Assignment (Due on 16/11/1999) Exercises 1.4, 1.8, 1.16, 1.39, 1.40

1. Assignment (Due on 16/11/1999) Exercises 1.4, 1.8, 1.16, 1.39, 1.40 Problems 1.16, 1.20, 1.27, 1.32, 1.33 marked with My office hours: 2:00 pm - 5:00 pm on Tuesdays

2. THERMODYNAMICS References ·Atkins, P.W., “Physical Chemistry”, Oxford University Press ·Castellan, G.W., “Physical Chemistry”, Addison Wesley ·Levine, I.R., “Physical Chemistry”, McGraw-Hill ·Laidler & Meiser, “Physical Chemistry”, Houghton Mifflin Co. ·Alberty, R.A. and Silbey, R., “Physical Chemistry”, Wiley

3. thermo – heat dynamics – changes definitions descriptions no proofs no violations Events,Experiments,Observations abstract generalize Laws of Thermodynamics: 0th, 1st, 2nd, 3rd Applications, Verifications

4. Note: ·does not worry about rate of changes (kinetics) but the states before and after the change ·not dealing with time

5. Classical Thermodynamics Marcoscopic observables T, P, V, … Statistical Thermodynamics Microscopic details dipole moment, molecular size, shape

6. Joule’s experiment T  mgh adiabatic wall (adiabatic process) U (energy change) = W (work) = mgh Thermometer w h

7. w Page 59 T  time interval of heating U = q (heat) Conclusion: work and heat has the same effect to system (internal energy change) * FIRST LAW: U = q + W

8. U: internal energy is a state function [ = Kinetic Energy (K.E.) + Potential Energy (P.E.) ] q: energy transfer by temp gradient W: force  distance E-potential  charge surface tension  distance pressure  volume First Law: The internal energy of an isolated system is constant

9. M M V2 V1 Convention: Positive: heat flows into system work done onto system Negative: heat flows out of system work done by system Pressure-volume Work P1 = P2

10. If weight unknown, but only properties of system are measured, how can we evaluate work? M M V2 P P V1 Assume the process is slow and steady, Pint = Pext

11. Free Expansion: Free expansion occurs when the external pressure is zero, i.e. there is no opposing force

12. Reversible change: a change that can be reversed by an infinitesimal modification of a variable. Quasi equilibrium process: Pint = Pext + dP (takes a long time to complete) infinitesimal at any time quasi equilibrium process

13. P 1 2 V Page 64-66 Example: P1 = 200kPa = P2 V1 = 0.04m3 V2 = 0.1m3 * what we have consider was isobaric expansion (constant pressure) other types of reversible expansion of a gas: isothermal, adiabatic

14. Page 65-66 Isothermal expansion remove sand slowly at the same time maintain temperature by heating slowly * P T V2 V1 V area under curve

15. Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3

16. V1 P1 T1 Adiabatic Reversible Expansion For this process PV = constant for ideal gas (proved later) (slightly larger than 1) V2 P2 T2

17. P a b d V Ex. V1 = 0.04m3, P1 = 200kPa, V2 = 0.1m3,  = 1.3 200kPa const d: pressure drop without volume change |Wa|>|Wb|>|Wc|>|Wd| isothermal adibatic

18. State Function Vs Path Function State function: depends only on position in the x,y plane e.g.: height (elevation) 300 1 B 2 200 100m Y X A

19. 2 481.13K 192.45K 3 Path Function: depends on which path is taken to reach destination from 1  2, difference of 300m (state function) but path A will require more effort. Internal energy is a state function, heat and work are path functions P 1 200kPa 0.04 0.1 V/m3

20. 5 moles of monoatomic gas

21. Multivariable Calculus Consider a 2 variable function or a surface in three dimensional plot z or  C’ C xc yc y x

22. At point C, there are two slopes orthogonal to each other in constant x or constant y direction C’ z C’ z C C x=xC y=yc y x

23. the change in  can be calculated as a sum of two parts: change in x direction and change in y direction partial derivative partial derivative w.r.t. x w.r.t. y

24. Joule’s second experiment Energy UU(V,T) ??? thermometer V2 V1 adiabatic wall At time zero, open valve

25. thermometer No temperature change adiabatic wall After time zero, V1 V1+V2 T=0  Q=0 W=0 no Pext U=0

26. Page 99 U=U(T) Energy is only a function of temperature for ideal gas * 0 (for ideal gas) CV is constant volume heat capacity

