Gibbs Free Energy and Chemical Potential. NC State University


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1 Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University
2 The internal energy expressed in terms of its natural variables We can use the combination of the first and second laws to derive an expression for the internal energy in terms of its natural variables. If we consider a reversible process: du = δq + δw δw = PdV (definition of work) δq = TdS (second law rearranged) Therefore, du = TdS  PdV This expression expresses the fact that the internal energy U has a T when the entropy changes and a slope P when the volume changes. We will use this expression to derive the P and T dependence of the free energy functions.
3 The Gibbs energy expressed in terms of its natural variables To find the natural variables for the Gibbs energy we begin with the internal energy: du = TdS  PdV and substitute into: dh = du + PdV + VdP to find: dh = TdS + VdP (S and P are natural variables of enthalpy) and using: dg = dh  TdS  SdT we find: dg = SdT + VdP (T and P are natural variables of G) Once again we see why G is so useful. Its natural variables are ones that we commonly experience: T and P.
4 The variation of the Gibbs energy with pressure We shown that dg = VdP  SdT. This differential can be used to determine both the pressure and temperature dependence of the free energy. At constant temperature: SdT = 0 and dg = VdP. The integrated form of this equation is: ΔG = P 2 P 1 VdP For one mole of an ideal gas we have: ΔG m = RT P 2 dp = RT ln P 2 P P 1 P 1 Note that we have expressed G as a molar quantity G m = G/n.
5 The variation of the Gibbs energy with pressure We can use the above expression to indicate the free energy at some pressure P relative to the pressure of the standard state P = 1 bar. G m T = G 0 T + RT ln P 1 bar G 0 (T) is the standard molar Gibb s free energy for a gas. As discussed above the standard molar Gibb s free energy is the free energy of one mole of the gas at 1 bar of pressure. The Gibb s free energy increases logarithmically with pressure. This is entirely an entropic effect. Note that the 1 bar can be omitted since we can write: RT ln P 1 bar = RT ln P RT ln 1 = RT ln P
6 The pressure dependence of G for liquids and solids If we are dealing with a liquid or a solid the molar volume is more or less a constant as a function of pressure. Actually, it depends on the isothermal compressibility, κ = 1/V( V/ P) T, but κ is very small. It is a number of the order 104 atm 1 for liquids and 106 atm 1 for solids. We have discussed the fact that the density of liquids is not strongly affected by pressure. The small value of κ is another way of saying the same thing. For our purposes p we can treat the volume as a constant and we obtain G T = G 0 m T + V m P 1
7 Question What is the isothermal compressibility κ = 1/V( V/ P) T for an ideal gas? A. κ = 1/V B. κ = 1/P C. κ = 1/T D. κ = nrt/p
8 Question What is the isothermal compressibility κ = 1/V( V/ P) T for an ideal gas? A. κ = 1/V B. κ = 1/P C. κ = 1/T D. κ = nrt/p
9 Question Compare the following two equations for the pressure dependence of the Gibbs free energy. Which one applies to the formation of diamond d from graphite? A. G m T = G 0 T + RT ln P 1 bar B. G m T = G 0 T + V m P 1
10 Question Compare the following two equations for the pressure dependence of the Gibbs free energy. Which one applies to the formation of diamond d from graphite? A. G m T = G 0 T + RT ln P 1 bar B. G m T = G 0 T + V m P 1
11 Systems with more than one component Up to this point we have derived state functions for pure systems. (The one exception is the entropy of mixing). However, in order for a chemical change to occur we must have more than one component present. We need generalize the methods to account for the presence of more than one type of molecule. In the introduction we stated that we would do this using a quantity called the chemical potential. ti The chemical potential is nothing more than the molar Gibbs free energy of a particular component. Formally we write it this way: G Rate of change of G as number of μ i = moles of i changes with all other ni T,P,j i variables held constant.
12 Example: a gas phase reaction Let s consider a gas phase reaction as an example. We will use a textbook tb example: N 2 O 4 (g) = 2 NO 2 (g) We know how to write the equilibrium constant for this reaction. 2 K = P NO 2 P N2 N 2 O 4 At constant T and P we will write the total Gibbs energy as: dg = μ NO2 dn NO2 + μ N2 O 4 dn N2 O 4 dg =2μ NO2 dn μ N2 O 4 dn We use the reaction stoichiometry to obtain the factor 2 for NO 2.
13 Definition of the Gibbs free energy change for chemical reaction We now define Δ rxn G: Δ rxn G = dg dn T,P This is Δ rxn G but it is not Δ rxn G o! Note that we will use Δ rxn G and ΔG interchangably. If we now apply the pressure dependence for one component, to a multicomponent t system: G m T = G 0 T + RT ln μ i T = μ i0 T + RT ln P 1 bar P i 1 bar These two expressions are essentially identical. The chemical potential, μ i, is nothing more than a molar free energy.
