1422-Chapt-15-Thermodynamics.good_notes (1).docx

8/11/2019 1422-Chapt-15-Thermodynamics.good_notes (1).docx

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Chapter 15

(not much on E)

Thermodynamics:

Enthalpy, Entropy

& Gibbs Free

Energy


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mo Thermodynamics: thermo = heat (energy)

dynamics = movement, motion

Some thermodynamic terms chemists use:

System: the portion of the universe that we areconsidering

open system: energy & matter can transfer

closed system: energy transfers only

isolated system: no transfers

Surroundings: everything else besides the system

Isothermal: a system that is kept at a constant

temperatureby adding or subtracting heat from the

surroundings.Heat Capacity: the amount of heat energy required to

raise the temperature of a certain amount of material by

1C (or 1 K).

Specif ic Heat Capacity: 1 g by 1C

Molar H eat Capacity: 1 mole by 1C


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mo Calorie: the amount of heat required to raise the

temperature of 1g of water by 1C.

specific heat of water = 1 cal/g C

1 calorie = 4.18 joules

Specific Heats and Molar Heat Capacities

Substance

Al

Specific Heat

(J/C

g Molar Heat (J/C

mol)

.0.380.450.84

.24.425.183.8

CuFe

CaCO3

Ethanol

WaterAir

2.43

4.181.00

112.0

75.3~ 29

important to: engineers chemists

EXAMPLE: How many joules of energy are needed raise

the temperature of an iron nail (7.0 g) from 25C to125C?The specific heat of iron is 0.45 J/CHeat energy = (specific heat)(mass)(T)

Heat energy = (0.45 J/C g)(7.0 g)(100C) = 315

J

g.

Note that T can be C or K, but NOT F. When jusbeing used

in a scientific formula it will usually be kelvin


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Problem: How much energy does it take to raise

the body temperature 2.5C (a fever of just over

103F) for someone who weighs 110 pounds (50 kg).

Assume an average body specific heat capacity of3 J/C.g.

Problem: What would be more effective at melting

a frozen pipehot water or a hair dryer (hot air

gun). Why?


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mo State Functions

System properties, such as pressure (P), volume (V),

and temperature (T) are called state functions.

The value of a state function depends only on the state

of the system and not on the way in which the system

came to be in that state.

A change in a state function describes a difference

between the two states. It is independent of the

process or pathway by which the change occurs.

For example, if we heat a block of iron from room

temperature to 100C, it is not important exactly how

we did it. Just on the initial state and the final state

conditions. For example, we could heat it to 150C,

then cool it to 100C. The path we take is unimportant,so long as the final temperature is 100C.

Miles per gallon for a car, is NOT a state function. It

depends highly on the path: acceleration, speed, wind,

tire inflation, hills, etc.

Most of the thermodynamic values we will discuss in

this chapter are state functions.


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mo Energy: " The capacity to do work

and/or transfer heat"

Forms of Energy:

Kinetic (Ekinetic = mv2)

Heat

Light (& Electromagnetic)

Electricity

Sound

Potential

Gravitational

Chemical

Nuclear - Matter (E = mc2)

WORK


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mo

First Law of Thermodynamics:

The total amount of energy (and

mass) in the universe is constant.

In any process energy can be

changed from one form to

another; but it can never be

created nor destroyed.

You can' t get something for

nothing


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mo

Enthalpy (Heats) of Reaction

The amount of heat released or absorbed by a

chemical reaction at constant pressure (as one woulddo in a laboratory) is called the enthalpy or heat or

reaction. We use the symbol H to indicate

enthalpy.

Sign notation (EXTREMELY IMPORTANT!!):

+H indicates that heat is being absorbed in the

reaction (it gets cold)

H indicates that heat is being given off in the

reaction (it gets hot) exothermic

endothermic

Standard Enthalpy = H ( is called a not)

Occurring under StandardConditions:

Pressure 1 atm (760 torr)

1.0 MConcentration

Temperature is notdefined or part of Standard

Conditions, but is often measured at 298 K (25C).


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mo Standard Enthalpy of Formation -- Hf

The amount of heat absorbed (endothermic) or

released (exothermic) in a reaction in which one

mole of a substance is formed from its elements in

their standard states, usually at 298 K (25C).

