Enthalpy vs entropy

The Measure of Enthalpy and Entropy and the difference between the two concepts.

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

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

3. ΔΔHH
vapvap and and ΔΔHH
fusfus
Last unit we looked at the change in Last unit we looked at the change in
enthalpy associated with a chemical enthalpy associated with a chemical
reaction and the energy flow as materials reaction and the energy flow as materials
(mostly water) are heated.(mostly water) are heated.
Today we’re going to extend that lesson to Today we’re going to extend that lesson to
include phase change.include phase change.

4. ΔΔHH
vapvap and and ΔΔHH
fusfus
If you put a thermometer in a glass of ice If you put a thermometer in a glass of ice
water, the temperature will read 273 K water, the temperature will read 273 K
until all the ice melts. Only after all the ice until all the ice melts. Only after all the ice
melts will the temperature begin its crawl melts will the temperature begin its crawl
up to room temperature.up to room temperature.
For similar reasons, the temperature of For 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?

5. ΔΔHH
vapvap and and ΔΔHH
fusfus
Melting ice and boiling water are both Melting ice and boiling water are both
endothermic processes and as such they endothermic processes and as such they
both require energy.both require energy.
If any of the water were to increase in If any of the water were to increase in
temperature that energy would temperature that energy would
immediately be transferred to the ice to immediately be transferred to the ice to
cause it to melt.cause it to melt.
For waterFor water
ΔΔHH
vapvap = 43.9 kJ/mol = 43.9 kJ/mol
ΔΔHH
fusfus = 6.02 kJ/mol = 6.02 kJ/mol

6. ΔΔHH
vapvap and and ΔΔHH
fusfus
Examples, Given (for water)Examples, Given (for water)
ΔΔHH
vapvap = 43.9 kJ/mol = 43.9 kJ/mol
ΔΔHH
fusfus = 6.02 kJ/mol = 6.02 kJ/mol
1.1.How much energy is required to vaporize How 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 melt How 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 raise How much energy would it take to raise
the temperature of ice at 268 K to 378 the temperature of ice at 268 K to 378
K? (getting harder)K? (getting harder)

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

8. A new lookA new look
If 90.0 g of water at 300. K are added to If 90.0 g of water at 300. K are added to
30.0 g of water at 350. K what will the 30.0 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

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

10. How to do itHow to do it
When the system reaches equilibrium all species When 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 to Next we find how much energy will be required to
melt all that ice. That’s easymelt all that ice. That’s easy
kJ 25.1
mol 1
kJ 6.02
18.0g
mol 1
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 now Since 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 is need 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)

11. SteamSteam
What happens when 1 mole of steam is brought What 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 this It 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
( ) ( )
kJ 2.57
J/kJ 1000
K 100.
gK
J
4.1818.0g
ΔH =
÷
÷
ø
ö
ç
ç
è
æ
·
=
This means that cooling a mole of water from This 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 the Add 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

12. Finishing upFinishing up
If 119.1 kJ are released when one mole of steam is cooled If 119.1 kJ are released when one mole of steam is cooled
to 273 K (as water) then how many moles of steam will be to 273 K (as water) then how many moles of steam will be
required to release the 25.1 kJ of energy required to melt required 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.
steam mol 0.211x
kJ 25.1
x
kJ 119.1
mol 1
=Þ=
K) (373
molK
atmL
0.08206mol) (0.211atm)V (1 ÷
ø
ö
ç
è
æ
·
·
=

13. 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 in Entropy can be thought of as the force in
the universe that pushes everything the universe that pushes everything
toward disorder and chaos.toward disorder and chaos.
The second law of thermodynamics The second law of thermodynamics
states: In any spontaneous process there states: In any spontaneous process there
is always an increase in the entropy of the is always an increase in the entropy of the
universe.universe.

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

15. Enthalpy vs. Entropy (Shiva)Enthalpy vs. Entropy (Shiva)
We know that the amount of energy in the We know that the amount of energy in the
universe is constant but the Entropy of the universe 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
ΔΔSS
univuniv = = ΔΔSS
syssys + + ΔΔSS
surrsurr
For any reaction the change in entropy of the universe is For any reaction the change in entropy of the universe is
going to be equal to the change in entropy of our system going to be equal to the change in entropy of our system
plus the change in entropy of the surroundings. If the plus the change in entropy of the surroundings. If the
change in entropy of the universe is positive change in entropy of the universe is positive  and the and the
reaction will happen spontaneously. If it’s negative reaction will happen spontaneously. If it’s negative 
and it will not happen, end of story.and it will not happen, end of story.

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

17. TemperatureTemperature
Entropy is always increasing but we can Entropy is always increasing but we can
manipulate entropy by changing the manipulate entropy by changing the
temperature.temperature.
What has more disorder, ice or liquid What has more disorder, ice or liquid
water?water?
Then why does ice ever form?Then why does ice ever form?
HH
22OO
(l)(l)  H H
22OO
(s)(s) ΔΔS = -# S = -# 
So how can this happen????????????So how can this happen????????????

18. Endo or exothermic?Endo or exothermic?
Is the freezing of water endothermic or Is the freezing of water endothermic or
exothermic?exothermic?
This release of energy from the system This release of energy from the system
makes the surroundings more disordered. makes the surroundings more disordered.
Hence, Hence, ΔΔSS
surrsurr = +# = +# 
ΔΔSS
univuniv = = ΔΔSS
surrsurr + + ΔΔSS
syssys
As long as As long as ΔΔSS
univuniv is + is +  and it may and it may
proceed.proceed.

