Thermodynamics: Spontaneous Processes, Entropy, and Free Energy

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Spontaneous Processes

• In thermodynamics, a spontaneous process is one that proceeds in a given direction without outside intervention.

• A nonspontaneous process only occurs for as long as energy is continually added to the system.

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Thermodynamics• The first law states that energy is neither

created or destroyed…the mass/energy of the universe is constant.

• The second law of thermodynamics states that the total entropy of the universe increases in any spontaneous process.

• Entropy (S) is a measure of the distribution of energy in a system at a specific temperature.

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Entropy and Microstates

• Quantum mechanics teaches that energy is not continuous at the atomic scale; only certain levels of energy are possible.

• The motion of molecules is quantized, which means different states are separated by specific energies.

• An energy state, also called an energy level, is an allowed value of energy.

• A microstate is a unique distribution of particles among energy levels.

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Motion… and beyond

• Three types of motion. Translational Rotational Vibrational

• As the temperature of a sample increases, the amount of motion increases.

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Energy States

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Not drawn to scale

OK, we have seen ‘energy levels’. What affects their spacings ! (…this is critical stuff)

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• Heavy systems (Br-Br)• Not constrained (gases)• Many particles

• light systems (H-H)• heavily constrained

(solids or liquids)• Few particles

Highlyentropic

lowentropy

These energy levels get populated by molecules…. We can have different ways of arranging the molecules across the energy states….

Imagine the number of permutations (W) for each cube ! (How many ways you can

arrange the colors)

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W=1 W=4

Statistical EntropyEntropy is related to the number of microstates by the following equation:

S = k ln(W)

S is entropyW is the number of microstates

k is Boltzmann’s constant (k = 1.38 x 10-23 J/K)

This equation indicates that entropy increases as the number of possible or alternative

microstates (W) increases.

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Let’s solve for W using a simple example of two different quantized systems….

• We have six quanta of total energy, 3 molecules per system, and two

different energy stackings.

0

6

0

3

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Distributions vsmicrostates

(permutations)

7 distributions 2 distributions

Entropy (Statistical) and Probability

S = k ln W

k = Boltzmann’s constant

= 1.38 x 10-23 J/K

W = The number of ways that the state can be achieved.

Ln(1) = 0

ln(220) = 20⋅ ln(2)

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Spring 2017 Exam Question

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Assume a heteronuclear diatomic molecule, AB, forms a 1-dimensional crystal by lining up along the x-axis. Also assume thateach molecule can only have one of six possible orientations,corresponding to atom A facing in either the + or – directions alongthe x-, y-, or z axes. If the molecules are arranged randomly in thesix directions, the molar entropy at absolute zero should be ?

A) Nk ln (6!)B) Nk ln (6)C) 0D) Nk ln (66)

Going back to gas expansion

Vi =Bi ⋅ v and Vf = Bf ⋅v

Wi =(Bi )N and Wf = =(Bf )N N is the molesof gas

microstates

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∆S = Sf – Si = k ln(Wf) – k ln(Wi)

= k ln(Wf/Wi)

Since V = Bv

Recall that Wi =(Bi )N and Wf = =(Bf )N

=

=

i

fN

i

f

BB

kNBB

k lnln

=

=

vVvV

kNvVvV

ki

fN

i

f

//

ln//

ln

=

i

f

VV

kN ln

Uff Da!Entropy

increases with

increases in volume

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Thermodynamic Entropy• Consider an isothermal process (a process

that takes place at a constant temperature). ∆S = qrev/T Qrev means the flow of heat as the process is

carried out reversibly in very small steps.∆Suniv = ∆Ssys + ∆Ssurr

• Spontaneous processes must result in an increase in the entropy of the universe.

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Spontaneity and Entropy

• If ∆Ssys > 0 and ∆Ssurr > 0, then ∆Suniv > 0

• If ∆Ssys < 0 then ∆Ssys + ∆Ssurr > 0 for ∆Suniv > 0

• If ∆Ssys > 0 and ∆Ssurr < 0 as long as ∆Ssys + ∆Ssurr > 0

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Third Law of Thermodynamics• Third Law of Thermodynamics - the entropy of a

perfect crystal is zero at absolute zero. S is explicitly known (=0) at 0 K, S values at other temps can

be calculated.

• Absolute entropy is the entropy change of a substance taken from S = 0 (at T = 0 K) to some other temperature.

• Standard molar entropy (So) is the absolute entropy of 1 mole of a substance in its standard state.

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Select Standard Molar Entropy Values (1 atm, 298 K)

Formula So

(J/(mol•K) Formula Name So

(J/(mol•K)Br2(g) 245.5 CH4(g) methane 186.2Br2(l) 152.2 CH3CH3(g) Ethane 229.5

Cdiamond(s) 2.4 CH3OH(g) Methanol 239.7Cgraphite(s) 5.7 CH3OH(l) 126.8

CO(g) 197.7 CH3CH2OH(g) Ethanol 282.6CO2(g) 213.8 CH3CH2OH(l) 160.7H2(g) 130.6 CH3CH2CH3(g) Propane 269.9N2(g) 191.5 CH3(CH2)2CH3(g) n-Butane 310.0O2(g) 205.0 CH3(CH2)2CH3(l) 231.0

H2O(g) 188.8 C6H6(g) Benzene 269.2H2O(l) 69.9 C6H6(l) 172.8NH3(g) 192.3 C12H22O11(s) Sucrose 360.2

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Trends• Ssolid < Sliquid < Sgas

H2O(g) is 188.8 J/(mol•K) and H2O(l) is 69.9 J/(mol•K)

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Compare with enthalpy plot (temp vs heat

added)

Other Trends

• Entropy increases as the complexity of

molecular structure increases.

