Thermodynamics B

Thermodynamics–Deals with the interconversion of heat an other forms of energy

First Law: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.

E = q + w

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.

CaCl2(s) Ca+2(aq) + 2 Cl - (aq) H2O H = - 82.8 kJ

NH4NO3(s) NH4+(aq) + NO3

- (aq) H2O 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)

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

So: So = Sof - So

i 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 0

Suniverse = Ssystem + Ssurroundings

Ssurroundings comes primarily from change in heat energy (q atconstant pressure) but also depends on the Temperature of the surroundings.

SsurrQsurr

T Hsys

T

Third Law of Thermodynamics

Entropy of a pure, perfectly crystalline substance is zero at absolute zero of temperature.

Why?At 0 K motion ceases, therefore, there is only onepossible structure and no randomness (perfect order)

S=k ln = number of arrangementsk = boltzmann constant

So, third law gives us Absolute Entropy

Determine if the entropy of the system:A) increases B) decreases C) stays the same

1. Benzene(s) --> benzene (l)A2. NaNO3(s) --> NaNO3 (aq) A

3. S(s) + O2(g) --> SO2(g) C4. MgCO3(s) --> MgO(s) + CO2(g)A

5. PCl3(l) + Cl2(g) --> PCl5(s)B6. 2 HgO(s) --> 2 Hg(l) + O2(g) A

Entropy and Physical Properties

1. S increases with mass: He < Ne < Ar

2. S increases with molecular complexity: CH$ < C2H6 < C3H8

3. S increases with softness of solid crystals4. 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 increase10. S increases as temperature increases

Which of the following will have the largest entropy?1) C3H8 (l)2) C3H8(g)3) C3H6(g)4) C4H10(l)5) C4H10(g)

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

Suniv = Ssys + Ssurr and it is not always possible to determine whatis happening in the system and the surroundings.

So, it is necessary to have another way to assess spontaneity that depends only on the system

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.

GT,P < 0 Spontaneous Reaction

G = H - T SGibbs - Helmholtz Equation

GT,P > 0 Non - Spontaneous Reaction

GT,P = 0 Reaction at Equilibrium

Gorxn

Standard Free Energy change for a reaction

Reactants and Products all in standard states

Standard States

Gas 1 atm PLiquid pure liquidSolid pure solidSolution 1 M solutionsElement Most stable form at 1 atm.

Gof

Standard free energy of formation

1 mole of a single product from elements in their standard states

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

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 spontaneousG = 0 EquilibriumG > 0 non spontaneous

How does this relate to standard conditions?

G = Go + RT ln Q Q = Reaction QuotientR = gas constantT = temp in K

G = Go + RT ln Q

G = 0 Reaction is ?At equilibrium

So: Q = ?K; the equilibrium constant

And: G = 0 = Go + RT ln K

Go = - RT ln KWhich gives us a relationship between equilibrium and thermodynamics.

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.

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 KSo: watch your units!!!

Problems: Given the following information for the reaction: 4 NO (g) 2 N2O (g) + O2(g) at 25oC Compound Ho

f kJ/mol So J/mol Gof kJ/mol

NO(g) 90.4 210.6 86.7 N2O (g) 81.5 220.0 103.6 O2 (g)

A: What is Go in kJ for the reaction?

Grxno Gf

o(products) Gfo(reactants)

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

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

= - 139.6 kJ

B: What is Ho in kJ for the reaction?

Hrxno Hf

o(products) Hfo(reactants)

Srxno So(products) So(reactants)

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

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

= - 198.6 kJ

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

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

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

SoH o Go

T

Go = Ho - T So

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

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

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

Srxno So(products) So(reactants)

= (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 + PV Ho = Eo + nRT

Eo = Ho - nRT

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

= - 196.1 kJ/mol

F: Is the reaction as written:

A) spontaneous B) non - spontaneous C) can’t tell

G: At what temperature will the reaction change direction?

Go = Ho - T So = 0

Ho = T So

TH o

So

198.6 kJ 0.200 kJ

= 993 K

H: What is the value of KGo = - RT ln K

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

ln K = 56.34 K = 2.95 x 10 24

A certain reaction gives off heat and increases in entropy. The reaction is?

A) spontaneous at all temperaturesB) non spontaneous at all temperaturesC) spontaneous at high temp and non spontaneous at lowD) spontaneous at low temp and non spontaneous at high

A certain reaction gives off heat and increases in entropy. The reaction is?

A) spontaneous at all temperaturesB) non spontaneous at all temperaturesC) spontaneous at high temp and non spontaneous at lowD) spontaneous at low temp and non spontaneous at high

G = H - T S

(-) -T(+)

G = - at all temp

So rxn spontaneous at all Temp

A reaction absorbs heat and increases in entropy.The reaction is?

A) spontaneous at all temperaturesB) non spontaneous at all temperaturesC) spontaneous at high temp and non spontaneous at lowD) spontaneous at low temp and non spontaneous at high

A reaction absorbs heat and increases in entropy.The reaction is?

A) spontaneous at all temperaturesB) non spontaneous at all temperaturesC) spontaneous at high temp and non spontaneous at lowD) spontaneous at low temp and non spontaneous at high

G = H - T S

(+) -

G = + at low temp, - at high tempSo: non spontaneous at low temp

spontaneous at high temp

T(+)