Second Law of Thermodynamics

Law of Disorder the disorder (or entropy) of a system tends to increase ENTROPY (S) Entropy is a measure of disorder Low entropy (S) = low disorder High entropy (S) = greater disorder Operates at the level of atoms and molecules hot metal block tends to cool gas spreads out as much as possible

Factors affecting Entropy A. Entropy increase as matter moves from a solid to a liquid to a gas Increasing Entropy B. Entropy increases when a substance is divided into parts Increasing Entropy

C. Entropy tends to increase in reactions in which the number of molecules increases Increasing Entropy D. Entropy increase with an increase in temperature

Entropy Changes in the System (∆S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [ +] - bS 0 (B) aS 0 (A) [+ ] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - The standard entropy of reaction (∆ S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = J/Kmol S 0 (O 2 ) = J/Kmol S 0 (CO 2 ) = J/Kmol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = – [ ] = J/Kmol

Entropy Changes in the System (∆S sys ) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, ∆S 0 > 0. If the total number of gas molecules diminishes, ∆ S 0 < 0. If there is no net change in the total number of gas molecules, then ∆ S 0 may be positive or negative BUT ∆ S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down, ∆ S is negative.

 S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: Gibbs Free Energy For a constant-temperature process:  G =  H sys -T  S sys Gibbs free energy (G)  G < 0 The reaction is spontaneous in the forward direction.  G > 0 reaction is spontaneous in the reverse direction. The reaction is non-spontaneous as written. The  G = 0 The reaction is at equilibrium.

 G =  H - T  S

aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - The standard free-energy of reaction (∆ G 0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn Standard free energy of formation (∆ G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f f

2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x– x–237.2 ] – [ 2x124.5 ] = kJ Is the reaction spontaneous at 25 0 C?  G 0 = kJ < 0 spontaneous

Recap: Signs of Thermodynamic Values NegativePositive Enthalpy (ΔH) ExothermicEndothermic Entropy (ΔS)Less disorderMore disorder Gibbs Free Energy (ΔG) SpontaneousNot spontaneous

Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/Kmol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium  G = 0 Q = K 0 =  G 0 + RT lnK  G 0 =  RT lnK