spontaneity entropy___free_energy

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


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


Kinetics depends on the pathway taken tells us the speed of the process depends on

activation energytemperatureconcentrationcatalysts


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


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


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


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


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?


Predict the sign of the entropy change solid sugar is added to water iodine vapor condenses onto a cold

surface forming crystals


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


A process is spontaneous if the Suniverse is positive

If the Suniverse is zero, there is no tendency for the reaction to occur


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


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


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


Signs of Entropy ChangesSsys Ssurr Suniv Spontaneous? + + - - + - - +


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


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


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


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


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


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


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)


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


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


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


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


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)


Entropy is a state function entropy does not depend on the

pathway taken Srxn = nSoproducts - nSoreactant


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


Calculate So forAl2O3(s) + 3 H2(g) --> 2 Al(s) + 3 H2O(g)Substance So (J/K.mol)Al2O3 51H2 131Al 28H2O 189


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


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


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)


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


Calculate Go for the reaction Cdia --> Cgr using the following data:

Cdia + O2 --> CO2(g) Go = -397 kJCgr + O2 --> CO2(g) Go

= -394 kJ


Calculate Go for the reaction2CH3OH + 3 O2--> 2 CO2 + 4 H2OSubstance Go

f(kJ/mol)CH3OH -163O2 0CO2 -394H2O -229


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


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


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?


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


Relationship between Go and Keq

Go Keq= 0 1< 0 >1> 0 < 1


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


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


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

spontaneity entropy___free_energy

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