- 72 Views
- Uploaded on
- Presentation posted in: General

Introduction to Entropy

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Introduction to Entropy

by Mike Roller

THIS MEANS THAT THINGS FALL.

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

Entropy: Time’s Arrow

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

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

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)

Probability of 1particle in left bulb= ½

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

" 3particlesall 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”).

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:

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

Everything in its place

No molecular motion

Solid

Liquid

Gas

vaporization

S

(qrev/T)

(J/K)

fusion

0

Temperature (K)

0

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

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

- VaporizationSgas > Sliquid
ΔHvaporization/Tvaporization = ΔSvaporization

- Increasing ngas in a reaction
- HeatingST2 > ST1 if T2 > T1
- Dissolving (usually)Ssolution > (Ssolvent + Ssolute)
- Molecular complexitymore bonds, more entropy
- Atomic complexitymore e-, protons, neutrons

- 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

Entropy and Gibbs Free Energy

by Mike Roller

- 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 = 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

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

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

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 = 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

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 kJSpontaneous

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

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

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

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

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

- 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