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Introduction to Entropy

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### Introduction to Entropy

### Entropy and Gibbs Free Energy

by Mike Roller

THEY FALL FROM HEIGHTS OF ENERGY AND STRUCTURED INFORMATION INTO MEANINGLESS, POWERLESS DISORDER.

MATTER IS ENERGY.

ENERGY IS INFORMATION.

EVERYTHING IS INFORMATION.

PHYSICS SAYS THAT STRUCTURES... BUILDINGS, SOCIETIES, IDEOLOGIES... WILL SEEK THEIR POINT OF LEAST ENERGY.

THIS IS CALLED ENTROPY.

Entropy (S)

= a measure of randomness or disorder

Second Law of Thermodynamics

occurs without outside intervention

In any spontaneous process, the entropy of the universe increases.

ΔSuniverse > 0

Another version of the 2nd Law:

Energy spontaneously spreads out if it has no outside resistance

Entropy measures the spontaneous dispersal of energy as a function of temperature

How much energy is spread out

How widely spread out it becomes

Entropy change = “energy dispersed”/T

Entropy of the Universe

ΔSuniverse = ΔSsystem + ΔSsurroundings

Positional disorder

Energetic disorder

ΔSuniverse > 0 spontaneous process

Both ΔSsys and ΔSsurr positive

Both ΔSsys and ΔSsurr negative

ΔSsys negative, ΔSsurr positive

ΔSsys positive, ΔSsurr negative

spontaneous process.

nonspontaneous process.

depends

depends

Entropy of the Surroundings(Energetic Disorder)

System

Entropy

Heat

ΔSsurr > 0

ΔHsys < 0

Surroundings

System

Surroundings

Heat

Entropy

ΔSsurr < 0

ΔHsys > 0

Low T large entropy change (surroundings)

High T small entropy change (surroundings)

Positional Disorder and Probability

Probability of 1 particle in left bulb = ½

" 2 particles both in left bulb = (½)(½) = ¼

" 3 particles all in left bulb = (½)(½)(½) = 1/8

" 4 " all " = (½)(½)(½)(½) = 1/16

" 10 " all " = (½)10 = 1/1024

" 20 " all " = (½)20 = 1/1048576

" a mole of " all " = (½)6.021023

The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

Entropy of the System: Positional Disorder

Ssolid < Sliquid << Sgas

Ludwig Boltzmann

Ordered

states

Low probability

(few ways)

Ludwig Boltzmann

Low S

Disordered

states

High probability

(many ways)

High S

Ssystem Positional disorder

S increases with increasing # of possible positions

The Third Law of Thermodynamics

The Third Law:

The entropy of a perfect crystal at 0 K is zero.

Everything in its place

No molecular motion

Entropy Curve

Solid

Liquid

Gas

vaporization

S

(qrev/T)

(J/K)

fusion

0

Temperature (K)

0

S° (absolute entropy) can be calculated for any substance

Entropy Increases with...

- Melting (fusion) Sliquid > Ssolid
ΔHfusion/Tfusion = ΔSfusion

- Vaporization Sgas > Sliquid
ΔHvaporization/Tvaporization = ΔSvaporization

- Increasing ngas in a reaction
- Heating ST2 > ST1 if T2 > T1
- Dissolving (usually) Ssolution > (Ssolvent + Ssolute)
- Molecular complexity more bonds, more entropy
- Atomic complexity more e-, protons, neutrons

Recap: Characteristics of Entropy

- S is a state function
- S is extensive (more stuff, more entropy)
- At 0 K, S = 0 (we can know absolute entropy)
- S > 0 for elements and compounds in their standard states
- ΔS°rxn = nS°products - nS°reactants
- Raise T increase S
- Increase ngas increase S
- More complex systems larger S

by Mike Roller

Entropy (S) Review

- ΔSuniverse > 0 for spontaneous processes
- ΔSuniverse = ΔSsystem + ΔSsurroundings

positional

energetic

- We can know the absolute entropy value for a substance
- S° values for elements & compounds in their standard states are tabulated (Appendix C, p. 1019)

- For any chemical reaction, we can calculate ΔS°rxn:
- ΔS°rxn = S°(products) - S°(reactants)

ΔSuniverse and Chemical Reactions

ΔSuniverse = ΔSsystem + ΔSsurroundings

For a system of reactants and products,

ΔSuniverse = ΔSrxn – ΔHrxn/T

- If ΔSuniverse > 0, the reaction is spontaneous
- If ΔSuniverse < 0, the reaction is not spontaneous
- The reverse reaction is spontaneous

- If ΔSuniverse = 0, the reaction is at equilibrium
- Neither the forward nor the reverse reaction is favored

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

Compound

C6H12O6(s)

O2(g)

