1. Copyright © Tyna L. Heise 2001 - 2002
Chapter Nineteen
Copyright © Tyna L. Heise
All Rights Reserved

2. Spontaneous Processes
Understanding and designing chemical reactions:
How rapidly does the reaction proceed?
- reaction rates
- controlled by a factor related to energy
- the lower the activation energy, the faster the reaction proceeds

3. Spontaneous Processes
How far toward completion will the reaction go?
- equilibrium constants
- depends on rates of forward and reverse reactions
- equilibrium should also be dependent on energy in some way due to dependence on reaction rates

4. Spontaneous Processes
Chemical thermodynamics is the relationship between equilibrium and energy.
First Law of Thermodynamics: for a reaction that occurs at constant pressure, the enthalpy change equals the heat transferred between the system and its surroundings Energy is conserved!!
* Enthalpy is important in helping us determine if a reaction will proceed!

5. Spontaneous Processes
energy is neither created nor destroyed in any process, energy can only be transferred or converted from one form to another
DE = q + w
a spontaneous process occurs without any outside intervention as energy is conserved

6. Spontaneous Processes

7. Spontaneous Processes
temperature is going to effect the spontaneity of a process
if discussing a phase change, at the temperature of the phase change, the phases compete for spontaniety, and neither is said to win over the other

8. Spontaneous Processes
Sample exercise: Under 1 atm pressure, CO2(s) sublimes at -78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at -100°C?

9. Spontaneous Processes
Sample exercise: Under 1 atm pressure, CO2(s) sublimes at -78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at -100°C?
*No, if the temp had been higher it would change phase spontaneously, but lower than the sublimation point favors the reverse, so the solid remains a solid.

10. Spontaneous Processes
Reversible and Irreversible Processes:
State functions: define a state and do not depend upon the pathway
temperature
internal energy
enthalpy

11. Spontaneous Processes
Reversible and Irreversible Processes:
Reversible processes is a unique way for a system to change its state, than go back to its state by following the exact same path but in the opposite direction
phase changes at constant temp
only one specific value of q (heat)
system in equilibrium

12. Spontaneous Processes
Reversible and Irreversible Processes:
Irreversible processes cannot be simply restored to their original state using the same path, it may be forced to go back, but by a different pathway
phase changes at different temps
two q values need to be established qforward and qreverse
any spontaneous reaction

13. Spontaneous Processes
Thermodynamics can tell us
direction of reaction
extent of reaction
NOT
speed of reaction

14. Entropy and the 2nd Law Spontaniety depends upon two factors
Enthalpy (DH): heat of reaction
exothermic normally spontaneous
endothermic normally NOT spontaneous
Entropy (DS): disorder of the system
natural law indicates reactions go in the direction that leads to more disorder

15. Entropy and the 2nd Law The Spontaneous Expansion of a Gas:
When the stopcock is opened, the gas will spontaneously flow to fill the empty half, but it WILL NOT flow backward without work being done on the system.

16. Entropy and the 2nd Law The Spontaneous Expansion of a Gas:
Gas expands because of the tendency for the molecules to ‘spread out’ among the different arrangements that they can take.

17. Entropy and the 2nd Law Entropy: measurement of randomness or chaos
melting ice
dissolving salts
The more disordered or random a system, the larger its entropy

18. Entropy and the 2nd Law
Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system:
a) CO2(s)  CO2(g)

19. Entropy and the 2nd Law
Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system:
a) CO2(s)  CO2(g)
Entropy increases

20. Entropy and the 2nd Law
Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system:
b) CaO(s) + CO2(g)  CaCO3(s)

21. Entropy and the 2nd Law
Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system:
b) CaO(s) + CO2(g)  CaCO3(s)
Entropy decreases

22. Entropy and the 2nd Law Entropy: measurement of randomness or chaos
state function
for a process that occurs at constant temperature, the entropy change is dependent on the heat transferred during the reverse of the reaction (qrev)
DS = qrev/T

23. Entropy and the 2nd Law
Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?

24. Entropy and the 2nd Law
Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?
DS = qrev/T

25. Entropy and the 2nd Law
Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?
DS = qrev/T qrev = DHvap
= 38.56kJ/mol
T = K

26. Entropy and the 2nd Law
Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?
DS = qrev/T kJ 1000 J 1 mol mol 1 kJ g
T = K

27. Entropy and the 2nd Law
Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?
DS = qrev/T J/g * 25.8 g = J
T = K

28. Entropy and the 2nd Law
Sample exercise: The normal boiling point of ethanol, C2H5OH, is 78.3°C, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?
DS = qrev/T = J/ K
= J/K

29. Entropy and the 2nd Law
Second law of Thermodynamics: In any reversible process, DSuniv = 0. In any irreversible reaction, DSuniv >0.
DSuniv = DSsys + DSsurr
In an isolated system, just the entropy of the system is considered.

30. Molecular Interpretation
On the microscopic level, the number of gas molecules can be directly related to the amount of entropy in a system.
The more gas molecules present, the higher the entropy value
a phase change that increases the number of gas molecules would increase entropy
a phase change that decreases the number of gas molecules would decrease entropy

31. Molecular Interpretation
Three moles of gas combine to form two moles of gas, thus decreasing the number of molecules.

