110c-lecture8 - Entropy Changes of Chemical Reactions Δ r...

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Unformatted text preview: Entropy Changes of Chemical Reactions ( Δ r S) a A + b B → y Y + z Z Δ r Sº = {y*Sº(Y) + z*Sº(Z)} – {a*Sº(A) + b*Sº(B)} Sº given in Table 21.2 or CRC Handbook H 2 (g) + ½ O 2 (g) → H 2 O(l) Δ r Sº = 1*Sº(H 2 O) – 1*Sº(H 2 ) – 1/2*Sº(O 2 ) = 1*(70 J K-1 mol-1 ) – (1)(130.7 J K-1 mol-1 ) – ½*(205.2 J K-1 mol-1 ) = -163.3 J K-1 mol-1 Negative Δ r Sº indicates the products are more ordered than reactants. (i.e. 2 gas phase reactants form only 1 condensed phase species) S(gas) > S(liquid) > S(solid) Entropy Trends: 1. S gas > S liq > S sol 2. S increases with mass (S He < S Ne < S Ar < S Kr ) 3. # of bonds (S(C 2 H 2 ) < S(C 2 H 4 ) < S(C 2 H 6 )) Helmholtz and Gibbs Free Energies (pp 881-893) Chapter 22: Problems 1, 8, 11, 19, 23, 24, 41, 54 Free energy functions (A and G) indicate direction of spontaneous process at T=const. “ Free energy ” means energy available to do maximal work. 1 st Law: Δ U < 0 or Δ H < 0 for mechanical system (ball rolls down hill). 2 nd Law: Δ S > 0 for spontaneous process of isolated system (mixing two gases). Helmholtz Free Energy (A): A ≡ U – TS or Δ A = Δ U – T Δ S Gibbs Free Energy (G): G ≡ H – TS or Δ G = Δ H – T Δ S Spontaneous Chemical Processes at const=T and P: Ba(OH) 2 (s) + 2 NH 4 NO 3 (s) → Ba(NO 3 ) 2 (s) + 2 H 2 O(l) + 2 NH 3 (aq) Δ r H > 0 (unfavorable) and Δ r S > 0 (favorable) which one wins tug-o-war? Δ r G = Δ r H -T Δ r S H 2 O(s) → H 2 O(l) at T=298 K and P=1 atm Δ fus H = +6.01 kJ/mol (unfavorable) and Δ fus S = +22 J/K (favorable) Δ fus G = Δ fus H – T Δ fus S = +6010 J/mol – (298 K)(22 J/K) = -546 J/mol Δ G < 0 ⇒ entropically driven Helmholtz Free Energy (A) Define a new state function (A): A ≡ U – TS Δ A decides interplay of Δ U or Δ S for predicting spontaneous process ( Δ A < 0). A is maximum work done by or to the system (A = (q rev + w rev ) – TS) dA = ( δ q rev + δ w rev ) – T δ q rev /T = δ w rev At const T and V: TdS- dU = SdT TdS- dU = ) TS- U ( d = dA TS U A − − = 1 st Law: dU = dq V 2 nd Law: dS ≥ δ q/T TdS ≥ δ q = dU 0 ≥ dU –TdS = dA ⇒ Δ A < 0 for spontaneous process at const T &V • For a spontaneous process, the Helmholtz Free Energy decreases ( dA < 0). • For a reversible process at equilibrium, dA= 0. • Helmholtz Free Energy (A) is a minimum at equilibrium at const T and V . For a finite and macroscopic change in a system at const T and V , ( ) ( ) S T U S T T S U TS U TS U A Δ − Δ = Δ − Δ − Δ = Δ − Δ = − Δ = Δ Helmholtz Free Energy Change ( Δ A) for Mixing Two Gases Volume and Temperature remain constant ( = Δ U and Δ T = 0). ( ) S T S T T S U TS U A Δ − = Δ − Δ − Δ = − Δ = Δ ....
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110c-lecture8 - Entropy Changes of Chemical Reactions Δ r...

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