chapter 19: chemical thermodynamics

Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics


Spontaneous processes

H2O (s) H2O (l)

at 25oC



CO2 (s) → CO2 (g)

at 25oC

‘dry ice’

at 25oC



rust

at 25oC

Fe (s) + O2 (g)Fe2O3 (s)



at 25oC 5 H2O (g) + 6 CO2 (g)C6H10O5 (s) + 6 O2 (g)

Burning Paper



H2O (s) → H2O (l) CO2 (s) → CO2 (g)

4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)

at 25oC

…happen “on their own”although they sometimes require an initial push to get going

…are “product favored”occur in a definite direction: towards the formation of product


H2O (s) H2O (l)

below 0oC

The direction of a spontaneous processes may depend on temperature


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

non-spontaneous

spontaneous

non-spontaneous

• 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


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)

S > 0 (positive)

S < 0 (negative)


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

In a gas, molecules are more randomly distributed


Larger molecules/atoms generally have a larger entropy

Larger molecules have more internal motion

Ssmall < Smedium < Slarge


Often, dissolving a solid or liquid will increase the entropy

dissolves

more disordered arrangementhigher entropy

lower entropy


Dissolving a gas in a liquid decreases the entropy

dissolves

overall more disordered arrangement:

higher entropy


The entropy of a substance increases with temperature


} size of molecules increases

} Sgas > Sliquid

} dissolving a gas in a liquid is accompanied bya lowering of the entropy

} dissolving a liquid in another liquid is accompanied by an increase in entropy


What is the sign of S for the following reactions?

FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g)solid solidgas gas

1 mol 2 mol

S > 0

Ba(OH)2 (s) → BaO (s) + H2O (g)

solid solid gas

S > 0

2 SO2 (g) + O2 (g) → 2 SO3 (g)

gas gas gas

2 mol 1 mol 2 mol

S < 0

Ag+ (aq) + Cl- (aq) → AgCl (s) S < 0

in solution insoluble


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 = [229.5] – [219.4 + 130.6] = -120.5 J/K

Do you expect S to be positive or negative?

So for elements are NOT zero


Which reactions are spontaneous?

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 Ho?

What is the sign of So?

Yes

Go = negative

Ho = negative (exothermic!)

So = positive


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

- + -

- - sign depends on T !low T => Go is negativehigh T => Go is positve

+ + sign depends on T !low T => Go is positivehigh T => Go is negative

+


e. I have no idea how to even start thinking about this

endothermic H > 0

Spontaneity: G must be negative - think about the signs of H and S



endothermic H > 0

number of gas molecules doubles S > 0



Go = Ho - TSo

Ho So

+ -

Go

- + -



+




endothermic H < 0

number gas molecules doubles S > 0


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. Does the the decomposition of water ever become spontaneous?

If the decomposition of water is spontaneous, thenthe formation of water is non-spontaneous:

2 H2 (g) + O2 (g) → 2 H2O (l) G > 0

i.e. is there a temperature at which G becomes > 0 (non-spontaneous)?



2 H2 (g) + O2 (g) → 2 H2O (l)


G = H - TS

H < 0S < 0


Go = Ho - TSo

Ho So

+ -

Go

- + -



+



2 H2 (g) + O2 (g) → 2 H2O (l)


G = H - TSH < 0S < 0

The decomposition of water becomes spontaneous at high temperatures!

The formation of water becomes non-spontaneous at high temperatures


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

CO (g) -110.5 -137.2 197.9


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

Ho =[2 x (-110.5] – [2 x 1.88 + 0]

2 C (s) + O2 (g) → 2 CO (g)

= -224.8 kJ

So = [2 x 197.9] – [2 x 2.43 + 205.0] = 185.9 J/K

Go = -224.8 kJ - 298K x 185.9 J/K1000J

1kJ = - 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 = [2 x (-137.2)] – [2 x 2.84 + 0] = -280.1 kJ

chapter 19: chemical thermodynamics

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