1 / 50

THERMODYNAMICS

THERMODYNAMICS. Chapter 19 . SPONTANEOUS PROCESS.  A process that occurs without ongoing outside intervention . Examples Nails rusting outdoors. 2H 2 (g) + O 2 (g) 2H 2 O(g). Ice melting at room temperature. Expansion of gas into an evacuated space.

sezja
Télécharger la présentation

THERMODYNAMICS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. THERMODYNAMICS Chapter 19

  2. SPONTANEOUS PROCESS  A process that occurs without ongoing outside intervention. • Examples • Nails rusting outdoors

  3. 2H2(g) + O2(g) 2H2O(g) • Ice melting at room temperature • Expansion of gas into an evacuated space • Formation of water from O2(g) and H2(g):

  4. T=0oC  water and ice in equilibrium H2O (l) H2O (s) Why are some processes spontaneous and others not? We know that temperature has an effect on the spontaneity of a process. e.g. T>0oC  ice melts  spontaneous at this temp. H2O (s)  H2O (l) T<0oC  water freezes  spontaneous at this temp. H2O (l)  H2O (s)

  5. Formation of water - SPONTANEOUS! 2H2(g) + O2(g) 2H2O(l) H = - 285.8 kJ.mol-1 Pt cat Exothermic processes tend to be spontaneous. Example Rusting of nail - SPONTANEOUS! 4Fe(s) + 3O2(g)  Fe2O3(s) H = - 822.2 kJ.mol-1

  6. However, the dissolution of ammonium nitrate is also spontaneous, but it is also endothermic. NH4NO3(s)  NH4+(aq) + NO3-(aq) H = +25.7 kJ.mol-1 So is: 2N2O5(s)  4NO2(g) + 2O2(g) H = +109.5 kJ.mol-1  a process does not have to be exothermic to be spontaneous.  something else besides sign of H must contribute to determining whether a process is spontaneous or not.

  7. That something else is: ENTROPY (S) extent of disorder! More disordered  larger entropy Entropy is a state function S = Sfinal - Sinitial Units: J K-1mol-1

  8. Entropy’s effects on the mind

  9. Examples of spontaneous processes where entropy increases: Dissolution of ammonium nitrate: NH4NO3(s)  NH4+(aq) + NO3-(aq) H = +25.7 kJ.mol-1 Decomposition of dinitrogen pentoxide: 2N2O5(s)  4NO2(g) + 2O2(g) H = +109.5 kJ.mol-1

  10. However, entropy does not always increase for a spontaneous process At room temperature: Spontaneous Non-spontaneous

  11. SECOND LAW OF THERMODYNAMICS The entropy of the universe increases in any spontaneous process. Suniverse = Ssystem + Ssurroundings Spontaneous process: Suniverse > 0 Process at equilibrium: Suniverse = 0 Thus Suniv is continually increasing! Suniv must increase during a spontaneous process, even if Ssyst decreases.

  12. Q: What is the connection between sausages and the second law of thermo? • A: Because of the 2nd law, you can put a pig into a machine and get sausage, but you can't put sausage into the machine and get the pig back.

  13. Suniv>0 For example: Rusting nail = spontaneous process 4Fe(s) + 3O2(g)  2Fe2O3(s) Ssyst<0 BUT reaction is exothermic, entropy of surroundings increases as heat is evolved by the system thereby increasing motion of molecules in the surroundings.  Ssurr>0 -ve +ve +ve Thus for Suniv=Ssyst+ Ssurr>0 Ssurr > Ssyst

  14. Special circumstance = Isolated system: • Does not exchange energy nor matter with surroundings Ssurr = 0 Spontaneous process: Ssyst > 0 Process at equilibrium: Ssyst = 0 Spontaneous  Thermodynamically favourable (Not necessarily occur at observable rate.) Thermodynamics  direction and extent of reaction, not speed.

  15. EXAMPLE • State whether the processes below are spontaneous, non-spontaneous or in equilibrium: • CO2 decomposes to form diamond and O2(g) • Water boiling at 100oC to produce steam in a closed container • Sodium chloride dissolves in water NON-SPONTANEOUS EQUILIBRIUM SPONTANEOUS

  16. MOLECULAR INTEPRETATION OF S Decrease in number of gaseous molecules decrease in S e.g. 2NO(g) + O2(g)  2NO2(g) 3 moles gas 2 moles gas

  17. Molecules have 3 types of motion: Translational motion - Entire molecule moves in a direction (gas > liquid > solid) Vibrational motion – within a molecule Rotational motion – “spinning” Greater the number of degrees of freedom  greater entropy

  18. Decrease in temperature  decrease in thermal energy  decrease in translational, vibrational and rotational motion  decrease in entropy As the temperature keeps decreasing, these motions “shut down”  reaches a point of perfect order.

  19. EXAMPLE • Which substance has the great entropy in each pair? Explain. • C2H5OH(l) or C2H5OH(g) • 2 moles of NO(g) or 1.5 moles of NO(g) • 1 mole O2(g) at STP or 1 mole NO2(g) at STP

  20. THIRD LAW OF THERMODYNAMICS The entropy of a pure crystalline substance at absolute zero is zero. S(0 K) = 0  perfect order

  21. Ginsberg's Theorem • (The modern statement of the three laws of thermodynamics) • 1. You can't win. • 2. You can't even break even. • 3. You can't get out of the game.

  22. Entropy increases for s  l  g

  23. EXAMPLE • Predict whether the entropy change of the system in each reaction is positive or negative. • CaCO3(s)  CaO(s) + CO2(g) • 2SO2(g) + O2(g)  2SO3(g) • N2(g) + O2(g)  2NO(g) • H2O(l) at 25oC  H2O(l) at 55oC +ve -ve 3 mol gas  2 mol gas Can’t predict, but it is close to zero ? 2 mol gas  2 mol gas +ve Increase thermal energy

  24. Standard molar entropy (So) = molar entropy for substances in their standard state • NOTE • So 0 for elements in their standard state • So(gas) > So(liquid) > So(solid) • So generally increases with increasing molar mass • So generally increases with increasing number of atoms in the formula of the substance

  25. Calculation of S for a reaction Stoichiometric coefficients (So from tabulated data)

  26. EXAMPLE Calculate So for the synthesis of ammonia from N2(g) and H2(g): N2(g) + 3H2(g)  2NH3(g) So/J.K-1.mol-1 N2(g) 191.5 H2(g) 130.6 NH3(g) 192.5

  27. Calculation of S for the surroundings For a process that occurs at constant temperature and pressure, the entropy change of the surroundings is: (T &P constant)

  28. GIBB’S FREE ENERGY (G)

  29. Hsys Suniv = Ssys - T Defined as: G = H – TS - state function - extensive property Suniv = Ssys + Ssurr At constant T and P:  - TSuniv = - TSsys + Hsys (at constant T & P) G = H – TS

  30. We know: Spontaneous process: Suniv > 0 Process at equilibrium: Suniv = 0 Therefore: Spontaneous process: Process at equilibrium: < 0 -TSuniv -TSuniv = 0 Gsyst = -TSuniv

  31. G = H – TS Spontaneity involves S H T Spontaneity is favoured by increasing S and H is large and negative.

  32. G allows us to predict whether a process is spontaneous or not (under constant temperature and pressure conditions): G < 0spontaneous in forward direction G > 0non-spontaneous in forward direction/spontaneous in reverse direction G = 0 at equilibrium

  33. But nothing about rate

  34. Standard free energy (Go) Go = Ho – TSo Standard states: Gas - 1 atm Solid - pure substance Liquid - pure liquid Solution - Concentration = 1M Gfo = 0 kJ/mol for elements in their standard states

  35. Tabulated data of Gfo can be used to calculate standard free energy change for a reaction as follows: Stoichiometric coefficients

  36. EXAMPLE The combustion of propane gas occurs as follows: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Using thermodynamic data for Go, calculate the standard free energy change for the reaction at 298 K. Gfo/kJ.mol-1 C3H8(g) -23.47 CO2(g) -394.4 H2O(g) -228.57 H2O(l) -237.13

  37. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Gfo/kJ.mol-1 C3H8(g) -23.47 CO2(g) -394.4 H2O(g) -228.57 H2O(l) -237.13 Go = [3(-394.4) + 4(-237.13)] – [(-23.47) – 5(0)] Go = -2108 kJ

  38. Free Energy and Temperature How is change in free energy affected by change in temperature? G = H – TS G = H - TS H S -TS - + + +at all temp - - at all temp + - - at high temp+at low temp + - + +at high temp - at low temp - + - -

  39. Note: For a spontaneous process the maximum useful work that can be done by the system: wmax = G “free energy” = energy available to do work

  40. EXAMPLE The combustion of propane gas occurs as follows at 298K: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2220 kJ.mol-1 a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho. b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Wasyour prediction correct?

  41. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2220 kJ.mol-1 a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho. Go = Ho – TSo -ve -ve 6 moles gas  3 moles gas  – TSo > 0 Ho – TSo will be less negative than Ho i.e. Go will be less negative than Ho

  42. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2220 kJ.mol-1 b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Wasyour prediction correct? Go = Ho – TSo Go = (-2220 kJ) – (298 K)(-374.46x10-3kJ.K-1.mol-1) Go = -2108 kJ.mol-1  Prediction was correct.

  43. EXAMPLE (TUT no. 5a) • At what temperature is the reaction below spontaneous? • AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s) • Ho = 851.5 kJ; So = 38.5 J K-1

  44. At what temperature is the reaction below spontaneous? AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s) Ho = 851.5 kJ; So = 38.5 J K-1 Go = Ho – TSo Assume H and S do not vary that much with temperature. Set G = 0 equilibrium 0 = (851.5 kJ) – T(38.5x10-3 kJ.K-1) T = 22117 K at equilibrium For spontaneous reaction: G < 0  T > 22117 K

  45. Free Energy and the equilibrium constant Recall:G = Change in Gibb’s free energy under standard conditions. G can be calculated from tabulated values. BUT most reactions do not occur under standard conditions. Calculate G under non-standard conditions: Q = reaction quotient R = gas constant = 8.314 J.K-1.mol-1

  46. Under standard conditions: (1 M, 1 atm) Q = 1  ln Q = 0  G = Go At equilibrium: G = 0 and Q = Keq  If Go < 0  ln Keq > 0  Keq > 1 i.e. the more negative Go, the larger K etc. Go < 0  Keq > 1 Go > 0  Keq < 1 Go = 0  Keq = 1 Also

  47. EXAMPLE Calculate K for the following reaction at 25oC: 2H2O(l) 2H2(g) + O2(g) Gfo/kJ.mol-1 H2O(g) -228.57 H2O(l) -237.13 Go = [2(0) + (0)] – [2(-237.13)] Go = 474.26 kJ.mol-1 474.26x103 J.mol-1 = -(8.314 J.K-1.mol-1)(298 K) lnK lnK = -191.4 K = 7.36x10-84

More Related