2. Thermodynamics vs. Kinetics
Domain of Kinetics
Rate of a reaction depends on the pathway from reactants to products.
Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.
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3. Spontaneous Processes and Entropy
Thermodynamics lets us predict the direction in which a process will occur but gives no information about the speed of the process.
A spontaneous process is one that occurs without outside intervention.
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4. What should happen to the gas when you open the valve?
CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
What should happen to the gas when you open the valve?
The gas should spread evenly throughout the two bulbs.
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5. The Expansion of An Ideal Gas Into an Evacuated Bulb
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6. Entropy
The driving force for a spontaneous process is an increase in the entropy of the universe.
A measure of molecular randomness or disorder.
A measure of how many micro states can exist.
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7. Entropy
Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state.
Nature spontaneously proceeds toward the states that have the highest probabilities of existing.
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8. The Microstates That Give a Particular Arrangement (State)

9. Positional Entropy
A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system.
Therefore: Ssolid < Sliquid << Sgas

10. Predict the sign of ΔS for each of the following, and explain:
CONCEPT CHECK!
Predict the sign of ΔS for each of the following, and explain:
The evaporation of alcohol
The freezing of water
Compressing an ideal gas at constant temperature
Heating an ideal gas at constant pressure
Dissolving NaCl in water + –
a) + (a liquid is turning into a gas)
b) - (more order in a solid than a liquid)
c) - (the volume of the container is decreasing)
d) + (the volume of the container is increasing)
e) + (there is less order as the salt dissociates and spreads throughout the water)
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11. Second Law of Thermodynamics
In any spontaneous process there is always an increase in the entropy of the universe.
The entropy of the universe is increasing.
The total energy of the universe is constant, but the entropy is increasing.
Suniverse = ΔSsystem + ΔSsurroundings
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12. ΔSsurr ΔSsurr = +; entropy of the universe increases
ΔSsurr = -; process is spontaneous in opposite direction
ΔSsurr = 0; process has no tendency to occur
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13. Entropy Changes in the Surroundings (ΔSsurr)
ΔSsurr is determined by flow of energy as heat
Exothermic process increases ΔSsurr
Important driving force for spontaneity
Endothermic process decreases ΔSsurr
Impact of transfer of energy as heat to or from the surroundings is greater at lower temperatures
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14. CONCEPT CHECK!
For the process A(l) A(s), which direction involves an increase in energy randomness? Positional randomness? Explain your answer. As temperature increases/decreases (answer for both), which takes precedence? Why? At what temperature is there a balance between energy randomness and positional randomness?
Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid).
As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored.
There is a balance at the melting point.
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15. ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow.
The magnitude of ΔSsurr depends on the temperature.
$50 to a millionaire or a high school student
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16. ΔSsurr
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17. ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH
If the reaction is exothermic:
ΔH has a negative sign
ΔSsurr is positive since heat flows into the surroundings
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19. Free Energy (G)
A process (at constant T and P) is spontaneous in the direction in which the free energy decreases.
Negative ΔG means positive ΔSuniv.
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20. All quantities refer to the system
Free Energy (G)
ΔG = ΔH – TΔS (at constant T and P)
All quantities refer to the system
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21. CONCEPT CHECK!
A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔSsurr ΔG Explain your answers. – +
As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium.
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22. Effect of ΔH and ΔS on Spontaneity
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23. Interactive Example 17.5 - Free Energy and Spontaneity
At what temperatures is the following process spontaneous at 1 atm?
What is the normal boiling point of liquid Br2?

24. Interactive Example 17.5 - Solution
The vaporization process will be spontaneous at all temperatures where ΔG°is negative
Note that ΔS°favors the vaporization process because of the increase in positional entropy, and ΔH°favors the opposite process, which is exothermic
These opposite tendencies will exactly balance at the boiling point of liquid Br2, since at this temperature liquid and gaseous Br2 are in equilibrium (ΔG°= 0)

25. Interactive Example 17.5 - Solution (Continued 1)
We can find this temperature by setting ΔG°= 0 in the following equation:

26. Interactive Example 17.5 - Solution (Continued 2)
At temperatures above 333 K, TΔS°has a larger magnitude than ΔH°, and ΔG°is negative
Above 333 K, the vaporization process is spontaneous
The opposite process occurs spontaneously below this temperature
At 333 K, liquid and gaseous Br2 coexist in equilibrium

27. Interactive Example 17.5 - Solution (Continued 3)
Summary of observations (the pressure is 1 atm in each case)
T > 333 K
The term ΔS°controls, and the increase in entropy when liquid Br2 is vaporized is dominant
T < 333 K
The process is spontaneous in the direction in which it is exothermic, and the term ΔH°controls

28. Interactive Example 17.5 - Solution (Continued 4)
T = 333 K
The opposing driving forces are just balanced (ΔH°= 0), and the liquid and gaseous phases of bromine coexist
This is the normal boiling point

29. Entropy Changes and Chemical Reactions
Positional probability determines the changes that occur in a chemical system
Fewer the molecules, fewer the possible configurations
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30. Figure 17.6 - Entropy of Water
H2O molecule can vibrate and rotate in several ways
Freedom of motion leads to a higher entropy for water 30

