chemical thermodynamics spontaneous processes reversible processes review first law second...

Chemical ThermodynamicsChemical Thermodynamics

SpontaneousProcesses

ReversibleProcesses

ReviewFirst Law

IrreversibleProcesses

Second Law Third LawEntropy

TemperatureDependence

GibbsFree Energy

EquilibriumConstant

04/19/23

• Thermodynamics is concerned with the question: can a reaction occur?

• First Law of Thermodynamics: energy is conserved.• Any process that occurs without outside intervention is

spontaneous.• When two eggs are dropped they spontaneously break.• The reverse reaction is not spontaneous.• We can conclude that a spontaneous process has a direction.

Spontaneous ProcessesSpontaneous ProcessesSpontaneous ProcessesSpontaneous Processes

• A process that is spontaneous in one direction is not spontaneous in the opposite direction.

• The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T > 0C, Water turning to ice is spontaneous at T < 0C.

Reversible and Irreversible Processes• A reversible process is one that can go back and forth

between states along the same path.


Reversible and Irreversible Processes

• Chemical systems in equilibrium are reversible.

• In any spontaneous process, the path between reactants and products is irreversible.

• Thermodynamics gives us the direction of a process. It cannot predict the speed at which the process will occur.


The Spontaneous Expansion of a Gas• Why do spontaneous processes occur?• Consider an initial state: two flasks connected by a closed

stopcock. One flask is evacuated and the other contains 1 atm of gas.

• The final state: two flasks connected by an open stopcock. Each flask contains gas at 0.5 atm.

• The expansion of the gas is isothermal (i.e. constant temperature). Therefore the gas does no work and heat is not transferred.

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof Thermodynamics


The Spontaneous Expansion of a Gas

• Why does the gas expand?

Entropy and the Entropy and the Second Law of Second Law of

ThermodynamicsThermodynamics

Entropy and the Entropy and the Second Law of Second Law of

ThermodynamicsThermodynamics

The Spontaneous Expansion of a Gas



Probability of having one player (out of four) getting 13 cards of same suit is:

111052.2497,347,688,39

1

Entropy• Entropy, S, is a measure of the disorder of a system.• Spontaneous reactions proceed to lower energy or higher

entropy.• In ice, the molecules are very well ordered because of the

H-bonds.• Therefore, ice has a low entropy.



Entropy• Generally, when an increase in entropy in one process is

associated with a decrease in entropy in another, the increase in entropy dominates.

• Entropy is a state function.

• For a system, S = Sfinal - Sinitial.

• If S > 0 the randomness increases, if S < 0 the order increases.



Entropy• Suppose a system changes reversibly between state 1 and

state 2. Then, the change in entropy is given by

– at constant T where qrev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.)



)(constant revsys T

Tq

S

The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9oC, and its molar enthalpy of fusion is 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? MW(Hg) = 200.59 g/mol

The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9oC, and its molar enthalpy of fusion is 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? MW(Hg) = 200.59 g/mol

-2.44 J K-1

The Second Law of Thermodynamics• The second law of thermodynamics explains why spontaneous

processes have a direction.• In any spontaneous process, the entropy of the universe

increases.

Suniv = Ssys + Ssurr

• Entropy is not conserved: Suniv is increasing.





Calculate ΔSsys , ΔSsurr , ΔSuniv for the reversible melting of 1 mole of Ice in a large, isothermal water bath at 0oC and 1 atm. Heat of Fusion for water is 6.01 kJ/mol.

Calculate ΔSsys , ΔSsurr , ΔSuniv for the reversible melting of 1 mole of Ice in a large, isothermal water bath at 0oC and 1 atm. Heat of Fusion for water is 6.01 kJ/mol.

Ssys = +22.0 J mol-1 K-1

• A gas is less ordered than a liquid that is less ordered than a solid.

• Any process that increases the number of gas molecules leads to an increase in entropy.

• When NO(g) reacts with O2(g) to form NO2(g), the total

number of gas molecules decreases, and the entropy decreases.