14 Question How should one think of the chemical potential, μ of component j? A. It is the potential energy of that component B. It is a molar free energy of that t component C. It is potential entropy of that component D. It is a molar entropy of that component
15 Question How should one think of the chemical potential, μ of component j? A. It is the potential energy of that component B. It is a molar free energy of that t component C. It is potential entropy of that component D. It is a molar entropy of that component Single component: G 0 P m T = G T + RT ln 1 bar Multiple components each have a μ j : μ i T = μ i0 T + RT ln P i 1 bar
16 Application of definitions to the chemical reaction We can write the Gibbs energy as: ΔG =2μ NO2 μ N2 O 4 and use the chemical potentials: 0 μ NO2 = μ NO2 + RT ln P NO2 μ 0 N2 O 4 = μ N2 O 4 + RT ln P N2 O 4 to obtain the following: 0 0 ΔG =2μ NO2 μ N2 O 4 +2RT ln P NO2 RT ln P N2 O 4 P NO ΔG = ΔG 0 + RT ln 2 2 P N2 O 4, ΔG 0 = 2μ 0 μ 0 NO2 N2 O 4
17 Note the significance of ΔG and ΔG o The change ΔG is the change in the Gibbs energy function. It has three possible ranges of value: ΔG < 0 (process is spontaneous) ΔG = 0 (system (y is at equilibrium) ΔG > 0 (reverse process is spontaneous) On the other hand ΔG o is the standard molar Gibbs energy change for the reaction. It is a constant for a given chemical reaction. We will develop these ideas for a general reaction later in the course. For now, let s consider the system at equilibrium. Equilibrium means ΔG = 0 so: P NO 2 0=ΔG 0 + RT ln K, ΔG 0 = RT ln K, K = P NO 2 P N2 O 4
18 Question Which statement is true at equilibrium if K = 2? A ΔG RT l (2) ΔG o 0 A. ΔG = RT ln(2), ΔG o = 0 B. ΔG = 0, ΔG o = RT ln(2) C. ΔG = ΔG o = RT ln(2) B. ΔG = 0, ΔG o = 0
19 Question Which statement is true at equilibrium if K = 2? A ΔG RT l (2) ΔG o 0 A. ΔG = RT ln(2), ΔG o = 0 B. ΔG = 0, ΔG o = RT ln(2) C. ΔG = ΔG o = RT ln(2) B. ΔG = 0, ΔG o = 0
20 Temperature dependence of ΔG o The van t Hoff equation We use two facts that we have derived to determine the temperature t dependence d of the free energy. Here are we are skipping some derivations that are not often used and deriving a very useful expression in biology. ΔG 0 = RT ln(k) ΔG 0 = ΔH 0 TΔS 0 ln(k) = ΔH0 RT + ΔS 0 R ln(k 0 2 ) ln(k 1 )= ΔH0 R T T 2 If we plot ln(k) vs 1/T the slope is ΔH o /R. This is a useful expression for determining the standard enthalpy change.
21 Example: the formation of diamond Graphite and diamond are two forms of carbon. Given that the free energy of formation of diamond d is: C(s, graphite) C(s, diamond) Δ r G o = kj/mol and the densities are: ρ(graphite) = 2.26 and ρ(diamond) = 3.51 calculate the pressure required to transform carbon into diamond. Solution: Graphite will be in equilibrium with diamond when: 0=ΔG 0 + ΔV m P 1 P = 1 ΔG0 = 1 ΔVm Mρd ρ M gr ΔG0
22 Example: the formation of diamond Plugging the values we find: 0=ΔG 0 + ΔV m P 1 P =1 ΔG 0 =1 ΔG 0 ΔV m M ρ M d ρ gr = J/mol kg/mol =1.5x 10 9 Pa = bar
23 Protein folding example: Two state t model k f U F k u Unfolded Folded K = [F]/[U] K = ff/(1ff) Fraction folded ff Fraction unfolded 1ff
24 Thermodynamic model ff = 1/(1+K) K = e ΔGo o /RT ff = 1/(1 + e ΔGo /RT ) ff = 1/(1 + e ΔHo /RT e ΔSo /R ) The temperature at which the protein is 50% folded can be defined as T m the melt temperature. At T o = 0orT = o o m, ΔG m ΔH /ΔS.
25 Equilibrium melt curves Mostly folded o o Mostly unfolded T m In this case: T m = 300 K = ΔH o /ΔS o
26 Van t Hoff plots Slope = ΔH o /R The standard method for obtaining the The standard method for obtaining the reaction enthalpy is a plot of ln K vs. 1/T
27 Question What determines the steepness of the equilibrium melt curve? A. The melt temperature B. The equilibrium constant C. The ratio of the enthalpy to the entropy D. The magnitude of the enthalpy and entropy
28 Question What determines the steepness of the equilibrium melt curve? A. The melt temperature B. The equilibrium constant C. The ratio of the enthalpy to the entropy D. The magnitude of the enthalpy and entropy
29 Question What can you say about the enthalpy of reaction for the process shown in the figure below? A. The reaction is exothermic. B. The reaction is endothermic. C. The reaction is spontaneous. D. None of the above.
30 Question What can you say about the enthalpy of reaction for the process shown in the figure below? A. The reaction is exothermic. B. The reaction is endothermic. C. The reaction is spontaneous. D. None of the above.
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