Why ?

Volume increases at constant T

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Entropy Differences in Allotropes

Limited Modes of Motion Additional Modes of Motion

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Calculating Entropy Changes• Entropy change for a reaction (∆Srxn) is

∆Srxn = ΣnSoproducts - ΣmSo

reactants

where n and m are the coefficients of the products and reactants in the balanced

equation.

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PracticeCalculate the standard entropy change

for the reaction

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)

Answer: - 619 J/mole*K

Remember that phase is

important…previous table

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Free Energy

• Free energy (G) is an indication of the energy available to do useful work.

• Free energy change (∆G) is the change in free energy of a process.

For spontaneous processes at a constant temperature and pressure ∆G < 0.

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Driving Forces for Spontaneous Chemical Processes

1. The formation of low energy products (exothermic processes; ∆H < 0)

2. The formation of products that have greater entropy than the reactants (∆S > 0).

• The free energy (G) relates enthalpy, entropy, and temperature for a process.

G = H - TS or ∆G = ∆H - T∆SAkin to

disposable income…

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Effects of ∆H, ∆S, and T on ∆G

∆H ∆S ∆G Spontaneity

<0 >0 Always <0 Always Spontaneous

<0 <0 <0 at low temp Spontaneous at Low Temperatures

>0 >0 <0 at high temp Spontaneous at High Temperatures

>0 <0 Always >0 Never Spontaneous

Let’s do ice melting….…lets make energy diagrams

(see board)

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Spontaneity’s Dependence on Temperature

<

…for melting of ice…

ΔH = 6.02 kJ/mole

Δ S = 0.0220 kJ/K*mole

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∆G = ∆H - T∆S

PracticeGiven the data below, calculate the temperature at which

the reaction becomes spontaneous. (∆H0 = -92 kJ/mol and ∆S0 = -199J/mol)

N2(g) + 3H2(g) ⇌ 2NH3(g)

http://www.galletti.de/e/index.htm

Dividing by -1 in an inequality requires a flip-flop of the inequality sign

T<462

°

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Calculation of Free Energies for Chemical Reactions

∆G0rxn = Σn∆G0

f, products - Σm∆G0f, reactants

The free energy for a chemical reaction indicates the maximum amount of energy that is free to do useful work.

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Practice

Calculate the free energy change for the following reaction using the ∆G0

f values in the appendix.

C12H22O11 (s) + 12O2 (g) ⇌ 12CO2 (g) + 11H2O (l)

-1543.8 kJ/mol 0 -394.4 -237.2

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Driving Reactions• Exergonic reactions are spontaneous

(∆G < 0).

• Endergonic reactions are nonspontaneous (∆G > 0).

…don’t forget about activation energy and

kinetics

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Thermodynamics of Equilibriums

• ∆G = ∆Go + RT ln(Q)

• At equilibrium: Q = K

∆G = ∆Go + RT ln(K) = 0

∆Go = -RT ln(K)

K = e-∆G /RTo

…based on thermodynamic

arguments

The Xo symbol represents

standard state

Free energy

N2O4(g) ⇌ 2 NO2(g)

No N2O4present

No NO2present

Mole fraction actually

⇌

2NO2

2N2O2

Combined (equilibrium) manifold – mixing !!!

Which manifold has the greatest entropy?

Relationship Between K and

∆G

K < 1

K > 1

K = 1

∆Go = -RT ln(K)

K = e-∆G /RT

Remember that K is at equilibrium

Shown to the left are three possible configurations (A, B, and C) for placing 4 atoms in two boxes. Which of the following processes is accompanied by the largest increase in entropy, ΔS?

A) A→ B B) B → C C) C →A

Appetizer Problems….many of these can be solved by using energy level diagrams

What is the W value for each state? S = k⋅ln(W)…compare S values among states

W = n!/(n1! ⋅n2!) …or think in terms of…

Consider the following possible gas phase reaction:

Which of the following is probably true for this reaction?

A) ΔH > 0 B) ΔS > 0 C) ΔG > 0

What can be said of ΔG° for the condensation of water vapor,

H2O(g) → H2O(l)

at 25 °C if the partial pressure of H2O(g) is 1.0 atm?

A) ΔG° > 0 B) ΔG° = 0 C) ΔG° < 0

To the left is a plot of vapor pressure versus temperature for the

vaporization of ethanol,

C2H5OH(l) → C2H5OH(g).

At which temperature is ΔG° = 0 for the vaporization of ethanol at

1.0 atm?

A) > 100 °C B) 100 °C C) < 100 °C

Which of the following plots shows the correct relationship between ΔG° (y-axis) and temperature (x-axis) for the sublimation of solid iodine to iodine vapor at 1.0 atm?

A) B) C)

The formation of ozone, O3(g), from molecular oxygen is an endothermic process, with ΔH° = 85 J/mole.

3 O2(g) 2 O3(g)

At what temperatures will the reaction proceed spontaneously if PO2 = PO3 = 1.0 atm?

A) High temperatures B) Low temperatures C) No temperatures