The Molecular Interpretation The Molecular Interpretation of Entropyof Entropy


HyperChem

• Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

• Entropy changes dramatically at a phase change.• As we heat a substance from absolute zero, the entropy

must increase.• If there are two different solid state forms of a substance,

then the entropy increases at the solid state phase change.



• Boiling corresponds to a much greater change in entropy than melting.

• Entropy will increase when– liquids or solutions are formed from solids,

– gases are formed from solids or liquids,

– the number of gas molecules increase,

– the temperature is increased.



ENTROPY OF THE UNIVERSEENTROPY OF THE UNIVERSE

Die Enegie der Welt ist Konstant;Die Enegie der Welt ist Konstant;

die entropie der Welt einendie entropie der Welt einen

Maximum ZuMaximum Zu

Entropy is the Arrow of TimeEntropy is the Arrow of Time

A Brief History of Time S. Hawking’s Grand Design: The Meaning of Life

http://www.brainpickings.org/index.php/2012/03/29/brian-cox-arrow-of-time/

http://www.brainpickings.org/index.php/2012/03/29/brian-cox-arrow-of-time/

http://www.youtube.com/watch?v=FnSEt2BCcRs

http://www.youtube.com/watch?v=FnSEt2BCcRs

• Absolute entropy can be determined from complicated measurements.

• Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H.

• Units: J/mol-K. Note units of H: kJ/mol.• Standard molar entropies of elements are not zero.• For a chemical reaction which produces n moles of

products from m moles of reactants:

Entropy Changes in Chemical ReactionsEntropy Changes in Chemical ReactionsEntropy Changes in Chemical ReactionsEntropy Changes in Chemical Reactions

reactantsproducts mSnSS

Calculate ΔS for the system, the surroundings, and the universe for the synthesis of ammonia at 298 K. [ΔHo

rxn = -92.38 kJ/mol]



Answer

• For a spontaneous reaction the entropy of the universe must increase.

• Reactions with large negative H values are spontaneous.• How do we balance S and H to predict whether a reaction

is spontaneous?• Gibbs free energy, G, of a state is

• For a process occurring at constant temperature

Gibbs Free EnergyGibbs Free EnergyGibbs Free EnergyGibbs Free Energy

TSHG

STHG

• There are three important conditions:– If G < 0 , forward reaction is spontaneous.– If G = 0 , reaction is at equilibrium.– If G > 0 , then forward reaction is not spontaneous. If G > 0,

work must be supplied from the surroundings to drive the reaction.

• For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.




Srxn = -198.7 J mol-1 K-1 . Calculate the Gibbs Free Energy change for the reaction.

-33.14 kJ/mol

Answer for Suniv

Standard Free-Energy Changes

• We can tabulate standard free-energies of formation, Gf (c.f. standard enthalpies of formation).

• Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and G = 0 for elements.

G for a process is given by


reactantsproducts ff GmGnG

Calculate ΔGorxn for the combustion of methane.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

ΔGof (kJ/mol): -50.8 0 -394.4 -228.6

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

ΔGof (kJ/mol): -50.8 0 -394.4 -228.6

-800.8 kJ/mol

Free Energy and TemperatureFree Energy and TemperatureFree Energy and TemperatureFree Energy and Temperature

• At equilibrium, G = 0 :

• From the above we can conclude:– If G < 0, then K > 1.

– If G = 0, then K = 1.

– If G > 0, then K < 1.

Free Energy and The Equilibrium ConstantFree Energy and The Equilibrium ConstantFree Energy and The Equilibrium ConstantFree Energy and The Equilibrium Constant

eqKRTG ln

Given ΔGo = -33.3 kJ/mol for the ammonia formation from nitrogen and hydrogen, calculate Keq at 25.0oC.

Given ΔGo = -33.3 kJ/mol for the ammonia formation from nitrogen and hydrogen, calculate Keq at 25.0oC.

Aurora + Entropy 6.82x105

Chemical ThermodynamicsChemical Thermodynamics

SpontaneousProcesses

ReversibleProcesses

ReviewFirst Law

IrreversibleProcesses

Second Law Third LawEntropy

TemperatureDependence

GibbsFree Energy

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)(constant revsys T

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S

wqE

0 sysG

0sys surruniv SSS

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