# Entropy, free energy and equilibrium by pptfiles

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```									Entropy, free energy and equilibrium
Spontaneity Entropy Free energy and equilibrium

Learning objectives
 Discuss what is meant by spontaneity  Discuss energy dispersal and its relevance to spontaneity  Describe the concept of a reversible process  Define entropy

Equilibrium
 At an equilibrium point, the system resists small disturbances (not necessarily large ones)
unstable

Locally stable

more stable

 At equilibrium, the rates of the forward and backward processes are equal

Spontaneity
 The tendency for a process to advance to equilibrium without external influence  Something that happens naturally is spontaneous  Any process will be spontaneous in one direction  The reverse is non-spontaneous  If work needs to be done, it is not spontaneous
 A rock naturally rolls down a hill - spontaneous  It must be pushed back up - nonspontaneous  A hot object naturally cools - spontaneous

Indicators of spontaneity
 What is the indicator of spontaneity?
 Heat evolved?

 But…endothermic reactions occur spontaneously as well (ice melting, salt dissolving)  Enthalpy is not an indicator of spontaneity, although most spontaneous processes are exothermic – energy is conserved not created  The amount of energy does not change in any process – but it is redistributed...

Various spontaneities: dispersal
 Matter disperses – gas fills a container, two liquids mix  Heat disperses – hot object cools on cold surface  Motion disperses – a ball stops bouncing  The reverses of these three well known processes never occur spontaneously

Reversibility
 A reversible process is one where the system and surroundings are changed to original values without any change  An irreversible process is one where the system and surroundings cannot be restored  A reversible process produces the maximum possible work

Reversibility and reality
 Reversibility only occurs when the system is in or almost at equilibrium – at an infinitesimal rate  In reality this does not obtain  Real processes produce less work than ideal processes  Spontaneous processes are irreversible  The reverse of a spontaneous process is nonspontaneous

Spontaneity and speed
 The speed of a reaction is not an indicator of its spontaneity.  Spontaneity is determined by the relative positions of the initial and final states (thermodynamic state functions)  Speed is determined by the pathway (kinetics)  Two independent regimes

Entropy – the mixing (distributing) link
 entropy measures distribution of energy over states  The more states available, the more entropy  It is a “state function” – depends only on initial and final states, not the pathway.  The entropy change for a process is  Systems move spontaneously to a state of greater entropy – greater distribution of energy  Disorder provides more states for energy distribution than ordered systems
 S  S final  S initial

Why do crystals form at all?
 Entropy is distribution of energy over microstates  Crystals are highly ordered arrangements
 Crystals should spontaneously fly apart to maximize disorder

 But...This view ignores energy of the lattice
 Energy input to break bonds corresponds to entropy decrease (localization of energy)

Don’t let them fool you
 A popular argument against evolution is that the formation of organized DNA molecules from a random soup of atoms and molecules contravenes Second Law  Just as crystals appear in a dish spontaneously so can DNA form from smaller units  N.B. Order (energy concentration) can appear spontaneously locally provided greater disorder (energy dispersal) is occurring elsewhere

Entropy and solubility
 Hydration increases entropy of the ions in the lattice
 Ions in solution have greater disorder

 but can decrease entropy of the solvent
 Solvent molecules now have greater order

 Excessive hydration by highly polarizing ions can reduce entropy of solvent
 CaSO4 is only slightly soluble

What will these socks ne’er be matched?
 Would you be stunned if the tumble dryer matched the socks?  Okay, you never match the socks anyway  Chaos in the sock drawer is natural  The same principles are applied to chemical change (sort of)

Chance meeting: entropy and probability
 Ordered states are less likely because there are fewer ways to obtain them
 Do our socks become matched spontaneously?  No, only one of many possible arrangements

 With only a few molecules the ordered state becomes massively less probable than a disordered state

Boltzmann and disorder
S  k ln W

 W is the number of possible arrangements of the state  k is Boltzmann’s constant =
 R/NA = 1.38x10-23 J/K

