Mr. Bartelt PresentsThe 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 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?

ΔΔHHvapvap and and ΔΔHHfusfus

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.

ΔΔHHvapvap and and ΔΔHHfusfus

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 until water, the temperature will read 273 K until all the ice melts. Only after all the ice 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?

ΔΔHHvapvap and and ΔΔHHfusfus

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 waterΔΔHHvapvap = 43.9 kJ/mol = 43.9 kJ/molΔΔHHfusfus = 6.02 kJ/mol = 6.02 kJ/mol

ΔΔHHvapvap and and 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 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 K? the temperature of ice at 268 K to 378 K? (getting harder)(getting harder)

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)

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

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

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 25.1mol 1

kJ 6.02

18.0g

mol 175.0g

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 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 we now need to find how much energy is released when one mole of steam is brought down to baseline (273 K as a liquid)mole of steam is brought down to baseline (273 K as a liquid)

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

J4.1818.0g

ΔH

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

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 0.211xkJ 25.1

x

kJ 119.1

mol 1

K) (373molK

atmL0.08206mol) (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 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 states: The second law of thermodynamics states: In any spontaneous process there is In any spontaneous process there is always an increase in the entropy of the always an increase in the entropy of the universe.universe.

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.

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

ΔΔSSunivuniv = = ΔΔSSsyssys + + ΔΔSSsurrsurr

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.

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 787Entropy 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.

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?

HH22OO(l)(l) H H22OO(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 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, ΔΔSSsurrsurr = +# = +#

ΔΔSSunivuniv = = ΔΔSSsurrsurr + + ΔΔSSsyssys

As long as As long as ΔΔSSunivuniv is + is + and it may and it may

proceed.proceed.

ΔΔSSunivuniv = = ΔΔSSsurrsurr + + ΔΔSSsyssys

The sign of The sign of ΔΔSSsyssys is easy to predict. If your system is easy to predict. If your system

gets more orderly, the gets more orderly, the ΔΔSSsys sys is negative is negative , but if it , but if it

gets more chaotic then gets more chaotic then ΔΔSSsyssys 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?

ΔΔSSsurrsurr is T dependent is T dependent

The above equations are important.The above equations are important. When water freezes When water freezes ΔΔSSsyssys 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 (ΔΔSSsyssys or or ΔΔSSsurrsurr) needs to dominate) needs to dominate

ΔΔSSsyssys is temperature dependent is temperature dependent At high tempertures At high tempertures ΔΔSSsys sys becomes small and weakbecomes small and weak

At low temperatures At low temperatures ΔΔSSsys sys is large and dominatesis large and dominates

T

ΔHΔSsurr syssurruniv ΔS ΔS ΔS

ΔΔSSsurrsurr is T dependent is T dependent

syssurruniv ΔS ΔS ΔS T

ΔHΔSsurr

0ΔS TΔH

Thus

ΔS TΔH - ST

ΔS T

ΔH- ΔS

ΔS ΔS ΔS

sys

sysuniv

sysuniv

syssurruniv

The reaction is spontaneous

Gibbs free energyGibbs free energyGibbs 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 = enthalpyT = 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(ΔΔSSunivuniv) = -) = -ΔΔGG

Let’s look at a few examplesLet’s look at a few examplesΔΔG = G = ΔΔH - T H - 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 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?

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

HH22OO(g)(g) H H22OO(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

J188.7S

Kmol

J69.96S

(g)2

(l)2

OH

OH

Calculating Calculating ΔΔGG ΔΔSSreactionreaction = 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 HH22OO(g)(g) H H22OO(l)(l)

ΔΔHHreactionreaction = = ΔΔHHproductsproducts – – ΔΔHHreactantsreactants

ΔΔHHreactionreaction = -286 –(-242) = - 44 kJ/mol = -286 –(-242) = - 44 kJ/mol

Kmol

kJ2.422H

Kmol

kJ286-H

(g)2

(l)2

OH

OH

Calculating Calculating 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 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

ΔΔG and Boiling point TempG and Boiling point TempΔΔG 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 H and H22OO(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

ΔΔSSreactionreaction = - 0.1187 kJ/mol = - 0.1187 kJ/mol ΔΔHHreactionreaction = - 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)