chapter 18: thermodynamics

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Chapter 18: Thermodynamics Renee Y. Becker Valencia Community College 1

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Chapter 18: Thermodynamics. Renee Y. Becker Valencia Community College. Thermodynamics. Thermodynamics: The study of interconversion of heat and other forms of energy Spontaneous process: Process that proceeds on its own without external influence (- G) - PowerPoint PPT Presentation

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Page 1: Chapter 18:  Thermodynamics

Chapter 18: Thermodynamics

Renee Y. BeckerValencia Community College

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Page 2: Chapter 18:  Thermodynamics

Thermodynamics

Thermodynamics: The study of interconversion of heat and other forms of energy

Spontaneous process: Process that proceeds on its own without external influence (-G)

Non-spontaneous: Needs continuous external influence to take place (+G)

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Page 3: Chapter 18:  Thermodynamics

Thermodynamics

Entropy (S): Disorder, molecular randomnessS = Sfinal - Sinitial

When disorder increases +SWhen disorder decreases -S

Enthalpy (H): Heat flowIn to the system +HOut of the system -H

Gibbs Free-Energy: Measure of spontaneityG = H-TS

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Page 4: Chapter 18:  Thermodynamics

Example 1: Predict the sign of S in the system for each of the

following

a)      H2O(g) H2O(l)

b)      I2(g) 2I-

(g)

c)      CaCO3(s) CaO(s) + CO2(g)

d) Ag+(aq) + Br-

(aq) AgBr(s) 4

Page 5: Chapter 18:  Thermodynamics

Example 2:

Which of the following reactions has an increase in entropy?

1. H2O(g) H2O(l)

2. H2O(l) H2O(g)

3. H2O(g) H2O(s) 5

Page 6: Chapter 18:  Thermodynamics

Entropy and temperature 3rd Law of Thermodynamics

a) The entropy of a perfectly ordered crystalline substance at 0 K is zero

b) As the temperature increases, the KE increases, Molecular motion increases, entropy increases

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Page 7: Chapter 18:  Thermodynamics

Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy, S

The entropy of one mole of the pure substance at 1 atm pressure and a specific temperature usually 25C

Standard Entropy of Reaction, SEntropy change for a chemical reaction

S = Sproducts - Sreactants

Based on 1 mole of substance so you have to multiply S by the number of moles present

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Page 8: Chapter 18:  Thermodynamics

Standard Entropy of Reaction, S

aA + bB cC + dD

S = [c S(C) + d S(D)] – [a S(A) + b S(B)]

Units: coefficients are moles

S = J/K molS = J/K

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Page 9: Chapter 18:  Thermodynamics

Example 3:

Calculate the standard entropy of reaction at 25C for the decomposition of calcium carbonate

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

Substance S (J/K mol)CaCO3(s) 92.9

CaO(s) 39.7

CO2(g) 213.6

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Page 10: Chapter 18:  Thermodynamics

Entropy and the Second Law of Thermodynamics 1st Law of Thermodynamics

In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant

2nd Law of ThermodynamicsIn any spontaneous process, the total entropy of a system and its surroundings always increases

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Page 11: Chapter 18:  Thermodynamics

Entropy and the Second Law of Thermodynamics

Stotal = Ssystem + Ssurroundings

if S > 0 spontaneousif S< 0 non spontaneous

if S = 0 equilibrium

Ssurr = -H / T

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Page 12: Chapter 18:  Thermodynamics

Entropy and the Second Law of Thermodynamics

a)  Exothermic reaction: H<0, because the surroundings gain heat (entropy increases), heat is lost from the

systemb) Endothermic reaction: H>0, surroundings lose heat

(entropy decreases), and system gains the heat

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Page 13: Chapter 18:  Thermodynamics

Example 4:

Which of the following reactions is endothermic?