CO2(g)

H2O(g)

ΔH°f (kJ/mol)

-1275

0

-393.5

-242

S° (J/mol K)

212

205

214

189

ΔSuniverse = ΔSrxn – ΔHrxn/T

ΔS°rxn = S°(products) - S°(reactants)

= [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))]

= [6(214) + 6(189)] – [(212) + 6(205)] J/K

ΔS°rxn= 976 J/K

ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)

= [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))]

= [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ

ΔH°rxn = -2538 kJ

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

Compound

C6H12O6(s)

O2(g)

CO2(g)

H2O(g)

ΔH°f (kJ/mol)

-1275

0

-393.5

-242

S° (J/mol K)

212

205

214

189

ΔSuniverse = ΔSrxn – ΔHrxn/T

ΔS°rxn = 976 J/K (per mole of glucose)

ΔH°rxn = -2538 kJ (per mole of glucose)

At 298 K,

ΔS°universe = 0.976 kJ/K – (-2538 kJ/298 K)

ΔS°universe = 9.5 kJ/K

Gibbs Free Energy (G)

ΔSuniverse = ΔS – ΔH/T

ΔG = ΔH – TΔS

Divide both sides by –T

-ΔG/T = -ΔH/T + ΔS

G = H – TS

At constant temperature,

ΔG = ΔH – TΔS

(system’s point of view)

–ΔGmeans+ΔSuniv

A process (at constant T, P) is spontaneous if free energy decreases

Josiah Gibbs

ΔG and Chemical Reactions

ΔG = ΔH – TΔS

- If ΔG < 0, the reaction is spontaneous
- If ΔG > 0, the reaction is not spontaneous
- The reverse reaction is spontaneous

- If ΔG = 0, the reaction is at equilibrium
- Neither the forward nor the reverse reaction is favored

- ΔG is an extensive state function

Ba(OH)2(s) + 2NH4Cl(s) BaCl2(s) + 2NH3(g) + 2 H2O(l)

ΔH°rxn = 50.0 kJ (per mole Ba(OH)2)

ΔS°rxn = 328 J/K (per mole Ba(OH)2)

ΔG = ΔH - TΔS

ΔG° = 50.0 kJ – 298 K(0.328 kJ/K)

ΔG° = – 47.7 kJ Spontaneous

At what T does the reaction stop being spontaneous?

The T where ΔG = 0.

ΔG = 0 = 50.0 kJ – T(0.328 J/K)

50.0 kJ = T(0.328 J/K)

T = 152 K

not spontaneous below 152 K

Effect of ΔH and ΔS on Spontaneity

ΔG = ΔH – TΔS

ΔG negative spontaneous reaction

ΔH

–

+

–

+

ΔS

+

+

–

–

- Spontaneous?
- Spontaneous at all temps
- Spontaneous at high temps
- Reverse reaction spontaneous at low temps

- Spontaneous at low temps
- Reverse reaction spontaneous at high temps

- Not spontaneous at any temp

Ways to Calculate ΔG°rxn

1.ΔG° = ΔG°f(products) - ΔG°f(reactants)

- ΔG°f = free energy change when forming 1 mole of compound from elements in their standard states
2.ΔG° = ΔH° - TΔS°

3.ΔG° can be calculated by combining ΔG° values for several reactions

- Just like with ΔH° and Hess’s Law

2H2(g) + O2(g) 2 H2O(g)

1.ΔG° = ΔG°f(products) - ΔG°f(reactants)

ΔG°f(O2(g)) = 0

ΔG°f(H2(g)) = 0

ΔG°f(H2O(g)) = -229 kJ/mol

ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ

2.ΔG° = ΔH° - TΔS°

ΔH° = -484 kJ

ΔS° = -89 J/K

ΔG° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ

2H2(g) + O2(g) 2 H2O(g)

3.ΔG° = combination of ΔG° from other reactions (like Hess’s Law)

2H2O(l) 2H2(g) + O2(g)ΔG°1 = 475 kJ

H2O(l) H2O(g)ΔG°2 = 8 kJ

ΔG° = - ΔG°1 + 2(ΔG°2)

ΔG° = -475 kJ + 16 kJ = -459 kJ

Method 1: -458 kJ

Method 2: -457 kJ

Method 3: -459 kJ

What is Free Energy, Really?

- NOT just “another form of energy”
- Free Energy is the energy available to do useful work
- If ΔG is negative, the system can do work (wmax = ΔG)
- If ΔG is positive, then ΔG is the work required to make the process happen
- Example: Photosynthesis
- 6 CO2 + 6 H2O C6H12O6 + 6 O2
- ΔG = 2870 kJ/mol of glucose at 25°C
- 2870 kJ of work is required to photosynthesize 1 mole of glucose

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