27. Kinetic Model for Gases Qualitatively: • Gases consists of spheres of negligible size, far apart from one other. • Particles in ceaseless random motion; no interactions except collisions

28. U V Constant Temperature U V T U T

29. * Energy is a function of Volume and temperature for real gases Interaction among molecules

30. Page 101 *Enthalpy Define H = U + PV   state function intensive variables locating the state Enthalpy is also a state function H = U + PV + VP

31. At constant pressure H = U+PV = U - W = Q H = QP constant pressure heating H is expressed as a functional of T and P for ideal gases (proved later)

32. Thermochemistry Heat transferred at constant volume qV = U Heat transferred at constant pessure qP = H Exothermic H = -ve Endothermic H = +ve

33. Standard states, standard conditions do not measure energies and enthalpies absolutely but only the differences, U or H The choice of standard state is purely a matter of convenience Analogy – differences in altitudes between 100 points and their elevation with respect to sea level

34. * What is the standard state ? The standard states of a substance at a specified temperature is its pure form at 1 bar

35. 25oC, 1 bar: the most stable forms of elements assign “zero enthalpy” Ho298 = 0 used for chemical reactions

36. Standard enthalpy of formation Standard enthalpy change for the formation of the compound from its elements in their reference states. Reference state of an element is its most stable state at the specified temperature & 1 bar C (s) + 2H2 (g)  CH4 (g) Hfo = -75 kJ 289K, 1 atm From the definition, Hfo for elements  0

37. Hess’s Law The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. Standard reaction enthalpy is the change in enthalpy when the reactants in their standard states change to products in their standard states.

38. Hess’ Law H1 H1 = H2 + H3 + H4 state function Hess’s law is a simple application of the first law of thermodynamics

39. e.g. C (s) + 2H2 (g)  CH4 (g) H1o = ? 298K, 1 atm C (s,graphite) + O2 (g)  CO2 (g) Ho = -393.7 kJ H2 (g) + ½O2(g)  H2O (l) Ho = -285.8 kJ CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) Ho = -890.4 kJ H1o = -393.7 + 2(-285.8) - (-890.4) = -75 kJ/mole

40. Heat of Reaction (Enthalpy of Reaction) Enthalpy change in a reaction, which may be obtained from Hfo of products and reactants Reactants  Products I stoichiometric coefficient, + ve products, - ve reactants E.g. CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)

41. Hfo/ kJ CH3Cl -83.7 HCl -92.0 Cl2 0 CH4 -75.3 Hro = (-83.7-92.0) - (-75.3+0) = -100.4 kJ Reactants Products elements elements

42. Page 83 2A + B  3C + D 0 = 3C + D - 2A - B Generally, 0 = J J J J denotes substances, J are the stoichiometric numbers *

43. Bond energy (enthalpy) Assumption – the strength of the bond is independent of the molecular environment in which the atom pair may occur. C (s,graphite) + 2H2 (g)  CH4(g)Ho = -75.4 kJ H2 (g)  2H (g) Ho = 435.3 kJ C (s,graphite)  C (g) Ho= 715.8 kJ  C (s,graphite) + 2H2 (g)  C (g) + 4H (g) Ho = 2(435.3)+715.8 = 1586.5 kJ C (g) + 4H (g)  CH4 (g) H = -75.4-1586.5 = -1661.9 kJ CH Bond enthalpy = 1661.9/4 kJ = 415.5 kJ

44. Temperature dependence of Hr P HrT CP,P R Hro CP,R 298 T

45. What is the enthalpy change for vaporization (enthalpy of vaporization) of water at 0oC? H2O (l)  H2O (g) Ho = -241.93 - (-286.1) = 44.01 kJmol-1 H2 = Ho + CP(T2-T1) assume CP,i constant wrt T H (273) = Ho(298) + CP(H2O,g) - CP (H2O,l)(273-298) = 44.10 - (33.59-75.33)(-25) = 43.0 kJ/mole

46. For ideal Gases:

47. 0 = R/P for ideal gas 0 for ideal gas

48. Prove PV = constant for adiabatic reversible expansion of an ideal gas 0 for ideal gas

49. Adiabatic expansion

50. Reversible vs Irreversible Non-spontaneous changes vs Spontaneous changes Reversibility vs Spontaneity First law does not predict direction of changes, cannot tell which process is spontaneous. Only U = Q + W