Physical Chemistry Lecture 18 Maxwell Relations Spontaneity under arbitrary conditions Constant U and V (∆S )U ,V ≥ 0 Constant T and V (∆A)T ,V ≤ 0 Constant T and P (∆G )T , P ≤ 0 Must calculate changes to make predictions Free-energy changes Helmholtz free energy, A dA = − SdT − PdV Gibbs free energy, G dG = − SdT + VdP Must know how entropy depends on temperature to evaluate free-energy ∂S ∂S change dS = dT + dV dS ∂T V ∂S = dT ∂T P ∂V T ∂S + dP ∂P T Evaluation by integration Evaluate changes of a function such as A or G in a process requires evaluating an integral of a derivative T2 ∆A = − ∫ S (T )dT − T1 T2 ∆G = − ∫ S (T )dT T1 V2 ∫ P(V )dV V1 + P2 ∫ V ( P)dP P1 Maxwell relation from Helmholtz energy dA = − SdT − PdV By definition Equality of crossed second partial derivatives of the state function, A ∂ ∂T ∂ ∂T ∂A ∂V T V (− P ) V ∂S ∂V T ∂ ∂A = ∂V ∂T V T ∂ (− S ) = T ∂V ∂P = ∂T V Maxwell relation from Gibbs energy dG = − SdT + VdP By definition Equality of crossed second partial derivatives of the state function, G ∂ ∂T ∂ ∂T ∂G ∂P T P (V ) P ∂S ∂P T ∂ ∂G = ∂P ∂T P T ∂ = (− S ) ∂P T ∂V = − ∂T P Evaluating entropy at a temperature By definition ∂S S (T , P) = S (Tref , P) + ∫ dT ∂T P Tref T T = S (Tref , P) + ∫ Tref C P (T ) dT T Determined by the way the heat capacity changes Must specify entropy at the reference temperature (Third Law) Evaluation by numerical integration ∆Si C P (Ti +1 ) C P (Ti ) + Ti +1 Ti (Ti +1 − Ti ) = 2 Summation gives change in entropy with temperature Can also use trapezoid to approximate area Heat Capacity of Platinum 0.1 0.09 0.08 0.07 CP (joule gm -1 K-1) Treat integral as a sum over small intervals Area of rectangle 0.06 0.05 0.04 0.03 0.02 0.01 0 0 10 20 30 40 50 T (K) 60 70 80 90 100 Evaluating entropy at a pressure By definition of the change in a state P function ∂S S (T , P) = S (T , Pref ) + ∫ ∂P Pref dP T ∂V ∫P ∂T P dT ref P = S (T , Pref ) − Evaluation by change of volume ∂S S (T , V ) = S (T , Vref ) + ∫ dV ∂V T Vref V ∂P = S (T , Vref ) + ∫ dT ∂T V Vref V Standard molar entropies at 298.15 K Evaluate entropies by numerical integration of heat capacities Solid Ag Sθ (J/K) 42.55 Liquid Br2 Sθ (J/K) Gas Sθ (J/K) 152.2 H2 130.7 AgCl 96.2 H2O 69.9 N2 191.6 Graphite 5.74 Hg 76.0 O2 205.1 Diamond 2.377 CH3OH 126.8 CO2 213.7 I2 116.1 C2H5OH 160.7 Ar 154.9 173. Xe 169.7 Hg2Cl2 196. C6H6 Internal energy derivatives For a reversible process dU = TdS − PdV Replace entropy differential by definition dU ∂S = T dT ∂T V ∂S + dV − PdV ∂V T ∂S = T dT ∂T V ∂S + T − P dV ∂V T ∂S = T dT ∂T V ∂P + T − P dV ∂T V The last substitution uses a Maxwell relation Indentifying partial derivatives Compare equation to standard form ∂U ∂T V ∂S = T ∂T V ∂U ∂V T ⇒ ∂P = T ∂T V ∂S ∂T V = CV T − P Gives evaluable forms for derivatives Other forms of work Stretching an elastic network dwelastic = fdl Moving charge against an electric potential dwelecrtic = VdQ Magnetizing a material dwmagnetic = BdM Creating a surface dwsurface = γdA Summary Free-energy changes specify spontaneity (dA)T,V ≤ 0 (dG)T,P ≤ 0 Must calculate the changes, ∆A and/or ∆G, to determine spontaneity of a process Maxwell relations provide connection between derivatives to allow proper integration