1. Concerned with the study of transformation of energy:
THERMODYNAMICS
Concerned with the study of transformation of energy:
Heat  work
LECTURER
DR. GABRIEL HAREWOOD
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2. Zeroth Law of Thermodynamics
The zeroth law states that if bodies A and B are separately in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other. OR
If A and B are in thermodynamic equilibrium, and B and C are in thermodynamic equilibrium, then A and C are also in thermodynamic equilibrium.

3. Thermodynamics: the 1st law
Spectroscopy handout
Thermodynamics: the 1st 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
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4. System and Surroundings
Spectroscopy handout
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
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5. Spectroscopy handout
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 from the surrounding
Exothermic: releases heat to the surrounding
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6. WORK 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
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7. Expansion Work
Expansion against zero external pressure (free expansion)
w = -pex.DV = 0 (external pressure = 0)
Reversible isothermal expansion
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
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8. 1st Law of Thermodynamics
Spectroscopy handout
1st 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?
INTERNAL ENERGY is a State Function
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Note: remind that deltaU in isothermal reversible expansion is now zero, as w is maximum, but so it heat absorbed
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 = qV
How do we measure heat?
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9. Bomb calorimetry Heat Capacity
Spectroscopy handout
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 DU.
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)
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10. Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it U or E.
2009, Prentice-Hall, Inc.

11. Spectroscopy handout
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12. 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)
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13. 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
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14. 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
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15. 3. Work (W) w = force × distance moved in direction of force i. e
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
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16. 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
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17. 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
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18. 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
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19. NB: qp  ∆U why. Only equal if no work is done i. e. ∆V = 0 i. e
NB: qp  ∆U why? Only equal if no work is done i.e. ∆V = 0 i.e. qv = ∆U
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20. Example: energy diagram
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21. WORK FOR A REVERSIBLE SYSTEM
Since work done by system = pex∆V (Vf > Vi)
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
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22. 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)
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23. i.e. V  0 (expansion work).
Enthalpy, H
Most reactions take place in an open vessel at constant pressure, pex. Volume can change during the reaction.
Enthalpy is the internal energy plus the product of pressure and volume.
H = U + pV
i.e. V  0 (expansion work).
Definition: H = qp i.e. heat supplied to the system at constant pressure.
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24. Spectroscopy handout
Enthalpy
Most reactions we investigate occur under conditions of constant PRESSURE (not Volume)
ENTHALPY: Heat of reaction at constant pressure!
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25. Use a “coffee-cup” calorimeter to measure it
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?
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26. 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
FROM THE DESK OF GRH
NB: Internal energy and Enthalpy are STATE FUNCTIONS

27. 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
Since work done = - pex V
H = (- pex V + q) +p V (pex= p)
H = ( -p V + q) + p V = q
 H = qp
FROM THE DESK OF GRH

28. 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
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29. STATE FUNCTION AND PATH FUNCTION
STATE Function: It depends only on the present state of the system, not on the path by which the system arrived at that state. 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).
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30. 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.
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31. C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l)
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.
FROM THE DESK OF GRH

32. H = U + (pV) = U + pi V + Vi p + p V NB: p = 1 bar, i. e
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
FROM THE DESK OF GRH
Endothermic reaction (q>0) results in an increase in enthalpy (DH>0)
Exothermic reaction (q<0) results in an decrease in enthalpy (DH<0)

33. 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(g) + 2 O2(g) → CO2(g) + 2H2O(l), H = kJmol-1
 U = – ((1 – 3)(8.314) 298)/ 1000
= – (-2)(8.314)(0.298) = kJ mol-1
FROM THE DESK OF GRH

34. 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)
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35. 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.
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36. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

37. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

38. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

39. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these equations is:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

40. Calculation of H We can use Hess’s law in this way:
H = nHf°products – mHf° reactants
where n and m are the stoichiometric coefficients.

41. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] – [( kJ) + (0 kJ)] = ( kJ) – ( kJ) = kJ

42. rxnHo = nDfHom(products) - nDfHom(reactants)
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
rxnHo = nDfHom(products) - nDfHom(reactants)
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43. Bond Energies eg. Calculate the 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
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
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44. Enthalpy Changes and Bond Energies
Energy is absorbed when bonds break. The energy required to break the bonds is absorbed from the surroundings.
If there was some way to figure out how much energy a single bond absorbed when broken, the enthalpy of reaction could be estimated by subtracting the bond energies for bonds formed from the total bond energies for bonds broken.
O2(g) 2O(g) H°=490.4 kJ
H2(g) 2H(g) H° =431.2 kJ
H2O(g)2H(g) + O(g) H°=915.6 kJ
We can estimate the bond enthalpies of O=O, H-H, and O-H as kJ/mol, kJ/mol, and kJ/mol, respectively.
2H2(g) + O2(g)  2H2O(g) H°= ?
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45. A calculation based on enthalpies of formation gave H° = -483.7 kJ
2H2(g) + O2(g)  2H2O(g)
moles of bonds broken
Energy absorbed
moles of bonds formed
Energy released
kJ each 862.4kJ kJ each kJ
kJ each 490.4kJ
___________________________________________
1352.7kJ kJ
H°= kJ = kJ.
(Remember that the minus sign means "energy released", so you add the bond energies for broken bonds and subtract energies for bonds formed to get the total energy.)
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A calculation based on enthalpies of formation gave H° = kJ
Bonds in a molecule influence each other, which means that bond energies aren't really additive. An O-H bond in a water molecule has a slightly different energy than an O-H bond in H2O2, because it's in a slightly different environment.
Reaction enthalpies calculated from bond energies are very rough approximations!

