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Thermodynamics

Thermodynamics. Three laws of thermodynamics Spontaneous processes Entropy Second and third law of thermodynamics Gibbs free energy Free energy of Chemical equilibrium. I. Three laws of thermodynamics. First law: conservation of energy Energy cannot be created or destroyed

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Thermodynamics

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  1. Thermodynamics • Three laws of thermodynamics • Spontaneous processes • Entropy • Second and third law of thermodynamics • Gibbs free energy • Free energy of Chemical equilibrium

  2. I. Three laws of thermodynamics • First law: conservation of energy • Energy cannot be created or destroyed • Energy can be transferred from one from to another DE = q + w • Second law • Deal with the spontaneity of chemical/physical processes • Third law • Determination of absolute value of entropy.

  3. II. Spontaneous chemical/physical processes • Spontaneous process: • process occurs by itself without outside assistance Ex. Which processes are spontaneous? A waterfall runs downhill spontaneous Iron exposed to oxygen and water forms rust spontaneous spontaneous A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes above 0 0C nonspontaneous At 1 atm, ice melts above 0 0C spontaneous Heat flows from a colder object to a hotter object nonspontaneous

  4. Example: A gas expands in an evacuated bulb spontaneous nonspontaneous

  5. What causes spontaneous processes? CH4(g) + 2O2(g) CO2(g) + 2H2O (l)DH0 = -890.4 kJ H+(aq) + OH-(aq) H2O (l)DH0 = -56.2 kJ H2O (s) H2O (l)DH0 = 6.01 kJ H2O NH4NO3(s) NH4+(aq) + NO3-(aq)DH0 = 25 kJ Does a decrease in enthalpy mean a spontaneous reaction? Spontaneous reactions • In general, decrease in enthalpy favors spontaneity of a reaction. • However, both enthalpy and entropy will affect the spontaneity of a chemical process.

  6. S order disorder S III. Entropy (S) • A measure of the randomness or disorder of a system. • A state function: properties that are determined by the state of the system, regardless of how that condition was achieved Ex. E, H, P, V, T, S DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0

  7. Distribution Microstates I. II. DS = k ln III. Wf Wi III. Entropy Ex. Number of ways to distribute 4 balls in two conpartment W = 1 • Related to the number of microstates. W = number of microstates S = k ln W W = 4 DS = Sf - Si W = 6 Wf> Wi thenDS > 0 Wf< Wi thenDS < 0

  8. Processes that lead to an increase in entropy (DS > 0)

  9. Entropy change for various processes • Temperature change As temperature goes up, entropy increases • Mixing of liquids/gases, dissolution Randomness increases, entropy increases • Phases changes Sgas > Sliquid > Ssolid • Chemical reactions If no. of moles of gases increases, then S increases.

  10. How does the entropy change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 600C to 800C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0)

  11. Standard entropy (So) • Absolute entropy of a substance at 25oC and 1 atm. • So is a positive quantity • Down the group (at the same physical state), So increases He < Ne < Ar < Kr HF < HCl < HBr < HI • Larger molecules, higher entropy CO2 > CO C2H6 > CH4

  12. IV. Second Law of Thermodynamics • Second Law of Thermodynamics • The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0

  13. The standard entropy of reaction (DS0rxn) is the entropy change for a reaction carried out at 1 atm and 250C. aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD DS0 DS0rxn = S n So(products) - S m So(reactants) rxn A. Entropy Changes in the System (DSsys) S : summation n : stoichiometric coefficient of product m : stoichiometric coefficient of reactant

  14. DS0 DS0 DS0 DS0 Your turn! What is the DSo for the following reaction at 250C? 1/2 N2(g) + 3/2 H2(g) NH3 (g) rxn rxn rxn rxn So(N2) = 191.5 J/K•mol So(H2) = 130.7 J/K•mol So(NH3) = 192.3 J/K•mol = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol = S0(NH3) – [1/2 x S0(N2) + 3/2 x S0 (H2)] = 192.3 – [1/2 x 191.5 + 3/2 x 130.7] = -99.5 J/K•mol = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] What is the DSo for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) Ex. So(CO) = 197.9 J/K•mol So(CO2) = 213.6 J/K•mol So(O2) = 205.0 J/K•mol

  15. (1) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) (2) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) (3) N2 (g) at 2 atm, 25oC N2 (g) at 2 atm, 0oC (4) N2 (g) at 2 atm, 25oC N2 (g) at 1 atm, 25oC (5) 2H2 (g) + O2 (g) 2 H2O (g) A. Entropy Changes in the System (DSsys) • When gases are involved in the reaction - If a reaction produces more gas than it uses, DSo > 0. - If a reaction produces less gas than it uses, DSo < 0. - If there is no change in the total number of gas, DSo will be small. Ex. Predict the sign of DS for the following reactions. DS ~ 0 DS < 0 DS < 0 DS > 0 DS < 0

  16. -DHsys Dssurr= T B. Entropy Changes in the Surroundings (DSsurr) Endothermic Process DSsurr < 0 Exothermic Process DSsurr > 0

  17. C. Third law of thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature (0 K). S = k ln W W = 1 S = 0

  18. DG = DHsys -TDSsys V. Gibbs free energy DSuniv = DSsys + DSsurr > 0 Spontaneous process: DSuniv = DSsys + DSsurr = 0 Equilibrium process: For a constant-temperature process: DH: change in enthalpy Gibbs free energy (G) DS: change in entropy DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

