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1 Thermodynamics and Chemical Equilibria

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1 Thermodynamics and Chemical Equilibria Classical Thermodynamics Thermodynamics and chemical equilibria • Key goals for today’s lecture: – Define: • Gibbs Free Energy (G) G = H - TS • Thermodyypnamics as a prediction as to the sp ontaneous nature of a • Lecture 2 8/27/09 chemical reaction A + B C • State Functions • Chapter 1 Voet, Voet and Pratt • First and Second Laws • Heat, work, internal energy, enthalpy, entropy, free energy, chemical potential, System, property, state, adiabatic, diathermal, reversible • Equilibrium constants Thermodynamic Definitions: Gibb’s Free Energy (What is Enthalpy and Entropy and their relationship to the First and Second Laws?) • Is a state function (a property of a system that depends only on the current state of the system and not its history) First, Let’s Define a System… • Gibb’s Free Energy is determined at constant T and P: a defined part of the universe a chemical reaction G = H - TS a btibacteria ΔG° = ΔH° -TΔS° a reaction vessel a metabolic pathway •The Gibbs free energy (G) of a system is defined by an enthalpy Surroundings: the rest of the universe term (H) (change of the total energy with the system), and the Open system: allows exchange of energy and matter entropy term (S) (change in the disorder) at temperature (T) Isolated system: no exchange of matter or energy. i.e. A perfect insulated box. First Law of Thermodynamics: Energy is Conserved Reversible and Irreversible Processes • Energy is neither created or destroyed. – In a chemical reaction, all the energy must be accounted for. • Reversible processes Fsurr • Equivalence of work w and energy (heat) q – Proceed infinitesimally out – Work (w) is defined as w = F x D (organized motion) of balance. – Heat (q) is a reflection of random molecular motions (heat) – Requires zero friction, epsilon heat gradients – Are hypothetical only • Heat q: – If q is positive reaction is endothermic system absorbs heat from surroundings –If q is negative exothermic system gives off heat. • Irreversible process – “Real world” process • Work w: Fsys – Have finite changes and P –If w is positive, the system does work ON the surroundings. loses. sys Remember sign conventions ΔE=q-w Piston displacement of thermodynamics from a ΔE=(-q)-(+w) DOES work(+w) on Fsys = Fsurr + dF steam engine surroundings Cylinder “Feels” hot as it looses heat (-q) to surroundings Control Volume: Defines the system boundary 1 ΔU = U -U = q - w Direction of heat flow by definition is most important. final initial Exothermic system releases heat = -q q = heat absorbed by the system from surroundings Endothermic system gains heat = +q If q is positive reaction is endothermic ΔU = a state function dependent on the current properties only. system absorbs heat from surroundings State Function → Any quantity whose value is If q is negative exothermic independent of its history. system gives off heat. ΔUispath independent while q and w are not state functions because they can be converted from one form of energy to the other. (excluding other forms of energy, e.g. electrical, light and nuclear energy, from this discussion.) Enthalpy (H) Enthalpy (H) At constant pressure w = PΔV + w1 When considering only pressure/volume work w1 = work from all means other than pressure-volume work. ΔH = qp -PΔV + PΔV = qp PΔV is also a state function. ΔH = qp when other work is 0 By removing* this type of energy from U, we get enthalpy or ‘to warm in’ qp is heat transferred at constant pressure. *remember the signs and direction In biological systems the differences between H = U + PV ΔUandΔH are negligible (e.g. volume changes) ΔH = ΔU + PΔV = qp -w + PΔV = qp -w1 The change in enthalpy in any hypothetical reaction Second Law and Entropy: pathway can be determined from the enthalpy Entropy is the arrow of time and in any cyclic process the entropy will either increase or remain the same. change in any other reaction pathway between the same products and reactants. G = H - TS This is a calorie (joule) “bean counting” • Entropy – measure the degree of randomness – drives it to the most probable state or maximum disorder • Lord Kelvin stated that it is impossible take heat from a hot Hot Cold reservoir and convert it to work without transferring heat to a cooler reservoir. • Entropy defines directionality of a ∆E Which way does heat travel? This directionality, is not mentioned in First Law 2 Entropy Gas on its own will expand to the available volume. • Two ways of formulating entropy. • Disorder increases – Probable distribution of energies: Number of