chapter 19: chemical thermodynamics spontaneous processes… …happen without outside help …are...

Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics

Spontaneous processes…

…happen without outside help

…are “product favored”


Spontaneous processes

at 25oC

H2O (s) → H2O (l)



CO2 (s) → CO2 (g)

at 25oC

‘dry ice’



4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)

at 25oC

rust



…occur in a definite direction: towards the formation of product

H2O (s) → H2O (l) CO2 (s) → CO2 (g)

4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)

at 25oC


The direction of a spontaneous processes may depend on temperature

at -10oC

H2O (l) → H2O (s)


H2O (l) → H2O (s)

CO2 (s) → CO2 (g)

2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)

at 25oC

• At a given temperature and pressure, processes arespontaneous only in one direction

• If a processes is spontaneous in one direction it is non-spontaneous in the other direction


Which reactions are spontaneous?

• many spontaneous reactions are exothermic (H < 0)

… but not all!

• Some reactions are endothermic (H > 0) and still spontaneous

NH4NO3 (s) → NH4+ (aq) + NO3

- (aq)

H > 0



Reactions proceed towards a more probable state

In general, the more probable state is associated

with more disorder


Entropy can be thought of as a measure of disorder

Ludwig Boltzmann (1844-1906)

S = k log W

k = 1.38 x 10-23 J / K

W = Wahrscheinlichkeit (probability)


The change in entropy for any process is:

S = Sfinal - Sinitial

What is the sign of S for the following processes at 25oC ?

H2O (s) → H2O (l)

CO2 (g) → CO2 (s)


Second Law of Thermodynamics

For any spontaneous process, the entropy of the universe increases

Souniverse = So

system + Sosurroundings > 0


Ssolid < Sliquid < Sgas

gasliquidsolid

Of all phase states, gases have the highest entropy


Larger Molecules generally have a larger entropy

Ssmall < Smedium < Slarge


Often, dissolving a solid or liquid will increase the entropy

dissolves


Dissolving a gas in a liquid decreases the entropy

dissolves


The entropy of a substance increases with temperature


S = k ln W

• W is a measure for how many different ways there areof arranging a molecule or an ensemble of molecules (the system)

• W reflects the number of microstates of a system

• The larger the possible number of microstates the higher the entropy


Ssolid < Sgas

gassolid

Of all phase states, gases have the highest entropy


Larger Molecules generally have a larger entropy

Ssmall < Slarge


Often, dissolving a solid or liquid will increase the entropy


Dissolving a gas in a liquid decreases the entropy


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What is the sign of S for the following reactions?

FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g)

Ba(OH)2 (s) → BaO (s) + H2O (g)

2 SO2 (g) + O2 (g) → 2 SO3 (g)

Ag+ (aq) + Cl- (aq) → AgCl (s)


For each of the following pairs, which substance has a highermolar entropy at 25oC ?

HCl (l) HCl (s)

C2H2 (g) C2H6 (g)

Li (s) Cs (s)

Pb2+ (aq) Pb (s)

O2 (g) O2 (aq) HCl (l) HBr (l)

CH3OH (l) CH3OH (aq)N2 (l) N2 (g)


Sorxn = Σ n So(products) – Σ m So(reactants)

If you know the standard molar entropies of reactants andproducts, you can calculate

S for a reaction:


substance So (J/K-mol)

H2 130.6

C2H4 (g) 219.4

C2H6 (g) 229.5

What is So for the following reaction?

C2H4 (g) + H2 (g) → C2H6 (g)

Sorxn =

Do you expect S to be positive or negative?

So for elements are NOT zero



We need to know the magnitudes of both S and H !

Go = Gibbs free energy…

… is a measure of the amount of “useful work” a system can perform

Go = Ho - TSo


Go = Ho - TSo

A reaction is spontaneous if Go is negative

J. Willard Gibbs(1839 – 1903)


2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)

The reaction of sodium metal with water:

Is the reaction spontaneous?

What is the sign of Go?

What is the sign of o?

What is the sign of So?


Go < 0 => reaction is spontaneous(“product favored”)

Go > 0 => reaction is non-spontaneous

Go = 0 => reaction is at equilibrium

“exergonic”

“endergonic”


Go = Ho - TSo

Ho So

+ -

Go

- +

- -

+ +


Go = Ho - TSo

H2O (s) → H2O (l) Go < 0

at 25oC (298K):

spontaneous:

Ho > 0

So > 0

So > 0 > Ho => Go < 0

but: if T becomes very small:

at 298K:

So > 0 < Ho => Go > 0


A diamond left behind in a burning house reacts according to

2 C (s) + O2 (g) → 2 CO (g)

What is the value of Go for the reaction at 298K ?

substance Hfo (kJ/mol) Gf

o (kJ/mol) So (J/K-mol)

O2 0 0 205.0

C (diamond, s) 1.88 2.84 2.43

C (graphite, s) 0 0 5.69

CO2 (g) -393.5 -394.4 213.6


There are two possible ways to calculate Go:

I) Go = Ho - TSo

II) Go = Σ n Gfo (products) – Σ m Gf

o (reactants)

calculate Go from Ho and So :


I) calculate Go from Ho and So : Go = Ho - TSo



O2 0 0 205.0

C (diamond, s) 1.88 2.84 2.43


CO (g) -110.5 -137.2 197.9

2 C (s) + O2 (g) → 2 CO (g)

Go = - 280.2 kJ




O2 0 0 205.0

C (diamond, s) 1.88 2.84 2.43


CO (g) -110.5 -137.2 197.9

2 C (s) + O2 (g) → 2 CO (g)

II) Go = Σ n Gfo (products) – Σ m Gf

o (reactants)

Go = -280.1 kJ


Consider the following reaction:

2 H2 (g) + O2 (g) → 2 H2O (l)

S = -326.3 J/K, H = -571.7 kJ , and G = -475.3 kJ at 25oC. At what temperature does the reaction become spontaneous in the

opposite direction?

G = H - TS

The reaction “switches” direction if H = TS , i.e. if G = 0

H < 0 S < 0 G < 0 at 298 K

G > 0 at high T


0 = H - TS

TS = H

T =ΔSΔH

x 1 kJ / 1000JT =

T = 1752.1 K

-571.7 kJ

-326.3 J/K

At temperatures greater than 1752.1 K liquid water will spontaneously decompose into H2 and O2 !!


At the normal melting point, the Gibbs free energies ofthe solid and liquid phase of any substance are equal:

H2O (s) → H2O (l)at 0oC

Go = 0

H2O (l) → H2O (g)at 100oC

Go = 0

At the normal boiling point, the Gibbs free energies ofthe liquid and gas phase of any substance are equal:


At a phase change: Go = 0

Go = Ho - TSo

0 = Ho - TSo

if we know Hofus (or Ho

vap) , we can

calculate

So at the melting (or boiling point):

Ho = TSo

So = Ho

T


Sofus

= Hofus

T

The entropy of melting for H2O

= 6.02 kJ/mol

273.15 K

= 22.0 J/mol-K

Sovap

= Hovap

T

The entropy of vaporization for H2O

= 40.7 kJ/mol

373.15 K

= 109 J/mol-K

units!melting temp

boiling temp

chapter 19: chemical thermodynamics spontaneous processes… …happen without outside help …are...

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