1. 1 Second Year Chemistry 1 st semester: Organic 1 st semester: Physical (2005-2006) December exams 2 nd : Analytical & Environmental 2 nd : Inorganic Summer exams Physical: 3 lecturers  8 topics Dónal Leech: four topics Thermodynamics Gases, Laws & Phases, Equilibrium

2. 2 Introduction Energetics and Equilibria What makes reactions “go”! This area of science is called THERMODYNAMICS Thermodynamics is expressed in a mathematical language BUT Don’t, initially anyway, get bogged down in the detail of the equations: try to picture the physical principle expressed in the equations We will develop ideas leading to one important Law, and explore practical applications along the way The Second Law of Thermodynamics

3. 3 Lecture Resources 12 lectures leading to four exam questions ( section A, you must answer two from this section ) Main Text: “ Elements of Physical Chemistry” Atkins & de Paula, 4 th Edition (Desk reserve) http://www.oup.com/uk/booksites/content/0199271836/ OTHERS. “Physical Chemistry” Atkins & de Paula, 7 th Edition or any other PChem textbook These notes available on NUI Galway web pages at http://www.nuigalway.ie/chem/degrees.htm See also excellent lecture notes from James Keeler, Cambridge, although topics are treated in a different running order than here.James Keeler, Cambridge

4. 4 Thermodynamics: the 1 st law The internal energy of an isolated system is constant Energy can neither be created nor destroyed only inter-converted Energy: capacity to do work Work: motion against an opposing force System: part of the universe in which we are interested Surroundings: where we make our observations (the universe) Boundary: separates above two

5. 5 System and Surroundings Systems Open: energy and matter exchanged Closed: energy exchanged Isolated: no exchange Diathermic wall: heat transfer permitted Adiabatic wall: no heat transfer

6. 6 Work and Heat Work (w): transfer of energy that changes motions of atoms in the surroundings in a uniform manner Heat (q): transfer of energy that changes motions of atoms in the surroundings in a chaotic manner Endothermic: absorbs heat Exothermic: releases heat

7. 7 Work Mechanical work can generally be described by dw = -F.dz Gravitational work (mg.dh) Electrical work ( .dq) Extension work (f.dl) Surface expansion work ( .d  ) As chemists we will concentrate on EXPANSION WORK (many chemical reactions produce gases)  w = - F.  z but p ex = F/A therefore w = - p ex.  V Expansion against constant external pressure

8. 8 Expansion Work In thermodynamics “reversible” means a process that can be reversed by an infinitesimal change of a variable. A system does maximum expansion work when the external pressure is equal to that of the system at every stage of the expansion Expansion against zero external pressure (free expansion) w = - p ex.  V = 0 (external pressure = 0) Reversible isothermal expansion

9. 9 Isothermal reversible expansion Exercise: Calculate the work done when 1.0 mol Ar(g) confined in a cylinder of volume 1.0 dm 3 at 25°C expands isothermally and reversibly to 2.0 dm 3.

10. 10

11. 11 1 st Law of Thermodynamics The internal energy of an isolated system is constant Energy can neither be created nor destroyed only inter-converted  U = q+w Exercise: A car battery is charged by supplying 250 kJ of energy to it as electrical work, but in the process it loses 25kJ of energy as heat to the surroundings. What is the change in internal energy of the battery? How do we measure heat? Use calorimetry. If we enclose our system in a constant volume container (no expansion), provided no other kind of work can be done, then w = 0.  U = q V INTERNAL ENERGY is a State Function

12. 12 Bomb calorimetry By measuring the change in Temperature of the water surrounding the bomb, and knowing the calorimeter heat capacity, C, we can determine the heat, and hence  U. Heat Capacity Amount of energy required to raise the temperature of a substance by 1°C (extensive property) For 1 mol of substance: molar heat capacity (intensive property) For 1g of substance: specific heat capacity (intensive property)

13. 13 Calorimeter calibration Can calibrate the calorimeter, if its heat capacity is unknown, by passing a known electrical current for a given time to give rise to a measured temperature change. Amperes.Volts.Sec = Coulombs.Volts = Joules Exercise: In an experiment to measure the heat released by the combustion of a fuel, the compound was burned in an oxygen atmosphere inside a calorimeter and the temperature rose by 2.78°C. When a current of 1.12 A from an 11.5 V source was passed through a heater in the same calorimeter for 162 s, the temperature rose by 5.11°C. What is the heat released by the combustion reaction?

