CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUMsnorthrup/chem1120/LectureNotesS11/Chap 18... · This...

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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 = q v q v = 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.

Transcript of CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUMsnorthrup/chem1120/LectureNotesS11/Chap 18... · This...

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 > ΔE. 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 = can a reaction occur naturally, that is, without outside intervention? Two

factors. 1. Reactions are favored by a decrease in energy of the system, i.e., when energy is

released. (downhill in energy)

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2. Reactions are favored by increased Entropy, that is, an increase in randomness, or dispersal of energy. Examples:

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 = a measure of the disorder or randomness of a system. 2. Second Law:

In spontaneous changes the total entropy of system plus surroundings increases. 3. ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 4. It is possible for entropy of a system to be decreasing (system becoming more

orderly): ΔSsystem < 0

but only if the entropy of the surroundings is increasing by an even greater amount. 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:

S = 0 for a pure, perfect crystal at T = 0 K 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 the system. 3. Gibbs Free Energy, G, provides the appropriate indicator.

G = H - TS

ΔG = ΔH - TΔS constant T and P processes 4. ΔG < 0 for spontaneous process, where ΔG is free energy change in system only.

G tends to go downhill

Now we have a new criterion for spontaneity, involving only system variables. ΔG has two factors now:

• ΔH - energy factor (or enthalpy) • TΔS - entropy factor

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° = standard Gibbs free energy of reaction. 2. ΔG° = -RT ln K

3. Also: K = e-ΔG°RT .

4. Note that:

ΔG° of rxn K Product Formation

if ΔG°< 0 K > 1 Products favored over reactants at equilib. if ΔG° = 0 K = 1 At equilib. [C]c[D]d... = [A]a[B]b... (very rare) if ΔG° > 0 K < 1 Reactants favored over products at equilib.

5. The free energy change of reaction occurring under non-standard-state conditions,

use: ΔG = ΔG° + RTlnQ 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:

ΔG° = ΔH° - TΔS°

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?

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NOTES: