Thermodynamics b
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Thermodynamics B. Thermodynamics Deals with the interconversion of heat an other forms of energy. Energy can be converted from one form to another, it can not be created or destroyed. i.e. The energy of the universe is constant. First Law:.  E = q + w.

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Thermodynamics B

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Thermodynamics b

Thermodynamics B


Thermodynamics b

  • Thermodynamics

    • Deals with the interconversion of heat an other forms of energy

Energy can be converted from one form to another, it

can not be created or destroyed. i.e. The energy of

the universe is constant.

First Law:

E = q + w


Thermodynamics b

Want to be able to predict if process or reaction will occur

spontaneously (without outside interference)

In many cases, spontaneous reactions are exothermic, but not always.

H2O

CaCl2(s)Ca+2(aq) + 2 Cl - (aq)

H = - 82.8 kJ

H2O

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

H = + 25.0 kJ

So: can’t use H to determine spontaneity.

New term: Entropy: (S)

Entropy is a state function

Entropy is a measure of the randomness or disorder of a system.

The greater the disorder, the greater the entropy

e.g. S(s) < S(l) << S(g)


Thermodynamics b

Different from enthalpy, can have standard absolute Entropy (So)

So: So = Sof - Soi

So always > 0

Second Law of Thermodynamics:

Entropy of the universe always increases for an irreversible

process or remains constant for a reversible process.

Suniverse = Ssystem + Ssurroundings

Ssurroundings comes primarily from change in heat energy (q at

constant pressure) but also depends on the Temperature of the

surroundings.


Thermodynamics b

Third Law of Thermodynamics

Entropy of a pure, perfectly crystalline substance is zero at

absolute zero of temperature.

At 0 K motion ceases, therefore, there is only one

possible structure and no randomness (perfect order)

Why?

  • = number of arrangements

    k = boltzmann constant

So, third law gives us Absolute Entropy


Thermodynamics b

Determine if the entropy of the system:

A) increasesB) decreasesC) stays the same

1. Benzene(s) --> benzene (l)

A

2. NaNO3(s) --> NaNO3 (aq)

A

3. S(s) + O2(g) --> SO2(g)

C

4. MgCO3(s) --> MgO(s) + CO2(g)

A

5. PCl3(l) + Cl2(g) --> PCl5(s)

B

6. 2 HgO(s) --> 2 Hg(l) + O2(g)

A


Thermodynamics b

Entropy and Physical Properties

  • S increases with mass:

    He < Ne < Ar

  • S increases with molecular complexity:

    CH$ < C2H6 < C3H8

  • S increases with softness of solid crystals

  • S is greater in metallic solids than in network solids:

    C (diamond) < C(graphite)

  • S increases when a solid melts

  • S increases when a solid or liquid vaporizes

  • S usually increases when a solid or liquid dissolves in water

  • S decreases when a gas dissolves in water

9. S increases as volume increase

10. S increases as temperature increases


Thermodynamics b

Which of the following will have the largest entropy?

C3H8 (l)

C3H8(g)

C3H6(g)

C4H10(l)

C4H10(g)


Thermodynamics b

In order to determine spontaneity, need to know Suniv > 0

Suniv = Ssys + Ssurr and it is not always possible to determine what

is happening in the system and the surroundings.

So, it is necessary to have another way to assess spontaneity that

depends only on the system


Thermodynamics b

Gibbs Free Energy (G)

G = H - TS where all the quantities are for the system

G is a state function

It is an Extensive property

There is NO absolute free energy, like enthalpy, we can

only measure G.

Gibbs - Helmholtz Equation

G = H - T S

GT,P < 0

Spontaneous Reaction

GT,P > 0

Non - Spontaneous Reaction

GT,P = 0

Reaction at Equilibrium


Thermodynamics b

Standard Free Energy change for a reaction

Gorxn

Reactants and Products all in standard states

Standard States

Gas1 atm P

Liquidpure liquid

Solidpure solid

Solution1 M solutions

ElementMost stable form at 1 atm.

Standard free energy of formation

Gof

1 mole of a single product from elements in their

standard states

By definition: Gof of an element in its standard state = 0


Thermodynamics b

G = H - T S

If all reactants and products are standard states then

Go = Ho - T So

Relationship for spontaneity based on non standard conditions

G < 0 spontaneous

G = 0 Equilibrium

G > 0 non spontaneous

How does this relate to

standard conditions?

Q = Reaction Quotient

G = Go + RT ln Q

R = gas constant

T = temp in K


Thermodynamics b

G = Go + RT ln Q

At equilibrium

G = 0 Reaction is ?

K; the equilibrium constant

So: Q = ?

And: G = 0 = Go + RT ln K

Which gives us a relationship between

equilibrium and thermodynamics.

Go = - RT ln K

Go < 0, K >>1 and the reaction at equilibrium has more products

than reactants and had to be spontaneous in the forward direction

to get there.

Go > 0, K << 1 and the reaction at equilibrium has more reactants

than products and had to be spontaneous in the reverse direction

to get there.


Thermodynamics b

Since the reactions are temperature dependent, you can find the

temperature when the reaction goes from spontaneous to

non spontaneous in the forward direction by setting Go = 0

Go = Ho - T So

Go = 0 = Ho - T So

Ho = T So

Units: Remember:  H in kcal/mol or kJ / mol

 G in kcal/ mol or kJ / mol

 S in cal / mol K or J/ mol K

So: watch your units!!!


Thermodynamics b

A: What is Go in kJ for the reaction?

= (2(N2O) + O2) - (4(NO))

= (2(103.6) + 0) - (4(86.7)

= - 139.6 kJ


Thermodynamics b

B: What is Ho in kJ for the reaction?

= (2(N2O) + O2) - (4(NO))

= (2(81.5) + 0) - (4(90.4))

= - 198.6 kJ

C: What is So in J/K for the reaction?

= (2(N2O) + O2) - (4(NO))

= (2(220.0) + ?) - (4(210.6))

Go = Ho - T So

= ((-198.6 kJ) - (-139.6 kJ))/ 298 K

= - 0.2000 kJ/K = - 200.0 J/ K


Thermodynamics b

D: What is So in J/K for O2(g)?

= (2(N2O) + O2) - (4(NO))

- 200.0 = (2(220.0) + O2) - (4(210.6))

So for O2(g) = 204.4 J/K

E: What is Eo for the reaction in kJ?

Ho = Eo + PV

Ho = Eo + nRT

Eo = Ho - nRT

= (-198.6) - ((3-4)(8.314/1000)(298))

= - 196.1 kJ/mol


Thermodynamics b

F: Is the reaction as written:

A) spontaneousB) non - spontaneousC) can’t tell

G: At what temperature will the reaction change direction?

= 0

Go = Ho - T So

Ho = T So

= 993 K

H: What is the value of K

Go = - RT ln K

- 139.6 kJ = - (8.314J/1000J/kJ)(298K)lnK

K = 2.95 x 10 24

ln K = 56.34


Thermodynamics b

A certain reaction gives off heat and increases in entropy.

The reaction is?

A) spontaneous at all temperatures

B) non spontaneous at all temperatures

C) spontaneous at high temp and non spontaneous at low

D) spontaneous at low temp and non spontaneous at high


Thermodynamics b

A certain reaction gives off heat and increases in entropy.

The reaction is?

A) spontaneous at all temperatures

B) non spontaneous at all temperatures

C) spontaneous at high temp and non spontaneous at low

D) spontaneous at low temp and non spontaneous at high

G = H - T S

(-) -

T(+)

- at all temp

G =

So rxn spontaneous at all Temp


Thermodynamics b

A reaction absorbs heat and increases in entropy.

The reaction is?

A) spontaneous at all temperatures

B) non spontaneous at all temperatures

C) spontaneous at high temp and non spontaneous at low

D) spontaneous at low temp and non spontaneous at high


Thermodynamics b

A reaction absorbs heat and increases in entropy.

The reaction is?

A) spontaneous at all temperatures

B) non spontaneous at all temperatures

C) spontaneous at high temp and non spontaneous at low

D) spontaneous at low temp and non spontaneous at high

G = H - T S

(+) -

T(+)

+ at low temp, - at high temp

So: non spontaneous at low temp

spontaneous at high temp

G =


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