CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUMsnorthrup/chem1120/LectureNotesS12/Chap 18 student.pdf1. U...
Transcript of CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUMsnorthrup/chem1120/LectureNotesS12/Chap 18 student.pdf1. U...
Chapter 18 Page 1
CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM
Part One: Heat Changes and Thermochemistry
This aspect of Thermodynamics was dealt with in Chapter 6. (Review) A. Statement of First Law. (Section 18.1)
1. U ≡ total internal energy possessed by a system. 2. q = heat absorbed by system during a process. 3. w = work done on system during a process. 4.
ΔU = q + w 5. Scenario:
w = work involved expanding against its surroundings
work of expansion is calculated as: w = -PΔV P = opposing pressure ΔV = volume change of system 6. If system is heated but not allowed to respond in any way, then:
ΔU = q no work is done
7. So we say:
ΔU = qv
qv = heat absorbed in a process in which system is not allowed to expand against its surroundings. (constant volume!!)
All energy input remains in the system.
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B. Enthalpy Changes. 1. Most processes we care about occur at constant Pressure, not constant Volume. 2. Example: Heating an open beaker of water. 3. qp = heat absorbed by system heated at constant pressure.
↑↑ expands slightly against surroundings, work is done Tiny amount of energy input qp does not end up increasing E
4. Enthalpy H = “heat content” of a system. ΔH = qp = enthalpy change = change in heat content of the system 5. ΔH slightly > ΔU. 6. Enthalpy and Energy are nearly synonymous, but ΔH is more directly measurable for
heat transfers under constant P conditions. ΔH = ΔU + PΔV
Part Two: Spontaneity of Reactions
A. Spontaneity =
1. Reactions are favored
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2. Reactions are favored by
a. expansion of a gas into a vacuum
Figure 18.6
b. dissolving of a crystal into solution
c. A pendulum in the atmosphere will always come to rest, never the opposite.
Figure 18.7
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d. Heat flow from high T region to low T region.
B. The Second Law of Thermodynamics.
1. Involves Entropy S = 2. Second Law: 3. 4. It is possible for entropy of a system to be decreasing (system becoming more
orderly): ΔSsystem < 0
but only if Example: Crystallization of an insoluble solid from solution is a decrease in disorder,
but still can occur spontaneously, because heat is released into the surroundings 5. It can be shown that the entropy change of a system in a process is always: ΔS > q/T q=heat flow into the system at temp T 6. The Third Law of Thermodynamics: i.e., a substance that is perfectly crystalline at 0 K has zero entropy.
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Therefore, there is an absolute zero on the entropy scale, unlike energy, and you can
find S of a sample substance, whereas you can’t find its U, only its ΔU
Figure 18.8
7. Hess’ Law can be used to calculate ΔS° of processes, using the data in Table 18.1 8. Entropy changes in a reaction - Entropy increases for:
a. a reaction in which a molecule is broken down into two or more smaller molecules.
b. a reaction in which there are an increasing number of moles of gas. c. A process in which a solid changes to liquid or a liquid changes to a gas.
C. Entropy of phase transitions:
€
ΔStrans =ΔHtrans
Ttrans
Part Three: Free Energy A. Free Energy Change, ΔG, and Spontaneity. (Section 18.4)
1. It is inconvenient to monitor S of universe as a criterion for spontaneity.
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2. More convenient to focus only on . 3. Gibbs Free Energy, G, provides the appropriate indicator.
4.
Now we have a new criterion for spontaneity, involving only system variables. ΔG has two factors now:
• •
5. Now we understand what we said about two factors before: ΔH negative (exothermic) helps ΔG to be negative ΔS positive (increased disorder) also helps ΔG to be negative 6. Dissolving NaCl(s) in H2O is endothermic:
ΔH positive but still is spontaneous because large positive ΔS term
-TΔS causes ΔG to be overall negative. ΔG = ΔH - TΔS (constant temperature and pressure)
ΔH = - ΔS = + Rxns are spontaneous at all temperatures. ΔH = - ΔS = - Rxns become spontaneous below a definite
temperature. ΔH = + ΔS = + Rxns become spontaneous above a definite temperature. ΔH = + ΔS = - Rxns are nonspontaneous at all temperatures.
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7. Can use Hess’ Law to calculate ΔGrxn° from standard free energies of formation, ΔGf°.
Therefore:
€
ΔGrxno = nΔGf
o
products∑ − nΔGf
o
reactants∑
just like we did for heats of reaction:
€
ΔHrxno = nΔHf
o
products∑ − nΔHf
o
reactants∑
And also:
€
ΔSrxno = nSo
products∑ − nSo
reactants∑
Part Four: Free Energy and the Equilibrium Constant
A. Relationship Between K and ΔG°. (Sect. 18.6) 1. ΔG° = 2. 3. 4. Note that:
ΔG° of rxn K Product Formation
if ΔG°< 0 K > 1 if ΔG° = 0 K = 1 if ΔG° > 0 K < 1
5. The free energy change of reaction occurring under non-standard-state conditions,
use: Where Q = reaction quotient.
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6. Interpretation of ΔG. Instantaneous indicator of which direction a reacting system will go spontaneously to
reach an equilibrium state. Dependent upon instantaneous value of Q. Also: ΔG = maximum useful work that can be performed by a process.
B. Change of ΔG° (and thus the Equilibrium Constant) with Temperature. (Section 18.7)
1. Can use the equation:
to quickly estimate how ΔG° changes with T if we simply plug in the values of ΔH° and ΔS° at 298 K, and put in T as the new temperature
2. Example problem: A chemical reaction has a ΔH° = -300 kJ and an entropy change of reaction ΔS° = -2.5 kJ/K. a. What is the ΔG° of reaction at 298 K? b. Is the reaction spontaneous at 298K? c. At what temperature will the reaction switch over to becoming spontaneous?