Chapter 20 - Spontaneous Change and Free Energyweb.unbc.ca/chemistry/chem101/CHEM101 Ch20...

Chapter 20 - Spontaneous Change and Free Energy

- the governing laws of the Universe are the three laws of thermodynamics

- these can be said in a number of ways but the best paraphrase that I know is:

"you can't win; you can't even break even;

and you can't quit the game"

"you can't win"Can never get more energy out of reaction than there is in the reaction. That is, the maximum amount of chemical energy available from a chemical reaction is the amount of energy put into creating the reactants in the first place.

"you can't break even"Some of the energy from a chemical reaction will be available to do work but some will be lost to entropy.

"you can't quit the game"The only perfect state is 'absolute zero' but as this is impossible to achieve, thermodynamics always applies to all systems.

20-1 The Meaning of Spontaneous Change

- a spontaneous process occurs when nothing interfers with a system - that is, no outside action is necessary for the process to proceed

- a non-spontaneous process requires the intervention of something outside of the system (i.e. heat or light)

Many examples of both:- cleaning a room - rusting metal- popping balloons - food going bad- leaking balloons - cooking

- some spontaneous processes can be reversed

i.e. 4Fe(s) + 3O2 (g) → 2Fe2O3 (s) - rusting iron2Fe2O3 (s) + 3C(s) → 4Fe(s) + 3CO2 (g) - refining iron

the former happens spontaneously in the presence of air and water; the latter requires the addition of heat and carbon

Two key points:

1) if a process is spontaneous then the reverse process has to be non-spontaneous.

(you can't go downhill both ways!)

2) both spontaneous and non-spontaneous reactions are possible or can occur

- it's just that non-spontaneous reactions require an external force of some sort

Can we make predictions about the direction of a reaction?Yes - if we understand the energy terms involved.

Enthalpy (symbol: H) is a measure of the thermicity of a chemical reaction

- exothermic - heat is given off- endothermic - heat is absorbed or taken in

P. Bertholet & J. Thomsen proposed that the direction of a reaction could be determined by its enthalpy alone

- a good first step as most exothermic reactions are spontaneous

However, not all!Consider the cooling bags that are used in sports. They get spontaneously colder and are endothermic. That is, the reaction draws in heat from the surroundings and yet, is spontaneous.

Note: recall Chapter 7 (pg. 221) where the book defines the terms "system" and "surroundings"

Refresh your memory of thermochemistry.

20-2 The Concept of Entropy

- thought experiment:

Consider to volumes or flasks joined by a stem (say A and B):

- with one atom in the system, it is possible for the atom to be in one container or the other (50:50 chance)

- with two atoms, there are other possibilities:Container A Container B

2 01 11 10 2

- subtle point - each atom could occupy each container and each would represent a unique configuration

- with three atoms, there are eight possibilities:AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB

which occur in a 1:3:3:1 ratio

Pascal's Triangle:1

1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 11 6 15 20 15 6 1

1 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 11 10 45 120 210 252 210 120 45 10 1

What has this to do with "entropy"?

- if we fill one of the vessels with gas and then open the tap between the two vessels, the molecules will redistribute themselves:

- but where did the energy to do this come from?- the molecules haven't changed- the temperature is the same

- it comes from maximizing the number of possible combinations- from increasing the number of possible arrangements of molecules between the chambers

The greater the number of configurations of the microscopic particles (atoms, ions, molecules) among the energy levels in a particular system, the greater the entropy of the system.

- the Boltzmann Equation quantifies this:

S = k(ln W)

- where 'S' is the symbol for entropy'k' is the Boltzmann constant, R/NA

'W' is the number of possible configurations or microstates

- this equation is a little unwieldy as we can never truly measure 'W' but conceptually, it is like tossing Avogadro's number of coins and monitoring the results for the distribution of heads and tails many, many times

- it's not that the particles couldn't rush back and refill the first container, it is that it is highly improbable that they will!

(Imagine tossing a coin and coming up with, say, 6.022 x1023 heads in a row! Many times!)

- one way to think about entropy is to view it as the randomness of the Universe - given a chance, random occurrences settle into the most probable state which is the most random configuration possible

- consider the consequences of tossing a coin six times- there are seven possible outcomes:

Heads Tails6 0 ³ only 1 way to do this!5 14 23 3 ³ but 20 ways to do this!2 41 50 6

- let's try it:

00100029123821110731115641723551618461110372228011910010

'05 #2'05 #1'04 #2'04 #1TailsHeads

Pascal's Triangle:1

1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 11 6 15 20 15 6 1

1 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 11 10 45 120 210 252 210 120 45 10 1 = 1024

As the number of samples increases, the distribution of ranges changes shape and narrows.

