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1 Physical Chemistry I (Basic Chemical Thermodynamics) [I] [II] [III] [IV] Introduction Energy(The First Law of Thermodynamics) Work Heat Capacity C p and C v Adiabatic Change Exact(=Perfect) Differential Heat of Chemical Reaction Entropy (The Second Law of Thermodynamics) Spontaneous Change Reversible Change and Irreversible Change Entropy 1) Randomness 2) Exact(=Perfect) Differential ds Calculation of Entropy Change 1) Adsorbed Heat by Reversible Change 2) Entropy Change accompany with Volume Change 3) S= f (T,P) Entropy of Mixing ΔS mix The Third Law Free Energy and Chemical Potential Helmholtz Free Energy and Gibbs Free Energy Exact(=Perfect) Differential da and dg Gibbs-Helmholtz Equation Properties of Mixtures; Partial Molar Quantities Partial Molar Free Energy; Chemical Potential a) Chemical Equilibrium b) Phase Equilibrium Thermodynamic Formulas Fundamental Relationship

2 A) 1 1mol A (A) B) 10 B (B) (A) (B) Isothermal compression, (B) (A) Isothermal expansion 0 (A) (B) (B) (A) 0

3 4 Heat of Solution ( ) 1 C 2 H 5 OH(l) +9H 2 O C 2 H 5 OH 9H 2 O 1mol 9mol 10%mol ΔH 1 = 7.021 kj/mol 2 C 2 H 5 OH(l) +4H 2 O C 2 H 5 OH 4H 2 O 1mol 9mol 20%mol ΔH 2 = 3.757 kj/mol ΔH 3) C 2 H 5 OH(4H 2 O) +5H 2 O C 2 H 5 OH 9H 2 O 20%mol 10%mol ΔH = ΔH 1 ΔH 2 = ( 7.021) ( 3.757) = 3.264 kj/mol 10%mol 20%mol 20%mol 10%mol EtOH ( ΔH) n w /n a = n H2O /n alc=etoh ΔH ΔH= ΔH (Integral Heat of Solution) (Diffrential Heat of Solution) H 2 O n w = n H2O >>0, EtOH 1mol EtOH

4 5218 J (A) 0 1atm, 22.4l (B) 0 10atm, 2.24l (A) (B) (A) 10atm 1atm W W =10 (22.4 2.24)=201.6(l atm) =201.6 1.013 10 2 = 20.422 (kj) =1 (2.24 22.4)= 20.16(l atm) =20.16 1.013 10 2 = 2.0422 (kj) 20.422 (kj) { 2.0422 (kj)}= 18.37 (kj) 334.72J(=80Cal 4.184) 1g A B (A) 10atm 1atm 18.37 (kj) 334.72J 54.9g AB (A)

5 Δs 10atm (A) (B) entropy Ideal gas R=0.08206(dm 3 atm K -1 mol -1 ) = 8.3145(J K -1 mol -1 ) A 0 1atm, 22.4 dm 3 (B) 0 10atm, 2.24 dm 3 ( ) Δs = nc V In( T 2 T 1 ) + nrin( V 2 V 1 ) (III-32) 1-a: Δs gas = nrin( V 2 V 1 )=1 8.3145 2.303log(2.24/22.4) =19.15 (-1)=-19.15(J K -1 ) (A) (B) entropy V B 1-b: W = PdV = nrt ( 1 V )dv = nrtin V B V A = -1 8.3145 2.303log(2.24/22.4) = 5218 ( 1)= +5218(kJ) ( ) V B V A V A

6 1-c: V B W = PdV = nrt ( 1 V )dv = nrtin V B V A V B V A V A U 1 =Q 1 + W 1 U 2 =Q 2 + W 2 ΔU=ΔQ ΔW ΔU =0 Q= W= PdV Δs gas = Q nrtin T = T V B V A = nrin V B V A = - 19.15(J K -1 ) Δs gas =( W)/T= ( 5238)/(273.2)= 19.15(J K -1 ) () 204.3 10 2 J 10atm W= PdV= 10 (2.24 22.4)=201.6 1.0132 10 2 =+204.3 10 2 (J) =+20.43(kJ) ΔU=0 20430J Δs ice =Δs surr =( W)/T= (+20430)/(273.2)= 74.8(J K -1 ) entropy 20430J 0 20430J Δ s total = Δs gas + Δs surr = 19.15+74.8=+55.6(J K -1 )