32. Molecular Interpretation
Degrees of freedom
creating new bonds decreases the freedom of movement atoms may have had.
3 degrees
motion in one direction, translational movement
vibrational movement
spinning, rotational movement

33. Molecular Interpretation
Figure 19.12

34. Molecular Interpretation
Sample exercise: Choose the substance with the greatest entropy in each case:
1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm.

35. Molecular Interpretation
Sample exercise: Choose the substance with the greatest entropy in each case:
1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C.

36. Molecular Interpretation
Sample exercise: Choose the substance with the greatest entropy in each case:
1 mol of H2(g) at STP or 1 mol of SO2(g) at STP.

37. Molecular Interpretation
Sample exercise: Choose the substance with the greatest entropy in each case:
1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.

38. Molecular Interpretation
Sample exercise: Predict whether is DS is positive or negative in each of the following processes:
HCl(g) + NH3(g)  NH4Cl(s)

39. Molecular Interpretation
Sample exercise: Predict whether is DS is positive or negative in each of the following processes:
2SO2(g) + O2(g)  2SO3(s)

40. Molecular Interpretation
Sample exercise: Predict whether is DS is positive or negative in each of the following processes:
cooling of nitrogen gas from 20°C to -50°C

41. Calculation of Entropy Changes
Entropy Calculations:
no easy method for measuring entropy
experimental measurements on the variation of heat capacity with temperature can give a value known as absolute entropy
zero point of reference for perfect crystalline solids
tabulated as molar quantities, J/mol-K

42. Calculation of Entropy Changes
Entropy differs from enthalpy
standard molar entropies are not 0
entropies of gases are greater than those of liquids and solids
entropies increase with molar mass
entropies increase with number of atoms in formula
DS° = nS(products) - mS(reactants)

43. Gibbs Free Energy
Spontaneity involves both thermodynamic concepts: entropy and enthalpy
DG = DH – TDS
G = Gibbs Free Energy
H = Enthalpy
T = Temperature (K)
S = Entropy

44. Gibbs Free Energy Gibbs Free Enegry
If DG is negative, the reaction is spontaneous and proceeds in the forward direction
If DG is zero, the reaction is at equilibrium
If DG is positive, the reaction is nonspontaneous and proceeds in the reverse direction

45. Gibbs Free Energy
Gibbs Free Enegry
State function
Table 19.3

46. Gibbs Free Energy
Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane:
CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g)

47. Gibbs Free Energy
Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane:
CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g)
( )

48. Gibbs Free Energy
Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane:
CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g)
( )
DG° = products - reactants

49. Gibbs Free Energy
Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane:
CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g)
( )
DG° = products – reactants
( ( )) – ( )

50. Gibbs Free Energy
Sample exercise: By using data from Appendix C, Calculate DG° at 298 K for the combustion of methane:
CH4(g) + 2O2(g) ®CO2(g) + 2H2O(g)
( )
DG° = products – reactants
( ( )) – ( )
-800.7

51. Gibbs Free Energy
Sample exercise: Consider the combustion of propane to form CO2(g) and H2O(g) at 298K. Would you expect DG° to be more negative or less negative than DH°?

52. Gibbs Free Energy
Sample exercise: Consider the combustion of propane to form CO2(g) and H2O(g) at 298K. Would you expect DG° to be more negative or less negative than DH°?
more negative, using Gibbs Free Enegry formula, more moles of gas being produced would be increasing entropy, +DS,

53. Free Energy and Temperature
Table 19.4

54. Free Energy and Temperature
Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate DH° and DS° at 298 K for the following reaction:
2SO2(g) + O2(g) ® 2SO3(g)

55. Free Energy and Temperature
Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate DH° and DS° at 298 K for the following reaction:
2SO2(g) + O2(g) ® 2SO3(g)
2(-296.9) (-395.2)
DH = 2(-395.2) - 2(-296.9) = kJ

56. Free Energy and Temperature
Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate DH° and DS° at 298 K for the following reaction:
2SO2(g) + O2(g) ® 2SO3(g)
2(248.5) (256.2)
DS = 2(256.2) – (2(248.5)+205.0) = J

57. Free Energy and Temperature
Sample exercise: Using the values obtained, estimate DG° at 400 K
DG = DH – TDS

58. Free Energy and Temperature
Sample exercise: Using the values obtained, estimate DG° at 400 K
DG = DH – TDS
= kJ – 400( kJ) =

59. Free Energy and the Equilibrium Constant
2 other important ways free energy is a powerful tool
Tabulate free energy under nonstandard conditions
Directly relate free energy to equilibrium constants

60. Free Energy and the Equilibrium Constant
Tabulate free energy under nonstandard conditions
DG = DG° + RTlnQ
R = J/mol-K
T = temperature K
Q = reaction quotient

61. Free Energy and the Equilibrium Constant
Sample exercise: Calculate DG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.
DG = DG° + RTlnQ

62. Free Energy and the Equilibrium Constant
Sample exercise: Calculate DG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.
DG = DG° + RTlnQ
2(-16.6) (298)ln(2.02/0.50(0.753))
-26.0 kJ

63. Free Energy and the Equilibrium Constant
At equilibrium, DG is equal to 0 so…
DG° = -RTlnKeq
DG° negative: K > 1
DG° zero : K = 1
DG° positive: K < 1
Keq = e-DG°/RT