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 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.
It may be slow or fast.
A gas fills its container uniformly. It never spontaneously collects at one end of the container.
Heat flow always occur from hot to a cooler one. The reverse process never occurs spontaneously.
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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. Calculate ΔH, ΔE, q, and w for the process you described above.
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.
Calculate ΔH, ΔE, q, and w for the process you described above.
All are equal to zero.
All are equal to zero. Since it is a constant temperature process, H = 0 and E = 0. The gas is working against zero pressure (evacuated bulb) so w = 0. E = q + w, so q = 0.
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6. 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.
c) Given your answer to part b, what is the driving force for the process?
Entropy
Some students are probably aware of the concept of entropy. This is a good introduction to the concept.
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7. The Expansion of An Ideal Gas Into an Evacuated Bulb
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8. Entropy
Spontaneous processes is an increase in property called entropy (s).
The driving force for a spontaneous process is an increase in the entropy of the universe.
It is a measure of molecular randomness or disorder.
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9. Entropy
Entropy is a 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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10. The Microstates That Give a Particular Arrangement (State)

11. Each configuration that gives a particular arrangement is called microstate.
Which arrangement is most likely to occur.
Table 17.2 shows relative probabilities of finding all the molecules in the left bulb.

12. 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
Question 17.31, 32

13. 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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14. 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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15. ΔSuniv ΔSuniv = +; entropy of the universe increases
ΔSuniv = -; process is spontaneous in opposite direction
ΔSuniv = 0; process has no tendency to occur, and the system is at equilibrium.
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16. 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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17. CONCEPT CHECK!
Describe the following as spontaneous/non-spontaneous/cannot tell, and explain.
A reaction that is:
Exothermic and becomes more positionally random
Spontaneous
Exothermic and becomes less positionally random
Cannot tell
Endothermic and becomes more positionally random
Endothermic and becomes less positionally random
Not spontaneous
Explain how temperature affects your answers.
a) Spontaneous (both driving forces are favorable). An example is the combustion of a hydrocarbon.
b) Cannot tell (exothermic is favorable, positional randomness is not). An example is the freezing of water, which becomes spontaneous as the temperature of water is decreased.
c) Cannot tell (positional randomness is favorable, endothermic is not). An example is the vaporization of water, which becomes spontaneous as the temperature of water is increased..
d) Not spontaneous (both driving forces are unfavorable).
Questions "a" and "d" are not affected by temperature. Choices "b" and "c" are explained above.

18. ΔSsurr
There are two important characteristics of the entropy changes that occur in the surroundings.
The sign of ΔSsurr depends on the direction of the heat flow. At constant temperature , an exothermic process in the system causes the heat flow into the surrounding. It increases the random motions and the entropy of the surroundings. ΔSsurr is positive
The magnitude of ΔSsurr depends on the temperature. The transfer of a given quantity of energy as a heat produces much greater percent change in the randomness of the surrounding at a low temperature than it does at a high temperature. Thus depends directly on the quantity of heat transferred and inversely proportional on the temperature.
In other words, the tendency for the system to lower its energy becomes a more important driving force at lower temperatures.
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19. ΔSsurr
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20. ΔSsurr
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21. ΔSsurr
We can express Δssurr in terms of the change in enthalpy ΔH for a process occurring at constant pressure. Heat flow (constant P) = change in enthalpy = ΔH Questions 17.33, 17.34
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23. Free energy (G)
Free energy (another thermodynamic term) is also related to spontaneity in dealing with temperature dependence of spontaneity.

24. Free Energy (G)
A process carried out at constant T and P is spontaneous if the ΔG is negative.
Negative (-ΔG) means positive (+Δsuniv).
So we have two functions that can be used to predict spontaneity.
i) Entropy of universe, which apply to all processes.
Ii) Free energy which can be used for processes carried out at constant temperature and pressure.
Question 17.37
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25. Free Energy (G) ΔG = ΔH – TΔS (at constant T and P)
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26. 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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27. EXERCISE!
The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.
ΔS = 132 J/K·mol
ΔSsurr = -132 J/K·mol
ΔG = 0 kJ/mol
S = 132 J/K·mol
Ssurr = -132 J/k·mol
G = 0 kJ/mol
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28. Spontaneous Reactions
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29. Effect of ΔH and ΔS on Spontaneity
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30. CONCEPT CHECK!
Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant.
Predict the signs of:
ΔH ΔSsurr ΔS ΔSuniv
Explain.
Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written.
Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive. –
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31. Fewer molecule means fewer possible configurations.
Entropy changes in the system are determined by positional probability.
Fewer molecule means fewer possible configurations.
Gases molecules generally possess more entropy than solid.
So whenever the products contains more moles of gases than reactant the entropy change is probably positive.
Question 17.41, 42

32. Third Law of Thermodynamics: In thermodynamics change in a certain function is usually important.
The change in enthalpy determines if the reaction is exothermic or endothermic at constant temperature.
The change in free energy determines if a process is spontaneous at constant temperature and pressure.
A perfect crystal represents the lowest possible entropy.
Law: The entropy of a perfect crystal at 0 K is zero.
The entropy of a substance increases with temperature.
Since S is zero for a perfect crystal at 0 K, the entropy value for a substance at a particular temperature can be calculated by knowing the temperature dependence entropy.
The standard entropy values (S0) of many common substances at 298 K and 1 atm. are listed in appendix 4.
Because the entropy is a state function of the system , the entropy change can be calculated by the following equation.
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33. 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.
Question 17.45,46
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
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34. 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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35. Standard Free Energy Change (ΔG°)
For a chemical reaction we are interested in the 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.
It is important to note that the ΔG° for a reaction is not measure directly.
ΔG° values for several reactions allow us to compare the relative tendency of these reactions . More negative the value of ΔG°, the further a reaction will go to the right to reach the equilibrium.
Question:17.53, 57, 59
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
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36. 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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37. 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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38. Free Energy and Pressure
G = G° + RT ln(P) or ΔG = ΔG° + RT ln(Q)
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39. Sketch graphs of: G vs. P H vs. P
CONCEPT CHECK!
Sketch graphs of:
G vs. P
H vs. P
ln(K) vs. 1/T (for both endothermic and exothermic cases)
G vs. P: A natural log graph (levels off as P increases).
H vs. P: no relationship (slope of zero).
lnK vs 1/T: endothermic - straight line, negative slope
exothermic - straight line, positive slope
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40. 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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41. 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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42. Change in Free Energy to Reach Equilibrium
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44. Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.
wmax = ΔG
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45. All real processes are irreversible.
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
All real processes are irreversible.
First law: You can’t win, you can only break even.
Second law: You can’t break even.
As we use energy, we degrade its usefulness.
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