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THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins

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  1. THERMODYNAMICS.Elements of Physical Chemistry. By P. Atkins Concerned with the study of transformation of energy: Heat  work

  2. CONSERVATION OF ENERGY – states that: • Energy can neither be created nor destroyed in chemical reactions. It can only be converted from one form to the other. • UNIVERSE • System – part of world have special interest in… • Surroundings – where we make our observations

  3. → → Example: ↔matter↔energy ↔ energy not mattermatter× Energy × Open system Closed system Isolated system

  4. If system is themally isolated called Adiabatic system eg: water in vacuum flask.

  5. WORK and HEAT • Work – transfer of energy to change height of the weight in surrounding eg: work to run a piston by a gas. • Heat – transfer of energy is a result of temperature difference between system and surrounding eg: HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l) - heat given off. If heat released to surroundings – exothermic. If heat absorbed by surroundings – endothermic.

  6. Example: Gasoline, 2, 2, 4 trimethylpentane CH3C(CH3)2CH2CH(CH3)CH3 + 25/2 O2→ 8CO2(g) +H2O(l) 5401 kJ of heat is released (exothermic) Where does heat come from? From internal energy, U of gasoline. Can represent chemical reaction: Uinitial = Ufinal + energy that leaves system (exothermic) Or Ui = Uf – energy that enters system (endothermic)

  7. Hence, FIRST LAW of THERMODYNAMICS(applied to a closed system) • The change in internal energy of a closed system is the energy that enters or leaves the system through boundaries as heat or work. i.e. • ∆U = q +w • ∆U = Uf – Ui • q – heat applied to system • W – work done on system • When energy leaves the system, ∆U = -ve i.e. decrease internal energy • When energy enter the system, ∆U = +ve i.e. added to internal energy

  8. Different types of energies: • Kinetic energy = ½ mv2 (chemical reaction) kinetic energy (KE)  k T (thermal energy)where k = Boltzmann constant • Potential energy (PE) = mgh – energy stored in bonds Now,U = KE + PE

  9. 3.Work (W) w = force × distance moved in direction of force i.e. w = mg × h = kg × m s-2 × m = kg m2 s-2 (m) (g) (h) 1 kg m2 s-2 = 1 Joule - Consider work – work against an opposing force, eg: external pressure, pex. Consider a piston

  10. Piston

  11. w = distance × opposing force w = h × (pex × A) = pex × hA Work done on system = pex × ∆V ∆V – change in volume (Vf – Vi) • Work done by system = -pex× ∆V Since U is decreased

  12. Example: C3H8(g) + 5 O2(g)→ 3CO2(g) + 4H2O(l)at 298 K  1 atm (1 atm = 101325 Pa), -2220 kJ = q What is the work done by the system? For an ideal gas; pV = nRT (p = pex) n – no. of moles R – gas constant T = temperature V – volume p = pressure

  13. V= nRT/p or Vi = niRT/pex 6 moles of gas: Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3 3 moles of gas: Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3 work done = -pex × (Vf – Vi) = -101325 (0.0734 – 0.1467) = +7432 J

  14. NB: work done = - pex (nfRT/pex – niRT/pex) = (nf – ni) RT Work done = -∆ngasRT i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J Can also calculate ∆U ∆U = q +w q = - 2220 kJ w = 7432.7 J = 7.43 kJ  ∆U = - 2220 + 7.43 = - 2212.6 kJ

  15. NB: qp ∆U why? Only equal if no work is done i.e.∆V = 0 i.e. qv = ∆U

  16. Example: energy diagram

  17. Since work done by system = pex∆V System at equilibrium when pex = pint (mechanical equilibrium) Change either pressure to get reversible work i.e. pex > pint or pint > pex at constant temperature by an infinitesimal change in either parameter

  18. For an infinitesimal change in volume, dV • Work done on system = pdV For ideal gas, pV = nRT p = nRT/ V  work =  p dV = nRT dV/ V = nRT ln (Vf/Vi) because dx/x = ln x • Work done by system= -nRT ln(Vf/Vi)

  19. Enthalpy, H • Most reactions take place in an open vessel at constant pressure, pex. Volume can change during the reaction i.e. V  0 (expansion work). Definition: H = qp i.e. heat supplied to the system at constant pressure.