Also called heat of formation.

Hf = 0 for any element in its standard state (the

natural elemental form at 1 atm or 1 M) at 298 K.

EXAMPLES:

grap e, s +O2(g) (g)H = 0 kJ/mol 0kJ/mol

.kJ/mol

rxn

product(one mole)elements in

theirstandard states

nega ve s gnheat released --exotherm ic rxn

H (CO2) = 393.5kJ/mol


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mo10g +(g) (g)

H = 0 kJ/mol 0kJ/mol

.mol

rxn

elements intheirstandardstates

product(twomoles

v e yto puton per molebasis!!

nega ve s gnheat released --

exotherm ic rxn H (H O) = 241.8kJ/mol

ote that I usually will nothave you calculate Hf onhomeworks or testsso you generally dont have to worry about normalizinanswer to a

per mole basis.Hess's Law -- Adding Reactions

The overall heat of reaction (Hrxn) is equal to the

sum of the Hf (products) minus the sum of the Hf

(reactants):

(# eqiv)H(reactants)

f

= eq v(products)rxn

Therefore, by knowing Hf of the reactants and

products, we can determine the Hrxn for any

reaction that involves these reactants and products.


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mo11EXAMPLE: CO2 is used in certain kinds of fireextinguishers

to put out simple fires. It works by smothering the fi" heavier" CO2 that replaces oxygen needed to maintfire.CO2 is not good, however, for more exotic electrical

chemical fires.g s

COg s

C s =kJ/mol

-kJ/mol

- mokJ/mol

TS TS

(# eqiv)H(reactants)

f

= eq v (products)rxn

(2 eqiv)(-602 kJ/mol) + (1

eqiv)(0 kJ/mol)

= rxn

(2 eqiv)(0 kJ/mol) + (1 eqiv)

393 kJ/mol)= -

kJ/mol)-

kJ/mol)rxn

= - mo +

393 kJ/mol

rxn

} highly exothermic

rxn !!

=kJ/mol

x

Therefore, Mg will " burn"CO2 !


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mo12You can also add two reactions together to get the Hfor

another new reaction:Calculate for the

following reaction:

HrxnH O l OH l =??4 5 nven ese wo reac ons an

thermod namic data:a+ 3O2(g)b) C H (g) +

3O (g)

g3H2O(l)2CO2(g) +

2H2O(l)

= -kJ/moln = -kJ/mol4 n

ow osolve:

s on e pro uc s e o e rsreaction -- so we want to5sw c equa on a aroun o ge a son the roduct side:5

g3H2O l

g +1367 kJ/mol

rxn

*

no e awhen we

chan es si n! ! !

reverse ereaction, H n

ow we can a e wo reac ons oge erto ive us the desirednereaction:

C2H5OH(l) + 3O2(g) H= +1367 kJ/mol

rxn

g3H2O l

+H = -1411kJ/molrxn

g3O

g2H2O l4

H = -44

kJ/molrxn

gH O l OH l4 5

If we have to multiply one (or more) of the reactions someconstant to get them to add correctly, then we also wohaveto multiply Hrxn for that reaction by the same amo


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mo14Problem: Calculate H for the followingrxnreactions given the following Hf values:

Hf (SO2, g) = 297kJ/mol

Hf (H2SO4, l) = 814

kJ/mol

Hf (H2O, l) = 286

kJ/mol

Hf (SO3, g) = 396kJ/mol

Hf (H2SO4, aq) = 90

kJ/mol

Hf (H2S, g) = 20

kJ/mol

a) S(s) + O2(g) SO2(g)

b) 2SO2(g) + O2(g)

c) SO3(g) + H2O(l)

d) 2H2S(g) + 3O2(g)

2SO3(g)

H2SO4(l)

2SO2(g) + 2H2O(l)


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mo15Internal Energy -- E

The internal energy, E, represents all the energy

contained within a material. It includes kinetic

energy (heat), intra- and intermolecular forces

(bond energies, electrostatic forces, van der Waals),

and any other forms of energy present. As with

enthalpy, H, the absolute value cant (or is extremely

difficult) to define.What we can track is the change in E:

E = EfinalEinital = Eproducts

Ereactants

A key relationship is:E = q + w

Where q = heat and w = work performed on or by

the system. Sign notations:

q= positive = heat added to system (adds energy)

q= negative = heat removed (removes energy)

w= positive = work done on system (adds energy

w= negative = work done by system (removes ener


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mo16The most common type of work involvespressure/volume changes: e.g., explosion of gasoline

vapors in an internal combustion engine. The

explosion creates a dramatic pressure and volumeincrease that pushes the piston and creates work.

