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# Ch. 20: Entropy and Free Energy - PowerPoint PPT Presentation

Ch. 20: Entropy and Free Energy. Thermodynamics: the science of energy transfer Objective: To learn how chemists predict when reactions will be product-favored vs. when they will be reactant-favored . Section 20.1. Ø Thermodynamics tells us NOTHING about the rate of reaction.

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Ch. 20: Entropy and Free Energy

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### Ch. 20: Entropy and Free Energy

• Thermodynamics: the science of energy transfer

• Objective: To learn how chemists predict when reactions will be product-favored vs. when they will be reactant-favored

### Section 20.1

• ØThermodynamics tells us NOTHING about the rate of reaction.

• The study of rates and why some reactions are fast and others are slow is called kinetics (Ch. 15.)

### Section 20.2 Entropy

• Entropy, S:Measure of dispersal or disorder.

• ØCan be measured with a calorimeter. Assumes in a perfect crystal at absolute zero, no disorder and S = 0.

• ØIf temperature change is very small, can calculate entropy change, DS = q/T (heat absorbed / T at which change occurs)

• ØSum of DS can give total entropy at any desired temperature. See Table 20.1

### Section 20.2 Entropy

• In general, the final state is more probable than the initial one if:

• (1)energy can be dispersed over a greater number of atoms and molecules (hot  cold)

• (2)the atoms and molecules can be more disordered (dissolving, diffusion of gas)

### Section 20.2 Entropy

• More specifically,

• (1)if energy and matter are both more dispersed, it is definitely product-favored

• (2)if only energy or matter is dispersed, then quantitative information is necessary to decide which effects are greater

• (3)if neither matter nor energy is more dispersed, then the process will be reactant-favored

### Entropy Examples (positive DS)

• Boiling water

• Melting ice

• Preparing solutions

• CaCO3 (s)  CaO (s) + CO2 (g)

### Entropy Examples (negative DS)

• Molecules of gas collecting

• Liquid converting to solid at room temp

• 2 CO (g) + O2 (g)  2 CO2 (g)

• Ag+ (aq) + Cl-(aq)  AgCl (s)

### Entropy Generalizations

• Sgas > S liquid > Ssolid

• Entropies of more complex molecules are larger than those of simpler molecules (Spropane > Sethane>Smethane)

• Entropies of ionic solids are higher when attraction between ions are weaker.

ØEntropy usually increases when a pure liquid or solid dissolves in a solvent.

Entropy increases when a dissolved gas escapes from a solution

### Laws of Thermodynamics

• First law: Total energy of the universe is a constant.

• Second law: Total entropy of the universe is always increasing.

• Third law: Entropy of a pure, perfectly formed crystalline substance at absolute zero = 0.

### Calculating DSosystem

• DSosystem =  So (products) -  So (reactants)

Can also relate surroundings to the system!

• DSosurroundings = q surroundings / T

= - DHsystem / T

### Calculating DSouniverse

• DSouniverse = DSosurroundings +DSosystem

• DSouniverse = - DHsystem / T +DSosystem

• Can use 2nd law to predict whether a reaction is product-favored or reactant-favored!

• The higher the temperature, the less important the enthalpy term is!

• Roald Hoffmann (1981 Nobel prize): “One amusing way to describe synthetic chemistry, the making of molecules that is at the intellectual and economic center of chemistry, is that it is the local defeat of entropy.”

### 20.3 Gibbs Free Energy

• DG is a measure of the maximum magnitude of the net useful work that can be obtained from a reaction!

### 20.3 Gibbs Free Energy

• DGsystem = - T DSuniverse

= DHsystem - TDSsystem

• DGosystem = DHosystem - T DSosystem

• DGorxn = DHorxn - T DSorxn

### 20.3 Gibbs Free Energy

• DGosystem or DGorxn If negative, then product-favored. If positive, then reactant-favored.

• DGoreaction =  Gfo (products) -  Gfo (reactants)

### 20.3 Gibbs Free Energy

• DG is a measure of the maximum magnitude of the net useful work that can be obtained from a reaction!

• Know the meaning of enthalpy-driven vs. entropy-driven reactions.

### 20.4 Thermodynamics and K

If not at standard conditions,

DG = DGo + RT ln Q

(Equilibrium is characterized by the inability to do work.)

At equilibrium, Q = K and DG = O

Therefore, substituting into previous equation gives 0 = DGo + RT ln K and

DGo = - RT ln K(can use Kp or Kc)

2020.5      Thermodynamics and Time

• First law: Total energy of the universe is a constant.

• Second law: Total entropy of the universe is always increasing.

• Third law: Entropy of a pure, perfectly formed crystalline substance at absolute zero = 0.

• Entropy : time’s arrow

• Absolutely MUST learn table in Chapter highlights!

### 20.4 Thermodynamics and K

• ØUnderstand relationship between DGo, K, and product-favored reactions!

• DGo<0 K>1product-favored

• DGo=0 K=1

• DGo>0 K<1reactant-favored