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Chapter 19 Spontaneous Change: Entropy and Free Energy. Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php. Spontaneous processes. We have a general idea of what we consider spontaneous to mean:

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chapter 19 spontaneous change entropy and free energy

Chapter 19Spontaneous Change: Entropy and Free Energy

Dr. Peter Warburton

peterw@mun.ca

http://www.chem.mun.ca/zcourses/1051.php

spontaneous processes
Spontaneous processes
  • We have a general idea of what we consider spontaneous to mean:
  • A spontaneous process WILL OCCUR in a system WITHOUT any outside action being performed on the system.
spontaneous processes3
Spontaneous processes

Object falling to earth is spontaneous

Ice melting above zero Celcius is spontatneous

spontaneous processes4
Spontaneous processes
  • Icewillmelt above zero Celcius.
  • We don’t have to DO anything!
  • Objectswillfallto earth.
  • We don’t have to DO anything!
spontaneous processes5
Spontaneous processes
  • Since we DON’T have to DO anything for these spontaneous processes to occur it APPEARS that an overall energy change from potential energy
  • to kinetic energy
  • IS SPONTANEOUS
non spontaneous processes
Non-spontaneous processes
  • We have a general idea of what we consider non-spontaneous to mean:
  • A non-spontaneous process WILL NOT OCCUR in a system UNTIL an outside action is performed on the system.
non spontaneous processes7
Non-spontaneous processes

Object rising from earth is

non-spontaneous

Ice freezing above zero Celcius is

non-spontatneous

non spontaneous processes8
Non-spontaneous processes
  • We canmakewater freeze above zero Celcius by increasing the pressure.
  • We can make an object rise from the earth by picking it up.
non spontaneous processes9
Non-spontaneous processes
  • Since we DO have to ACT for these non-spontaneous processes to occur it APPEARS that an overall energy change from kinetic energy
  • to potential energy
  • IS NON-SPONTANEOUS
chemistry and spontaneity
Chemistry and spontaneity
  • We know there are chemical processes that are spontaneousbecause we can put the chemical system together and reactants become products without us having to do anything.
  • H3O+ (aq) + OH- (aq)  2 H2O (l)
chemistry and non spontaneity
Chemistry and non-spontaneity
  • We know there are chemical processes that are non-spontaneousbecause we can put the chemical system together and reactants DO NOT become products.
  • The system we put together stays like it is
  • UNTIL WE CHANGE SOMETHING!
  • 2 H2O (l)  2 H2 (g) + O2 (g)
spontaneous vs non spontaneous
Spontaneous vs. non-spontaneous
  • It is obvious by the examples we’ve looked at that the opposite of every spontaneous process is a non-spontaneous process.
  • In chemical systems we’ve seen that if we put a chemical system together a reaction occursuntil the system reaches equilibrium.
  • Whether the forward reaction or the reverse reactiondominates depends on which of the two reactions is spontaneous at those conditions!
spontaneity and energy
Spontaneity and energy
  • In our examples it APPEARED that the spontaneous process ALWAYS takes a system to a lower potential energy.
spontaneity and energy15
Spontaneity and energy
  • If this were true all exothermic processes would be spontaneous and all endothermic processes would be non-spontaneous.
  • THIS ISN’T TRUE!
  • H2O
  • NH4NO3 (s)  NH4+ (aq) + NO3- (aq)
  • is spontaneous even though
  • DH = +25.7 kJ
recall the first law
Recall the First Law
  • The First Law of thermodynamics stated that the energy of an ISOLATED system is constant.
  • What’s the largest ISOLATED system we can think of?
  • It’s the UNIVERSE!
  • The energy of the universe is constant!
recall the first law17
Recall the First Law
  • On the universal scale, there is no overall change in energy, and so lower energy CANNOT be the only requirement for spontaneity.
  • There must be something else as well!
further proof lower energy isn t enough
Further proof lower energy isn’t enough
  • If an ideal gas expands into a vacuum at a constant temperature, then
  • no work is done
  • and
  • no heat is transferred
further proof lower energy isn t enough19
Further proof lower energy isn’t enough
  • No work done and no heat transferred means
  • NO OVERALL CHANGE in energy
  • of the system
further proof lower energy isn t enough20
Further proof lower energy isn’t enough
  • This spontaneous process has no overall change in energy!
entropy
Entropy
  • Entropy (from Greek, meaning
  • “in transformation”)
  • is a thermodynamic property
  • that relates
  • the distribution of the total energy of the system
  • to the available energy levels of the particles.
entropy22
Entropy
  • A general way to envision entropy is
  • “differing ways to move”
  • Consider mountain climbers on a mountain. Two factors affect the distribution of mountain climbers on a mountain:
  • Total energy of all the climbers
  • and how many places can you stop on the mountain
consider hungry mountain climbers
Consider hungry mountain climbers
  • Hungry mountain climbers have
  • little total energy amongst themselves to climb a mountain, so most of them are near the bottom, while some are distributed on the lower parts of the mountain.