 The entropy is proportional to the natural log of the number of arrangements of the state

Entropy of a disordered system
 An ordered arrangement has W = 1, S = 0  Entropy of one mole of disordered molecules
S  k ln W  k ln 2
NA

 kN

A

ln 2  R ln 2

 S = 5.76 J/K

Entropy and gas expansion
 There is only one possible way for all the gas molecules to fill A and leave B empty. There are ways 2 N for NA molecules to occupy A and B  Entropy associated with gas mixing  Entropy associated with gas expansion: Doubling the volume doubles the number of positions (microstates) for distribution of energy
A

 S  R ln

V final V initial

Making sense of units and definition of entropy
 Units of entropy are J/K  How do these units connect to disorder and probability?  Disorder is not entropy  Disorder increases the number of microstates available  Clausius definition of entropy is: Change in entropy = (heat supplied)/temperature q  S sys  T

Work and gas expansion
 Work associated with isothermal reversible V  expansion of gas w   nR T ln  
2 rev

 V1 

 Isothermal means that ΔE = 0 = qrev + wrev  But...
V  q rev   w rev  nR T ln  2   V1 
 S sys  q rev T  V  nR T ln  2   V1  T V   nR ln  2   V1 

 Equivalent to result obtained from consideration of arrangements

Les Regles du Jeu (Rules of the game) Thermodynamics is the Law
 First Law: The total energy of a system and its surroundings is constant in any process  E sys  q  w  Second Law: In any spontaneous process, the total entropy of a system and its surroundings increases  S tot  0

Third Law of Thermodynamics
 The entropy of a perfectly ordered crystalline substance at 0K is zero
 Increasing T causes increase in entropy through molecular motion (rotational, vibrational and translational), and changes of state

Entropy and temperature
 Disorder and motion  Greater motion corresponds to greater number of microstates – entropy increases with T

Entropy of a system increases with T
 Increasing T increases entropy through greater molecular motion
 In a solid an increased number of vibrational energy states – more ways to distribute energy

 Phase changes cause step change because of increased number of microstates in less condensed phase

Standard molar entropy
 S° The entropy of one mole of the pure substance at 1 atm pressure and a specified temperature, usually 25°C
 Determined experimentally from heat capacity measurements

Comparison of different substances
 Gases have highest values  Solids have the lowest values

Standard entropy of reaction
S  S
o o o products

S

o reactants

In the reaction N2O4 = 2NO2
S  2S
o NO 2

S

o N 2O4

 Products have more particles than reactants  Predicting entropy change from chemical equation by counting particles

Entropy: connecting the microscopic to the macroscopic
 Microscopic:
 Measure of microstates and disorder

 Macroscopic:
 Indicator of spontaneous process

Three results

 S tot   S sys   S surr

 Stotal > 0 the process is spontaneous

 Stotal < 0 the process is nonspontaneous
 Stotal = 0 the process is at equilibrium

Surroundings
 Entropy change for the system is obtained from the entropies of the initial and final states  What about the surroundings?  At constant pressure, the entropy change in the surroundings is related to the enthalpy change for the system
 S surr    H T

Enthalpy change of system determines entropy change of surroundings
 Heat released by the system increases the disorder of the surroundings.  The effect of this is modulated by the temperature:
 At low temperature the effect is much more significant  At high temperature, where there is already considerable disorder, the effect is muted – the difference between tossing a rock into a calm pool (low T) and a storm-tossed ocean (high T)

Land of the Free... Energy
 Since Stotal is obtained from the Ssystem and the ΔHsystem, everything can be written in terms of the system:  Gibbs free energy
G  H  TS  State function, depends only on initial and final states G  H  TS

