Mr. Bartelt Presents
The greatest fight in the universe
Enthalpy vs. Entropy

EnthalpyEnthalpy
 We already know about enthalpy.We already know about enthalpy.
Enthalpy is related to the amount ofEnthalpy is related to the amount of
energy that is lost or gained by a systemenergy that is lost or gained by a system
during a normal chemical process.during a normal chemical process.
 We have seen both endothermic andWe have seen both endothermic and
exothermic processes in lab. Which doexothermic processes in lab. Which do
you think is favored by nature?you think is favored by nature?
 Why?Why?

ΔΔHHvapvap andand ΔΔHHfusfus
 Last unit we looked at the change inLast unit we looked at the change in
enthalpy associated with a chemicalenthalpy associated with a chemical
reaction and the energy flow as materialsreaction and the energy flow as materials
(mostly water) are heated.(mostly water) are heated.
 Today we’re going to extend that lesson toToday we’re going to extend that lesson to
include phase change.include phase change.

ΔΔHHvapvap andand ΔΔHHfusfus
 If you put a thermometer in a glass of iceIf you put a thermometer in a glass of ice
water, the temperature will read 273 Kwater, the temperature will read 273 K
until all the ice melts. Only after all the iceuntil all the ice melts. Only after all the ice
melts will the temperature begin its crawlmelts will the temperature begin its crawl
up to room temperature.up to room temperature.
 For similar reasons, the temperature ofFor similar reasons, the temperature of
boiling water will never exceed 373 K.boiling water will never exceed 373 K.
 Why is this?Why is this?

ΔΔHHvapvap andand ΔΔHHfusfus
 Melting ice and boiling water are bothMelting ice and boiling water are both
endothermic processes and as such theyendothermic processes and as such they
both require energy.both require energy.
 If any of the water were to increase inIf any of the water were to increase in
temperature that energy wouldtemperature that energy would
immediately be transferred to the ice toimmediately be transferred to the ice to
cause it to melt.cause it to melt.
 For waterFor water
ΔΔHHvapvap = 43.9 kJ/mol= 43.9 kJ/mol
ΔΔHHfusfus = 6.02 kJ/mol= 6.02 kJ/mol

ΔΔHHvapvap andand ΔΔHHfusfus
 Examples, Given (for water)Examples, Given (for water)
 ΔΔHHvapvap = 43.9 kJ/mol= 43.9 kJ/mol
 ΔΔHHfusfus = 6.02 kJ/mol= 6.02 kJ/mol
1.1. How much energy is required to vaporizeHow much energy is required to vaporize
10.0 g of water? (easy)10.0 g of water? (easy)
2.2. How much energy is required to meltHow much energy is required to melt
10.0 g of water? (easy)10.0 g of water? (easy)
3.3. How much energy would it take to raiseHow much energy would it take to raise
the temperature of ice at 268 K to 378the temperature of ice at 268 K to 378
K? (getting harder)K? (getting harder)

Hard problemsHard problems
 How much boiling water would you needHow much boiling water would you need
to add to a cup with 100.0 g of ice andto add to a cup with 100.0 g of ice and
100.0 g of water at equilibrium to get all100.0 g of water at equilibrium to get all
the ice to melt? (hard)the ice to melt? (hard)
 If you doubled that amount of water whatIf you doubled that amount of water what
would the final temperature be? (hard)would the final temperature be? (hard)

A new lookA new look
 If 90.0 g of water at 300. K are added toIf 90.0 g of water at 300. K are added to
30.0 g of water at 350. K what will the30.0 g of water at 350. K what will the
final temperature be?final temperature be?
It sounds hard, but there’s a trick:It sounds hard, but there’s a trick:
Just make a weighted average:Just make a weighted average:
(90.0*300+30.0*350)/120 = 312.5(90.0*300+30.0*350)/120 = 312.5

The baseline approachThe baseline approach
 Some problems will require a moreSome problems will require a more
sophisticated approachsophisticated approach
 How many liters of steam at 1 atm andHow many liters of steam at 1 atm and
373 K will it take to melt exactly 75.0 g of373 K will it take to melt exactly 75.0 g of
ice at STP?ice at STP?
 It sounds really hard, but it can be workIt sounds really hard, but it can be work
quite simply with the baseline approachquite simply with the baseline approach

