Lecture 28 Thermodynamics: Gibbs Free Energy, Equilibrium Constants and the Entropy Change for a Bimolecular Reaction

Transcription

1 Physical Principles in Biology Biology 3550 Fall 2017 Lecture 28 Thermodynamics: Gibbs Free Energy, Equilibrium Constants and the Entropy Change for a Bimolecular Reaction Monday, 6 November c David P. Goldenberg University of Utah goldenberg@biology.utah.edu

2 Enthalpy (H) Two consistent definitions for H at constant pressure: E minus work due to volume change H = E + P V = E w p If there is no volume change, then w p = 0, and H = E. The heat absorbed during the constant pressure process: H = q p Chemical and biochemical reactions are usually studied under conditions of constant (or nearly constant) pressure, and V is usually very small. H can be experimentally measured.

3 The Gibbs Free Energy (G) G = H TS sys For a process at constant temperature: G = H T S sys For a process at constant temperature and constant pressure: G = T S univ The Gibbs free energy is used to apply the second law. If G < 0, the process will be spontaneous. If G > 0, the reverse process will be spontaneous. If G = 0, the system and the surroundings are at equilibrium.

4 The Equilibrium Constant for a Chemical Reaction k f A B kr Rate expression: d[a] dt At equilibrium: d[a] dt = k f [A] + k r [B] = k f [A] + k r [B] = 0 [B] eq [A] eq = k f k r = K eq

5 A Slightly More Complicated Reaction 2 A k f B kr Rate expression: d[a] = k dt f [A] 2 + k r [B] At equilibrium: d[a] = k dt f [A] 2 + k r [B] = 0 [B] eq [A] 2 eq = k f k r = K eq

6 The Equilibrium Constant and Free Energy Change for a Chemical Reaction A B K eq = [B] eq [A] eq At equilibrium, a little bit of work can push the reaction in either direction. This defines the condition where G = 0 Definition of G for a chemical reaction: The free energy change for converting one mole of reactants to one mole of products, at specified concentrations, without significantly changing concentrations(!). Implies a very, very large volume! Or, a mechanism that restores concentrations continuously.

7 The Relationship Between Free Energy and the Equilibrium Constant A B K eq = [B] eq [A] eq G depends on concentrations of A and B. If If [B] [A] < K eq, the reaction will try to shift towards B, G < 0 [B] [A] > K eq, the reaction will try to shift back to A, G > 0 Define standard concentrations to specify G: Concentrations of all species are 1 M for solutions or 1 atm for gasses. Free energy change at these concentrations is called G.

8 The Relationship Between Free Energy and the Equilibrium Constant If reactants and products are at equilibrium concentrations: G = 0 If reactants and products are at standard state concentrations (1 M): G G G = RT ln [B] eq [A] eq = RT ln K eq G and K eq convey essentially the same information. For other concentrations: G = G + RT ln [B] [A]

9 The Relationship Between Free Energy and the Equilibrium Constant G = RT ln K eq A B K eq = [B] eq [A] eq If K eq > 1: Reaction favors B over A. G < 0 If K eq < 1: Reaction favors A over B. G > 0 If K eq = 1: A and B have equal free energies. G = 0

10 Reactions with More than One Reactant or Product A reaction with two reactants and two products: aa + bb cc + dd a, b, c and d represent stoichiometry The equilibrium constant: K eq = [C]c eq [D]d eq [A] a eq [B]b eq Products over reactants! (Mrs. Freeman) The reaction quotient: Q = [C]c [D] d [A] a [B] b When concentrations are not necessarily at equilibrium. G = G + RT ln Q

11 Clicker Question #1 For the reaction: A + B C What are the units of the equilibrium constant? 1 M 2 No units 3 M 1

12 From Two Molecules to One A reaction with two reactants and one products: A + B C The equilibrium constant: K eq = [C] eq [A] eq [B] eq Some problems: The equilibrium constant has units of inverse concentration. The numerical value of K eq will depend on the units that we use for concentration What happens when we try to calculate G? G = RT ln K eq Logarithms are functions of pure numbers. How do we take a logarithm of a value with units?

13 From Two Molecules to One A + B C Defining standard states is important! Standard state concentration is usually defined as 1 M For thermodynamic purposes, refine the definition of the equilibrium constant: ( [C]eq /1 M ) K eq = ( [A]eq /1 M )( [B] eq /1 M ) Now, K eq is dimensionless, and we don t have to feel guilty about taking the logarithm. But, value of G depends on choice of standard state concentrations. Why?

14 The Entropy Change for a Bimolecular Reaction A + B C Assume a volume made up of N c little cubes: The number of ways to place a molecule of A and a molecule of B in the volume: Ω A,B = N 2 c The number of ways to place a molecule of C in the volume: Ω C = N c The entropy change S = k ln(ω C ) k ln(ω A,B ) = k ln(n c ) k ln(nc 2 ) ( ) Nc = k ln = k ln(n c ) N 2 c The entropy decreases!

15 The Entropy Change for a Bimolecular Reaction From previous slide: S = k ln(n c ) In this case, the number of little cubes doesn t cancel out, and the size of the cubes matters. A plausible estimate: Make the cubes about the right size to hold one or two small molecules, V c = (1 nm) 3 = 1 nm 3 Specify the reaction volume, V in liters. N c = V (L) ( ) 1 nm 10 3 m nm V (L) L m

16 The Entropy Change for a Bimolecular Reaction From previous slide: S = k ln(n c ) = k ln(v (L) ) For one mole each of A and B converted to C: S = N α k ln(v (L) ) = R ln(v (L) ) For V = 1 L: S = R ln(10 24 ) = 8.31 J/K 53 = 460 J/K

17 Clicker Question #2 If the reaction volume is doubled for the reaction A + B C, How will the entropy change be affected? 1 S will be unaffected. 2 S will become more negative. 3 S will become less negative. S = k ln(n c ) = k ln(v (L) )

18 A Bimolecular Reaction A + B C S = 460 J/K Total free energy change for the reaction: G = H T S Contribution of entropy change to the free energy change at 298 K: T S = 298 K 460 J/K = 140 kj/mol This is a large, unfavorable contribution to G. For the reaction to be favorable, there must be other factors that make the total free energy change negative. What factors might do that?