1. N2O4(g) 2 NO2(g) H = +57.1 kJ

2. 2 NO2(g) N2O4(g) H = -57.1 kJ

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Page 14: Chapter 18:  Thermodynamics

Free-Energy

Free energy, GG = H – TSG = H - TS

if G < 0 spontaneous Only direct relationship that if G > 0 nonspontaneous always holds up!if G = 0 equilibrium

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Page 15: Chapter 18:  Thermodynamics

Example 5: Consider the decomposition of gaseous N2O4

 

N2O4(g) 2 NO2(g) H = +57.1 kJ

S = +175.8 J/K

a) Is this reaction spontaneous under standard-state conditions at 25C?

b) Estimate the temperature at which the reaction becomes spontaneous

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Page 16: Chapter 18:  Thermodynamics

Standard Free-Energy Changes for Reactions 1. Standard State Conditions: Solids, liquids, and

gases in pure form at 1 atm pressure, Solutes at 1M concentration, specified temperature, usually 25 celsius

2. Standard Free Energy Change, G: The change in free energy that occurs when reactants in their SS are converted to products in their SS

3. G = H - TS G = H - TS

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Page 17: Chapter 18:  Thermodynamics

Example 6:

Consider the thermal decomposition of calcium carbonate

CaCO3(s) CaO(s) + CO2(g) H = 178.3 kJ

S = 160.4 J/K a) Calculate the standard free energy change for this

reaction at 25Cb)    Will a mixture of solid CaCO3, CaO, and gaseous CO2 at 1 atm pressure react spontaneously at 25C?c) Assuming that H and S are independent of temperature, estimate the temperature at which the reaction becomes spontaneous

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Page 18: Chapter 18:  Thermodynamics

Standard Free Energies of Formation 1. Standard Free Energy of Formation, Gf

The free energy change for formation of one mole of the substance in its standard state from the most stable form of its constituent elements in their standard states

2. Gf measures the substances thermodynamic stability with respect to its constituent elements

3. -Gf are stable and do not decompose to their constituent elements under standard state conditions

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Page 19: Chapter 18:  Thermodynamics

Standard Free Energies of Formation

4. +Gf are thermodynamically unstable with respect to their constituents elementsa) There is no point in trying to synthesize a substance that has a +Gf because it would degrade into it’s constituents

b) You would need to synthesize it at different temperatures and or pressures or start with different starting materials that has a reaction with a -Gf

5. G = Gf(products) - Gf(reactants)

6. General reaction: aA +bB cC + dDG = [cGf(C) + dGf(D)] – [aGf(A) + bGf(B)]

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Page 20: Chapter 18:  Thermodynamics

Example 7:

Calculate the standard free energy change for the reaction of calcium carbide with water. Might this reaction be used for the synthesis of acetylene (C2H2)?

 

CaC2(s) + 2 H2O(l) C2H2(g) + Ca(OH)2(s)

Gf (CaC2) = -64.8 kJ/mol

Gf (H2O(l)) = -237.2 kJ/mol

Gf (C2H2) = 209.2 kJ/mol

Gf (Ca(OH)2) = -898.6 kJ/mol20

Page 21: Chapter 18:  Thermodynamics

Free Energy Changes and Composition of the Reaction Mixture Standard state conditions are unrealistic, the reaction

itself will change the temperature and pressure, so what can we use to calculate the free energy change under non-standard state conditions?

G = G + RT ln Q

R = gas constant T = temperature in Kelvins

Q = reaction quotient (Qc or Qp) 21

Page 22: Chapter 18:  Thermodynamics

Example 8:

Calculate the Free energy change for the formation of ethylene (C2H4) from carbon and hydrogen at 25C when the partial pressures are 100 atm H2 and 0.10 atm C2H4

2 C(s) + 2 H2(g) C2H4(g) G = 68.1 kJ/mol

 Is the reaction spontaneous in the forward or reverse

direction?

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Page 23: Chapter 18:  Thermodynamics

Free Energy and Chemical Equilibrium 1. When the RM is mostly reactants

Q<<1 RT lnQ <<0 G<0

2. When the RM is mostly products Q>>1 RT lnQ >>0 G>0

3. G= -RT ln K

K = equilibrium constant Kc or Kp

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Page 24: Chapter 18:  Thermodynamics

Example 9:

Calculate the Kp at 25C for the reaction

 CaCO3(s) CaO(s) + CO2(g)

G= 130.5 kJ/mol

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Page 25: Chapter 18:  Thermodynamics

Example 10:

Which of the following can always predict the spontaneity of a reaction?

1. H

2. S

3. G

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