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Second Law of Thermodynamics

Second Law of Thermodynamics. 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. Law of Disorder the disorder (or entropy) of a system tends to increase.

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Second Law of Thermodynamics

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  1. Second Law of Thermodynamics

  2. 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 Law of Disorder the disorder (or entropy) of a system tends to increase • 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. The standard entropy of reaction(∆ S0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aS0(A) bS0(B) [ + cS0(C) dS0(D) [ = - ] + ] aA + bBcC+ dD - S mS0(reactants) S nS0(products) = DS0 DS0 DS0 DS0 rxn rxn rxn rxn What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol Entropy Changes in the System (∆Ssys) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol

  6. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Entropy Changes in the System (∆Ssys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, ∆S0> 0. • If the total number of gas molecules diminishes, ∆ S0< 0. • If there is no net change in the total number of gas molecules, then ∆ S0may be positive or negative BUT ∆ S0will be a small number. The total number of gas molecules goes down, ∆ Sis negative.

  7. Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 reaction is spontaneous in the reverse direction.The reaction is non-spontaneous as written. The DG = 0 The reaction is at equilibrium.

  8. DG = DH - TDS

  9. The standard free-energy of reaction (∆ G0) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bBcC+ dD - [ + ] [ + ] = - mDG0 (reactants) S S = f Standard free energy of formation (∆ G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DG0 DG0 rxn rxn f DG0of any element in its stable form is zero. f nDG0 (products) dDG0 (D) cDG0 (C) aDG0 (A) bDG0 (B) f f f f f f

  10. - mDG0 (reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ = 12DG0 (CO2) 2DG0 (C6H6) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? Is the reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 spontaneous

  11. Recap: Signs of Thermodynamic Values

  12. Gibbs Free Energy and Chemical Equilibrium DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium Q = K DG = 0 0 = DG0 + RT lnK DG0 = -RT lnK

  13. DG0 = -RT lnK

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