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Spontaneity, Entropy & Free EnergyFirst Law of Thermodynamics
Basically the law of conservation of energyenergy can be neither created nor destroyedi.e., the energy of the universe is constant
• the total energy is constant• energy can be interchanged
– e.g. potential energy (stored in chemical bonds) can be converted to thermal energy in a chemical reaction
– CH4 + O2 --> CO2 + H2O + energy Doesn’t tell us why a reaction proceeds in a
particular direction
Spontaneity, Entropy & Free Energy
Spontaneous Processes and Entropy Spontaneous processes occurs without
outside intervention Spontaneous processes can be fast or
slow
Spontaneity, Entropy & Free Energy
Thermodynamics lets us predict whether a process will
occur tells us the direction a reaction will go only considers the initial and final states does not require knowledge of the
pathway taken for a reaction
Spontaneity, Entropy & Free Energy
Kinetics depends on the pathway taken tells us the speed of the process depends on
activation energytemperatureconcentrationcatalysts
Spontaneity, Entropy & Free Energy
Spontaneous Processes a ball rolls downhill, but the ball never
spontaneously rolls uphill steel rusts, but the rust never spontaneously
forms iron and oxygen a gas fills its container, but a gas will never
spontaneously collect in one corner of the container.
Water spontaneously freezes at temperatures below 0o C
Spontaneity, Entropy & Free Energy
What thermodynamic principle explains why these processes occur in one direction?
The driving force for a spontaneous reaction is an increase in the entropy of the universe
Spontaneity, Entropy & Free Energy
Entropy Symbol: S A measure of randomness or disorder The natural progression is from order to
disorder It is natural for disorder to increase Entropy is a thermodynamic function Describes the number of arrangements that
are available to a system in a given state
Spontaneity, Entropy & Free Energy
Entropy The greater the number of possible
arrangements, the greater the entropy of a system, i.e., there is a large positional probability.
The positional probability or the entropy increases as a solid changes from a liquid or as a liquid changes to a gas
Spontaneity, Entropy & Free Energy
Ssolid < Sliquid < SgasChoose the substance with the higher
positional entropy: CO2(s) or CO2(g)? N2(g) at 1 atm and 25oC or N2(g) at .010 atm
and 25oC?
Spontaneity, Entropy & Free Energy
Predict the sign of the entropy change solid sugar is added to water iodine vapor condenses onto a cold
surface forming crystals
Spontaneity, Entropy & Free Energy
Second Law of Thermodynamics The entropy of the universe is
increasing The universe is made up of the system
and the surroundings Suniverse = Ssystem + Ssurroundings
Spontaneity, Entropy & Free Energy
A process is spontaneous if the Suniverse is positive
If the Suniverse is zero, there is no tendency for the reaction to occur
Spontaneity, Entropy & Free Energy
The effect of temperature on spontaneity H2O(l) --> H2O(g) water is the system, everything else is the
surroundings Ssystem increases, i.e. Ssystem is positive,
because there are more positions for the water molecules in the gas state than in the liquid state
Spontaneity, Entropy & Free Energy
What happens to the surrounding? Heat leaves the surroundings, entering the
system to cause the liquid molecules to vaporize
When heat leaves the surroundings, the motion of the molecules of the surroundings decrease, which results in a decrease in the entropy of the surroundings
Ssurroundings is negative
Spontaneity, Entropy & Free Energy
Sign of S depends on the heat flow Exothermic Rxn: Ssurr >0 Endothermic Rxn: Ssurr< 0
Magnitude of S is determined by the temperature Ssurr = - H
Spontaneity, Entropy & Free Energy
Signs of Entropy ChangesSsys Ssurr Suniv Spontaneous? + + - - + - - +
Spontaneity, Entropy & Free Energy
Free Energy aka Gibbs Free Energy G another thermodynamic function
related to spontaneityG = H - TSfor a process that occurs at constant
temperature (i.e. for the system):G = H - TS
Spontaneity, Entropy & Free Energy
How does the free energy related to spontaneity?G = H - TSG = - H + S (remember, - H = Ssurr ) T T T G = Ssurr + Ssys (remember, Ssurr + Ssys = Suniv) T-G = Suniv T
Spontaneity, Entropy & Free Energy
Suniv > 0 for a spontaneous reaction G < 0 for a spontaneous reaction G > 0 for a nonspontaneous reaction
Useful to look at G because many chemical reactions take place under constant pressure and temperature
Spontaneity, Entropy & Free Energy
H2O(s) --> H2O(l) Ho = 6.03 x 103J/mole So = 22.1 J/K.mole Calculate G, Ssurr, and Suniv at -10oC,
0oC, and 10oC
Spontaneity, Entropy & Free Energy
For the melting of ice Ssys and Ssurr oppose each other spontaneity will depend on temperature So is positive because of the increase
in positional entropy when the ice melts Ssurr is negative because the reaction is
endothermic
Spontaneity, Entropy & Free Energy
At what temperatures is Br2(l) --> Br2(g) spontaneous?
What is the normal boiling point of Br2?