  20. Properties of enthalpy • Enthalpy is the sum of internal energy and the product of pV of that substance. i.e H = U + pV (p = pex) • Some properties of H

  21. Hi = Ui + pVi Hf = Uf + pVf Hf – Hi = Uf – Ui +p(Vf – Vi) or H = U + p V

  22. Since work done = - pexV H = (- pex V + q) +pV (pex= p) • H = ( -p V + q) + p V = q •  H = qp

  23. suppose p and V are not constant? • H = U + ( pV) expands to: • H = U + pi V + Vi P + (P) (V) • i.e. H under all conditions. • When p = 0 get back H = U + pi V  U + p V • When V = 0: H = U + Vi p

  24. Enthalpy is a state function.

  25. NB: work and heat depend on the path taken and are written as lower case w and q. Hence, w and q are path functions. The state functions are written with upper case. eg: U, H, T and p (IUPAC convention).

  26. Standard States • By IUPAC conventions as the pure form of the substance at 1 bar pressure (1 bar = 100,000 Pa). What about temperature? • By convention define temperature as 298 K but could be at any temperature.

  27. Example: C3H8(g) + 5 O2(g)→ 3CO2(g) + 4H2O(l) at 1 bar pressure, qp = - 2220 kJmol-1. • Since substances are in the pure form then can write H = - 2220 kJ mol-1 at 298 K  represents the standard state.

  28. H2(g)→ H(g) + H(g), H diss = +436kJmol-1 H2O(l)→ H2O(g), Hvap = +44.0 kJmol-1 Calculate U for the following reaction: CH4(l) + 2 O2(g)→ CO2(g) + 2H2O(l), H = - 881.1kJmol-1

  29. H = U + (pV) = U + pi V + Vi p + p V NB: p = 1 bar, i.e. p = 0  H =U + pi V Since -pi V = - nRT, U = H - nRT

  30. calculation  U = - 881.1 – ((1 – 2)(8.314) 298)/ 1000 = - 881.1 – (-1)(8.314)(0.298) = - 2182.99 kJ mol-1

  31. STANDARD ENTHALPY OF FORMATION, Hf • Defined as standard enthalpy of reaction when substance is formed from its elements in their reference state. • Reference state is the most stable form of element at 1 bar atmosphere at a given temperature eg. At 298 K Carbon = Cgraphite Hydrogen = H2(g) Mercury = Hg(l) Oxygen = O2(g) Nitrogen = N2(g)

  32. NB: Hf of element = 0 in reference state • Can apply these to thermochemical calculations eg. Can compare thermodynamic stability of substances in their standard state. • From tables of Hf can calculate H f rxn for any reaction.

  33. Eg. C3H8 (g) + 5O2(g)→ 3CO2(g) + 4H2O(l) • Calculate Hrxn given that: Hf of C3H8(g) = - 103.9 kJ mol-1 Hf of O2(g) = 0 (reference state) Hf of CO2(g) = - 393.5 kJ mol-1 Hf of H2O(l) = - 285.8 kJ mol-1 Hrxn = n H (products)- n H(reactants)

  34. Hf(products) = 3  (- 393.5) + 4  (- 285.8) = - 1180.5 -1143.2 = - 2323.7 kJ mol-1 Hf(reactants) = - 103.9 + 5  0 = - 103.9 kJ mol-1 Hrxn = - 2323.7 – (- 103.9) = - 2219.8 kJ mol-1 = - 2220 kJ mol-1

  35. Answer same as before. Eq. is valid. • Suppose: solid → gas (sublimation) • Process is: solid → liquid → gas • Hsub = Hmelt + Hvap • Ie. H ( indirect route) = . H ( direct route)