If one has a constant volume situation, then no

pressure/volume work will be done and w = 0.

So under constant volume conditions:

E = qThe change in internal energy is, therefore, equal to

the amount of heat added or removed from the

material (system).


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mo17Relationship Between E & H

E = q (at constant volume and temperature)

H = q (at constant pressure and temperature)

The difference is that volume changes occur for H athat typically involves work of some type. Remembethat significant volume changes only occur when gassare involved. So we only need to be concerned about

volume work when there is a change in the amount oproduced in a chemical reaction.

The relationship between H and E is defined,

therefore, as:

H = E + (n)RT

Where R = gas constant; T = temperature in kelvin,and:

n = equivalents (moles) of product gas

equivalents (moles) of reactant gas

If n = 0, then H = E.

But even when n 0, the PV work component is ussmall. See example in textbook (pages 574-575).


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mo18Entropy

The final state of a system is more energetically

favorable if:

1. Energy can be dispersed over a greater number

and variety of molecules.

2. The particles of the system can be more

dispersed (more disordered).

The dispersal of energy and matter is described by

the thermodynamic state function entropy, S.

The greater the dispersal of energy or matter in a

system, the higher is its entropy. The greater the

disorder (dispersal of energy and matter, both in

space and in variety) the higher the entropy.Adding heat to a material increases the disorder.

-structure

-disordered

-disordered


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mo19Unlike H, entropy can be defined exactly becauseof the Third Law of Thermodynamics:

Third Law of Thermodynamics: Any

pure crystalline substance at a

temperature of absolute zero (0.0 K) has

an entropy of zero (S = 0.0 J/Kmol).

Sign notation (EXTREMELY IMPORTANT!!):

+S indicates that entropy is increasing in the

reaction or transformation (it's getting more

disordered -- mother nature likes)

S indicates that entropy is decreasing in thereaction or transformation (it's getting less

disordered {more ordered} -- mother nature doesn't

like, but it does happen)


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mo20Qualitative " Rules" About Entropy:

1) Entropy increases as one goes from a solid to a

liquid, or more dramatically, a liquid to a gas.

0d s

0

15

0

10

0

50

uid

p asetransi t ions

re K2) Entropy increases if a solid or liquid is dissolved

in a solvent.

3) Entropy increases as the number of particles

(molecules) in a system increases:

N2O4(g) 2NO2(g)

S= 304 J/K(1

mole)

S= 480 J/K (2

moles)These first 3 above are most important forevaluating Srxn.


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mo21The rules below are mainly for comparing theentropy of individual molecules or materials.

4) The Entropy of any material increases with

increasing temperature

5) Entropy increases as the mass of a molecule

increases

S(Cl2(g)) > S(F2(g))

S= 165 J/K mol S= 158 J/Kmol

6) Entropy is higher for weakly bonded compounds

than for compounds with very strong covalent

bonds

S(graphite) > S(diamond)

S= 5.7 J/K mol S= 2.4 J/Kmol

Note that for individual molecules(materials) the

higher the entropy, the more likely the molecule will

want to fall apart to produce a number of smallermolecules.


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7) Entropy increases as the complexity (# of atoms,

# of heavier atoms, etc.) of a molecule increases

ntropy o a er es o Gaseous

H drocarbons

S = 186 J/KmolHe a

ne

S = 201 J/Kmol

S = 220 J/Kmol

ce y en e

Hy e

ne

S = 230 J/Kmol

S = 270 J/KmolH

H H ane

C

H

C

HH ropane


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mo23What are the biggest factors for evaluating Srxn for

a chemical rxn?

