Chapter 19 Thermodynamics And Equilibrium. I. THERMODYNAMICS A) Definition of Thermodynamics - It is the study of energy changes that accompany physical.

Chapter 19

Thermodynamics And Equilibrium

I. THERMODYNAMICS

A) Definition of Thermodynamics - It is the study of energy changes that accompany physical and chemical changes. B) Most of the energy produced in modern society comes from chemical reactions.

1) The burning of petroleum products - gasoline, fuel oil2) The burning of natural gas3) The burning of coal

C) Thermodynamics is based on our experience with how nature behaves. Our study of nature has resulted in Three Laws of Thermodynamics.D) You studied the first law previously, I hope. 1) Energy is conserved in physical and chemical changes. 2) The total quantity of energy in the universe is assumed to be a constant. 3) Energy is converted from one form to another while the total remains the same.

4) You should have studied Hess's Law and H in conjunction with the first law. (The change in enthalpy is the heat of reaction at constant pressure.)

II. THE SECOND LAW OF THERMODYNAMICS AND

SPONTANEITY

A) One of the basic goals of chemistry is to predict whether or not a reaction will occur when substances are brought together; that is, whether the reaction is or is not spontaneous.

A spontaneous process is a physical or chemical change that occurs by itself. It can start by itself, or once started it continues by itself without further exertion.

B) THIS DOES NOT MEAN RAPIDLY.

Remember rate or speed of reaction is the concern of kinetics. It could be a fraction of a second or 100 years before a measurable amount of products are formed and still be called a spontaneous reaction.

C) Some examples of spontaneous processes:

1) Iron rusts in moist air.

2) Ice melts at 25oC3) Natural gas continues to burn once it is lit. 4) Water runs downhill. 5) A wound clock (watch) runs down.

6) Heat flows from hotter to cooler objects. 7) Hydrogen burns in oxygen to form water.

8) Sodium reacts with water.

D) An early criterion for predicting whether or not a process was spontaneous was that it had to be exothermic.

1) If this were true, the only information we would need to predict whether or not a process was spontaneous is whether H is

___________________________ .

2) But just looking at the set of spontaneous processes above, we see an endothermic one. Which one is that?

H2O(s) ----> H2O(l) at 25oC

H = + 6.0 kJ/molOther spontaneous processes are:

H2O(l) ----> H2O(g) at 100oC H = + 40.7 kJ/mol

CaCO3(s) ----> CaO(s) + CO2(g) at 1100oC H = + 178 kJ/mol

3) The sign of H is not enough information to predict accurately whether or not a process is spontaneous.

III. ENTROPY - THE THERMODYNAMIC QUANTITY CENTRAL TO THE SECOND LAW OF THERMODYNAMICS

A) Entropy, symbol S, is a thermodynamic quantity that is a measure of the randomness or disorder in a system.

John's Backpack

B) An example of a spontaneous change:

The distribution of the "gas" between the two flasks has been a spontaneous change.

For mixing of two ideal gases, after a certain amount of time, each gas will be evenly distributed between the two flasks. The mixing will be a spontaneous change. There has been no temperature change, no changes in energy, and no outside stirring was required.

C) How do we explain this phenomenon?

1) There must be a natural tendency to mixed-up-ness. A higher probability is associated with the mixed state compared with the unmixed state.

2) The final mixed state is more probable than the separated state.

3) The driving force of the process is the tendency toward mixed-up-ness. The random motion of the molecules tends to produce disorder.

IV. THE SECOND LAW OF THERMODYNAMICS

A) The total entropy of a system and its surroundings always increases for a spontaneous process. The entropy of the universe increases in a spontaneous process.

B) The driving force for a spontaneous process is an increase in the entropy of the universe. This is based on experience, but has not been proven rigorously.

C) Two fundamental laws of nature are particularly important in thermodynamics:

1) Systems tend to attain a state of minimum potential energy.

2) Systems tend toward a state of maximum disorder.

D) For a given substance, the ___________ is the state of lowest entropy, the __________ is intermediate, and the ______________ is the highest entropy state.

E) Why should a substance spontaneously freeze at a temperature below its melting point? Doesn't this appear to violate the 2nd law? It seems to be spontaneous yet there is a decrease in the entropy of the system, the water is going from a higher to a lower entropy state.

F) All entropy effects must be considered.

1) When two ideal gases mix, there is no exchange of matter or energy between the system and the surroundings. The only entropy effect is the increase in the isolated system itself.

2) Usually the system is not isolated from the surroundings.

3) When a liquid freezes, the heat of fusion is involved. What is that?______________ .

This heat is absorbed by the surroundings and the motion of the molecules of the surroundings _______________ , therefore the entropy of the surroundings __________ .

The decrease in the entropy of the system is more than offset by the increase in the entropy of the surroundings so there is a net _________ in the randomness of the universe.

4) Mathematically we can express the Second Law of Thermodynamics as follows:

For a spontaneous process:

Suniverse = Ssystem + Ssurroundings must be greater than zero.

For an equilibrium process:

Suniverse = Ssystem + Ssurroundings must be equal to zero.

5) It is unfortunate, but it is not always easy to assess the change in randomness of the surroundings.

V. THE THIRD LAW OF THERMODYNAMICS

A) THE ENTROPY OF A PERFECT CRYSTALLINE SOLID AT 0 K IS ZERO (0).

B) Why should that seem reasonable?

C) Since we have the above definition we can obtain entropies, S's, of pure substances.

D) This can be contrasted to enthalpies for which we could obtain only what?

E) Another word of caution is necessary. So at standard state conditions has a numerical value for a pure element. Under these conditions, H = 0.

F) The standard entropy of a substance or ion, also called its absolute entropy, So, is the entropy value for the standard state of the species. Table 18.1 gives the standard entropies of various substances at 25oC and 1 atm. See also appendix C.

G) Example: For the reaction below, calculate S.

Hg(l) + ½ O2(g) ---> HgO(s)

The first thing I want you to do is guess the sign for S.

It appears that the entropy is decreasing. The system is going from a random to a more ordered system. You should guess that the sign of S for the reaction is ___________.

The Standard Entropy Change is calculated from the following:

Soreaction = So

products - Soreactants

Sor = So

HgO(s) - [SoHg(l) + 1/2 So

O 2(g)]

Sor = 1 mol(70.27 J/molK) - [1 mol(76.027

J/molK) + 1/2 mol(205.0 J/molK)]

Sor = -108.3 J/K

The negative sign indicates that the entropy associated with the reaction is decreasing. We predicted this earlier.H) The entropy usually increases in the following situations:

1) A reaction in which a molecule is broken into two or more smaller molecules or atoms. 2) A reaction in which there is an increase in the number of moles of gas. 3) A process in which a solid changes to a liquid or gas or a liquid changes to a gas.

VI. GIBBS FREE ENERGY

A) Since it is difficult to assess all of the entropy changes in the system and surroundings, the entropy increasing criterion for spontaneity is often difficult to obtain. Our goal is to have a criterion for spontaneity prediction that focuses only on the system because it is easy __________ .

B) Our criterion for spontaneity which relies only on our understanding of what is happening to the system is related to the store of useful work in a system. What do we mean by useful work? C) Theoretically, spontaneous reactions can be used to obtain useful work.

D) The maximum useful work which can be obtained in driving a turbine or work produced by a battery is the change in the Free Energy, G.

There is a relationship among the enthalpy change, H, the entropy change, S, and the change in free energy, G.

G = H - S

The T must be in Kelvins.

E) A process or reaction will occur spontaneously at constant T and P only if it is capable of doing useful work, i.e. G is negative. The free energy of the system must decrease

F) If G = zero, the system is at equilibrium. G) If G is positive the reaction as written is not spontaneous, but the reverse reaction is spontaneous.

H) Look at G = H - S

1) H reflects the heat energy transferred between the system and surroundings.

2) S is a measure of the change in the randomness of the system. The energy consumed or evolved with respect to the change in order of a system.

G = H – TS 3) The more negative H is, the more likely the reaction will be spontaneous. This reflects the drive toward minimum energy. Also, H is always in kJ/mol and S is in J/molK. BE CAREFUL This has caused more lost points than I care to remember. 4) If the reaction is exothermic (H is -) and proceeds with an increase in entropy (S is +) the reaction is spontaneous regardless of the temperature. Why?___________________

5) If H is + ( what kind of reaction?) and S is - (the reaction proceeds with a decrease in entropy), then G must be ________________ no matter what the temperature. The reaction must be __________________.

6) If H is -, the reaction is _____________ and S is -, then G is - at low temperatures, reaction is _______________ . G is ____________ at high temperatures and _______________ .

7) If H is + (_____________________), and S is +, then G is ______ at low T and _______________________. G is ______ at high T and ______ ___________________ as written.

REMEMBER SPONTANEOUS DOES NOT MEAN _____________.

8) Free Energy (G ) is the link between the drive toward minimum energy and maximum entropy.

I) Calculation of G from Go= Ho – So

Calculate Go for the following reaction:

Hg(l) + ½ O2(g) -----> HgO(s)

Hfo for HgO(s) = -90.79 kJ/mol

So for the reaction is - 108.3 J/molK

Go = -90.79 kJ - (-32.3 kJ) = -58.5 kJ

The reaction is spontaneous even though the entropy is decreasing. H is large enough to overcome this decrease in entropy.

J) Gof is the standard free energy of

formation. This is the free energy change that occurs when one mole of substance is formed from its elements in their most stable states at 1 atm and at a specified temperature, usually 25oC, 298 K.

1) The Gof of an element in its standard state

is zero.

