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

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

5. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) - heat given off.
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. w = force × distance moved in direction of force
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 = Pa), 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 × × 298)/ = m3
3 moles of gas:
Vf = (3 × × 298)/ = m3
work done = -pex × (Vf – Vi)
= ( – ) = J

14. NB: work done = - pex (nfRT/pex – niRT/pex)
= (nf – ni) RT
Work done = -∆ngasRT
i.e. work done = - (3 – 6) × × 298 = J
Can also calculate ∆U
∆U = q +w q = kJ
w = J = 7.43 kJ
 ∆U = = 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. Hf – Hi = Uf – Ui +p(Vf – Vi)
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 = kJmol-1.
Since substances are in the pure form then can write
H = 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 = kJmol-1
Calculate U for the following reaction:
CH4(l) + 2 O2(g) → CO2(g) + 2H2O(l), H = kJmol-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 = – ((1 – 2)(8.314) 298)/ 1000
= – (-1)(8.314)(0.298) = 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) = kJ mol-1
Hf of O2(g) = 0 (reference state)
Hf of CO2(g) = kJ mol-1
Hf of H2O(l) = kJ mol-1
Hrxn = n H (products)- n H(reactants)

34. Hf(products) = 3  (- 393.5) + 4  (- 285.8)
= = kJ mol-1
Hf(reactants) =  0 = kJ mol-1
Hrxn = – ( ) = kJ mol = 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 = kJmol-1…………………………….(1)
CO (g) +1/2 O2 (g) →CO 2(g), Hcomb = 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= –
(-283.0) = 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)
= ( 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. pV diagram
An irreversible path can be followed: Look at pV diagram again.

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 perfect gas 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 x (298)/202650
= x 10-2 m3
Vf = 1 x x 298/ = x 10-2 m3
therefore, w = -pex (Vf- Vi)
= ( ) x 10-2 = J
U = q + w; for a perfect gas U = 0
therefore q = -w and
q = -(-1239) = J

51. Work done along adiabatic path
ie q = 0 , no heat enters or leaves the system.
Since U = q + w and q =0
U = w
When a gas expands adiabatically, it cools.
Can show that: pVγ = constant, where ( Cp/Cv =γ )
and: piViγ = pfVfγ and since:
-p dV = Cv dT

52. Work done for adiabatic path = Cv (Tf- Ti)
For n mol of gas: w = n Cv (Tf –Ti)
Since piViγ = pfVf γ
piViγ/Ti = pfVf γ/ Tf
 Tf = Ti(Vi/Vf)γ-1
 w =n Cv{ ( Ti( Vi/ Vf)γ-1 – Ti}
An adiabatic pathway is much steeper than pV = constant pathway.

53. Summary
piVi = pfVf for both reversible and irreversible
Isothermal processes.
For ideal gas: For T =0, U = 0, and H=0
For reversible adiabatic ideal gas processes:
q=0 , pVγ = constant and
Work done = n Cv{ ( Ti( Vi/ Vf)γ-1 – Ti}
piViγ = pfVfγ for both reversible and irreversible adiabatic ideal gas.

54. 2nd Law of Thermodynamics
Introduce entropy, S (state function) to explain spontaneous
change ie have a natural tendency to occur- the apparent
driving force of spontaneous change is the tendency of energy
and matter to become disordered. That is, S increases on disordering.
2nd law – the entropy of the universe tends to increase.

55. Entropy S = qrev /T ( J K-1) at equilibrium
Sisolated system > 0 spontaneous change
Sisolated system < 0 non-spontaneous change
Sisolated system = 0 equilibrium

56. Properties of S If a perfect gas expands isothermally from
Vi to Vf then since U = q + w = 0
 q = -w ie
qrev = -wrev and
wrev = - nRT ln ( Vf/Vi)
At eqlb., S =qrev/T = - qrev/T = nRln (Vf/Vi)
ie S = n R ln (Vf/Vi)
Implies that S ≠ 0 ( strange!)
Must consider the surroundings.

57. Surroundings Stotal = Ssystem + Ssurroundings
At constant temperature surroundings give heat to the system to maintain temperature.
 surroundings is equal in magnitude to heat gained or loss but of opposite sign to make
S = 0 as required at eqlb.