1) phase change 2) change in # of molecules

Problem: For the following reactions, is the entropy

of the reaction increasing or decreasing?

a) Ag+(aq) + Cl-(aq)

b) H2CO3(aq)

AgCl(s)

H2O + CO

2(g)

c) Ni(s) + 4CO(g)

d) H2O(s)

Ni(CO)4(l)

H2O(l)

e) graphite diamond

f) 2Na(s) + 2H2O

g) H2S(g) + O2(g)

h) 2H2O(l)

2Na+(aq) + 2OH (aq) +H2(g)

-

H2O(l) + SO(g)

2H2(g) + O2(g)

i) CO2(g) + CaO(s)

) CaCl2(s) + 6H2O(l)

CaCO3(s)

CaCl26H2O(s)

k) 2NO2(g) N2O4(g)


24/40

Just as with enthalpies, one can calculate entropies

of reaction.

(# eqiv)S(reactants)

S = (# eqiv)S

(products)rxn

EXAMPLE:

g s +CO

g s +C s

S = 32 J/Kmol 215

J/Kmol J/Kmol J/Kmol

TS TS

(# eqiv)S(reactants)

S = (# eqiv)S

(products)rxn

= eq v mo + eq v

(2 eqiv)(32 J/Kmol) + (1 eqiv

J/Kmol)

rxn

=

J/K.mol) J/K.mol)rxn

}S = 218J/K.mol

entropy s ecreas ng(reaction is becomingmore ordered

rxn


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mo25Spontaneous Processes

A process that takes place without the net input of

energy from an external source is said to be

spontaneous (notinstantaneous).

1) Rxn of sodium metal with water:

-2Na(s) + 2H2O 2Na+(aq) + 2OH (aq) +

H2(g)

2) Combustion rxns:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

2H2O(l)2H2(g) + O2(g)

3) Expansion of a gas into a vacuum

xCO2(g) yCO2(s) + zCO2(g) (x = y + z)

4) A salt dissolving into solution:

NH4NO3(s) + H2O(l) NH4+(aq) + NO3-(aq)


26/40

mo26Second Law of Thermodynamics: In

any spontaneous process the entropy

of the universeincreases

Suniverse = Ssystem + Ssurroundings

Second Law (variant): in trying to do

work, you alwayslose energy to the

surroundings.

You can' t even break even!

Neither entropy (S) or enthalpy (H)

alone can tell us whether a chemical

reaction will be spontaneous or not.

An obvious (?) conclusion is that one

needs to use some combination of the two.


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mo27Gibbs Free Energy

The combination of entropy, temperature

and enthalpy explains whether a reaction is

going to be spontaneous or not. The symbol

Gis used to define the Free Energyof a

system. Since this was discovered by J.

Willard Gibbs it is also called the Gibbs

Free Energy. "Free" energy refers to the

amount of energy available to do work once

you have paid your price to entropy. Note

that this is not given simply by H, the heat

energy released in a reaction.

G = H TS

When G is negative, it indicates that a

reaction or process is spontaneous. A

positive G indicates a non-spontaneous

reaction.


28/40

mo28

G = H TS

S

+

G = negative G = ??

spontaneous

spontaneousat all

temperaturesat hightemperatures

- +0 H

G = ?? G = positive

spontaneousat lowtemperatures

non-spontaneousat alltemperatures

-

Spontaneous = exoergic (energy releasing)

Non-spontaneous = endoergic (energy releasing)


29/40

mo29Remember that entropies are

given in units of J/Kmol while

enthalpies and free energies are in

kJ/mol.

DON'T forget to convert all units to

kJ or J when using both S and H

in the same equation! !


30/40

mo30G vs. G: Standard vs. Non-Standard Condition

Remember that the (not) on G indicates that th

numerical value of G is based on the reaction atstandard conditions (1 M solution concentration, 1 atgas pressure). Temperature is NOT part of standardconditions!

As soon as one has a concentration different than 1 M1 atm pressure, the not goes away and one has G

Consider the reaction:

Initial: 1 atm 1 atm 1 atm

2SO2(g) + O2(g) 2SO3(g)

Grxn = 142 kJ/mol

The Grxnof 142 kJ/mol is for when each gas is pr

with a concentration of 1 atm. This indicates that thereaction under these conditions will proceed to makeproducts (spontaneous).