Goreaction = Go

products - Goreactants

3 O2(g) + 6 C(gr) + 6 H2(g) ----> C6H12O6(s)

Gof = - 910.56 kJ/mol

For the reaction below, calculate the Go reaction:

2 NO(g) + O2(g) ---> 2 NO2(g)

Goreaction = Go

products - Goreactants

Goreaction = 2 mol(Go

fNO2) - [2 mol(GofNO)

+ 1mol(GofO2)]

Goreaction = 2mol(51 kJ/mol) - [2 mol(86.6

kJ/mol) + 1mol(0)] Go

reaction = ___________

The reaction is __________________________ under standard state conditions.

VII. FREE ENERGY AND EQUILIBRIUM

A) A system at constant temperature and pressure will proceed spontaneously in the direction that lowers the free energy. This is why reactions proceed to equilibrium. The equilibrium position represents the lowest free energy value available to a particular reaction system.

B) For a reaction at conditions which are not standard state, the following equation has been derived:

G = Go + RT ln Q

What is Q?

For the general reaction

A(g) + B(g) ---> 2 C (g)

R = 8.31 J/molK

Go is the free energy change for reactants and products at 298K and 1 atm.

G is the free energy change for the reaction at specified pressures of reactants and products.

For the reaction:

CO(g) + 2 H2(g) ---> CH3OH(l)

Calculate G at 25oC when the pressure of CO is 5.0 atm and that of H2 is 3.0 atm.

G = Go + RT ln Q

We first have to compute Go from the thermodynamic tables.

Goreaction = Go

products - Goreactants

Goreaction = [(1 mol)(-166.2 kJ/mol)] - [(1 mol)

(-137.2 kJ/mol) + (2 mol)(0 kJ/mol)]

Goreaction = ______________

G = Go + RT ln Q

G = - 2.90 X 104 J + (-9.5 X 103 J) = ______ J

G = ______ kJ

Therefore the reaction is ________________ .

We can use the equation to find rG, the instantaneous rate of change of the Gibb's function at specific concentrations (more correctly activities) and you will use these values to predict whether or not a precipitate will occur in your experiment.

For the reaction:

Ag+(aq) + Cl-

(aq) ---> AgCl(s)

You will want to know if a precipitate forms upon the addition of 5 mL each of 2M AgNO3 and 2 M HCl.

You will use the equation

rG = Go + RT ln Q

The first thing you will have to do is calculate Go from values in the table. For the above reaction that is - 55.6 kJ.

To compute rG, we need to take into account the concentrations of the reactants, to calculate the reaction quotient, Q.

Since we are mixing equal volumes of 2 M solutions, the concentrations to use in Q are not 2 M. We have doubled the volume by mixing so we have halved the concentrations of the ions. Thus the [Ag+] and [Cl-] are now 1 M. Make certain you understand this point as you did not quite get it in the solubility chapter.

Thus, rG = - 55.6 kJ Why?

C) G's relationship with Kp

G = Go + RT ln Q

at equilibrium G = 0; Q = Kp

Then 0 = Go + RT ln Kp

Go = - RT ln Kp

for Go to be -, the ln Kp must be +, Kp > 1

for Go to be +, the ln Kp must be -, Kp < 1.

D) If Go is a large negative number (more negative than about -10 kJ), reactants are transformed almost entirely to products when equilibrium is reached.

F) If Go is between - 10 and + 10 kJ the reaction gives an equilibrium mixture with significant amounts of both reactants and products.

E) If Go is a large positive number (more positive than about +10 kJ), the reactants do not give significant amounts of products when equilibrium is reached.

G) Calculate Kp for the following reaction at 298 K: 2 C2H2(g) + 5 O2(g) ---> 4 CO2(g) + 2 H2O(g)

Gorxn = [4 mol(Go

f CO2(g) + 2 mol(Gof

H2O(g)] - [2 mol(Gof C2H2(g) + 5 mol(Go

f O2(g)]

Gorxn = [4 mol(-394.4kJ/mol) + 2 mol

(-228.6kJ/mol)] - [2mol(209kJ/mol) + 5 mol (0)]

Gorxn = [-1578 -457.2]kJ - [+418]kJ =

- 2453 kJGo

rxn = - RT ln Kp

- 2453 kJ = -2.49 kJ ln Kp

taking the antiln of such a big number will result in an error message on some calculators. If that is the case, you will have to go to log base 10.

2.303 log Kp = ln Kp = 985

What does this mean?

This method also works for net ionic equations – problem 18.52 in the text.

VIII. APPLICATION OF G TO PHASE CHANGES - MELTING AND BOILING

A) At the melting point and boiling point we have an equilibrium situation for change of phase, therefore G = 0.

G = H - S

0 = H - S

H = TS

B) The boiling point of CHCl3 is 61.7oC. The Hvap is 31.4 kJ/mol. Calculate the S of vaporization.

C) Estimate the melting point of benzene in oC, if Hfus is 10.9 kJ/mol and the Sfus is 39.1 J/molK.