Few energy levels can be reached!

consider well fed mountain climbers
Consider well fed mountain climbers
  • Well-fed mountain climbers have
  • more total energy amongst themselves to climb a mountain, so the climbers will be more spread out on the whole mountain.

More energy levels can be reached!

temperature and total energy
Temperature and total energy
  • The total energy shared by molecules is related to the temperature.
  • A given number of molecules at a low temperature (less total energy) have less“differing ways to move”than the same number of molecules at a high temperature (more total energy).
higher mountain means more places to stop
Higher mountain means more places to stop
  • If a well-fed mountain climber tries to climb Signal Hill, they will most likely reach the top. They have only a few places to stop (levels)because Signal Hill is a small mountain.
  • The same well-fed mountain climber on Mount Everest has a greater number of places to stop(levels) because it is a larger mountain.
volume and energy levels
Volume and energy levels
  • The number of levels of energy distribution of molecules is related to the volume.
  • A given number of molecules in a small volume have less“differing ways to move”than the same number of molecules in a larger volume.
entropy28
Entropy
  • The greater the number of “differing ways to move” molecules can take amongst the
  • available energy levels of a system
  • of a given state (defined by temperature and volume, and number of molecules),
  • the greater the entropy of the system.
expansion into vacuum
Expansion into vacuum
  • A gas expands into a vacuum because the increased volume allows for a greater number of“differing ways to move” for the molecules, even if the temperature is the same.
expansion into vacuum30
Expansion into vacuum
  • That is, the entropy increases when the gas is allowed to expand into a vacuum.
  • Entropy increase plays a role in spontaneity!
entropy is a state function
Entropy is a state function
  • Entropy, S, is a state function like enthalpy or internal energy.
  • The entropy of a system DEPENDS ONLY on the current state
  • (n, T, V, etc.) of the system, and NOT how the system GOT TO BE in that state.
boltzmann equation and entropy
Boltzmann equation and entropy

More available energy levels when the size of a box increases – like expanding a gas into a vacuum – ENTROPY INCREASES!

More energy levelsare accessible when the temperature increases – ENTROPY INCREASES!

change in entropy is a state function
Change in entropy is a state function
  • Because entropy is a state function, then change in entropy DS is ALSO a state function.
  • The difference in entropy between two states ONLY depends on the entropy of the initial and final states, and NOT the path taken to get there.
hess s law
Hess’s Law
  • Recall Hess’s Law – as long as we get from the same initial state to the same final statethen DH will be the same regardless of the steps we add together.
  • Change in entropy DS will work exactly the same way! As long as we get from the same initial state to the same final statethen DS will be the same regardless of the steps we add together.
boltzmann equation and entropy35
Boltzmann equation and entropy
  • n, T, V help define the number of states(number of available energy levels) the system can have.
  • The many “different ways to move” of molecules in a particular state are called microstates.
  • Hopefully it makes sense that more total states should automatically mean more total microstates.
  • The number of microstates is often symbolized by W.
playing cards
Say we have a deck of 52 playing cards.

Choosing oneplaying card is a state.

If we choose the first card out of the pack, there are 52 microstatesfor thisfirst state.

The second card (second state)we choose has 51 microstates, and so on.

Playing cards
playing cards37
Overall there are

W =52!  8 x 1067

possible distributions

(total microstates) for

52 playing cards!

Playing cards
playing cards and coin flips
If we flip 52 coins(a coin is one state), with two possible microstates(heads or tails) each, there are

W =252 = 4.5 x 1015

possibledistributions.

(total microstates)

A deck of 52 playing cards has greater entropy than 52 coins!

Playing cards and coin flips
boltzmann equation and entropy39
Boltzmann equation and entropy
  • Ludwig Boltzmann formulated the relationship between the number of microstates (W) and the entropy (S).
  • S = k ln W
  • The constant k is the Boltzmann constant which has a value equal to the gas constant R divided by Avagadro’s number NA
boltzmann equation and entropy40
Boltzmann equation and entropy
  • S = k ln W
  • where k = R / NA
  • k = (8.3145 JK-1mol-1) / (6.022 x 1023mol-1)
  • k = 1.381 x 10-23 JK-1
  • We can see the units for entropy will be Joules per Kelvin(JK-1)
measuring entropy change
Measuring entropy change
  • From the
  • units for entropy(JK-1)
  • we get an idea of how we might measure entropy change DS
  • It must involve some sort ofenergy changerelative to thetemperature change!
measuring entropy change42
Measuring entropy change
  • DS = qrev / T
  • The change in entropy is the heat involved in areversible process at a constant temperature.
heat is not a state function
Heat IS NOT a state function
  • Since heat IS NOT a state function we need a reversible process to make it ACT LIKE a state function.
reversible processes
Reversible processes
  • In a reversible process a change in one direction is exactly equal and opposite to the change we see if we do the change in the reverse direction.
  • In reality it is impossible to make a reversible process without making an infinite number of infinitesimally small changes.
reversible processes45
Reversible processes
  • We can however imagine the process is done reversibly and calculate the heat involved in it, so we can calculate the reversible entropy change that could be involved in a process.

DSrev = qrev / T

endothermic increases in entropy
Endothermic increases in entropy

In these three processes the molecules gain greater “differing ability to move.”

The molecules occupy more available microstates at the given temperature, and so the entropy increases in all three processes!

generally entropy increases when
Generally entropy increases when…
  • …we go from solid to liquid.
  • …we go from solid or liquid to gas.
  • …we increase the amount of gas in a reaction.
  • …we increase the temperature.
  • …we allow gas to expand against a vacuum.
  • …we mix gases, liquids, or otherwise make solutions of most types.
problem
Problem
  • Predict whether entropy increases, decreases, or we’re uncertain for the following processes or reactions:

Answers: a) decreases b) increases c) uncertain d) increases

evaluating entropy and entropy changes
Evaluating entropy and entropy changes
  • Phase transitions – In phase transitions the heat change does occur reversibly, so we can use the formula
  • DSrev = qrev / T
  • to calculate the entropy change. In this case the heat is the enthalpy of the phase transition and the temperature is the transition temperature
  • DS = DHtr / Ttr
evaluating entropy and entropy changes50
Evaluating entropy and entropy changes
  • Phase transitions – For water going from ice (solid) to liquid, DHfus = 6.02 kJmol-1 at the melting point (transition temp.) of 273.15 K (0 C)
  • DSfus = DHfus / Tmp
  • DSfus = 6.02 kJmol-1 / 273.15 K
  • DSfus = 22.0 JK-1mol-1
evaluating entropy and entropy changes51
Evaluating entropy and entropy changes
  • Phase transitions – For water going from liquid to gas, DHvap = 40.7 kJmol-1 at the boiling point (transition temp.) of 373.15 K (100 C)
  • DSvap = DHvap / Tbp
  • DSvap = 40.7 kJmol-1 / 373.15 K
  • DSvap = 109 JK-1mol-1
problem52
Problem
  • What is the standard molar entropy of vapourisation DSvap for CCl2F2 if its boiling point is -29.79 C and DHvap = 20.2 kJmol-1?

Answer: DSvap = 83.0 JK-1mol-1

problem53
Problem
  • The entropy change for the transition from solid rhombic sulphur to solid monoclinic sulphur at 95.5 C is DStr = 1.09 JK-1mol-1. What is the standard molar enthalpy change DHtr for this transition?

Answer: DHtr = 402 Jmol-1

absolute entropies
Absolute entropies
  • Say we imagine a system of molecules that has no total energy. At this zero-point energy there can ONLY be ONEpossible distribution of microstates, as no molecule has the energy to occupy a higher energy level.
  • The entropy CAN NEVER get smaller than its value in this situation, so we define the entropy S of this situation as ZERO.
absolute entropies55
Absolute entropies
  • This kind of imagining is the Third Law of Thermodynamics which states that
  • The entropy of a pure perfect crystal at 0K is zero.
  • At conditions other than at absolute zero, our entropy is that of the perfect system (zero)PLUS any entropy changes that come changing temperature and/or volume.
  • These are absolute entropies!
standard molar entropies
Standard molar entropies
  • One mole of a substance in its standard state will have an absolute entropy that we often call the standard molar entropy S.
  • These are usually tabulated at 298.15 K
  • In a chemical process we can then use these standard molar entropies to calculate the entropy change in the process.
standard molar entropies58
Standard molar entropies
  • DS = [SnpS(products) - SnrS(reactants)]
  • Hopefully this looks somewhat familiar!
  • We have seen a special treatment of Hess’s Law in Chem 1050 where
  • DH = [SnpDHf(products) - SnrDHf(reactants)]
  • We can do something similar with ANY thermodynamic property thatIS A STATE FUNCTION!
standard molar entropies59

Enthalpies of formation ARE NOT absolute!

Standard molar entropies
  • DS = [SnpS(products) –
  • SnrS(reactants)]
  • DH = [SnpDHf(products) –
  • SnrDHf(reactants)]
problem60
Problem
  • Use the data given to calculate the standard molar entropy change for the synthesis of ammonia from its elements.
  • N2 (g) + 3 H2 (g)  2 NH3 (g)
  • S298 for N2 = 191.6 JK-1mol-1
  • S298 for H2 = 130.7 JK-1mol-1
  • S298 for NH3 = 192.5 JK-1mol-1

Answer: -198.7 JK-1  mol-1 (per mole of rxn)

problem61
Problem
  • N2O3 is an unstable oxide that readily decomposes. The decomposition of 1.00 mol N2O3 to nitrogen monoxide and nitrogen dioxide at 25 C is accompanied by the entropy change DS = 138.5 JK-1mol-1. What is the standard molar entropy of N2O3 (g) at 25 C?
  • S298 for NO (g) = 210.8 JK-1mol-1
  • S298 for NO2 (g) = 240.1 JK-1mol-1

Answer: 312.4 JK-1 mol-1

the second law of thermodynamics
The second law of thermodynamics
  • We’ve seen that entropyMUST play a role in spontaneity, because the total energy of the universe doesn’t change.
  • We could say that an entropy increaseleads to spontaneity, but we have to be careful.
the second law of thermodynamics63

-ve since “freezing” is the reverse of “fusion” (like we do in Hess’s Law)

The second law of thermodynamics
  • Ice freezing below 0 Celcius is spontaneous, but the entropy of the water decreases in the process!
  • DSfreeze = -DHfus / Tmp
  • SinceDHfusandTmpare+ve,
  • thenDSfreeze is –ve!
the second law of thermodynamics64
The second law of thermodynamics
  • The water is ONLY the system.
  • The rest of the universe (the surroundings) must experience an opposite heat change as it takes the heat the freezing water gave off (a +ve DH for the surroundings), which means
  • the entropy of the REST OF THE UNIVERSEINCREASES in the process of water freezing!
the second law of thermodynamics65
The second law of thermodynamics
  • The total entropy change in any process is the entropy change for the systemPLUS
  • the entropy change for the surroundings
  • DSuniverse= DStotal =DSsys+DSsurr
  • Now we can connect entropy and spontaneity!
the second law of thermodynamics66
The second law of thermodynamics
  • In any spontaneous process
  • the entropy of the universe INCREASES.
  • DSuniverse =DSsys+DSsurr> 0
  • This is the Second Law of Thermodynamics!
water freezing
Water freezing
  • So while water freezing below zero Celciusdecreases the entropy of the system, the heat given off to the surroundingsincreases the entropyof the surroundingsto a greater extent.
  • The total entropy change of the universe is positiveand the process of water freezing below 0 Celcius is spontaneous!
free energy
Free energy
  • We’ve seen entropy increases when molecules have more ways to distribute themselves amongst the
  • energy levels.
free energy69
Free energy
  • However, some of the energy a molecule uses to put itself at a higher energy levelCAN NO LONGER be used to to do work becausedoing work would put the molecule backat a lower energy level, which would automatically decrease entropy.
  • The energy is NOT free
  • (or available) to be used!
free energy70

DS = DHrev / T

Reversible since the rest of the universe is SO BIG

then

TDS = DH

Free energy

DSuniverse =DSsys+DSsurr

  • TDSuniverse =TDSsys+ TDSsurr
  • TDSuniverse =TDSsys+DHsurr
free energy71
Free energy
  • TDSuniverse =TDSsys+DHsurr
  • TDSuniverse =TDSsys-DHsys

Since DHsys = -DHsurr

free energy72
Free energy

TDSuniverse =TDSsys – DHsys

-TDSuniverse =DHsys- TDSsys

  • DG =DHsys– TDSsys
  • DG is the free energy
  • (Gibbs free energy)
free energy73
Free energy
  • DG = -TDSuniv
  • For a spontaneous processDSuniv > 0, which means for a spontaneous processDG < 0!
free energy74
Free energy
  • DG<0 is spontaneous
  • DG>0 is non-spontaneous
  • DG=0 is at equilibrium
problem76
Problem

Predict the spontaneity at low and high temperatures for:

N2 (g) + 3 H2 (g)  2 NH3 (g) DH = -92.22 kJ

2 C (s) + 2 H2 (g)  C2H4 (g) DH = 52.26 kJ

problem77
Problem

N2 (g) + 3 H2 (g)  2 NH3 (g) DH = -92.22 kJ

Spontaneous @ low T and nonspontaneous @ high T

2 C (s) + 2 H2 (g)  C2H4 (g) DH = 52.26 kJ

Nonspontaneous @ all T

standard free energy change d g
Standard free energy change DG
  • Just like we can have a standard enthalpy change DH for chemicals, we can also define the standard free energy change…
  • DG =DH- TDS
standard free energy of formation d g f
Standard free energy of formation DGf
  • Standard enthalpies of formation DHf of elements in their standard states are zero:
  • DHrxn = [Snp DHf(products) –
  • Snr DHf(reactants)]
  • Standard free energies of formation DGf of elements in their standard states are zero:
  • DG = [Snp DGf(products) –
  • Snr DGf(reactants)]
problem80
Problem
  • Calculate DG at 298.15 K for the reaction
  • 4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s)
  • by using the two following sets of data. Compare your answers.
  • a) DH = -1648 kJ mol-1
  • and DS = -549.3 JK-1 mol-1
  • b) DGf (Fe) = 0 kJmol-1
  • DGf (O2) = 0 kJmol-1
  • DGf (Fe2O3) = -742.2 kJmol-1
problem answer
Problem answer
  • a) DG = -1484 kJmol-1
  • b) DG = -1484.4 kJmol-1
  • The answers are the same because free energy is a state function.
free energy and equilibrium
Free energy and equilibrium
  • We’ve already seen that
  • DG < 0 is spontaneous
  • DG > 0 is non-spontaneous
  • One process in a written reaction is spontaneous while the reverse is not, as long as DG≠0. Therefore
  • DG=0 is at equilibrium
water and steam
Water and steam
  • H2O (l, 1 atm)  H2O (g, 1 atm)
  • DG373.15 K = 0 kJmol-1
  • The system is at equilibrium at 1 atm (standard conditions) and at the boiling point temperature!
water and steam84
Water and steam
  • H2O (l, 1 atm)  H2O (g, 1 atm)
  • DG298.15 K = 8.590 kJmol-1
  • The system is not at equilibrium at 1 atm (standard conditions) and at the room temperature!
water and steam85
Water and steam
  • H2O (l, 1 atm)  H2O (g, 1 atm)
  • DG298.15 K = 8.590 kJmol-1
  • The forward process is non-spontaneous (DG > 0) so the reverse process is spontaneous and condensation occurs.
water and steam86
Water and steam
  • H2O (l, 0.03126 atm) 
  • H2O (g, 0.03126 atm)
  • DG298.15 K = 0 kJ
  • The system is at equilibrium at 0.03126 atm (non-standard conditions) and at the room temperature!
water and steam87
Water and steam
  • H2O (l, 0.03126 atm) 
  • H2O (g, 0.03126 atm)
  • DG298.15 K = 0 kJmol-1
  • Water CAN evaporate at room temperature,
  • just not to give an equilibrium pressure of 1 atm!
non standard conditions
Non-standard conditions
  • As we’ve seen with the previous water example, our interest in an equilibrium system is often at non-standard conditions,
  • so knowingDG
  • is usually not as useful as
  • knowing DG.
non standard conditions89
Non-standard conditions
  • For an ideal gas DH does not change if pressure changes, so at all non-standard conditionsDH = DH.
  • For an ideal gas DS does change if pressure changes (expansion into vacuum shows us this!), so at all non-standard conditionsDS ≠DS.
non standard free energy
Non-standard free energy
  • Because of these facts, the non-standardfree energy change is
  • DG = DH- TDS
  • But the standardfree energy change is
  • DG = DH- TDS
non standard free energy91
Non-standard free energy
  • The difference between standard and non-standardfree energy is totally due to the difference in entropy change between the standard and non-standard conditions
  • DG - DG = - T(DS-DS)
  • DG = DG+ T(DS-DS)
boltzmann distribution
Boltzmann distribution
  • By the Boltzmann distribution
  • S = R ln W for one mole of particles
  • DG = DG +T(R ln W - R ln W)
  • DG = DG +T(R ln W / W)
  • We are comparing a real system with a standard one. We are dealing with activities!
boltzmann distribution93
Boltzmann distribution
  • DG = DG +RT ln Qeq
  • In the comparison, we are looking at a reaction quotient!
  • In other words, we are looking at how our non-standard system is different from the system we haveat standard conditions!
boltzmann distribution94
Boltzmann distribution
  • DG = DG +RT ln Qeq
  • If our system
  • is at equilibrium
  • then Qeq = Keq
  • and DG = 0
boltzmann distribution95
Boltzmann distribution
  • DG = DG +RT ln Keq= 0
  • which means
  • DG =-RT ln Keq
equilibrium constant
Equilibrium constant
  • The thermodynamic equilibrium constantfor a reaction is directly related to the standard free energy change!
  • DG =-RT ln Keq
  • Keq= e-DG/RT
equilibrium constant97
Equilibrium constant

