

=(-G-TS)/T² = -H/T²

 d(G/T)/d(1/T)=H

 ΔG(T₂)/T₂ – ΔG(T₁)/T₁ = ΔH[1/T₂-1/T₁]

Gibbs-Helmoltz equation

 dG = VdP-SdT = VdP

 G(P₂)-G(P₁) = ∫VdP

 For a solid or liquid: G(P₂) –G(P₁) = V(P2- P₁)

 For a gas molecule: G(P₂)-G(P₁) = ∫nRT/P

= nRT lnP₂/P₁

Pressure dependence of Gibb’s

Free Energy