Entropy Change:- an entropy change is based on two measurable quantities:

Temperature and heat

)S = qrev

T

- and as 'S' is a state function, so is )S

- the proportionality to heat comes from the available energy levels - the more energy, the more levels occupied

- it is hard to know how entropy will change for a given reaction - not always obvious- usually, entropy parallels the changes in the number of particles and the volume of that they have available

(more volume = more possible arrangements)

for example, consider the melting of ice- this is endothermic

- it absorbs heat from the surroundings- but it is also spontaneous

- driven by entropy- rigid structure gives way to liquid with more degrees of freedom

- the molecules have gained some rotational and translational freedom- more energy levels ö more entropy- but the volume has decreased!

general "Rules" are provided on pg. 788entropy increases:

- solids form pure liquids or solutions- solids or liquids are converted to gases- the number of molecules of gas increase- the temperature of a substance increases

but there are always exceptions!

20.3 Evaluating Entropy and Entropy Change

- in a few simple cases, it is possible to evaluate )S- it is also possible to evaluate absolute entropy, S

phase transitions- in an equilibrium, the exchange of heat can be carried out reversibly and the heat must equal the entropy

)Str = )Htr

Ttrwhere 'tr' means transition (i.e. fusion or vaporization)

- from the heat of fusion and the normal melting point, for example, we can work out the entropy:

H2O(s, 1 atm) W H2O(l, 1 atm) )Hfus = 6020 J/mol at 273K

therefore, )Sfus = (6020 J/mol)/273.15K = 22.04 J/K@mol

- for vapourization,

H2O(l, 1 atm) W H2O(g 1 atm) )Hvap = 40,700 J/mol at 373K

therefore, )Svap = (40,700 J/mol)/373.15K = 109.1 J/K@mol

Trauton's Rule:- says that for many liquids, the standard entropy of vaporization is:

)SEvap = )HE

vap ≈ 87 J/mol@KTbp

- while this is not true for water (109.1 J/mol@K) because of hydrogen bonding, it is useful for many other compounds

- note that this also implies that )HEvap increases with

increasing boiling point - makes sense

Absolute Entropies- the entropy of a pure, perfect crystal at 0K is zero

- this is the third law of thermodynamics- everything is above absolute zero so

everything has entropy- the third law gives us a zero point against which to make measurements

- the entropy of one mole of substance in its standard state is called the "standard molar entropy", SE

- given in Appendix D -

- to calculate the entropy change in a reaction:

)SE = G<pSE(products) - G<rSE(reactants)

- where G means "the sum of" and <p and <r are the stoichiometric coefficients for the product and reactant species, respectively

i.e. 2CO(g) + O2 (g) → 2CO2 (g)

)SE = 2SECO2 - 2SE

CO - SEO2

= 2(213.7) - 2(197.7) - (205.1) = -173.1 J/mol@K

- note that this represents an entropy decrease which is consistent with having fewer molecules on the product side of the reaction than on the reactant side

- also, that SE for CO2 is 213.7 compared to only197.7 for CO with the difference arising from having more vibrationally active modes and that SE for O2 is not zero despite the fact that it is a pure substance

- in general, the more complex a molecule is, the larger the molar entropy of the substance

20.4 Criteria for Spontaneous Change: The Second Law

If spontaneity is tied to increasing entropy, then how do we explain the freezing of ice at -10EC?- this is definitely a spontaneous process, but it involves a decrease in entropy (a loss of both vibrational and translational freedom)- the way out is to realize that we must consider the entropy of both the system and the surroundings:

)Stotal = )Suniverse = )Ssystem + )Ssurroundings

and it is )Suniverse that must be greater than zero

That is, the entropy of the Universe must increase for any spontaneous reaction

- this is one way of stating the second law of thermodynamics

From this, if:)Ssystem is +ve and )Ssurroundings is +ve

then the reaction will always be spontaneous)Ssystem is -ve and )Ssurroundings is -ve

then the reaction will always be non-spontaneousand otherwise, it is ambiguous

Free EnergyIf we are to use )Suniverse = )Ssys + )Ssurr

then we have a problem because )Ssurr is actually impossible to measure. (We can't really make statements about the whole Universe outside of the system!)

In Section 7-6, we saw that qsurr = -qsys = -)Hsys

and from this, )Ssurr = -)Hsys

T

by substituting this into our equation and multiplying through by -T, we get:

-T)Suniverse = )Hsys - T)Ssurr

we redefine the left hand side to be a new function, G, that is the "Gibbs Free Energy" and we get:

Free energy change: )G = )H - T)SFree energy: G = H - TS

now we have a criteria to define spontaneity under all circumstances

if )G < 0 then the process is spontaneous as writtenif )G = 0 then the process is at equilibriumif )G > 0 then the process is not spontaneous as written

Applying free energy analysis to reactions gives:)H )S )G- + - spontaneous at all temperatures- - - " at low temperatures- - + " at high temperatures+ + - " at high temperatures+ + + " at low temperatures+ - + not spontaneous at any temperature

now we have a useful tool for determining whether or not a reaction will proceed from a thermodynamic point of view (remembering that the kinetics are important, too)

and we know why molecules decompose at high temperatures - eventually T)S will always win and relative to their constituent atoms, molecules have a positive entropy

