Introduction to Entropy

Introduction to Entropy

by Mike Roller

Entropy (S)= a measure of randomness or 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 MEANS THAT THINGS FALL.

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

THIS IS CALLED ENTROPY.

Entropy: Time’s Arrow

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

How much energy is spread out How widely spread out it becomesEntropy change = “energy dispersed”/T

Second Law of Thermodynamicsoccurs without outside intervention

Entropy of the UniverseΔSuniverse = 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)

SystemHeat Entropy

Surroundings

SystemHeat Entropy

Surroundings

TΔH

ΔS syssurr

Low T large entropy change (surroundings)High T small entropy change (surroundings)

ΔHsys < 0

ΔHsys > 0

ΔSsurr > 0

ΔSsurr < 0

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”).

Ssolid < Sliquid << Sgas

Entropy of the System: Positional Disorder

Ludwig Boltzmann Orderedstates

Disorderedstates

Low probability(few ways)

High probability(many ways)

Low S

High S

Ssystem Positional disorder

S increases with increasing # of possible positions

Ludwig Boltzmann

The Third Law:The entropy of a perfect

crystal at 0 K is zero.

Everything in its placeNo molecular motion

The Third Law of Thermodynamics

Entropy Curve

Solid GasLiquid

S(qrev/T)(J/K)

Temperature (K)0

0

fusion

vaporization

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

Entropy and Gibbs Free Energy

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)

CompoundC6H12O6(s)

O2(g)

CO2(g)

H2O(g)

ΔH°f (kJ/mol)-1275

0-393.5-242

S° (J/mol K)212205214189

ΔSuniverse = ΔSrxn – ΔHrxn/TΔS°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)

CompoundC6H12O6(s)

O2(g)

CO2(g)

H2O(g)

ΔH°f (kJ/mol)-1275

0-393.5-242

S° (J/mol K)212205214189

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

–ΔG means +ΔSuniv

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

Gibbs Free Energy (G)

Josiah Gibbs

G = H – TSAt constant temperature,

ΔG = ΔH – TΔS(system’s point of view)

ΔG = ΔH – TΔSDivide both sides by –T

-ΔG/T = -ΔH/T + ΔS

ΔSuniverse = ΔS – ΔH/T

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

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

ΔG = ΔH – TΔSΔG negative spontaneous reaction

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

Ways to Calculate ΔG°rxn

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 kJΔS° = -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 kJMethod 2: -457 kJMethod 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