Equilibrium and Energy PDF

 Equilibrium and Energy Equilibrium Energy Defined by equilibrium First law: H constants, Keq Second law: S Define structure of systems G = H – TS Stability when system at Equilibrium when G = 0 equilibrium...

 Equilibrium and Energy Equilibrium Energy Defined by equilibrium First law: H constants, Keq Second law: S Define structure of systems G = H – TS Stability when system at Equilibrium when G = 0 equilibrium Spontaneous change when Instability when system not G < 0 at equilibrium Must be a relationship between G and Keq At equilibrium G° = −RTlnKeq  Points about G° = −RTlnKeq True at equilibrium If not at equilibrium, need G − G° = RTlnKeq G is the free energy for the system; G° free energy for the process when at equilibrium Note: penultimate step of derivation gives (Partial) pressure(s) for gases, equivalent to concentrations of components for other phases. Replacement by concentrations, e.g., [C]c, gives equations for Keq G° for the isomerization of glucose-6-phosphate (G6P) to fructose- 6-phosphate (F6P) is 2.1 kJ mol−1. What concentrations of G6P and F6P will be present at equilibrium at 25 °C starting with 0.1 M G6P? 0.0428 − 0.428x = x 0.0428 = 1.428x x = 0.03M = [F6P] [G6P] = 0.1M − 0.03M = 0.07M  Tutorial: G, Keq problem The concentration of a solution of G3P was initially 0.05 M. Isomerase was added. After the mixture came to equilibrium at 25 °C, the concentration of G3P was 0.002 M. Calculate ΔG° for the reaction. What are the implications of the values of ΔG° and Keq for the reaction as given above? Energy and equilibrium of systems, so far… From the 1st law: enthalpy H = qp= U + PV Concerns energy changes to the system during processes; lowering enthalpy preferred From the 2nd law: entropy S = q Concerns changes to the disorder of the system during processes; increasing entropy preferred Need a thermodynamic concept which captures both of these  Criterion for the direction of spontaneous change For any spontaneous process at constant T and P A spontaneous process implies irreversible change, i.e: 𝑞𝑖𝑟𝑟𝑒𝑣 ∆𝑆 > or qirrev < TS 𝑇 q = H (const. T, P) So H < TS or H − TS < 0 Process will continue as long as H − TS < 0 (Gibbs) Free Energy Require: H − TS < 0 Define G = H − TS J. Willard Gibbs (1839-1903) Process will proceed as long G < 0 G known as the Gibbs free energy G is a thermodynamic state function  G = H − TS If G < 0, preferred direction of process Arrives at equilibrium when G = 0 If the H term dominates, have an enthalpy controlled process (H large and –ve) If the TS terms dominates, have an entropy controlled process  Enthalpy vs. entropy E.g. analysis of Mg2+ using complexation by EDTA EDTA tetra sodium salt Widely used method for analysis of Mg ions in aqueous solution Note: Mg2+ ion highly solvated by water Method used pharmaceutically, in pharmacology, biochemistry, etc. Note, process goes essentially to completion i.e., fully to the right hand side as drawn above  Enthalpy vs. entropy Can determine the relative contributions of enthalpy and entropy to the process. Hence can determine the thermodynamic driving forces for the process and… …infer the molecule-level processes Find that H small and +ve But that S is significant and +ve Release of five H2O molecules and two NaCl, increased disorder (more components generated) Overall G –ve, process spontaneous left to right as drawn  Standard states of free energies G°: value of G for a pure substance under 1 atmosphere of pressure (at some specified temperature) G°: change in free energy when 1 mole of products in their standard states are converted to 1 mole of products in their standard states  Using ΔG values, example (ATP, ADP and AMP: key molecules in biochemical energy transfer) ? (−30.5) − (−31.1) = 0.6 kJ mol−1  Tutorial: G = H − TS 1. ATP → ADP + Pi G° = −30.5 kJ mol−1 (from thermal analysis) H° = −20.1 kJ mol−1 Calculate S° = (all at 37 °C) Comment on the significance of the values of G, H and S 2. H° and S° for the unfolding of a protein are 250.8 kJ mol−1 and 752 J K−1 mol −1 respectively. Above what temperature (in °C) will unfolding of the protein be spontaneous? Comment on the values.