31. Entropy Changes in Reactions That Involve Gaseous Molecules
Change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products
If the number of product molecules is greater than the number of reactant molecules:
Positional entropy increases
ΔS is positive
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32. Interactive Example 17.6 - Predicting the Sign of ΔS°
Predict the sign of ΔS°for each of the following reactions
Thermal decomposition of solid calcium carbonate
Oxidation of SO2 in air

33. Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero.
The entropy of a substance increases with temperature.
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34. Standard Entropy Values (S°)
Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
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35. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information:
EXERCISE!
Calculate ΔS° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
S° (J/K·mol)
Na(s)
H2O(l)
NaOH(aq)
H2(g)
ΔS°= –11 J/K
[2(50) + 131] – [2(51) + 2(70)] = –11 J/K
ΔS°= –11 J/K
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36. Standard Free Energy Change (ΔG°)
The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
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37. CONCEPT CHECK!
A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS° ΔG° Explain.
The reaction is exothermic, more ordered, and spontaneous.
Thus, the sign of H is negative; S is negative; and G is negative.
– – –
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38. Consider the following system at equilibrium at 25°C.
CONCEPT CHECK!
Consider the following system at equilibrium at 25°C.
PCl3(g) + Cl2(g) PCl5(g)
ΔG° = −92.50 kJ
What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain.
The ratio will decrease.
S is negative (unfavorable) yet the reaction is spontaneous (G is negative). Thus, H must be negative (exothermic, favorable). Thus, as the temperature is increased, the reaction proceeds to the left, decreasing the ratio of partial pressure of PCl5 to the partial pressure of PCl3.
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39. Interactive Example 17.11 - Calculating ΔG°
Methanol is a high-octane fuel used in high-performance racing engines
Calculate ΔG °for the following reaction:
The following energies of formation are provided: 39

40. Interactive Example 17.11 - Solution
Use the following equation:

41. Interactive Example 17.11 - Solution (Continued)
The large magnitude and the negative sign of ΔG°indicate that this reaction is very favourable thermodynamically

42. Free Energy and Pressure
System under constant P and T proceeds spontaneously in the direction that lowers its free energy
Free energy of a reaction system changes as the reaction proceeds
Dependent on the pressure of a gas or on the concentration of species in solution
Equilibrium - Point where free energy value is at its lowest
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43. Free Energy and Pressure (Continued 1)
For ideal gases:
Enthalpy is not pressure-dependent
Entropy depends on pressure due to its dependence on volume
At a given temperature for 1 mole of ideal gas:
Slarge volume > Ssmall volume
Or,
Slow pressure > Shigh pressure
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44. Free Energy and Pressure (Continued 2)
G° - Free energy of a gas at 1 atm
G - Free energy of the gas at a pressure of P atm
R - Universal gas constant
T - Temperature in Kelvin
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45. Free Energy and Pressure (Continued 3)
Q - Reaction quotient
T - Temperature in Kelvin
R - Universal gas constant ( J/K·mol)
ΔG°- Free energy change at 1 atm
ΔG - Free energy change at specified pressures
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46. Interactive Example 17.13 - Calculating ΔG°
One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases
Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol

47. Interactive Example 17.13 - Solution
To calculate ΔG for this process, use the following equation:
First compute ΔGₒ from standard free energies of formation

48. Interactive Example 17.13 - Solution (Continued 1)
One might call this the value of ΔG°for one round of the reaction or for 1 mole of the reaction
Thus, the ΔG°value might better be written as –2.9×104 J/mol of reaction, or –2.9×104 J/mol rxn
Use this value to calculate the value of ΔG

49. Interactive Example 17.13 - Solution (Continued 2)
ΔG°= – 2.9×104 J/mol rxn
R = J/K·mol
T = = 298 K
Note that the pure liquid methanol is not included in the calculation of Q

50. Interactive Example 17.13 - Solution (Continued 3)

51. Interactive Example 17.13 - Solution (Continued 4)
Note that ΔG is significantly more negative than ΔG°, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm
This result can be expected from Le Châtelier’s principle

52. The Meaning of ΔG for a Chemical Reaction
A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.
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53. The equilibrium point occurs at the lowest value of free energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
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54. Change in Free Energy to Reach Equilibrium
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56. Interactive Example 17.15 - Free Energy and Equilibrium II56

57. Interactive Example 17.15 - Solution
Calculate ΔG° from ΔG° = ΔH°– TΔS°

58. Interactive Example 17.15 - Solution (Continued 1)
Therefore,
ΔG° = ΔH°– TΔS°
= (– 1.652×106 J) – (298 K) (– 543 J/K)
= – ×106 J

59. Interactive Example 17.15 - Solution (Continued 2)
Therefore,
K = e601
This is a very large equilibrium constant
The rusting of iron is clearly very favourable from a thermodynamic point of view

60. Reversible and Irreversible Processes
Reversible process: Universe is exactly the same as it was before a cyclic process
Irreversible process: Universe is different after a cyclic process
All real processes are irreversible
Characteristics of a real cyclic process
Work is changed to heat
Entropy of the universe increases
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