University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2014

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1 Lecture 11 07/18/14 University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2014 A. he Helmholt Free Energy and Reversible Work he entropy change S provides an absolutely general criterion for assessing the spontaneous direction for physical processes. However the Second Law specifies that S universe 0 so a entropy calculation must include S system and S surr. Suppose a system is isolated meaning it cannot exchange energy or matter with its surroundings. If this system is not at equilibrium it must undergo transport processes which are irreversible process that establish equilibrium. ransport occurs until equilibrium is established and an indication that an isolated system is at equilibrium is that it s entropy is maximum. his isolated system cannot decreases its entropy because to do so would be to violate the Second Law. Suppose a system is closed meaning it cannot matter with its surroundings but it can exchange energy in the forms of heat and work. If the system and the surroundings are not at equilibrium again irreversible transport processes must establish equilibrium. ransport occurs until equilibrium is established and an indication that the entropy of the surroundings and the system is maximum. he universe cannot decrease its entropy because to do so would be to violate the Second Law. Suppose a closed system is not at equilibrium. It exchanges heat and work with the surroundings. Now the surroundings are massive compared to the system and thus any exchange of energy with the system will not perturb the surroundings much from equilibrium. herefore the surrounding exchange heat reversibly or dssurr =. But the system is not at equilibrium and does not exchange heat reversibly. his irrev means that dssys because = irrev. he best we can say is dssys > because qrev > qirrev Once the system reaches equilibrium any slight perturbation from equilibrium results rev in an entropy change dssys =. What is always true is ds. From the First Law dusys = + dw. From the Second Law ds. Combining the two du ds + dw. For a reversible exchange of heat and P-V work du = rev + dwrev = ds PdV Now consider the state function U-S. he differential change in this state function is d ( U S ) = du ds Sd = δ q PdV ds Sd (11.1)

2 If the process is reversible then =ds. If the process is irreversible <ds. herefore is it always safe to say that d ( U S ) Sd PdV (11.2). And at constant V and (i.e. d=dv=0) equatyion 11.2 reduces to d( U S) 0 (11.3) his means that the state function A=U-S decreases via irreversible processes until it reaches a minimum when equilibrium is reached. Note that although A the Helmholt energy has units of Joules A is not a conserved quantity. he Helmholt energy is also called the work function or the reversible work because for a reversible change at constant temperature (d=0) da = Sd + δ wrev = dwrev B. Gibbs Energy: non-pv Work and Applications at Constant P and For calculating the spontaneous direction of processes occurring at constant P and the Gibbs Free energy G is used : G=H-S=U+PV-S Analogous behavior to A: dg = d( U + PV ) d( S) = du + PdV + VdP ds Sd = δ q PdV + PdV + VdP ds Sd (11.4) = δ q + VdP ds Sd If the process is reversible then δ qrev = ds and so dg = ds + VdP ds Sd = Sd + VdP (11.5) At constant P and for reversible changes with only PV work: dg=0 If the change is irreversible δ qirrev < δ qrev = ds and then dg < Sd + VdP (11.6) Similarly at constant P and dg<0. It is therefore certain that for both irreversible and reversible changes at constant P and : dg 0 (11.7) Equation 11.7 is basically the criteria for a system being at equilibrium or undergoing an irreversible change. he Gibbs energy or Gibbs function is also used to quantify reversible work at constant P and. Assume in addition to PV there are other work terms including electrical work (to move charge in a potential gradient) osmotic work (to move mass in a concentration gradient) etc. dg = dh d ( S ) = d ( U + PV ) d ( S ) = δqrev + δwrev + PdV + VdP Sd ds (11.8) = ds + δ wother PdV + PdV + VdP Sd ds = δ w + VdP Sd other Note at constant P and (i.e. dp=d=0) dg = dw other (11.9)

3 G = w other means the Gibbs free energy change G measures the amount of non-p-v work that a physical process can produce if it is conducted reversibly. Because the reversible work is the maximum work that can be produced G measures the upper limit of work that can be obtained from a spontaneous process at constant P and. Example for a electrochemical cell G = welec = ni Ε where w elec is the reversible electrical work n is the number of moles of electrons transported by the electrochemical cell I is Faraday s constant and Ε is the potential difference between the half cell. If no work other than P-V work is performed then dg = VdP Sd At constant (i.e. d=0) dg=vdp. Example: Calculate the free energy change when 1 mole of an ideal gas changes its pressure from 1 atm to 10 atm. P 2 2 dg = VdP dg = G2 G1 = G = VdP 1 = 10atm = 10atm nr F = = = H G P dp nr dp nr ln P = 1atm = 1atm = b gb. / gb gln =. 1mole 8 31J mole K 300K kJ he absolute standard free energy of a gas with a pressure of one atmosphere is designated G 0. 0 F P G = G + nr H G I 0 ln G nr P atm K J = + lnb g 1 Standard Free Energy Changes Definition: he standard molar free energy of formation G 0 f is the free energy change required to form one mole of a pure compound in its standard state from its constituent elements all in their standard states Gf = H f S f Example: Cbsgr g+ O2bgg CO2 bgg 0 0 H = b1moleg H f bco2g = kj S = S CO g S O g S O s gr b g fc 2 h fc 2b gh fc 2b g b gb g = 1mole J / mole K = 2. 86J / K G = H S = kj 298K 2. 86J / K = kJ f b gb g he Gibbs Equations All thermodynamic relationships can be derived from six basic equations h I KJ

4 du = ds PdV H = U + PV A = U S U H (11.10) G = H S CVm = CPm = applicable to closed system reversible PV work only. he Gibbs equations are expressions for dh da and dg that are analogous to du = ds PdV for closed systems undergoing reversible changes and PV work only. hey are: du = ds PdV; dh = ds + VdP (11.11) da = Sd PdV; dg = Sd + VdP It is the basic strategy of thermodynamics to provide any change in any state function U H G A or S resulting from a change in any state variable V or P in terms of a few easily measured properties. As it turns out all state functions S U changes as a result of changes in state variables (e.g. etc. ) can be P expressed in terms of P V and only three properties: H 1 V 1 V C Pm β κ = = = (11.12) V V P he first property C P m is just the heat capacity at constant pressure. he second property β is called the thermal expansivity which measures the degree to which the volume of a material changes with temperature. he third property κ is called the isothermal compressibility and measures the degree to which the volume of a material changes with pressure. he Gibbs equations are combined with the exact differential expressions for du dh da and dg to give the Maxwell relations. he exact differential expressions are U U du = ds PdV = ds + dv S S (11.13) P = S S and similarly using dh da and dg we get the three other Maxwell relations V S P S V = = = (11.14) P S S P.Only the last two relations S P S V = = (11.15) P are of practical use.

5 U Example: Determine the differential in terms of state variables P V and/or H 1 V 1 V and the three properties: C Pm β κ = = = V V P. Solution: U S V du = ds PdV = P S P du = P = P Given We have already shown that V P V P β = 1 = = V V P κ P U P β = P= P κ H Similarly for example = V β + V P

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