19. ΔΔSS
univuniv = = ΔΔSS
surrsurr + + ΔΔSS
syssys
The sign of The sign of ΔΔSS
syssys is easy to predict. If your system is easy to predict. If your system
gets more orderly, the gets more orderly, the ΔΔSS
sys sys is negative is negative , but if it , but if it
gets more chaotic then gets more chaotic then ΔΔSS
syssys is positive is positive ..
But the system is only half the picture, the But 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 the When ice melts, it throws energy into the
surroundings and makes them more chaotic surroundings and makes them more chaotic ..
So the important question is:So the important question is:
What causes freezing to take place at certain What causes freezing to take place at certain
temperatures and not at other temperatures?temperatures and not at other temperatures?

20. ΔΔSS
surrsurr is T dependent is T dependent
The above equations are important.The above equations are important.
When water freezes When water freezes ΔΔSS
syssys becomes more ordered becomes more ordered

But the process is exothermic, which makes the But the process is exothermic, which makes the
surroundings more chaotic surroundings more chaotic 
One term (One term (ΔΔSS
syssys or or ΔΔSS
surrsurr) needs to dominate) needs to dominate
ΔΔSS
syssys is temperature dependent is temperature dependent
At high tempertures At high tempertures ΔΔSS
sys sys becomes small and weakbecomes small and weak
At low temperatures At low temperatures ΔΔSS
sys sys is large and dominatesis large and dominates
T
ΔH
ΔS
surr-=
syssurruniv ΔS ΔS ΔS +=

21. ΔΔSS
surrsurr is T dependent is T dependent
syssurruniv ΔS ΔS ΔS +=
T
ΔH
ΔS
surr-=
( ) ( )
0ΔS TΔH
Thus
ΔS TΔH - ST
ΔS
T
ΔH
- ΔS
ΔS ΔS ΔS
sys
sysuniv
sysuniv
syssurruniv
<-
-=D
+=
+=
The reaction is spontaneous

22. 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 - T H - T ΔΔSS
Note: that if Note: that if ΔΔG is negative, the reaction will G is negative, the reaction will
occur spontaneously.occur spontaneously.
-(-(ΔΔH - T H - T ΔΔS) = T(S) = T(ΔΔSS
univuniv) = -) = -ΔΔGG

23. Let’s look at a few examplesLet’s look at a few examples
ΔΔG = G = ΔΔH - T H - T ΔΔSS
ΔΔHHΔΔSSΔΔGGSpontaneous ?Spontaneous ?
- - + + - -  YES!!!!!!!YES!!!!!!!
+ + - - + +  NO!!!!!!!!NO!!!!!!!!
- - - -  ??Depends on TDepends on T
+ + + + ??Depends on TDepends on T

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

25. Calculating Calculating ΔΔSS
ΔΔS is calculated in a similar fashion to S 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
HH
22OO
(g)(g)  H H
22OO
(l)(l)
ΔΔSS
reactionreaction = = ΔΔSS
productsproducts – – ΔΔSS
reactantsreactants (Just like (Just like ΔΔH)H)
ΔΔSS
reactionreaction = 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
·
=
·
=

26. Calculating Calculating ΔΔGG
ΔΔSS
reactionreaction = 188.7 – 69.96 = -118.7 = 188.7 – 69.96 = -118.7 
Now that we have Now that we have ΔΔS all we need it T and S all we need it T and ΔΔH H
and we can find and we can find ΔΔG usingG using
 HH
22OO
(g)(g)  H H
22OO
(l)(l)
ΔΔHH
reactionreaction = = ΔΔHH
productsproducts – – ΔΔHH
reactantsreactants
ΔΔHH
reactionreaction = -286 –(-242) = - 44 kJ/mol = -286 –(-242) = - 44 kJ/mol 
Kmol
kJ
2.422H
Kmol
kJ
286-H
(g)2
(l)2
OH
OH
·
-=
·
=

27. Calculating Calculating ΔΔG (cont…)G (cont…)
ΔΔSS
reactionreaction = - 118.7 J/mol = - 118.7 J/mol  -0.1187 kJ/mol -0.1187 kJ/mol
ΔΔHH
reactionreaction = - 44 kJ/mol = - 44 kJ/mol 
Now we plug and chug. If T = 300 K we would Now 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 - T H - 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

28. 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 H
22OO
(g)(g) and H and H
22OO
(l)(l) are in are in
equilibrium, hence the reaction is spontanious in equilibrium, hence the reaction is spontanious in
both directions or in neither (take your pick)both directions or in neither (take your pick)
Regardless Regardless ΔΔG = 0G = 0
ΔΔSS
reactionreaction = - 0.1187 kJ/mol = - 0.1187 kJ/mol 
ΔΔHH
reactionreaction = - 44 kJ/mol = - 44 kJ/mol 
ΔΔG = G = ΔΔH - T H - T ΔΔSS
0 = -44 – (T)(-0.1187)0 = -44 – (T)(-0.1187)
T=371 K T=371 K  98ºC (pretty close to water’s BP) 98ºC (pretty close to water’s BP)