DG  0

Keq  1

DG >> 0

then Keq is very small

DG << 0

then Keq is very large

predicting reaction direction
Predicting reaction direction
  • DG < 0 means the forward reaction is spontaneous at the given non-standard conditions
  • DG < 0 means the forward reaction is spontaneous at standard conditions and so K > 1
predicting reaction direction100
Predicting reaction direction
  • DG = 0 means the system is at equilibrium at the given non-standard conditions
  • DG = 0 means the system is at equilibrium at standard conditions and so Keq = 1 which only occurs at one specific T!
predicting reaction direction101
Predicting reaction direction
  • DG > 0 means the forward reaction is non-spontaneous at the given non-standard conditions
  • DG > 0 means the forward reaction is non-spontaneous at standard conditions and so K < 1
predicting reaction direction102
Predicting reaction direction
  • DG = DG
  • ONLY at standard conditions!
thermodynamic equilibrium constant k eq
Thermodynamic equilibrium constant Keq
  • The thermodynamic equilibrium constantKeqis expressed in terms of activities, which are unitless quantities.
  • Activities relate properties like concentration or pressure compared to a standard property value, like 1 M for concentration or 1 bar for pressure.
thermodynamic equilibrium constant k eq104
Thermodynamic equilibrium constant Keq

a A + b B  c C + d D

where ax = [X] / c0 (c0 is a standard concentration of 1 M)

or ax = Px / P0 (P0 is a standard pressure of 1 atm)

Note that ax = 1 for pure solids and liquids

free energy and equilibrium constants
Free energy and equilibrium constants
  • DG = -RT ln Keq
  • IS ALWAYS TRUE
  • DG = -RT ln Kc
  • and
  • DG = -RT ln Kp
  • DO NOT have to be true!
problem106
Problem
  • Write thermodynamic equilibrium constant expressionsfor each of the following reactionsand relate them to Kc and Kp where appropriate:
  • a) Si (s) + 2 Cl2 (g)  SiCl4 (g)
  • b) Cl2 (g) + H2O (l)  HOCl (aq) + H+ (aq) + Cl- (aq)
problem answer107
Problem answer
  • a) Si (s) + 2 Cl2 (g)  SiCl4 (g)
  • b) Cl2 (g) + H2O (l)  HOCl (aq) + H+ (aq) + Cl- (aq)
problem108
Problem
  • Use the given data todetermine if the following reaction is spontaneous at standard conditions at 298.15 K:
  • N2O4 (g)  2 NO2 (g)
  • DGf (N2O4) = 97.89 kJmol-1
  • DGf (NO2) = 51.31 kJmol-1