HokJ/mol So = 93.0 J/K.mol
Spontaneity, Entropy & Free Energy
Entropy Changes in Chemical Reactions Just like physical changes, entropy
changes in the surroundings are determined by heat flow
Entropy changes in the system are determined by positional entropy (the change in the number of possible arrangements)
Spontaneity, Entropy & Free Energy
N2 (g) + 3 H2(g) --> 2 NH3 (g) The entropy of the this system
decreases because four reactant molecules form two product
moleculesthere are less independent units in the
systemless positional disorder, i.e. fewer possible
configurations
Spontaneity, Entropy & Free Energy
When a reaction involves gaseous molecules: the change in positional entropy is
determined by the relative numbers of molecules of gaseous reactants and products
I.e., if you have more product molecules than reactant molecules, S will be positive
Spontaneity, Entropy & Free Energy
In thermodynamics, the change in a function is usually what is important usually we can’t assign an absolute
value to a function like enthalpy or free energy
we can usually determine the change in enthalpy and free energy
Spontaneity, Entropy & Free Energy
We can assign absolute entropy values, i.e., we can find S
A perfect crystal at 0 K, while unattainable, represents a standard all molecular motion stops all particles are in their place the entropy of a perfect crystal at 0
K is zero = third law of thermodynamics
Spontaneity, Entropy & Free Energy
Increase the temperature of our perfect crystal molecular motion increases disorder increases entropy varies with temperature See thermodynamic tables for So values
(at 298 K and 1 atm)
Spontaneity, Entropy & Free Energy
Entropy is a state function entropy does not depend on the
pathway taken Srxn = nSoproducts - nSoreactant
Spontaneity, Entropy & Free Energy
Calculate So at 25oC for 2NiS(s) + 3 O2(g) --> 2 SO2(g) + 2 NiO(s)Substance So(J/K.mol)SO2 248NiO 38O2 205NiS 53
Spontaneity, Entropy & Free Energy
Calculate So forAl2O3(s) + 3 H2(g) --> 2 Al(s) + 3 H2O(g)Substance So (J/K.mol)Al2O3 51H2 131Al 28H2O 189
Spontaneity, Entropy & Free Energy
What did you expect the So to be?Why is it large and positive?
H2O is nonlinear and triatomicH2O has many rotational and vibrational motions
H2 is linear and diatomicH2 has less rotational and vibrational motions
The more complex the molecule, the higher the So
Spontaneity, Entropy & Free Energy
Free Energy and Chemical ReactionsStandard Free Energy Change
Go
the change in the free energy that occurs if the reactants in their standard states are changed to products in their standard states
can’t be measured directly calculate from other values allows us to predict the tendency for a reaction
to go
Spontaneity, Entropy & Free Energy
How do we calculate Go? Go = Ho - TSo (for a reaction carried
out at constant temperature) Use Hess’ Law Use Go
f (standard free energy of formation)Go = nGo
f (products) - nGof (reactants)
Spontaneity, Entropy & Free Energy
Calculate Go for the reaction at 25oC2SO2(g) + O2(g) --> 2 SO3(g)Substance Ho
f(kJ/mol) So (J/K.mol)SO2(g) -297 248SO3 -396 257O2 0 205
Spontaneity, Entropy & Free Energy
Calculate Go for the reaction Cdia --> Cgr using the following data:
Cdia + O2 --> CO2(g) Go = -397 kJCgr + O2 --> CO2(g) Go
= -394 kJ
Spontaneity, Entropy & Free Energy
Calculate Go for the reaction2CH3OH + 3 O2--> 2 CO2 + 4 H2OSubstance Go
f(kJ/mol)CH3OH -163O2 0CO2 -394H2O -229
Spontaneity, Entropy & Free Energy
The dependence of free energy on pressure How does pressure affect enthalpy and entropy?
Pressure does not affect enthalpyPressure does affect entropy because
pressure depends on the volume• 1 mole of a gas at 10.0 L has more positions
available than 1 mole of a gas at 1.0 L• Slarge volume > Ssmall volume
• Slow pressure > Shigh pressure
Spontaneity, Entropy & Free Energy
Given that G = Go + RTln(P) where G is the free energy at some P (not necessarily 1
atm) where Go is the free energy at 1 atm
Ex: N2(g) + 3 H2(g) --> 2 NH3(g)(lots of equations…lots of equations…)
G = Go + RT ln Q Q is the reaction quotient (from the law of mass action) T is the temperature in K R is the gas constant, 8.3145 J/mol.K
Spontaneity, Entropy & Free Energy
Calculate G at 25o C for the reaction CO(g) + 2 H2(g) --> CH3OH where carbon monoxide is 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.
What does the answer tell us about this reaction under these conditions?
Spontaneity, Entropy & Free Energy
Free Energy and Equilibrium Equilibrium occurs at the lowest value of
free energy available to the reaction system, i.e., when G = 0
At equilibrium, G = 0, Q = Keq soG = 0 = Go + RT ln Keq
Go = - RT ln Keq
Use this equation to find Keq given Go, or to find Go
given Keq
Spontaneity, Entropy & Free Energy
Relationship between Go and Keq
Go Keq= 0 1< 0 >1> 0 < 1
Spontaneity, Entropy & Free Energy
For N2 + 3 H2 --> 2 NH3, Go = - 33.3 kJ per mole of N2 consumed at 25oC. Predict the direction in which the reaction will shift to reach equilibriuma. PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00 x 10-2 atmb. PNH3 = 1.00 atm, PN2 = 1.00 atm, PH2 = 1.00 atm
Spontaneity, Entropy & Free Energy
4Fe + 3 O2 <====> 2Fe2O3 Calculate the equilibrium constant using the following information:
Substance Hof (kJ/mol) So(J/K.mol)
Fe2O3 -826 90Fe 0 27O2 0 205
Spontaneity, Entropy & Free Energy
Keq and temperatureWe used Le Chatelier’s Principle to
determine how Keq would change when temperature changes
Use G to determine the new Keq at a new temperature Go = -RT ln K = Ho - TSo
ln K = - Ho . 1 + So
R T R