  19. Ex. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) mDHo (reactants) - S S = f DS0 DH0 DHo rxn rxn rxn [12x(–393.5) + 6x(–285.8) ] – [2 x 49.04 ] = -6534.9 kJ = nDHo (products) = [12 x S0(CO2)+ 6 x S0(H2O)] – [2 S0(C6H6) + 15 x S0(O2)] = -438.0 J/K f What is the standard free-energy change for the following reaction at 25 0C? DGorxn = DHorxn -TDSorxn DSorxn = S n So(products) - S m So(reactants) DGorxn = -6534.9 kJ - 298.15 K x(- 438.0 J/K) x (1 kJ/1000J) (spontaneous process) = -6404.3

  20. aA + bB cC + dD - [ + ] [ + ] = mDG0 (reactants) - S S = f • Standard free energy of formation (DG0f) • The free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. • DG0f of any element in its stable form is zero. DG0 DG0 rxn rxn f nDG0 (products) aDG0 (A) dDG0 (D) bDG0 (B) cDG0 (C) f f f f f A. Standard free-energy of reaction (DG0rxn) • DG0rxn: The free-energy change for a reaction when it occurs under standard-state conditions.

  21. Ex. 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) ] – [ 2 x 124.5 ] = -6405 kJ = 12DG 0 (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? DG 0f (CO2)= -394.4 kJ/mol DG 0f (H2O)= -237.2 kJ/mol DG 0f (C6H6)= 124.5 kJ/mol Is the reaction spontaneous at 25 0C? spontaneous DG0rxn = -6405 kJ < 0

  22. B. Temperature and direction of spontaneous change Ex. At 25oC, H for a reaction is positive and S is negative. At what temperature(s) will the reaction be spontaneous? (a) at all temperatures (b) higher temperatures (c) lower temperatures (d) the reaction cannot be spontaneous

  23. Ex. CH3OH (l) CH3OH (g) Estimate the normal boiling point of methanol. CH3OH (l) : DH0f = -238.6 kJ/mol, DS0 =127 J/K.mol CH3OH (g) : DH0f = -201.2 kJ/mol, DS0 =238 J/K.mol At boiling temperature(Tb), gas and liquid are at equilibrium DG0rxn=0 DH0rxn-TbDS0rxn=0 DHorxn= DHof (CH3OH,g)-DHof (CH3OH,l)= DHovap =(-201.2)-(-238.6) = 37.4 kJ/mol DSorxn= DSovap = 238-127 = 111 J/K.mol Tb= DH0vap/DS0vap= 37.4 x103/111 = 337 K = 64oC Cp to expt 65oC

  24. CaCO3(s) CaO (s) + CO2(g) Temperature and Spontaneity of Chemical Reactions From thermodynamic table DH0 = 177.8 kJ DS0 = 160.5 J/K DG0 = DH0 – TDS0 At 25 0C, DG0 = 130.0 kJ 835 0C DG0 = 0 at

  25. H2O (l) H2O (g) 40.79 kJ = DS = 373 K DH T C. Gibbs Free Energy and Phase Transitions At equilibrium DG0 = 0 = DH0 – TDSo = 109 J/K

  26. Ex. Estimate DGo for the following reaction at 298K and 500 K. N2O4(g) 2 NO2(g) N2O4(g) : DH0f = 9.66 kJ/mol, DS0 =304.3 J/K.mol NO2(g) : DH0f = 33.85 kJ/mol, DS0 =240.5 J/K.mol DG0rxn= DH0rxn-TDS0rxn DHorx= 2 DHof (NO2)-DHof (N2O4)=2 x(33.85)-(9.66)= 58.04 kJ DSorxn = 2 x 240.5- 304.3 = 176.7 J/K Reaction will be spontaneous at __________ temperature high At 298 K DG0rxn= 58.04 kJ- 298Kx 176.7 J/Kx(1 kJ/1000J)=5.38 kJ At 500 K DG0rxn= 58.04 kJ- 500Kx 176.7 J/Kx(1 kJ/1000J)=-30.31 kJ At what temperature will the reaction be at equilibrium? DG0rxn= 58.04 kJ- T x 176.7 J/Kx(1 kJ/1000J) =0 T=328.5 K

  27. DG = DG0 + RT lnQ DG0 = - RT lnK VI. Gibbs free energy and chemical equilibrium Under non-standard state condition 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

  28. Ex. o K = e- DG / RT Calculate KP and DGat 298 K for the Haber reaction when you start with 1.0 atm N2, 2.0 atm H2, and 4.0 atm NH3.DG0rxn=-33.3 kJ. N2(g) + 3 H2 (g) = 2 NH3 (g) DG0 = - RT lnK lnK = -DG0 / RT K = exp(-(-33.3 x103)/8.314/298)=e13.44 = 6.87 x 105 DG = DG0 + RT ln Q Q= PNH32/(PH23PN2) = 2.0 Q <K, spontaneous DG = DG0 + RT ln Q = -33.3 kJ+ 8.314 J/K.mol x 298 K x ln(2.0) = -31.6 kJ

  29. As reaction proceeds

  30. DG0 = - RT lnK

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