  36. Hess’ Law • - the standard enthalpy of a reaction is the sum of the standard enthalpies of the reaction into which the overall reaction may be divided. Eg. • C (g) + ½ O2(g)→ CO (g) , Hcomb =? at 298K

  37. From thermochemical data: • C (g) +O2(g)→ CO 2(g)Hcomb=-393.5 kJmol-1…………………………….(1) • CO (g) +1/2 O2 (g) →CO 2(g), Hcomb = -283.0 kJ mol-1……………………. (2) • Subtract 2 from 1 to give: • C (g) + O2 (g) – CO (g) – 1/2 O2(g) → CO2(g) – CO2(g) •  C (g) + ½ O 2(g) → CO (g) , Hcomb= -393.5 – • (-283.0) = - 110.5 kJ mol-1

  38. Bond Energies • eg. C-H bond enthalpy in CH4 • CH4(g)→ C (g) + 4 H (g) , at 298K. • Need: Hf of CH 4 (g) =- 75 kJ mol-1 • Hf of H (g) = 218 kJ mol-1 • Hf of C (g) = 713 kJ mol-1

  39. Hdiss =  nHf(products)-  nHf ( reactants) • = 713 + ( 4x 218) – (- 75) = 1660 kJ • mol-1 • Since have 4 bonds : C-H = 1660/4 = 415 • kJ mol-1

  40. Variation of H with temperature Suppose do reaction at 400 K, need to know Hf at 298 K for comparison with literature value. How? • As temp.î HmÎ ie. Hm T •  Hm = Cp,m  T where Cp,m is the molar • heat capacity at constant pressure.

  41. Cp,m = Hm/ T = J mol-1/ K • = J K-1 mol-1 •   HT2 =  HT1 +  Cp ( T2 - T1) • Kirchoff’s equation. • and •  Cp = n Cp(products)- nCp(reactants) • For a wide temperature range: Cp ∫ dT between T1 and T2. • Hence : qp = Cp( T2- T1) or H = Cp T and.

  42. qv = Cv ( T2 – T1) or CvT = U • ie. Cp = H / T ; Cv =U /T • For small changes: • Cp = dH / dT ; Cv = du / dT • For an ideal gas: H = U + p V • For I mol: dH/dT = dU/dT + R •  Cp = Cv + R • Cp / Cv = γ ( Greek gamma)

  43. Work done along isothermal paths • Reversible and Irreversible paths • ie T =0 ( isothermal) • pV = nRT= constant • Boyle’s Law : piVi =pf Vf • Can be shown on plot:

  44. pV diagram Pivi pV= nRT = constant Pfvf

  45. Work done = -( nRT)∫ dV/V = - nRT ln (Vf/Vi) • Equation is valid only if : piVi=pfVf and therefore: Vf/Vi = pi/pf and • Work done = -( nRT) ln (pi/pf) and follows the path shown.

  46. An irreversible path can be followed: Look at pV diagram again. pV diagram

  47. An Ideal or Perfect Gas • NB For an ideal gas, u = 0 Because: U  KE + PE  k T + PE (stored in bonds) Ideal gas has no interaction between molecules (no bonds broken or formed)

  48. Therefore u = 0 at T = 0 Also H = 0 since (pV) = 0 ie no work done This applies only for an ideal gas and NOT a chemical reaction.

  49. Calculation • eg. A system consisting of 1mole of perfectgas at 2 atm and 298 K is made to expand isothermally by suddenly reducing the pressure to 1 atm. Calculate the work done and the heat that flows in or out of the system.

  50. w = -pex V = pex(Vf -Vi) Vi = nRT/pi = 1 x 8.314 x (298)/202650 = 1.223 x 10-2 m3 Vf = 1 x 8.314 x 298/101325 = 2.445 x 10-2 m3 therefore, w = -pex (Vf- Vi) = -101325(2.445-1.223) x 10-2 = -1239 J U = q + w; for a perfect gas U = 0 therefore q = -w and q = -(-1239) = +1239 J