Significance of ΔG
 S tot   S sys   S surr

 But from before,
 S tot   S sys   H sys T

T  S tot  T  S sys   H sys

 So…

T  S tot    G

ΔG is indicator of spontaneity
 G   T  S tot

1. ΔG < 0 reaction always spontaneous 2. ΔG > 0 reaction always nonspontaneous 3. ΔG = 0 reaction at equilibrium The Gibbs Free Energy is a measure of the total entropy change

Four possible conditions
ΔH + + ΔS + + ΔG - Or + +
Spontaneity

Example
2NO2 = N2 + 2O2
N2 + 3H2 = 2NH3

Spontaneous at all T

Spontaneous at low T, nonspontaneous at high T Nonspontaneous at all T

3 O2 = 2 O 3

- Or +

Spontaneous at high T, nonspontaneous at low T

2HgO = Hg + O2

Standard free energies
 Solids liquids and gases in pure form at 1atm pressure  Solution at 1 M concentration  Standard temperature usually 25°C  Standard free energy change: ΔG° Change in free energy that occurs when reactants in standard states are converted to products in their standard states

ΔG° as a predictor of reactions
 Consider the reaction

N 2 ( g )  3 H 2 ( g )  2 NH 3 ( g )
    We want to calculate ΔG° Need ΔH° and ΔS° ΔH° is equal to ΔH°(formation) for NH3 ΔS° comes from the S° values for the reactants and product ΔH° = 2 x -46.1 kJ/mol = -92.2 kJ/mol ΔS° = -198.7J/mol K ΔG°(25°C) = -92.2 – 298x-198.7x.001 kJ/mol = -33.0 kJ/mol

Standard free energy of formation
 The free energy of formation of one mole of the substance in its standard state from the most stable forms of the elements in their standard states  Thus the ΔG°f for NH3 is given by -33.0/2 kJ/mol = -16.5 kJ/mol
 Elements in the standard state have ΔG°f = 0

Standard free energy of formation of some common compounds
Substance

Formula

ΔGfº/kJ /mol

Substance
Nitrogen dioxide Water

Formula

ΔGfº/kJ/ mol

Acetylene

C2H2
NH3 CO2 C2H4

209.2
-16.5 -394.4 68.1

NO2
H2O C C

51.3
-237.2 2.9 0

Ammonia

Carbon dioxide
Ethylene

Diamond

Graphite

The importance of the state
 Using reactants in different states will require modification to calculation for ΔG°  Consider graphite and diamond – two forms of carbon. Is it perhaps surprising that diamond is less stable than graphite? ΔG°f (diamond) = 2.9 kJ/mol  Free energy change for conversion of diamond into CO2 is larger than for conversion of graphite into CO2

ΔG°f and stability
 Many common compounds are unstable with respect to their elements – NO2, C2H4 and C2H2
 They will spontaneously decompose to the elements according to thermodynamics  However, they are generally regarded as stable  Kinetic barriers prevent rapid decomposition

Thermodynamic functions and chemical processes
 The foolish man builds his process on a foundation without ΔGfº  ΔG°f calculations can save a lot of unnecessary work  Indicates favourability of a reaction under standard conditions  If unfavourable, need to either modify the conditions or explore alternative synthesis pathways
 Example:  The formation of NO is not favoured from N2 and O2;  but it is favoured by reaction of O2 with NH3

Accounting for actual conditions
 In most reactions, the reactants and products are not in standard states

 G   G  RT ln Q
o

Q is the reaction quotient – similar in form to K
 Pressures for gases  Concentrations for liquids

N 2 ( g )  3 H 2 ( g )  2 NH 3 ( g )

Q 

p

2 NH

3

pN2 p

3 H2

Free energy and equilibrium
 Q «1, ΔG < 0 Drives towards products  Q » 1, ΔG > 0 Drives back towards reactants  At equilibrium, ΔG = 0

ΔG° = -RT ln K

Relationship between ΔGfº and K
ΔGº Ln K <0 >0 =0 >0 <0 =0 K >1 <1 1
Comment

Mainly products

Mainly reactants

Even ratio of reactants and products

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