CHEMICAL THERMODYNAMICS AP Chemistry Chapter 19 Notes

CHEMICAL

THERMODYNAMICS

AP Chemistry

Chapter 19 Notes

REVIEW1st Law of

Thermodynamics

Energy in the universe is conserved

REVIEW

Path Functionvalue depends on how process takes

place (i.e. q, w)

heat

q = mCpT

REVIEWState Function

value independent of path - is a defined reference or zero

point (i.e. H)

Enthalpy - Enthalpy - HHzero pointzero point

HHff00 of of elementselements in in

natural form at natural form at 252500CC

Enthalpy - H

types:

Hrxn = nHf0(prod) -

nHf0(react)

Enthalpy - H

types:

Hfus , Hvap

ENTROPY “S”ENTROPY “S”a measure of a measure of

randomness or randomness or disorder of a disorder of a

systemsystem

Entropy is NOT conserved. The universe seeks

maximum disorder.

2nd Law of Thermodynamics

for a spontaneousprocess, entropy

increases

Entropy - state functionEntropy - state function

zero reference: S = 0zero reference: S = 0

for a perfect crystal at for a perfect crystal at absolute zeroabsolute zero

Ssolid < Sliquid << Sgas

S0 = standard entropyof elements & cmpdsat P = 1 atm; T = 250C

units = J/K mol

[Appendix L, text]

Calculate S using Hess’ Law

S(rxn) = nS0(prod) - nS0(react)

Example:

Ca(s) + C(gr) + 3/2 O2 CaCO3(s)

Suniv = Ssys + Ssurr

if Suniv > 0 process spontaneous

Suniv = Ssys + Ssurr

if Suniv < 0 process ?

Suniv = Ssys + Ssurr

if Suniv = 0 process ?

to relate S and H

consider

H2O(s) H2O(l)

where the water is the system & everything

else is the surroundings

Temperature Dependence

S = J/mol (1/T)

Ssurr = -H/T

Ssys Ssurr Suniv spon

+ +

Ssys Ssurr Suniv spon

+ + + Y

- -

Ssys Ssurr Suniv spon

+ + + Y

- - - N

+ -

Ssys Ssurr Suniv spon

+ + + Y

- - - N

+ - ? D

- +

Ssys Ssurr Suniv spon

+ + + Y

- - - N

+ - ? D

- + ? D

change phase

change phase0change phase T

HS

For a phase change:

GIBB’S FREEENERGY

G

most abstract of thermodynamic state

functions

w1 w = reversible

work

w2

P

V

Definition:

G = w + PVw: reversible workPV: pressure/volume work

isothermal, reversible path

G = w + PV + VP

at constant PP = 0 so VP = 0

G = w + PV

G = w + PV + VPat constant PG = w + PV

at constant VV = 0 so PV = 0G = w = useful work

G cannot be G cannot be measuredmeasured

must measure must measure G G over a processover a process

ZERO REFERENCEG = 0 for elements in

stable form under Standard

Thermodynamic Conditions

T = 25oC ; P = 1 atm

Gf0 = standard Free

Energy of Formation from the elements

Appendix L, text

G follows Hess’ Law:

G0 (rxn) = nGf

0(p) - nGf0(r)

Summary of Laws ofThermodynamics

Zeroth Law:Heat Gain = Heat Loss

Summary of Laws ofThermodynamics

First Law:Law of Conservation

of Energy

Summary of Laws ofThermodynamics

Second Law:Defines Entropy

Summary of Laws ofThermodynamics

Third Law:Defines Absolute Zero

GIBB’S HELMHOLTZEQUATION

G H

T

combine

G = -aT + H

G, H are state functions, thus “a” must be a state function

G = -S(T) + H

Gibb’s HelmholtzEquation

G = H - TS

Units on the State Functions

HkJ

mol

SJ

K mol

GkJ

mol

G H T S

GT

HT

S

SST

Gsurr

surrST

H but

univST

G

univsurr SSS

but

thus, a process is spontaneousif and only if

G is negative

Spontaneitycontrolled by

enthalpy(minimum energy)

Sponaneitycontrolled by

enthalpyentropy

(maximum disorder)

Sponaneitycontrolled by

enthalpyentropy

both

Predict Spontaneity

IF H(-) and S(+)

G = -H - T(+S)

G < 0, => spontaneous

Predict Spontaneity

IF H(+) and S(-)

G = +H - T(-S)

G > 0, => NOT spontan

Summary of Spontaneity

H S G Spont. - + - yes + - + no + + + or - ? - - + or - ?

Uses of theGibb’s Helmholtz

Equation

1. Find the molar entropy of formation for ammonia.

2. Elemental boron, in thin fibers, can be made from a boron halide:

BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)

Calculate: Calculate: HH00, , SS00 and and GG00..

Spontaneous?Spontaneous?

Driving force?Driving force?

3. Using thermodynamic information, determine the boiling point of bromine.

ThermodynamicDefinition ofEquilibrium

Geq = 0

by definition

G = H - TS&

H = E + PV

thus,

G = E + PV - TS

take derivative of both sides

dG = dE + PdV + VdP - TdS - SdT

for a reversible process

TdS = q

derivative used for state function

whilepartial derivative used

for path function

if the only work is PVwork of expansion

PdV = w

First Law ofThermodynamics

E = q - w

dE = q - w = 0

thus

q = wor

TdS = PdV

by substitution

dG = 0 + PdV + VdP - PdV - SdTor

dG = VdP - SdT

let’s assume we havea gaseous system atequilibrium, therefore,examine Kp at thatconstant temperature

at constant T

SdT = 0

thusdG = VdP

dG V dP

VnRT

P

G

G

P

P

1

2

1

2

but

dG nRTdPP

G

G

P

P

1

2

1

2

dG nRTdPP

G G nRTPP

G

G

P

P

1

2

1

2

2 12

1

ln

G G RTPP

n

1 21

2

ln

if “condition 2” isat standard thermodynamic conditions, thenG2 = G0 and P2 = 1 atm

thus

G G RT Pn 0 ln

determine G for

aA + bB cC + dD

where all are gases

G = Gprod - Greact

= cGC + dGD - aGA - bGB

but

cG cG RT PC Cc 0 ln

and likewise for the others

cG dG aG bGC D A B0 0 0 0

Grxn0

and

G G RTP P

P PrxnC

cD

d

Aa

Bb

0 ln( ) ( )

( ) ( )

( ) ( )

( ) ( )?

P P

P PC

cD

d

Aa

Bb

but

Thus,

G = G0 + (RT) ln Q

But

aA + bB cC + dD

G = 0

thus

G RT Krxn P0 ln

or in general

G RT Krxn eq0 ln

THERMODYNAMICSTHERMODYNAMICS

&&

EQUILIBRIUMEQUILIBRIUM

3. Using thermodynamic information, determine

the boiling point of bromine.

FACT: Product-favored FACT: Product-favored

systems have Ksystems have Keqeq > 1. > 1.

Thermodynamics and KThermodynamics and Keqeq

Therefore, both Therefore, both

∆G˚∆G˚rxnrxn and K and Keqeq are are

related to reaction related to reaction

favorability.favorability.

Thermodynamics and KThermodynamics and Keqeq

KKeqeq is related to reaction is related to reaction

favorability and thus to ∆Gfavorability and thus to ∆Goorxnrxn..

The larger the value of K the The larger the value of K the more negative the value of more negative the value of

∆G∆Goorxnrxn

Thermodynamics and KThermodynamics and Keqeq

∆∆GGoorxnrxn= - RT lnK= - RT lnK

where R = 8.314 J/K•molwhere R = 8.314 J/K•mol

Thermodynamics and KThermodynamics and Keqeq

Calculate K for the reactionCalculate K for the reaction

NN22OO44 2 NO2 NO22

∆∆GGoorxnrxn = +4.8 kJ = +4.8 kJ

K = 0.14K = 0.14

When ∆GWhen ∆G00rxnrxn > 0, then K < 1 > 0, then K < 1

∆Gorxn = - RT lnK

Thermodynamics and Keq

∆∆G, ∆G˚, and KG, ∆G˚, and Keqeq

∆∆G is change in free energy at G is change in free energy at non-standard conditions.non-standard conditions.

∆∆G is related to ∆G˚G is related to ∆G˚∆∆G = ∆G˚ + RT ln Q G = ∆G˚ + RT ln Q

where Q = reaction quotient where Q = reaction quotient

∆∆G, ∆G˚, and KG, ∆G˚, and Keqeq

When Q < K or Q > K, reaction is When Q < K or Q > K, reaction is spontaneous.spontaneous.

When Q = K reaction is at equilibriumWhen Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibriumWhen ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln KTherefore, ∆G˚ = - RT ln K

Product favored Product favored reactionreaction

–∆–∆GGoo and K > 1 and K > 1

In this caseIn this case

∆ ∆GGrxnrxn is < ∆G is < ∆Goorxnrxn , so , so

state with both state with both reactants and reactants and

products present products present is MORE STABLE is MORE STABLE

than complete than complete conversion.conversion.

∆∆G, ∆G˚, G, ∆G˚, andand K Keqeq

Product-favored reaction. Product-favored reaction.

2 NO2 NO22 N N22OO44

∆∆GGoorxnrxn = – 4.8 kJ = – 4.8 kJ

Here ∆GHere ∆Grxnrxn is less than ∆G is less than ∆Goorxnrxn , so the , so the

state with both reactants and state with both reactants and products present is more stable products present is more stable than complete conversion.than complete conversion.

∆G, ∆G˚, and Keq

∆∆GGoorxnrxn is the change in free is the change in free

energy when reactants energy when reactants

convert COMPLETELY to convert COMPLETELY to

products.products.

Thermodynamics and KThermodynamics and Keqeq

OverviewOverview

KKeqeq is related to reaction is related to reaction

favorability.favorability.

When ∆GWhen ∆Goorxnrxn < 0, reaction < 0, reaction

moves energetically moves energetically

“downhill”“downhill”

4. For the following reaction, calculate the temperature at which the reactants are favored.

)g(NH)g(H2

3)g(N

2

1322

THERMODYNAMICSTHERMODYNAMICSOFOF

CHEMICALCHEMICALREACTIONSREACTIONS

5. How much useful work can be obtained from an engine fueled

with 75.0 L of hydrogen at 10 C at 25 atm?

6. The reaction to split water into hydrogen and oxygen can be promoted by firstreacting silver with water.

2 Ag(s) + H2O(g) Ag2O(s) + H2(g)

Ag2O(s) 2 Ag(s) + 1/2 O2(g)

Calculate H0, S0 and G0 for each reaction.

Combine the reactions and calculate H0 and G0 for the combination.

Is the combination spontaneous?

At what temperature does the second reaction become spontaneous?

7. The conversion of coal into hydrogen for fuel is:

C(s) + H2O(g) CO(g) + H2(g)

Calculate G0 and Kp at 250C.

Is the reaction spontaneous?

At what temperature does the reaction become spontaneous?

Calculate the temperature at which K = 1.0 x 10-4.

8. The generation of nitric acid in the upper atmosphere might destroy the ozone layer by: NO(g) + O3(g) NO2(g) + O2(g)

Calculate G0 (reaction)and K at 250C.