14. 14 Enthalpy Most reactions we investigate occur under conditions of constant PRESSURE (not Volume) ENTHALPY: Heat of reaction at constant pressure! Use a “coffee-cup” calorimeter to measure it Heat capacity Exercise: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffee- cup calorimeter, the temperature increases from 21°C to 27.5°C. What is the enthalpy change, if the density is 1g/mL and specific heat 4.18 J/g.K?

15. 15 Perfect gas enthalpy Use intensive property of molar enthalpy and internal energy At 25°C, RT = 2.5 kJ/mol Thermicity-Revision Endothermic reaction ( q >0) results in an increase in enthalpy (  H >0) Exothermic reaction ( q <0) results in an increase in enthalpy (  H <0) NB: Internal energy and Enthalpy are STATE FUNCTIONS

16. 16 Temperature variation of enthalpy Convenient empirical expression to use for heat capacity is: Exercise: What is the change in molar enthalpy of N 2 when it is heated from 25°C to 100 °C, given that:

17. 17 Relation between heat capacities

18. 18 Thermochemistry Chemists report data for a set of standard conditions: The standard state of a substance (°) is the pure substance at exactly 1 bar It is conventional (though not obligatory) to report data for a T of 298.15K Standard enthalpies of phase transition Energy that must be supplied (or is evolved) as heat, at constant pressure, per mole of molecules that undergo the phase transition under standard conditions (pure phases), denoted  H° Note: the enthalpy change of a reverse transition is the negative of the enthalpy change of the forward transition

19. 19 H°H°

20. 20 Sublimation l Direct conversion of a solid to a vapour The enthalpy change of an overall process is the sum of the enthalpy changes for the steps into which it may be divided

21. 21 Enthalpies of ionisation (kJ/mol)  ion H°(T)= Ionisation energy(0) + (5/2)RT (see Atkins & de Paula, Table 3.2)

22. 22 Problems  Ethanol is brought to the boil at 1 atm. When the electric current of 0.682 A from a 12.0 V supply is passed for 500 s through a heating coil immersed in the boiling liquid, it is found that the temperature remains constant but 4.33 g of ethanol is vapourised. What is the enthalpy of vapourisation of ethanol at its boiling point at 1 atm?  Calculate the standard enthalpy of sublimation of ice at 0°C given that  fus H° is 6.01 kJ/mol and  vap H° is 45.07 kJ/mol, both at 0°C.   sub H° for Mg at 25°C is 148 kJ/mol. How much energy as heat must be supplied to 1.00 g of solid magnesium metal to produce a gas composed of Mg 2+ ions and electrons?

23. 23 Bond enthalpies (kJ/mol)

24. 24 Problem  Estimate the standard reaction enthalpy for the formation of liquid methanol from its elements as 25°C

25. 25 Enthalpies of combustion Enthalpies (heats) of combustion: complete reaction of compounds with oxygen. Measure using a bomb calorimeter. Most chemical reactions used for the production of heat are combustion reactions. The energy released when 1g of material is combusted is its Fuel Value. Since all heats of combustion are exothermic, fuel values are reported as positive.  Most of the energy our body needs comes from fats and carbohydrates.  Carbohydrates are broken down in the intestines to glucose. Glucose is transported in the blood to cells where it is oxidized to produce CO 2, H 2 O and energy:  C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l)  c H°=-2816 kJ  The breakdown of fats also produces CO 2 and H 2 O  Any excess energy in the body is stored as fats

26. 26 Heats of formation If one mole of the compound is formed under standard conditions from its elements in their reference state then the resulting enthalpy change is said to be the standard molar enthalpy (Heat) of formation,  f H° where the subscript indicates this. The reference state is the most stable form under the prevailing conditions.

27. 27 Hess’s Law To evaluate unknown heats of reaction The standard enthalpy of a reaction is the sum of the standard enthalpies for the reactions into which the overall reaction may be divided  rxn H o =  f H o m(products) -  f H o m(reactants)

28. 28 Variation of  r H° with T  r H°(T 2 ) =  r H°(T 1 ) +  r C p °(T 2 -T 1 ) If heat capacity is temperature dependent, we need to integrate over the temperature range  r C p ° =  C p,m ° (products) -  C p,m ° (reactants) Kirchoff’s Law

29. 29 Thermodynamics: the 2 nd law Deals with the direction of spontaneous change (no work required to bring it about) Kelvin Statement No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work Impossible!