way of • N identical molecules in a bulb, open the stop cock you arranging N particles in ni groups get 2N equally probable ways that N molecules can – Carnot cycle: Examining the efficiency of a idealized be distributed in the bulb. reversible cycle and realize that even “perfect” process cannot convert 100% heat into work. – Implies qlow> 0 – ie. Efficiency < 1 The process of diffusion of a gas from the left bulb initially W2 = 1 and Entropy (S) S = 0 to Right (N/2) + Left (N/2) at equilibrium gives a : measure the degree of randomness ΔS that is (+) with a constant energy process such S = kblnW that ΔU = 0 while ΔS>0 Each molecule has an inherent amount of energy which drives it to the most probable state or maximum disorder. This means if no energy flows into the bulbs from the kB or Boltzman’s constant equates the arrangement outside expansion will cool the gas! Conservation of Energy probability to calories (joules) per mole. says that the increase in Entropy is the same as the decrease in thermal (kinetic) energy of the molecules!! Entropy is a state function and as such its value depends only on parameters that describe a state of matter. It is difficult or (impossible) to count the number of At constant pressure we have changes in qp (Enthalpy) and arrangements or the most probable state! changes in order - disorder (Entropy) So how do we measure entropy? A spontaneous process gives up energy and becomes more disordered. final dq ΔS ≥ ∫ G = H - TS inital T It takes 80 kcal/mol of heat to change ice at zero °C to water at zero °C Describes the total usable energy of a system, thus a change from one state to another produces: 80,000 = 293 ev’s or entropy units 273 A Reversible process means at equilibrium during the change. This is ΔG = ΔH - TΔS = qp - TΔS impossible but makes the calculations easier but for irreversible process q ΔS ≥ If ΔG is negative, the process is spontaneous T 3 ΔS ΔH Equilibrium Constants and ΔG +-All favorable at all temperature Now if we are at equilibrium or ΔG = 0 and ΔG° is the spontaneous free energy of the reaction in the standard state. Then: --Enthalpy favored. [C]c[D]d Spontaneous at o ΔG = 0 = ΔG + RTln a b temperature below [A] [B] T=ΔH ΔS [C]c[D]d ΔG o = − RTln ++Entropy driven, [A]a [B]b enthalpy opposed. OR Spontaneous at Temperatures above ΔG o = −RTlnK T=ΔH eq ΔS o o o -+Non-spontaneous i.e. ΔG = ∑ΔGf (products) - ∑ΔG f (reactants) STP So what does ΔGo really mean? Standard Temperature and Pressure and at 1M concentration. c d o [C] [D] We calculate ΔG’s under these conditions. ΔG = ΔG + RTln [A]a [B]b aA + bB cC + dD ΔG We can calculate a G for each component __ __ __ __ (1) ΔG = cGc + d Gd - a Ga - bG b o ΔGo (2) Ga - G a = RTln[A] combining (1) and (2) c d Keq (3) o [C] [D] If Keq = 1 then ΔG = 0 ΔG = ΔG + RTln a b [A] [B] o ΔG equates to how far Keq varies from 1!! o o The Variation of Keq with ΔG at 25 C o The van’t Hoff Relationship Keq ΔG 6 - (kJ·mole-1 Keq can vary from 10 to 10 • Methodology of finding ΔH and ΔS from experimental data. 6 or more!!! 106 -34.3 4 10 -22.8 ⎛ ⎞ 102 -11.4 ΔG° ΔH°−TΔS° lnKeq =− =−⎜ ⎟ 101 -5.7 RT ⎝ RT ⎠ o ΔG is a method to calculate two 0 10 0.0 1 ⎛ ΔH°⎞ Van’t Hoff plot reactions whose Keq’s are different: 10-1 5.7 lnKeq = ⎜ ΔS°− ⎟ however the initial products and 10-2 11.4 R ⎝ T ⎠ - ΔHo reactants maybe far from their 10-4 22.8 = Slope eq R equilibrium concentrations 10-6 34.3 ΔSo c d lnK = Intercept R [C] [D] - H o 1 So Must be used Δ ⎛ ⎞ Δ so a b lnK eq = ⎜ ⎟ + [A] [B] R ⎝ T ⎠ R 1 T 4 Standard State for Biochemistry o o Most times ′ ΔG = ΔG or However, species with either H O or H+ requires consideration. 2 ΔG° versus ΔG°' Unit Activity For A + B C + D + nH2O n 25 oC o [C][D][H2O] o [C][D] ΔG = - RTln ΔG′ = - RTlnK′eq = - RTln [A][B] [A][B] pH = 7.0 (not 0, as used in chemistry) This is because water is at unity. Water is 55.5 M [H2O] is taken as 1, however, if water is in the Keq equation then [H2O] = 55.5 and for 1 mol of H2O formed: o o The prime indicates Biochemical standard state ΔG′ = ΔG + nRTln[H2O] o o o -1 ' ΔG′ = ΔG + 9.96 kJ • mol K eq ΔG ' Coupled Reactions As long as the overall pathway is exergonic, it will operate in a forward manner. A + B C + D ΔG1 (1) D + E F +G ΔG2 (2) ΔG ≥ 0 reaction 1 will not occur as written. If 1 Thus, the free energy of ATP hydrolysis, a is sufficiently exergonic so However, if ΔG 2 ΔG1 + ΔG 2 ≤ 0 highly exergonic reaction, is harnessed to Then the combined reactions will be favorable through drive many otherwise endergonic biological the common intermediate D processes to completion!! A + B + E C + F + G ΔG3 ΔG3 = ΔG1 + ΔG 2 ≤ 0 Questions: Units A.
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