As the reactants start reacting, however, theirconcentrations decrease (SO2twice as fast as O2) an

G turns into G and becomes less negative.

When G = 0 the reaction has reached equilibrium.Although for this rxn, SO2is probably the limiting

reagent (not enough present to complete the rxn).


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mo32Just as with enthalpies and entropies, one cancalculate free energies of reaction.

(# eqiv)

G(reactants)f

= eq v

(products)

rxn

EXAMPLE:

g s +CO (g)

g s +C(s)

G = 0kJ/mol

-kJ/mol

- mokJ/mol

TS S

(# mol) G(reactants)

f

= mo (products)rxn

= mo - mo + mo m (2 mol)(0 kJ/mol) + (1 mol)(

kJ/mol)

rxn

G = (-1140kJ)

-rxn = - +

394 kJrxn

} rxn!highly exothermicrxn !!

=kJrxn

ompare o w c was - orthe same rxn.

rxn

e mssng o energy wen oENTROPY!


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mo33Example: To make iron, a steel mill takes Fe2O3 (rust

or iron ore) and reacts it with coke (a complex, impure

form of carbon) to make iron and CO2. Based on the

data below, this is a non-spontaneous reaction at roomtemperature, but it becomes spontaneous at higher

temperatures. Assuming that H and S do not

change much with temperature, calculate the temp-

erature above which the reaction becomes spontaneous

(i.e., Grxn = 0).

Hrxn= +465 kJ/mol

Srxn= +552 J/molK (or0.552 kJ/molK

Grxn= +301 kJ/mol (at 298 K)

Grxn = Hrxn TSrxn

as we raise the temperature, G will eventuallyreach 0 andthen go negative & spontaneous, so let G = 0solve forT, the temperature at which this will happen:0 = Hrxn TSrxn

rearrang ng o so ve orgives:

/T = (Hrxn) (Srxn)

T = (465 kJ/mol) / (0.552 kJ/molK)

T = 842 K

(above this temperature Grxn will be negat ivewill have

a spontaneous reaction)


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mo34Problem: Calculate Grxn for the following.

Gf (SO2, g) = 300kJ/mol

Gf (SO3, g) = 371kJ/mol

Gf (H2SO4, l) = 690 kJ/mol Gf (H2SO4, aq) 742 kJ/molGf (H2O, l) = 237kJ/mol

Gf (H2S, g) = 34kJ/mol

a) S(s) + O2(g) SO2(g)

b) 2SO2(g) + O2(g)

c) SO3(g) + H2O(l)

d) 2H2S(g) + 3O2(g)

2SO3(g)

H2SO4(l)

2SO2(g) + 2H2O(l)


35/40

mo35Comparisons of Hrxn and Grxn

S(s) + O2(g) SO2(g)

Hrxn = 297 kJ/molSrxn = 11 J/mol K

Grxn = 300 kJ/mol

2SO2(g) + O2(g) 2SO3(g)

Hrxn = 198 kJ/mol

Grxn = 142 kJ/mol

SO3(g) + H2O(l) H2SO4(l)

Hrxn = 132 kJ/mol

Grxn = 82 kJ/mol

2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l)

Hrxn = 1126 kJ/mol

Grxn = 1006 kJ/mol


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mo36Entropy of Fusion and Vaporization

While the entropy of a substance increases steadily

with increasing temperature, there is a considerable

ump in the entropy at a phase transition:

250

o as

150

100

50

d

phasetransi t ions

Temperature(K)This jump at the melting point is called the entropy

of fusion, Sfusion, and as you might expect, it is

related to the enthalpy (or heat) of fusion, Hfusion:

HfusionSfusion

Tm


37/40


38/40

Problem: Calculate the boiling point for ethanol

(CH3CH2OH) from the data in the thermodynamic

tables.

(literature value = 78.5C)


39/40

mo39


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mo40

, , ,Davis, Peck & Stanley. Thomson Brooks/ColePublisher.

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1422-Chapt-15-Thermodynamics.good_notes (1).docx