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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.

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Presentation on theme: "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."— Presentation transcript:

1 Second Law of Thermodynamics

2 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

3 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

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

5 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) = 197.9 J/Kmol S 0 (O 2 ) = 205.0 J/Kmol S 0 (CO 2 ) = 213.6 J/Kmol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/Kmol

6 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.

7  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.

8  G =  H - T  S

9 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

10 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–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous

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

12 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

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