How to do itHow to do it
 When the system reaches equilibrium all speciesWhen the system reaches equilibrium all species
will be liquid water at 273 K. That is our baseline.will be liquid water at 273 K. That is our baseline.
 Next we find how much energy will be required toNext we find how much energy will be required to
melt all that ice. That’s easymelt all that ice. That’s easy
kJ25.1
mol1
kJ6.02
18.0g
mol1
75.0g =••
 That’s how much energy it takes to get our ice to the baseline.That’s how much energy it takes to get our ice to the baseline.
 Since heat gained by ice is going to equal heat lost by steam we nowSince heat gained by ice is going to equal heat lost by steam we now
need to find how much energy is released when one mole of steam isneed to find how much energy is released when one mole of steam is
brought down to baseline (273 K as a liquid)brought down to baseline (273 K as a liquid)

SteamSteam
 What happens when 1 mole of steam is broughtWhat happens when 1 mole of steam is brought
down to 273 K and remains a liquid?down to 273 K and remains a liquid?
1.1. It condenses: 43.9 kJ will be released when thisIt condenses: 43.9 kJ will be released when this
happenshappens
2.2. It cools from 373 K to 273 KIt cools from 373 K to 273 K
( ) ( )
kJ2.57
J/kJ1000
K100.
gK
J
4.1818.0g
ΔH =






•
=
 This means that cooling a mole of water fromThis means that cooling a mole of water from
boiling to freezing releases 75.2 kJ of energy.boiling to freezing releases 75.2 kJ of energy.
 Add that to the heat released from theAdd that to the heat released from the
condensation process and you get 43.9+75.2 =condensation process and you get 43.9+75.2 =
119.1 kJ/mol119.1 kJ/mol

Finishing upFinishing up
 If 119.1 kJ are released when one mole of steam is cooledIf 119.1 kJ are released when one mole of steam is cooled
to 273 K (as water) then how many moles of steam will beto 273 K (as water) then how many moles of steam will be
required to release the 25.1 kJ of energy required to meltrequired to release the 25.1 kJ of energy required to melt
our 75.0 g of ice?our 75.0 g of ice?
 Easy ratio:Easy ratio:
 Finally use PV=nRT to solve for the volume in liters.Finally use PV=nRT to solve for the volume in liters.
steammol0.211x
kJ25.1
x
kJ119.1
mol1
=⇒=
K)(373
molK
atmL
0.08206mol)(0.211atm)V(1 





•
•
=

EntropyEntropy
 If you thought that enthalpy was abstract,If you thought that enthalpy was abstract,
welcome to Entropy.welcome to Entropy.
 Entropy can be thought of as the force inEntropy can be thought of as the force in
the universe that pushes everythingthe universe that pushes everything
toward disorder and chaos.toward disorder and chaos.
 The second law of thermodynamicsThe second law of thermodynamics
states: In any spontaneous process therestates: In any spontaneous process there
is always an increase in the entropy of theis always an increase in the entropy of the
universe.universe.

Enter ShivaEnter Shiva
 Shiva is the Hindo lord ofShiva is the Hindo lord of
chaos. He/she (inchaos. He/she (in
Hindoism he/she is aHindoism he/she is a
hermaphrodite) is thehermaphrodite) is the
“lord of chaos”.“lord of chaos”.
Whenever a processWhenever a process
produces an increase inproduces an increase in
Entropy Shiva is happy.Entropy Shiva is happy.

Enthalpy vs. Entropy (Shiva)Enthalpy vs. Entropy (Shiva)
 We know that the amount of energy in theWe know that the amount of energy in the
universe is constant but the Entropy of theuniverse is constant but the Entropy of the
Universe is always increasing.Universe is always increasing.
 Shiva is happy =Shiva is happy = 
Shiva’s lawShiva’s law
ΔΔSSunivuniv == ΔΔSSsyssys ++ ΔΔSSsurrsurr
For any reaction the change in entropy of the universe isFor any reaction the change in entropy of the universe is
going to be equal to the change in entropy of our systemgoing to be equal to the change in entropy of our system
plus the change in entropy of the surroundings. If theplus the change in entropy of the surroundings. If the
change in entropy of the universe is positivechange in entropy of the universe is positive  and theand the
reaction will happen spontaneously. If it’s negativereaction will happen spontaneously. If it’s negative 
and it will not happen, end of story.and it will not happen, end of story.

What makesWhat makes 
 Disorder!!! A gas his more scattered andDisorder!!! A gas his more scattered and
spread outspread out 
 Look on page 787Look on page 787
 Entropy is increased if there are moreEntropy is increased if there are more
possible arrangements at the end of thepossible arrangements at the end of the
reaction than beforereaction than before
 The mores spread out the better. Shiva isThe mores spread out the better. Shiva is
eagerly awaiting theeagerly awaiting the
heat death of the universeheat death of the universe, when, when
maximum entropy is achieved. This willmaximum entropy is achieved. This will
take a while, fear not.take a while, fear not.

TemperatureTemperature
 Entropy is always increasing but we canEntropy is always increasing but we can
manipulate entropy by changing themanipulate entropy by changing the
temperature.temperature.
 What has more disorder, ice or liquidWhat has more disorder, ice or liquid
water?water?
 Then why does ice ever form?Then why does ice ever form?
HH22OO(l)(l)  HH22OO(s)(s) ΔΔS = -#S = -# 
 So how can this happen????????????So how can this happen????????????

Endo or exothermic?Endo or exothermic?
 Is the freezing of water endothermic orIs the freezing of water endothermic or
exothermic?exothermic?
 This release of energy from the systemThis release of energy from the system
makes the surroundings more disordered.makes the surroundings more disordered.
Hence,Hence, ΔΔSSsurrsurr = +#= +# 
ΔΔSSunivuniv == ΔΔSSsurrsurr ++ ΔΔSSsyssys
 As long asAs long as ΔΔSSunivuniv is +is +  and it mayand it may
proceed.proceed.

ΔΔSSunivuniv == ΔΔSSsurrsurr ++ ΔΔSSsyssys
 The sign ofThe sign of ΔΔSSsyssys is easy to predict. If your systemis easy to predict. If your system
gets more orderly, thegets more orderly, the ΔΔSSsyssys is negativeis negative , but if it, but if it
gets more chaotic thengets more chaotic then ΔΔSSsyssys is positiveis positive ..
 But the system is only half the picture, theBut the system is only half the picture, the
surroundings are important as well.surroundings are important as well.
 When ice melts, it throws energy into theWhen ice melts, it throws energy into the
surroundings and makes them more chaoticsurroundings and makes them more chaotic ..
 So the important question is:So the important question is:
What causes freezing to take place at certainWhat causes freezing to take place at certain
temperatures and not at other temperatures?temperatures and not at other temperatures?

ΔΔSSsurrsurr is T dependentis T dependent
 The above equations are important.The above equations are important.
 When water freezesWhen water freezes ΔΔSSsyssys becomes more orderedbecomes more ordered

 But the process is exothermic, which makes theBut the process is exothermic, which makes the
surroundings more chaoticsurroundings more chaotic 
 One term (One term (ΔΔSSsyssys oror ΔΔSSsurrsurr) needs to dominate) needs to dominate
 ΔΔSSsyssys is temperature dependentis temperature dependent
 At high temperturesAt high tempertures ΔΔSSsyssys becomes small and weakbecomes small and weak

T
ΔH
ΔSsurr −= syssurruniv ΔSΔSΔS +=

ΔΔSSsurrsurr is T dependentis T dependent
syssurruniv ΔSΔSΔS +=
T
ΔH
ΔSsurr −=
( ) ( )
0ΔSTΔH
Thus
ΔSTΔH-ST
ΔS
T
ΔH
-ΔS
ΔSΔSΔS
sys
sysuniv
sysuniv
syssurruniv
<−
−=∆
+=
+=
The reaction is spontaneous

Gibbs free energyGibbs free energy
 Gibbs free energy is defined asGibbs free energy is defined as
G=H-TS (does this look familiar?)G=H-TS (does this look familiar?)
Where H = enthalpyWhere H = enthalpy
T = temp (K)T = temp (K)
S = entropyS = entropy
 The free energy equation can be rewritten:The free energy equation can be rewritten:
ΔΔG =G = ΔΔH - TH - T ΔΔSS
Note: that ifNote: that if ΔΔG is negative, the reaction willG is negative, the reaction will
occur spontaneously.occur spontaneously.
-(-(ΔΔH - TH - T ΔΔS) = T(S) = T(ΔΔSSunivuniv) = -) = -ΔΔGG

Let’s look at a few examplesLet’s look at a few examples
ΔΔG =G = ΔΔH - TH - T ΔΔSS
ΔΔHH ΔΔSS ΔΔGG Spontaneous ?Spontaneous ?
--  ++  --  YES!!!!!!!YES!!!!!!!
++  --  ++  NO!!!!!!!!NO!!!!!!!!
--  --  ?? Depends on TDepends on T
++  ++  ?? Depends on TDepends on T

CondensationCondensation
 When water condenses your systemWhen water condenses your system
becomes more organizedbecomes more organized 
 However, condensation is exothermic andHowever, condensation is exothermic and
releases energy which makes thereleases energy which makes the
surroundings more disorderedsurroundings more disordered 
 This means that temperature determines ifThis means that temperature determines if
water will freezewater will freeze
 But how do we calculateBut how do we calculate ΔΔS?S?

CalculatingCalculating ΔΔSS
 ΔΔS is calculated in a similar fashion toS is calculated in a similar fashion to ΔΔH.H.
 ΔΔS is looked up in the back of the bookS is looked up in the back of the book
 Back to the example of the freezing waterBack to the example of the freezing water
HH22OO(g)(g)  HH22OO(l)(l)
ΔΔSSreactionreaction == ΔΔSSproductsproducts –– ΔΔSSreactantsreactants (Just like(Just like ΔΔH)H)
ΔΔSSreactionreaction = 69.96 – 188.7 = -118.7= 69.96 – 188.7 = -118.7 JJ/mol*K/mol*K 
NOTE:NOTE: ΔΔS is in Joules NOT kJS is in Joules NOT kJ
Kmol
J
188.7S
Kmol
J
69.96S
(g)2
(l)2
OH
OH
•
=
•
=

CalculatingCalculating ΔΔGG
 ΔΔSSreactionreaction = 188.7 – 69.96 = -118.7= 188.7 – 69.96 = -118.7 
 Now that we haveNow that we have ΔΔS all we need it T andS all we need it T and ΔΔHH
and we can findand we can find ΔΔG usingG using
 HH22OO(g)(g)  HH22OO(l)(l)
ΔΔHHreactionreaction == ΔΔHHproductsproducts –– ΔΔHHreactantsreactants
ΔΔHHreactionreaction = -286 –(-242) = - 44 kJ/mol= -286 –(-242) = - 44 kJ/mol 
Kmol
kJ
2.422H
Kmol
kJ
286-H
(g)2
(l)2
OH
OH
•
−=
•
=

CalculatingCalculating ΔΔG (cont…)G (cont…)
 ΔΔSSreactionreaction = - 118.7 J/mol= - 118.7 J/mol  -0.1187 kJ/mol-0.1187 kJ/mol
 ΔΔHHreactionreaction = - 44 kJ/mol= - 44 kJ/mol 
 Now we plug and chug. If T = 300 K we wouldNow we plug and chug. If T = 300 K we would
expect condensation to take place spontaneously.expect condensation to take place spontaneously.
ΔΔG =G = ΔΔH - TH - T ΔΔSS
ΔΔG = -44 –(300)(-0.1187)G = -44 –(300)(-0.1187)
ΔΔG = -8.39 kJ/molG = -8.39 kJ/mol
 But what about @ 444 K???But what about @ 444 K???
ΔΔG = -44 –(444)(-0.1187)G = -44 –(444)(-0.1187)
ΔΔG = +8.7 kJ/molG = +8.7 kJ/mol

ΔΔG and Boiling point TempG and Boiling point Temp
 ΔΔG can be used to predict boiling point.G can be used to predict boiling point.
At the boiling point HAt the boiling point H22OO(g)(g) and Hand H22OO(l)(l) are inare in
equilibrium, hence the reaction is spontanious inequilibrium, hence the reaction is spontanious in
both directions or in neither (take your pick)both directions or in neither (take your pick)
RegardlessRegardless ΔΔG = 0G = 0
 ΔΔSSreactionreaction = - 0.1187 kJ/mol= - 0.1187 kJ/mol 
 ΔΔHHreactionreaction = - 44 kJ/mol= - 44 kJ/mol 
ΔΔG =G = ΔΔH - TH - T ΔΔSS
0 = -44 – (T)(-0.1187)0 = -44 – (T)(-0.1187)
T=371 KT=371 K  98ºC (pretty close to water’s BP)98ºC (pretty close to water’s BP)