Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved. E = q + w.

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Thermodynam ics Chapter 18

Transcript of Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved. E = q + w.

Page 1: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

ThermodynamicsChapter 18

Page 2: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

1st Law of Thermodynamics

Energy is conserved.

E = q + w

Page 3: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

SPONTANEOUS: occur without any outside intervention

Example: drop an egg

The REVERSE is not spontaneous!!

Page 4: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

REVERSIBLE PROCESS: change can be restored to its’ original state by exactly reversing the change.

Example: ice water at 0o C

Page 5: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

IRREVERSIBLE PROCESS: cannot simply be reversed to original state.

Example: gas expanding

Page 6: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

Processes in which the disorder of the system increases tend to occur

spontaneously.

Ex: gas expanding, ice melting, salt dissolving

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ENTROPY: (S) the change in disorder. (Change in randomness)

The more disorder, the larger the entropy.

S = Sfinal - Sinitial

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S = > 0 when the final state is in more disorder

S = < 0 when the final state is more ordered than original

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H2O (l) H2O (s)

Ag+(aq) + Cl-

(aq) AgCl(s)

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- a solid melts

- a liquid vaporizes

- a solid dissolves in water

- a gas liquefies

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For a process at constant temperature, the entropy change is the value of qrev divided by the absolute

temperature.S = qrev/T

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Example: Calculate the entropy change when 1 mol of water is

converted into 1 mol of steam at 1 atm pressure. (Hvap = 40.67

kJ/mol)

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(1 mole)(40.67 kJ/ mol)(1000 J/1 kJ)373 K

S = 109 J/K

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The normal freezing point of mercury is -38.9oC, an its

molar enthalpy of fusion is Hfus = 2.331 kJ/mol. What is the entropy change when 50.0

g of Hg(l) freezes at the normal freezing point?

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-2.48 J/KThe answer is negative because the

process brings more order

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The normal boiling point of ethanol, is 78.3oC and its molar

enthalpy of vaporization is 38.56 kJ/mol. What is the change in

entropy when 25.8 g of C2H5OH(g) condenses to liquid at

the normal boiling point?

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-61.4 J/K

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The element gallium, Ga, freezes at 29.8oC, and its enthalpy of fusion is 5.59

kJ/mol. Calculate the value of S for the

freezing of 90.0 g of Ga(l).

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S = -23.8 J/K

Page 20: Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

2nd Law of Thermodynamics: In any reversible process, Suniverse = 0. In any

irreversible (spontaneous) process, Suniverse > 0.

Suniverse = Ssystem + Ssurroundings

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On a Molecular Level

TRANSLATIONAL MOTION: movement of molecules

VIBRATIONAL MOTION: the movement of atoms within the molecule.

ROTATIONAL MOTION: the molecules spinning

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Increasing Temperature Increases Entropy

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3rd Law of Thermodynamics: the entropy of a pure crystalline

substance at absolute zero is zero. S(0K) = 0

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In general, the entropy increases when:

Liquids or solutions are formed from solids

Gases are formed from either solids or liquids

The number of molecules of gas increases during a chemical reaction.

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CaCO3(s) CaO(s) + CO2(g)

N2(g) + 3H2(g) 2NH3(g)

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Standard molar entropies: (So) absolute entropies for substances in their standard

state. (J/mol-K)

1. Unlike enthalpies of formation, the standard molar entropies of elements are not zero.

2. The So of gases are greater than those of liquids and solids.

3. The So generally increases with increasing molar mass.

4. The So generally increase with the number of atoms in the formula.

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So = nSo(products) - mSo(reactants)

Calculate So for the synthesis of ammonia:

N2(g) + 3H2(g) 2NH3(g)

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So = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K) + (3

mol)(130.6 J/mol-K)] =

-198.3 J/K

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Using Appendix C, calculate the standard entropy change, for the

following reaction:

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

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180.4 J/K

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C2H4(g) + H2(g) C2H6(g)

NH3(g) + HCl(g) NH4Cl(s)

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-120.5 J/K

-284.6 J/K

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GIBBS FREE ENERGY

The spontaneity of a reaction involves both enthalpy and entropy. The relationship is

known as free energy.

G = H - TS

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1. If G is negative, the reaction is spontaneous in the forward direction.

2. If G is zero, the reaction is at equilibrium.

3. If G is positive, the reaction in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. However, the reverse reaction will be spontaneous.

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G

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Go = nGfo(products) - mGf

o(reactants)

N2(g) + 3H2(g) 2NH3(g)

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-33.32 kJ

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CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

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CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

-800.7 kJ

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Assuming no change for Ho and So, what happens to Go

with an increase in temperature?

N2(g) + 3H2(g) 2NH3(g)

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Calculate G at 298K for a reaction mixture that

consists of 1.0 atm N2, 3.0 atm H2 and 0.50 atm NH3.

N2(g) + 3H2(g) 2NH3(g)

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Use standard free energies of formation to calculate the equilibrium constant

K at 25oC for the reaction involved in the Haber

process.