7 U = Q+W du = TdS-PdV H = U+PV dh = TdS+VdP A = U-TS da = -SdT-PdV (IV-8) G = H-TS dg = -SdT + VdP (IV-12) Euler s relationship T V T P S V S P S S T T = P S = V S = P T P V = V T V P Maxwell s relationship (IV-16) (IV-17) (III-30) (III-39) Maxwell equation du = TdS-PdV U V U V T T = T S V = T P T V V P = T P T P V P (III-29) H P H P T T dh = TdS+VdP = T S P T = T V T + V = T V T P P + V + V (III-30) da = -SdT-PdV dg = -SdT + VdP (IV-8) (IV-12) A V A T G P G T T V T P = P = S = V = S da = -SdT-PdV (IV-8) dg = -SdT + VdP (IV-12) (IV-18) (IV-19) (IV-20) (IV-21)

8 U S U V H S H T V S T S = T = P = T = V du = TdS-PdV dh = TdS+VdP 1) x =f (y, z) 2) 3) dx = x y 1 y x x y z z Z = x y y z dy + x z x z z x y y = 1 dz (S-1) (II-32) (S-2) 4) dx = Kdy + Ldz K z y = L y z Eulers reciprocity relation (II-24) U T H T 1 V V P V T C V C P P 1 V V P α T β α : thermal expansion coefficient β : compressibility (II-14) (II-15) (S-3) (S-4) Gibbs-Helmholtz Equation ΔA =ΔU TΔS ΔG =ΔH TΔS P ΔG/T T ( ) ΔA =ΔU + T ΔA T ( ) ΔG =ΔH + T ΔG T (IV-24) (IV-25)

9 ΔG T T ( ) ΔG = 1 T ΔG + 1 T ( ΔG) 2 T T = T ΔG T 2 = ΔH T 2 Properties of Mixtures; Partial Molar Quantities X (V, U, H, S, A, G) X = f (T,P, n 1, n 2, - - - - - - - n i - - - -) (IV-26) dx = X dt + X dp + X T P, n 1, n 2, P dn T, n 1, n 2, n 1 +... + X 1 n i T, P, n 2, T, P, n j, dn i +... (IV-27) where n j ; n i n X X n i (IV-28) i T, P, n j T, P dx = X 1 dn 1 + X 2 dn 2 +... +X i dn i +... (IV-29) X = X 1 n 1 + X 2 n 2 +... +X i n i +... (IV-30) X = V; V i = V (IV-31) n i T, P, n j X = S; S i = S n i T, P, n j (IV-32) X = H; H i = H n i T, P, n j (IV-33) X = G; G i = G n i T, P, n j (IV-34) Partial Molar Enthalpy Change

10 ( ) ΔH i = ΔH n i T, P, n j Partial Molar Free Energy ; Chemical Potential G i = G μ n i i dg = -SdT+VdP+μ 1 dn 1 + μ 2 dn 2 + + μ i dn i (IV-35) (IV-36) G = μ 1 n 1 + μ 2 n 2 + + μ i n i (IV-37) a) (Chemical Equilibrium) ( μ 1,n i ) react ( ) = μ 1,n i aa + bb+.. react Ideal Gas Mixture productt ll + mm+.. Product eq. (III-42) ΔS MIX = R n i InX i 2 (n 1, n 2 ) S 0 1,S 0 2 ; ΔS MIX = ( n 1 S 1 + n 2 S 2 ) n 1 S 0 0 0 0 ( 1 + n 2 S 2 )= n 1 ( S 1 S 1 )+ n 2 ( S 2 S 2 ) ( ) S 1 S 0 1 =ΔS 1 = ΔS MIX n 1 ( ) S 2 S 0 2 =ΔS 2 = ΔS MIX n 2 T, P, n 2 T, P, n 1 ΔH MIX = 0 ΔG MIX =ΔH MIX TΔS MIX (IV-38) (IV-39) ΔG MIX = - TΔS MIX ΔG MIX = TΔS MIX = T( R n i InX i )= RT n i InX i (IV-40)

11 ΔG 1 = G 1 G 0 1 = G MIX n 1 ΔG 2 = G 2 G 0 2 = G MIX n 2 = RTInX 1 T, P, n 2 = RTInX 2 T, P, n 1 (IV-41) (IV-41) μ 1 = μ 1 0 + RTInX 1 μ 2 = μ 2 0 + RTInX 2 (IV-42) Dolton X 1 = P 1 P X 2 = P 2 P P 1, P 2 P = P 1 = 1atm 1atm 0 μ = μ μ 0 : μ i = μ i 0 + RTInX i μ i = μ i 0 + RTInP i

12 Clapeyron-Clausius equation / G 1 = G 2 dg 1 = dg 2 dg = VdP SdT dg 1 = V 1 dp S 1 dt dg 2 = V 2 dp S 2 dt V 1 dp S 1 dt = = V 2 dp S 2 dt T (K), H b dp dt = S 2 S 1 V 2 V 1 S 2 S 1 =ΔS = H b T dp dt = H b ( ) TV 2 V 1 Clapeyron equation V 2 >>V 1 V 2 V 1 V 2 1700 dp dt = H b TV 2 = PH b RT 2 dp P 1 dt = H b RT 2 dinp dt = H b RT 2 Clapeyron-Clausius equation H b H b (A) InP = H b RT + const. (P 1, T 1 ) (P 2, T 2 ) In P 2 = H b 1 1 P 1 R T 2 T 1 (B) (C)

13 (Phase Rule) C α, β, γ, P (C - 1) P (C - 1)P μ α 1 = μ β p 1 == μ 1 μ α 2 = μ β p 2 == μ 2 C C μ α c = μ β p c == μ c P : C : P μ A = μ 0 A +RTInC A (P - 1) C F = 2+ (C 1)P (P 1)C = 2 + C P 1) C = 1, 2 P = 2 F = 2 + C P F = 1

14 2) C = 2, P = 2 F = 2 + C P F = 2 3) α- β C = 1, 2 P = 2 F = 2 + C P F = 1 96 4) 3 C = 1, P = 3 F = 2 + C P F = 0 0.0075 (273.16K) 0.006atm 71 c t b x t : b t : c t : t : x t : b t : c t : t : x

15 U(S,V) c U = U(S,V,n 1,n 2,,n c ) U(S,V,n i ) n i i U du = U ds + U dv + S V, n i V S, n i = TdS PdV + μ i dn i c i =1 c i =1 U dn n i = TdS PdV + i S, V, n j i c i =1 U dn n i i S, V, n j i μ i chemical potential μ i = U n i S, V, n j i dn i 2 IV.1

16 2 IV.2 IV.3 IV.3 1 3 2 2 2 T(X g ) (X l ) *X g X 2 X 2 X l X 2 T 1 T 2 2 T T 2 X 2 2 2

17 IV.4 (critical solution temperature) (a)(b) (c) (a) (b) (c) 2 2 1 2 2 2 2 IV.5 [G/(n 1 +n 2 )] 2 2 2 (X B, X B ) 2 2 2 2 2 2 1 IV.4(a) 2 y, z y, z (tie line)2 2 2 IV.1 2 2 1 2

18 IV.6 IV.7 b 50 (E, eutectic point) 2 3 1 (e) 2 (solid solution) Ni-Cu 2 IV.8 IV.3 IV.6 IV.8

19 (S 1,S 2 ) 2 (S A,S B ) IV.9 2 IV.6 2 IV.10 1 2 (congruent melting compound, C ) 2 (congruent melting point) 2 3 ( r = 1)

20 2 / ( ) 2 X 2 n g, n l P 2 (n g + n l )X 2 = n g X g 2 + n l l X 2 X 2 X 2 g X 2 l X 2 = n l n g Gas P Liquid g l 0 X 2 X 2 X 2 2