Answer: DG= 4.73 kJmol-1, not spontaneous at standard conditions.

problem109
Problem
  • Based on the problem of the previous slide, determine which direction the reaction will go in if 0.5 bar of each gas is placed in an evacuated container:
  • N2O4 (g)  2 NO2 (g)

Answer: Since DG= 4.73 kJmol-1, then Keq = 0.15 = Kp. Since Qp = 0.5 the reaction should go from right to left.

problem110
Problem
  • Determine the equilibrium constant at 298.15 K for the following reaction using the given data:
  • AgI (s)  Ag+ (aq) + I- (aq)
  • DGf (AgI) = -66.19 kJmol-1
  • DGf (Ag+) = 77.11 kJmol-1
  • DGf (I-) = -51.57 kJmol-1
problem answer111
Problem answer

Since DG= 91.73 kJmol-1, then Keq = 8.5 x 10-17

= Kc = Ksp.

If we compare to the Ksp value in Table 18.1 (8.5 x 10-17) we see we are definitely in the right ballpark!

d g and k eq are functions of t
DG and Keq are functions of T
  • We’ve already seen that equilibrium constants change with temperature.
  • Why?
  • Keq = e-DG/RT
  • and
  • DG =DH- TDS
d g and k eq are functions of t113
DG and Keq are functions of T
  • DG = -RT ln Keq =DH- TDS
  • ln Keq=-(DH/RT)+ (DS/R)
problem114
Problem
  • At what temperature will the following reaction have Kp = 1.50 x 102?
  • 2 NO (g) + O2 (g)  2 NO2 (g)
  • DH = -114.1 kJmol-1
  • and DS = -146.5 J K-1mol-1

Answer: T = 606 K

d g and k eq are functions of t115
DG and Keq are functions of T
  • DG = -RT ln Keq =DH- TDS
  • ln Keq=-(DH/RT)+ (DS/R)
  • IfDHandDS are constant over a temperature range
  • then
d g and k eq are functions of t116
DG and Keq are functions of T
  • ln K1=-(DH/RT1)+ (DS/R)
  • minus

ln K2=-(DH/RT2)+ (DS/R)

ln K1 -ln K2 =

-(DH/RT1)- (-DH/RT2)

d g and k eq are functions of t117
DG and Keq are functions of T
  • ln K1 -ln K2 =

[-DH/R] [(1/T1)- (1/T2)]

OR

  • ln [K1/K2]=

[-DH/R] [(1/T1)- (1/T2)]

van t hoff equation
van’t Hoff equation
  • The van’t Hoff equation relates equilibrium constants to temperatures
  • ln [K1/K2]=[-DH/R] [(1/T1)- (1/T2)]
  • It looks very similar to the Arrhenius equationand so a plot ofln K versus 1/Tshould give a straight line with a slope of[-DH/R]
slide119

Plot of ln Kp versus 1/T for the reaction.

slope = -DH / R, so for this reaction

DH = -180 kJmol-1

2 SO2 (g) + O2 (g) 2 SO3 (g)

problem120
Problem
  • The following equilibrium constant data have been determined for the reaction
  • H2 (g) + I2 (g)  2 HI (g)
  • Kp = 50.0@ 448 C

Kp = 66.9@ 350 C

Estimate DH for the reaction.

Answer: DH = -11.1 kJmol-1.

coupled reactions
Coupled reactions
  • If we have a non-spontaneous (DG > 0) reaction with a product that appears as a reactant in a different reaction that is spontaneous (DG < 0) then we can couple the two reactions(do them in the same container at the same time) to drive the non-spontaneous reaction!
coupled reactions122
Coupled reactions
  • This should make sense because if we have two equilibria in one container, one with a small K(like Ksp) and one with a large K(like Kf for complex formation), then the coupled reaction that is the sum of the two reactions has a K value that is larger than Ksp!
  • We have made the non-spontaneous reaction occur!