46. FROM THE DESK OF GRH

47. 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. increase Hm increase
ie. Hm  T
 Hm = Cp,m  T where Cp,m is the molar
heat capacity at constant pressure.
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48. Variation of DrH° with T
Kirchoff’s Law
DrH°(T2) = DrH°(T1) + DrCp°(T2-T1)
DrCp° = S nCp,m°(products) - S nCp,m°(reactants)
FROM THE DESK OF GRH
If heat capacity is temperature dependent, we need to integrate over the temperature range

49. Work done along isothermal paths
Reversible and Irreversible paths
ie T = o ( isothermal)
pV = nRT= constant
Boyle’s Law : piVi =pf Vf
Can be shown on plot:
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50. pV Diagram Pivi pV= nRT = constant Pfvf REVERSIBLE
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Pfvf

51. 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.
FROM THE DESK OF GRH

52. 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) 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.
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53. therefore, w = -pex (Vf- Vi) = -101325(2.445-1.223) x 10-2 = -1239 J
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.
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
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54. Relation between heat capacities
Cp / Cv = γ ( Greek gamma)
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55. Work done along adiabatic path
ie q = 0 , no heat enters or leaves the system.
Since U = q + w and q =0
U = w
No heat leaves the system if no work is done.
For a perfect gas expanding adiabatically, a decrease in T is expected because work is done.
This implies the the kinetic energy of the molecules decrease as work is done. So the T falls.

56. 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.
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57. 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
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2nd Law: The Entropy of an isolated system increases in the course of spontaneous change
DStot>0

58. The Second Law (cont.) the system and surroundings to predict if a
Three possibilities:
If DSuniv > 0…..process is spontaneous
If DSuniv < 0…..process is spontaneous in opposite direction.
If DSuniv = 0….equilibrium
• Here’s the catch: We need to know DS for both
the system and surroundings to predict if a
reaction will be spontaneous!

59. 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 = - wrev/T = nRln (Vf/Vi)
ie S = n R ln (Vf/Vi)
Implies that S ≠ 0 ( strange!)
Must consider the surroundings.
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60. Heat transferred = qP,surr = - qP,system= -DHsys
The Second Law (cont.)
Consider a reaction driven by heat flow from the surroundings at constant P.
Exothermic Process: DSsurr = heat/T
Endothermic Process: DSsurr = -heat/T
Heat transferred = qP,surr = - qP,system= -DHsys

61. DSsurr = -DH/T = -778 kJ/298K = -2.6 kJ/K
Example
For the following reaction at 298 K:
Sb4O6(s) + 6C(s) Sb(s) + 6CO2(g) DH = 778 kJ
What is DSsurr?
DSsurr = -DH/T = -778 kJ/298K = -2.6 kJ/K

62. 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.
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63. S = Cv ∫ dT /T between Ti and Tf S = Cv ln ( Tf/ Ti )
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
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64. The Third Law
Recall, in determining enthalpies we had standard state values to use. Does the same thing exist for entropy?
The third law: The entropy of a perfect crystal at 0K is zero.
The third law provides the reference state for use in calculating absolute entropies.

65. What is a Perfect Crystal?
Perfect crystal at 0 K
Crystal deforms at T > 0 K

66. 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
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67. Calculation
eg. 2H2 (g) + O2( g) → 2H2O( l ), H = kJ mol-1 The 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
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68. 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.
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69. ie. S > 0, G < 0 for spontaneous change ;
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)
FROM THE DESK OF GRH

70. 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 dw = -pdV  dG = Vdp - SdT
FROM THE DESK OF GRH

71. For chemical Reactions:
G = n G (products) -  n G (reactants)
and
Grxn = Hrxn - T Srxn
FROM THE DESK OF GRH

72. 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.
FROM THE DESK OF GRH

73. Equilibrium diagram
FROM THE DESK OF GRH

74. For non-spontaneous rxn. GB > GA, G is + ve
FROM THE DESK OF GRH

75. Signs of Thermodynamic Values
Negative
Positive
Enthalpy (ΔH)
Exothermic
Endothermic
Entropy (ΔS)
Less disorder
More disorder
Gibbs Free Energy (ΔG)
Spontaneous
Not spontaneous

76. 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.
FROM THE DESK OF GRH

77. 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 I2(s)= 0
FROM THE DESK OF GRH

78. 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
FROM THE DESK OF GRH

79. 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
FROM THE DESK OF GRH

80. 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.
FROM THE DESK OF GRH

81. Effect of temperature on K (van’t Hoff equation)
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 )
FROM THE DESK OF GRH

82. Summary 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
FROM THE DESK OF GRH