30. 30 Entropy The apparent driving force for spontaneous change is the dispersal of energy A thermodynamic state function, Entropy, S, is a measure of the dispersal of energy (molecular disorder) of a system 2 nd Law: The Entropy of an isolated system increases in the course of spontaneous change  S tot >0

31. 31 Thermodynamic definition of S  Concentrates on the change in entropy:  S = q rev /T Can use this equation to quantify entropy changes. We will see later (3 rd & 4 th year) a statistical description of entropy S = k lnW (Boltzmann formula)

32. 32 Heat Engines All heat engines have a hot region “source” and a cold region “sink”: some energy must be discarded into the cold sink as heat and not used to do work

33. 33 Expansion entropy  Intuitively can guess that entropy increases with gas expansion.  Thermodynamic definition allows us to quantify this increase Recall that: w = -nRT ln (V f /V i ) BUT q rev = -w (  U = 0 for isothermal processes)  S = nR ln (V f /V i ) Note: independent of T Also: Because S is a state function, get the same value for an irreversible expansion

34. 34 Heating Entropy

35. 35 Entropy of phase transition  Entropy of fusion  Entropy of vapourisation Trouton’s rule The entropy of vapourisation is approximately the same (85 J/K.mol) for all non-polar liquids

36. 36 Phase transitions  To evaluate entropies of transition at T other than the transition temperature Entropy of vapourisation of water at 25°C?  Sum of  S for heating from 25°C to 100°C,  S for vapourisation at 100°C, and  S for cooling vapour from 100°C to 25°C. Try it! (+118 J/K.mol).

37. 37 Entropy changes in the surroundings  S tot =  S sys +  S sur  S tot =  S sys – q/T Example: Water freezing to ice. Entropy change of system is -22 J/K.mol, and heat evolved is -6.01 kJ/mol. Entropy change in surroundings must be positive for this process to occur spontaneously. Check this for different temperatures. Note that  S tot = 0 at equilibrium

38. 38 Spontaneity of water freezing  At 5°C:  S tot = -22 JK -1 mol -1 – (6,010Jmol -1 /278K)  = -0.38 JK -1 mol -1  At -5°C:  S tot = -22 JK -1 mol -1 – (6,010Jmol -1 /268K)  = +0.43 JK -1 mol -1  At 0°C:  S tot = -22 JK -1 mol -1 – (6,010Jmol -1 /273K)  = 0.01 JK -1 mol -1  To find transition temperature, set  S tot = 0 and solve for T. 273.18 K (slight error because of rounding of entropy and heats).  S tot =  S sys - q sys /T

39. 39 Problem  Typical person heats the surroundings at a rate of 100W (=J/s). Estimate entropy change in one day at 20°C.  q sur = 86,400 s × 100 J/s   S sur = q sur /T = (86,400 × 100 J)/293 K = 2.95 × 10 4 J/K

40. 40 3 rd Law  Entropy of sulfur phase transition is 1.09 J/K.mol.  Consider plot at left. Subtract entropy for phase transition (to give plot at right)  T=0 intercept is the same. Entropies of all perfectly crystalline substances are the same at T=0.

41. 41 Absolute and standard molar entropies (S and S 0 m ) Absolute entropies can be determined by integration of areas under heat capacity/T as a function of T, and including entropies of phase transitions. Standard molar entropies are the molar entropies of substances at 1bar pressure (and usually 298 K)

42. 42 Standard molar entropies

43. 43 Standard reaction entropies  Difference in molar entropy between products and reactants in their standard states is called the standard reaction entropy and can be expressed (like enthalpy) as:  Note: absolute entropies, S, and standard molar entropies, S 0 m, are discussed in section 4.7 of the textbook  rxn S o =  S o m(products) -  S o m(reactants)

44. 44 Spontaneity of reactions Consider the reaction: 2H 2 (g) + O 2 (g)  2H 2 O (l)  r S 0 = 2(70 J/K.mol) –[2(131 J/K.mol) + (205 J/K.mol)] = -327 J/K.mol But this reaction is spontaneous (explosive even!) When considering the implications of entropy, we must always consider the total change of the system and its surroundings  r H 0 = -572 kJ/mol. Therefore  r S sur = +1920 J/K.mol  r S tot is positive +1590 J/K.mol (spontaneous reaction!).

45. 45 Gibbs Energy  Introduced by J.W. Gibbs to combine the calculations of 2 entropies, into one.J.W. Gibbs  Because  S tot =  S –  H/T (constant T and P)  Introduce G = H – TS (Gibbs “free” energy)  Then  G =  H – T  S (constant T)  So that  G = – T  S tot (constant T and P)  G =  H – T  S In a spontaneous change at constant temperature and pressure, the Gibbs energy decreases

46. 46 Maximum non-expansion work  Can derive ( see box 4.5 in textbook ) that  G = w’ max  Example: formation of water: enthalpy -286kJ, free energy -237kJ  Example: suppose a small bird has a mass of 30 g. What is the minimum mass of glucose that it must consume to fly to a branch 10 m above the ground? (  G for oxidation of glucose to carbon dioxide and water is -2828 kJ at 25°C) Exercise: A human brain operates at about 25 W (J/s). What mass of glucose must be consumed to sustain that power for 1 hour?

47. 47 Problems solved w’ = (30 × 10 -3 kg) × (9.81 m s -2 ) × 10 m Note 1J = 1kg m 2 s -2 n = 2.943 J/(2828 × 10 3 J/mol) m = nM = (1.04 × 10 -6 mol) × 180 g/mol = 1.9 × 10 -4 g Answer 2: 5.7g

48. 48 Phase Equilibria Phase transitions Changes in phase without a change in chemical composition Gibbs Energy is at the centre of the discussion of transitions Molar Gibbs energy G m = G/n Depends on the phase of the substance A substance has a spontaneous tendency to change into a phase with the lowest molar Gibbs energy

49. 49 Variation of G with pressure  We can derive (see derivation 5.1 in textbook) that  G m = V m  p  Therefore  G m >0 when  p>0  Can usually ignore pressure dependence of G for condensed states  Can derive that, for a gas:  G m = RT ln(p f /p i )

50. 50 Proof-go back to fundamental definitions G = H – TS; H = U + pV; dU = dq + dw For an infinitesimal change in G: G + dG = H + dH – (T + dT)(S + dS) = H + dH – TS – TdS – SdT – dTdS dG = dH – TdS – SdT Also can write: dH = dU + pdV + Vdp dU = TdS – pdV (dS = dqrev/T and dw = -pdV) dG = Vdp – SdT Master Equations

51. 51 Variation of G with temperature  G m = -S m  T Can help us to understand why transitions occur The transition temperature is the temperature when the molar Gibbs energy of the two phases are equal. The two phases are in EQUILIBIRIUM at this temperature

52. 52 Phase diagrams  Map showing conditions of T and p at which various phases are thermodynamically stable  At any point on the phase boundaries, the phases are in dynamic equilibrium

53. 53 Location of phase boundaries  Clapeyron equation (see derivation 5.4)  Clausius-Clapeyron equation (derivation 5.5) Constant is  vap S/R

54. 54 Derivations dG m = V m dp – S m dT dG m (1) = dG m (2) V m (1)dp – S m (1)dT = V m (2)dp – S m (2)dT {V m (2) – V m (1)}dp = {S m (2) – S m (1)}dT  trs V dp =  trs S dT T  trs V dp =  trs H dT dp/dT =  trs H/(T  trs V)

55. 55 Derivations: liquid-vapour transitions dp/dT =  vap H/(T  vap V) ≈  vap H/{T V m (g)} =  vap H/{T (RT/p)} (dp/p)/dT =  vap H/(RT 2 ) (d lnp)/dT =  vap H/(RT 2 ) + constant

56. 56 Using the equation  The vapour pressure of mercury is 160 mPa at 20°C. What is its vapour pressure at 50°C given that its enthalpy of vapourisation is 59.3 kJ/mol?  The vapour pressure of pyridine is 50.0 kPa at 365.7 K and the normal boiling point is 388.4 K. What is the enthalpy of vapourisation of pyridine?  Estimate the normal and standard boiling point of benzene given that its vapour pressure is 20.0kPa at 35°C and 50.0kPa at 58.8°C. Remember:  BP: temperature at which the vapour pressure of the liquid is equal to the prevailing atmospheric pressure.  At 1atm pressure: Normal Boiling Point (100°C for water)  At 1bar pressure: Standard Boiling Point (99.6°C for water; 1bar=0.987atm, 1atm = 1.01325bar)

57. 57 Summary l Thermodynamics tells which way a process will go Internal energy of an isolated system is constant (work and heat). We looked at expansion work (reversible and irreversible). Thermochemistry usually deals with heat at constant pressure, which is the enthalpy. Spontaneous processes are accompanied by an increase in the entropy (disorder?) of the universe Gibbs free energy decreases in a spontaneous process