20.5 Standard Free Energy Change, )GE

- the standard free energy of formation, )GE, is the free energy change for a reaction in which a substance in its standard state is formed from its elements in their reference forms in their standard state

- as in Section 7.8, this leads to a value of zero for the free energy of formation of the elements in their reference form at a pressure of 1 bar

We can calculate )GE from the data in Appendix Dnote: )G changes sign if the reaction is reversed

)G is the sum of the individual steps

)GE = G<pGE(products) - G<rGE(reactants)

where the symbols have the same meanings as beforeExample:

2CO(g) + O2 (g) → 2CO2 (g) )HE = -566 kJ/mol)SE = -173.1 J/mol@K

)GE = )HE - T)SE = -566 - (298.15x-173.1)= -514.4 kJ

or: )GE = 2()GECO2) - 2()GECO) - ()GEO2)= 2(-394.4) - 2(-137.2) - (0) = -514.4 kJ

20.6 Free Energy Change and Equilibrium

note that if )G < 0 means the process is spontaneous and )G > 0 means that the process is non-spontaneous, then at some point, as the temperature increases, )G must be 0 and equilibrium must be achieved.

Relationship of )GE to )G for Non-standard Conditions- )GE is the free energy at the "standard state"- it only applies at one temperature and pressure, and hence, is of limited use

- if we want to describe a reaction at any temperature, then we want non-standard conditions

)G = )GE + RTlnQ

where Q is the "reaction quotient" - our instantaneous Keq value - the concentrations at our non-standard state

this equation can be used to determine spontaneity under any conditions of composition but, at equilibrium, we have Q = Keq and )G = 0, so,

)GE = -RTlnKeq

if we know )GE at any temperature, then we know Keqand vice versa i.e. at 298 K, )GE Keq

+200 9.1 x10-36

+50 1.7 x10-9

0 1.0-200 1.1 x1035

Further, the meaning of free energy change:

)G )GE

< 0 spontaneous spontaneous for reactants andproducts in their standard state (Keq>1)

= 0 equilibrium at equilibrium for reactants and products in their standard state (Keq=1)

>0 non-spontaneous non-spontaneous for reactants and products in their standard state (Keq<1)

Thermodynamic Equilibrium Constants: Activities

definition of activity:

α = the effective concentration of a substance in the systemthe effective concentration of that substance in a

standard reference state

i.e. for concentrations of solutions, the reference state could be 1.0 M and for a 0.1M solution, we would have:

α = 0.1M = 0.11.0M

- hence, we can use logarithms on activities as they are unitless quantities- mostly, as we have been doing, we assume that activity equals concentration

- when we write an equilibrium expression in terms of activities, it is called a "thermodynamic equilibrium constant"

- using the thermodynamic equilibrium constant, we can convert Keq to )GE and back again

Example:Determine the equilibrium constant at 298.15K for the dissolution of calcium carbonate in an acidic solution.

equilibrium: CaCO3 (s) + 2H+(aq) → Ca2+

(aq) + CO2 (g) + H2O(l)

)GE = G<pGE(products) - G<rGE(reactants)= (-553.6 + -394.4 + -237.1) - (-1129 + 2(0)) = -56.1

and lnKeq = (-)GE/RT) = 56.1 kJ/mol = 22.63(8.3145 J/mol@K)(298.15K)

Keq = e22.63 = 6.73 x109

20.7 )GE and Keq as Functions of Temperature- not surprisingly, )GE is temperature dependent- we can use this to calculate Keq at various temperatures:

- RTlnKeq = )GE = )HE - T)SE

- we can also work out the relationship between )HE and )SE by measuring the change in Keq with temperature

lnKeq = -)HE + )SERT R

- if we assume that )HE and )SE are constants over the temperature range then this is a straight line

Figure 20-12

- we can also manipulate the equation to give:

ln K2 = -)HE (T2-1 - T1

-1)K1 R

- which is in the same form as the Clausius-ClapeyronEquation:

ln P2 = -)Hvap (T2-1 - T1

-1)P1 R

- and the Arrhenius equation:ln k2 = -)Eact (T2

-1 - T1-1)

k1 R - all energy processes!

20.8 Coupled Reactions- just because a reaction is non-spontaneous, doesn't mean it won't go- we have seen two ways to drive a reaction

1) change the reaction conditions to ones that make the reaction spontaneous

2) carry out the reaction by electrolysis

- a third method is to use a coupled reaction- to combine a non-spontaneous reaction with a spontaneous reaction so that the overall reaction is spontanteous

Example:Cu2O(s) → 2Cu(s) + ½O2 (g) )GE673K = +125kJ

which means it wouldn't go by itself

however,Cu2O(s) → 2Cu(s) + ½O2 (g) )GE673K = +125 kJ

C(s) + ½O2 (g) → CO(g) )GE673K = -175 kJCu2O(s) + C(s) → 2Cu(s) + CO(g) )GE673K = -50 kJ

- the overall reaction is spontaneous at 673K and is the sort of reaction that is used to refine copper ore to metal