21 entropy eq.(iii-37) T2 C T2 Δ = P S dt = C Pd ln T T1 T T1 20 80g100 60g entropy t 80 (t 20) = 60 (100 t) t = 54.3 = 327.5 K =Cal/g K=4.184J/g K 1mol C p =4.184 18=75.3J K 1 mol 1 80g 327.5K entropy T2 80 327. 5 80 327. 5 ΔS1 = nc P ln = 75. 3 ln = 75. 3 2. 303log T1 18 293. 2 18 293. 2 J [J] = 770. 74 0. 04805 = + 37. 03 = [K][mol] [K] 60g 100 =373.2K 327.5K entropy T2 60 327. 5 60 327. 5 ΔS 2 = nc P ln = 75. 3 ln = 75. 3 2. 303log T1 18 373. 2 18 373. 2 = 32. 8 entropy Δ 2 S over all = ΔS 1 + ΔS = 37.03 32.8 = 4.23 [J/K] > 0 ΔS > 0

22 [IV] Free Energy and Chemical Potential Helmholtz Free Energy and Gibbs Free Energy [Energy or (Enthalpy) and Entropy] U or H ( ), S ( ) A U TS Helmholtz Free energy at constant volume G H TS Gibbs Free energy at constant pressure (IV-1) (IV-2) A or G: ΔA=ΔU TΔS (IV-3) ΔG=ΔH TΔS (IV-4) H (ΔH<0) G (ΔG<0) S (ΔS>0) G G ΔG = 0 Available capacity U A H, S G

23 ΔA = A 2 A 1 ΔG = G 2 G 1 (ΔA) T,V = 0 (ΔG) T,P = 0 (Equilibrium) ΔU = Q + W = TΔS+ W ΔH = ΔU + PΔV = TΔS+ W + PΔV ΔS (IV-3) ΔA=ΔU TΔS ΔA=ΔU TΔS = TΔS+ W TΔS = W (IV-4) : ΔG = ΔH TΔS ΔG = ΔH TΔS = TΔS+ W + PΔV TΔS = W + PΔV (IV-5) (IV-6) (IV-5) A ; Work function (Energy Dimension ) (IV-6) G ; Thermodynamic function ( G ) (Free energy) G 1 = G 2 (ΔG) T,P = 0 ΔG = ΔH TΔS ΔH > 0 = TΔS ΔH ΔG<0 eq. (IV-6) ΔG = W + PΔV W : PΔV : = ΔG<0, ΔG = 0 ΔG = ΔH TΔS

24 = Cell emf (electromotive force) Cell Chemical Reaction Voltaic cell (reversible cell ) Zn electrode 1 Cu electrode dil. H 2 SO 4 Daniell Cell = Reversible cell Zn ZnSO 4 aq. CuSO 4 aq. Cu Cu Zn, emf = 1.1V Zn + Cu 2+ Zn 2+ + Cu 1.1V Zn + Cu 2+ Zn 2+ + Cu 1.1V Cu 2+ Zn + Cu 2+ Zn 2+ + Cu Zn Zn 1.1V 1.1V emf W= ΔG = nfε

25 F = N A e = 6.022 10 23 (mol 1 ) 1.602 10 19(C) =9.648 10 4 (C mol 1 ) 9.65 10 4 Where: F: Faraday constant N A : Avogadro constant e: elementary charge () electric element 1[VC] =1[J] ΔG = nfε (IV-7) = 2 9.648 10 4 (mol 1 )(C) 1.1(V)= 21.23 10 4 = 213.4 (kj/mol) 1.1V emf Daniell Cell 213.4(kJ/mol) Zn + Cu 2+ 213.4(kJ/mol) Zn 2+ + Cu 213.4(kJ/mol) Daniell Cell 1M CuSO 4 Zn ΔH = 218(kJ/mol) Zn + Cu 2+ Zn 2+ + Cu ΔG = 213.4(kJ/mol) at 25 ΔG = ΔH TΔS cell ΔS cell = (ΔH ΔG )/T=[ 218 ( 213.4)]/298.2 = 0.0191(kJ/K)= 19.1 (J/K mol) 218kJ ΔH = 218 kj mol ΔS sur =ΔH/T= 218000/298.2=731.1(J/K mol) ΔS total = ΔS cell +ΔS sur = 19.1 + 731.1=712(J/K mol) > 0

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