Chem131_Chapter5

7/28/2019 Chem131_Chapter5

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STANDARD THERMODYNAMIC FUNCTIONS

aA+ bB cC+ dD

Chemical reaction

A + bB = cC+ dD

Equilibrium

Need to define the zeros from which things are measured!(e.g., sea level, center of earth)

Standard sates of pure substances(denoted by superscript )

Solid and liquid: P = 1 bar and temperature = T

Gas: P = 1 bar, temperature = T, ideal gas

V

m,200

H

m,200

H

200molar quantity

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Standard enthalpy of a reaction

aA1 + bA2 cA3 + dA4

H

rxn,T H

T H

HT = cH

m,T(A3) + dH

m,T(A4) aH

m,T(A1) bHm,T(A2)

HT = 3H

m,T(A3) + 4H

m,T(A4) + 1H

m,T(A1) + 2Hm,T(A2)

H

T =i

iH

m,T(A

i)2


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H

T =

i

iH

m,T(Ai)

molar enthalpy of Ai in its standard state at temperature TH

m,T(Ai) =

U

m,T(Ai) = molar internal energy of Ai in its standard state at temperature T

Not measurable

Not interesting

Only changes matter!

H

T

tabulated, but not for every reaction...

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Standard enthalpy of formation

(tabulated, at least for some temperatures)

fH

T =

is the enthalpy change associated with the process offorming the substance of interest starting from theseparated components in the standard states of theirmost stable forms (temperature = T and pressure = 1 bar)

6 C(graphite, 298K, P) + 3 H2(ideal gas, 298K, P) C6H6(liquid, 298K, P

)

How can we calculate the enthalpy of formation of benzene at T = 298 K?

How can we calculate the enthalpy of formation of formaldehyde at T = 307 K?

C(graphite, 307K, P) + H2(ideal gas, 307K, P)

+

1

2 O2(ideal gas, 307K, P

)

H2CO(ideal gas, 307K, P

)

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6 C(graphite, 298K, P) + 3 H2(ideal gas, 298K, P) C6H6(liquid, 298K, P

)

How can we calculate the enthalpy of formation of benzene at T = 298 K?

How can we calculate the enthalpy of formation of formaldehyde at T = 307 K?

C(graphite, 307K, P) + H2(ideal gas, 307K, P)

+1

2O2(ideal gas, 307K, P

) H2CO(ideal gas, 307K, P)

fH

T = 0 for each element in its reference form

fH

T = 0 for graphite (at both 298 K and 307 K

fH

T = 0 for H2 (at both 298 K and 307 K

fH

T = 0 for O2 at 307 K

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aA1 + bA2 cA3 + dA4

H

T =

i

iH

m,T(Ai) =

i

ifH

T(Ai)

Reactants in standard states

at temperature T

Products in standard states

at temperature T

H

T

Elements in standard states

at temperature T

fH

T(reactants) fH

T(products)

H

T = fH

T(products)fH

T(reactants)6


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How to determine the enthalpy of formation

fH

T =fH

T(step 1)

+fH

T(step 2)

+

fH

T(step 3)+fH

T(step 4)

+fH

T(step 5)

+fH

T(step 6)

6 steps & use the properties of a state function

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Step 1: Enthalpy change associated with the hypothetical transformation from anideal gas at T and 1 bar to a real gas at T and 1 bar

Need to calculate Hreal from Hideal

H= Hideal Hreal = Ha +Hb +Hc

Real gas at T and P = 0 bar Ideal gas at T and P = 0 bar

Ha

Hb

Hc

Real gas at T and Po Ideal gas at T and PoH= Hideal Hreal

T = constant

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Real gas at T and Po Ideal gas at T and Po


H= Hideal Hreal


HaHb

Hc

T = constant

Hb = Ub + PVb Hb=

Ub

Ub = intermolecular interactions 0 P 0when

check the behavior ofreal gases according to

Boyles and Charleslaws

Hb = 010


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H= Hideal Hreal


HaHb

Hc

T = constant

HP

T

= VTVHc

Hc =?

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H= Hideal Hreal


HaHb

Hc

HP

T

= VTV

T = constant

Hc = 0

Hc

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Step 1: Enthalpy change associated with the hypothetical transformation from anideal gas at T and 1 bar to a real gas at T and 1 bar



Need to calculate Hreal from Hideal

H= Hideal Hreal

Ha

Hb

Hc

T = constant

H= 0

P

(V TV)dP

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Step 2: Enthalpy change associated with mixing the different substances at T and 1bar (more important for gases)

CHAPTER 6

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Step 3: Enthalpy change associated with bringing the substances from T and Po toT and P of interest

H(step3) =

i

Hi(T, P T, P)

Hi = T

T

CP,idT+ P

P(Vi TVii)dP

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Step 4: Enthalpy change associated with the reaction that leads to the formation ofthe compound of interest from the mixed elements at T and P.

Use of a calorimeter(see book for a description of how it works)

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Step 5: Enthalpy change associated with bringing the substances from T and P atwhich they are formed to the standard conditions T and Po.

H(step5) =

i

Hi(T, P T, P)

Hi = T

TCP,idT+

P

P(Vi TVii)dP

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Step 6: Enthalpy change associated with the hypothetical transformation from areal gas at T and 1 bar to an ideal gas at T and 1 bar

Inverse process of step 1(now done for the newly formed gas species)

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Relation between Ho and Uo

H= U+ PV H= U+ PVconstant pressure

aA1 + bA2 cA3 + dA4

V

= cV

m,A3+ dV

m,A4 aV

m,A1 bV

m,A2

H= U

+ P

V

volume change of solids and

liquids is small

V

i,gases

iV

m,i

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H= U

+ P

V

aA1 + bA2 cA3 + dA4

V

= cV

m,A3+ dV

m,A4 aV

m,A1 bV

m,A2

V

i,gases

iV

m,i

V

m,i=

RT

P

V

m,i = ngRT

P

i,gases

i = ng

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H = U

+ PV

aA1 + bA2 cA3 + dA4

H

= U

+ngRT

H

U

if there are no gases

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Hesss law

Often the reaction that forms a compound from its elements cannot be done.

Ho cannot be determined directly2C(graphite) + 3H2(g) C2H6(g)

2H6(g) +7

2O2(g) 2CO2(g) + 3H2O(l)

(graphite) + O2(g) CO2(g)

H2(g) +1

2O2(g) H2O(l)

H

1

H

2

H

3

(1)

(2)

(3)

2C(graphite) + 3H2(g) C2H6(g) (1) + 2 (2) + 3 (3)

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Hesss law

Often the reaction that forms a compound from its elements cannot be done.

Ho cannot be determined directly2C(graphite) + 3H2(g) C2H6(g)

2H6(g) +7

2O2(g) 2CO2(g) + 3H2O(l)

(graphite) + O2(g) CO2(g)

H2(g) +1

2O2(g) H2O(l)

H

1

H

2

H

3

(1)

(2)

(3)

H

= H

1+ 2H

2+ 3H

3

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Temperature dependence ofHo

We know Ho at T1 and we would like to know Ho at T2: dH

dT=?

dH

dT=

i

i

dH

m,i

dT

dH

dT=

i

iC

P,m,i = C

P

H

T2H

T1=

T2

T1

C

PdT

Kirchhoffs Law

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S O C C O S


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Temperature dependence ofHo

H

T2H

T1=

T2

T1

C

PdT

Kirchhoffs Law

H

T2H

T1 C

P(T2 T1)

C

P,m = a+ bT+ cT2+ dT

3

IfT is small:

Otherwise use Kirchhoffs law with:

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Conventional entropies

A lot of bookkeeping... in the end we are interested inS

(1) Choose an arbitrary value (=0) for entropy of each element in a chosenreference state (T=0, Po)

(2) Tabulate Sfor any change from elements in the chosen reference state tothe desired substance in a standard state

Entropy reference state

Pure element in its stable condensed form (solid or liquid) at 1 bar in the limit T 0 K

Arbitrary definition

S

m,0 = limT0

S

m,T = 0

pure element in its stable condensed form

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How about So at T > 0 K?

If there is a phase change...

S= S

T S

0=

T

0

Cp

TdT

S=

Tpc0

Cp

TdT+Spc +

TTpc

Cp

TdT

How about the conventional entropy of a compound?

Its complicated!

S=qrev

T

a chemical reaction is an irreversible process!

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Third law of thermodynamics

Measurements ofG for several reactions led Nernst (1907) to postulate

limT0

G

T

P

= 0

G

T

P

= S

G

T

P

= S

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Measurements ofG for several reactions led Nernst (1907) to postulate

limT0

G

T

P

= 0

G

T

P

= S

G

T

P

= S

limT0

S = 0

Important

not true when the substances involved in the reaction are not in internal equilibriume.g., supercritical liquids

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For any isothermal process that involves only substances in internal equilibrium,the entropy change goes to zero as T goes to zero

limT0

S = 0

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How to calculate conventional entropies

A+B C

S

= 0 S

= 0 S

= 0

element element compound

T 0

T = 0 S=?

S= Sm,T2 =Tfus0

Cp,m(solid)T

dT+

fusH

m

T+

T2Tfus

Cp,m(liquid)T

dT+ . . .

there may be otherphase changes

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S= Sm,T2 =

Tfus0

Cp,m(solid)

TdT+

fusH

m

T+

T2Tfus

Cp,m(liquid)

TdT+ . . .

If gases are involved we need to calculate the entropy change associated to thehypothetical transformation from an ideal gas at P = 1 bar and T to a real gas at P =

1 bar and T

S(ideal,T,P) S(real,T,P)



T = constantSa

Sb

Sc

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S(ideal,T,P

)

S(real,T,P

)



T = constantSa

Sb

Sc

SP

T

=

VT

P

For steps a and c

Sb = 0In the limit T0, the entropy of ideal and real gasesbecome equal (statistical mechanics, Chem 132)

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Standard enthalpy of reaction

S

T =

i

iS

m,T,i

usually tabulated at T = 298 K

ST2= S

T1+

T2

T1

CP

TdT+phase change contributions

At different T,

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Standard Gibbs energy of reaction

aA+ bB cC+ dD

pure separated products intheir standard states at T

pure separated reactants intheir standard states at T

G

T =

i

iG

m,T,i G

T =

i

ifG

T,i

fG

T = fH

T,i TfS

T,i

enthalpies of formation from So and 3rd law of thermodynamics

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Chem131_Chapter5