58. Rem: dq = Cv dT and
dS = dqrev / T
 dS = Cv dT/ T and
S = Cv ∫ dT /T between Ti and Tf
S = Cv ln ( Tf/ Ti )
When Tf/ Ti > 1 , S is +ve
eg. L → G , S is +ve
S → L , S is +ve and since qp = H
Smelt = Hmelt / Tmelt and
Svap = Hvap / Tvap

59. Third Law of Thermodynamics
eg. Standard molar entropy, SmThe entropy of a perfectly crystalline substance is zero at T = 0
Sm/ J K-1 at 298 K
ice
water NB. Increasing disorder
water vapour 189
For Chemical Reactions:
Srxn =  n S (products) -  n S ( reactants)
eg. 2H2 (g) O2( g) → 2H2O( l ), H = kJ mol-1

60. Calculation
Ie surroundings take up + 572kJ mol-1 of heat
Srxn = 2S(H2Ol) - (2 S (H2g ) + S (O2g) )
= JK-1 mol-1 ( strange!! for a spontaneous reaction; for this S is + ve. ).
Why? Must consider S of the surroundings also.
S total = S system + S surroundings
S surroundings = + 572kJ mol-1/ 298K = x 103JK-1 mol-1
 S total =( JK-1mol-1) x 103 = 1.59 x 103 JK-1 mol-1
Hence for a spontaneous change, S > 0

61. Free Energy, G Is a state function. Energy to do useful work.
Properties
Since Stotal = Ssystem + Ssurroundings
Stotal = S - H/T at const. T&p
Multiply by -T and rearrange to give:
-TStotal = - T S + H and since G = - T Stotal
ie. G = H - T S
Hence for a spontaneous change: since S is + ve, G = -ve.

62. Free energy ie. S > 0, G < 0 for spontaneous change ;
at equilibrium, G = 0.
Can show that : (dG)T,p = dwrev ( maximum work)
 G = w (maximum)

63. Properties of G G = H - T S dG = dH – TdS – SdT H = U + pV
dH = dU + pdV + Vdp
Hence: dG = dU + pdV + Vdp – TdS – SdT
dG = - dw + dq + pdV + Vdp – TdS – SdT
 dG = Vdp - SdT

64. For chemical Reactions:
G = n G (products) -  n G (reactants)
and
Grxn = Hrxn - T Srxn

65. Relation between Grxn and position of equilibrium
Consider the reaction: A = B
Grxn = GB - GA
If GA> GB , Grxn is – ve ( spontaneous rxn)
At equilibrium, Grxn = 0.
ie. Not all A is converted into B; stops at equilibrium point.

66. Equilibrium diagram

67. For non-spontaneous rxn. GB > GA, G is + ve

68. Gas phase reactions Consider the reaction in the gas phase:
N2(g) + 3H2(g)→ 2NH3(g)
Q =( pNH3 / p)2 /( ( pN2/ p) (pH2/ p)3 ) where :
Q = rxn quotient ; p = partial pressure and p = standard pressure = 1 bar
Q is dimensionless because units of partial pressure cancelled by p .
At equilibrium:
Qeqlb = K = (( pNH3 / p)2 / ( pN2/ p) (pH2/ p)3 )eqlb

69. Activity ( effective concentration)
Define: aJ = pJ / p where a = activity or effective concentration.
For a perfect gas: aJ = pJ / p
For pure liquids and solids , aJ = 1
For solutions at low concentration: aJ = J mol dm-3
K = a2NH3 / aN2 a3H2
Generally for a reaction:
aA + bB → cC + dD
K = Qeqlb = ( acC adD / aaA abB ) eqlb = Equilibrium constant

70. Relation of G with K Can show that: Grxn = Grxn + RT ln K
At eqlb., Grxn = 0
 Grxn = - RT ln K
Hence can find K for any reaction from thermodynamic data.

71. Can also show that:
ln K = - G / RT
K = e - G / RT eg
H2 (g) + I2 (s) = 2HI (g) , Hf HI = kJ mol-1 at 298K; Hf H2 =0 ; Hf I 29(s)= 0

72. calculation Grxn = 2 x 1.7 = + 3.40 kJ mol-1
ln K = x 103 J mol-1 / J K-1 mol-1 x 298K =
ie. K = e – = 0.25
ie. p2 HI / pH2 p = ( rem. p = 1 bar; p 2 / p = p )
 p2 HI = pH2 x 0.25 bar

73. Example: relation between Kp and K
Consider the reaction:
N2 (g) + 3H2 (g) = 2NH3 (g)
Kp =( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3
and
K = [( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3] eqlb
 Kp = K (p)2 in this case. ( Rem: (p)2 / (p)4 = p -2

74. For K >> 1 ie products predominate at eqlb. ~ 103
K<< 1 ie reactants predominate at eqlb. ~ 10-3
K ~ 1 ie products and reactants in similar amounts.

75. Effect of temperature on K
Since  Grxn = - RT ln K = Hrxn - TSrxn
ln K = - Grxn / RT = - Hrxn/RT + Srxn/R
 ln K1 = - Grxn / RT1 = - Hrxn/RT1 + Srxn/R
ln K2 = - Grxn / RT2 = - Hrxn/ RT2 + Srxn/ R
ln K1 – ln K2 = - Hrxn / R ( 1/ T1 - 1/ T2 ) 0r
ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation