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Ch 18 Free Energy and Thermodynamics:

Homework: Read Ch 18, Work out sample/practice exercises in the sections as you read,

Ch 18: 27, 31, 33, 41, 43, 47, 51, 55, 61, 63, 67, 71, 77, 87

Check for MasteringChemistry due dates.

Turn in your Research topic by the due date

Chemical Thermodynamics:

Chemical Thermodynamics studies energy relationships between Enthalpy,

Entropy, Equilibrium, and Gibbs free energy. Thermodynamics can predict the

spontaneous direction of a reaction under specified conditions, but not the rate of

the reaction.

Thermodynamics vs. Kinetics:

Which one deals with state functions, independent of path?

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Spontaneous Processes:

Spontaneous process: A spontaneous process is one that proceeds in a specific

direction on its own without any outside assistance under specified conditions (T,

P). The products are considered thermodynamically more stable. The reverse

reaction is nonspontaneous. Some reactions are always spontaneous, others are

never spontaneous and many reactions can be either, depending on conditions. Most

spontaneous reactions favor exothermic processes (-H) and increased entropy

(+S). Often one factor may override the other. Equilibrium reactions can shift to

be spontaneous in the forward or reverse direction with just slight changes in the

environment/conditions. A spontaneous process always moves a reaction mixture

toward equilibrium.

Reversible process will restore all components to the original conditions and states

with no net change to the system or surroundings, an example that is reversible is

the melting or freezing of ice at 1 atm and 0°C. Reversible processes only occur for

situations at equilibrium, Suniv = .

Irreversible process will not restore to original, an example is to mix equal portions

of hot and cold liquid together that end with an in between final temperature. Any

spontaneous process is irreversible,Suniv > .

HISTORY: Carnot Engine: In 1824 a French Engineer, Sadi Carnot published an

analysis of factors that determine efficiency of a steam engine. An ideal engine that

converts all the energy content of fuel to useable work is impossible because heat is

always lost to the surroundings. This study led to engineering more efficient

engines and was the first study that led to the discipline thermodynamics. Rudolph

Clausius: Forty years later the ratio of qp/T (entropy) was found to have a special

significance and was comparable to energy.

Ideal Engine: An ideal engine that has maximum efficiency operates under ideal

conditions in which all processes are reversible.

The Energy Tax: Every energy transition results in a “loss” of energy, heating up

the surroundings. An ideal engine is not possible.

Heat Tax: Fewer steps are generally more efficient.

The heating of your home with natural gas directly (1 step) is much more efficient

than heating with electricity, which may take 5 or more steps.

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Entropy, S:

Second Law: Entropy Principle: Every spontaneous change increases the total

entropy change of the universe. This does not mean local decreases in

entropy cannot take place. For the entropy of a system to decrease, the

entropy of the surroundings must increase more. Suniv > The

universe tends toward greater randomness.

for reversible process Suniv = 0

Third Law: Zero Entropy Established: The entropy of a pure, perfect, crystalline

substance (no disorder) is zero at the temperature of absolute zero, 0

Kelvin.

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Entropy: Entropy is a state function that measures randomness, ways of being.

This includes all possible motions: Vibration, rotation, translation.

S= qp/T = H/T

S = klnW (entropy = Boltzmann’s constant, 1.38 x 10-23 J/K time

the natural log of the ways of being, number of microstates)

S= Sfinal – Sinital (S, entropy values can be found in the Appendix)

S= klnW final – klnW inital = kln(Wfinal/Winitial)

Entropy has units of J/mol K and can be found in Appendix C.

Be aware that the units for entropy are in J and not kJ!

Predicting the sign for S: (+ more ways, favorable)

1) Phase changes: s l g… H2O (l) H2O (g) (+S)

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2) Temp: warming… +S O2 (g) 25°C O2 (g) 80°C

3) Volume: expanding… +S 10 ml gas 1.0 liter of same gas

4) Mixing: dissolving ... +S NaCl (s) + H2O (l) NaCl (aq)

5) Rearrange to create more ways of being…

+S H2 (g) + Cl2 (g) 2 HCl (g)

6) # parts: increase… +S 10 g rock 10 g sand

7) More moles gas… +S N2O4 (g) 2 NO2 (g)

Example 1:

a) What is the entropy of a system with only 1 microstate?

S = klnW

b) What sign is S for the following:

Temperature decreases

Volume increases

gas liquid

increase the moles of gas

Dissolving sugar in hot tea

N2 (g) + 3H2 (g) 2 NH3 (g)

Ag+(aq) + Cl-1(aq) AgCl (s)

c) Circle the one with the higher entropy in each of the following pairs.

1 mole O2 (g) ; 25°C or 1 mole O2 (g) ; 50°C

1 mole N2O4 gas or 2 mole NO2 gas

NaCl (s) or NaCl (aq)

1 mol N2 gas in 22.4 L or 1 mol N2 gas in 2.2 L

1 mol H2O (l) or 1 mol H2O (s)

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Solving Srxn: Just like enthalpy, entropy can be solved using the Appendix values.

S°rxn= nS°products - nS°reactants

Example 2: S° for CO2 (g) = 213.8 J/mol K

a) Predict the sign of the standard entropy change for the reaction…

CH4(g) + 2 O2 (g) CO2 (g)+ 2 H2O (l)

b) Calculate S° using the appendix values.

Example 3:

a) Predict the sign of the entropy change for the vaporization of water…

b) Calculate S° using the appendix values.

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Trouton’s Law: Most liquids follow Trouton’s Law- the molar heat of entropy of

vaporization is approximately 88 J/mol K; Svap= 88+/- 5J/mol K

Example 4:

Most liquids follow Trouton’s rule. Normal boiling points and enthalpies of

vaporization of several organic liquids are in the table.

Substance Normal Boiling Pt

(˚C)

Hvap (kJ/mol)

Acetone, (CH3)2CO 56.1 29.1

Dimethyl ether,

(CH3)2O

-24.8 21.5

Ethanol, C2H5OH 78.4 38.6

Octane, C8H18 125.6 34.4

Pyridine, C5H5N 115.3 35.1

a) Estimate Svap for each liquid using … Svap= Hvap/Tbp

Acetone, (CH3)2CO

Dimethyl ether, (CH3)2O

Ethanol, C2H5OH

Octane, C8H18

Pyridine, C5H5N

b) Which do not follow Trouton’s Law?

c) Explain the exceptions using intermolecular forces?

d) Should water follow Trouton’s Law? Check the accuracy of your

conclusion using the Appendix.

e) Chlorobenzene boils at 131.5˚C, estimate the Hvap for C6H5Cl.

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Example 5: Estimate the normal boiling point in degrees Celsius of CCl4. Use the

appendix values.

Gibbs Free Energy:

Gibbs Free Energy is the maximum amount of useful energy obtainable in the form

of work. It uses both enthalpy and entropy to predict spontaneity.

Gf˚ , Gibbs Free Energy can be found in the Appendix

G = H - TS (remember to convert S into kJ for consistency)

G°rxn= nG°f (products) - nG°f(reactants)

G under nonstandard conditions is a reliable indicator of spontaneity.

G > 0; a positive free energy is nonspontaneous

G= 0; a zero free energy is at equilibrium

G< 0; a negative free energy is spontaneous

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Example 6:

Predict the sign and calculate the Gibbs free energy change under standard

conditions (25˚C) for the reaction… Use the appendix values.

CH4(g) + 2 O2 (g) CO2 (g)+ 2 H2O (l)

a) From this equation G°rxn= nG°f (products) - nG°f (reactants)

b) From this equation G°rxn= H°rxn - TS°rxn

Gibbs Free Energy and Temperature: Hf° andS° data do not significantly change with varying temperatures.

H

(kJ/mol)

S

(kJ/mol)

G = H - TS Spontaneity conditions

_ + _ Spontaneous at all Temps

-196 +126 _ 2H2O2 (l) 2 H2O (l) + O2 (g)

+ _ + Nonspontaneous at all Temps

+285 -137 + 3 O2 (g) 2 O3 (g)

_ _ + or - Spontaneous at low Temps

Nonspontaneous at high Temps

-233 -424 Equil T = ? 2H2S (g) + SO2 (g)

2 H2O (l) + 3 S (s)

+ + + or - Spontaneous at high Temps

Nonspontaneous at low Temps

+176 +285 Equil T = ? NH4Cl (s) NH3(g) + HCl (g)

The equilibrium temperature can be found for the bottom two conditions by setting

G to zero; H = TS; = H /S (use kJ, T is in Kelvin)

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Example 7:

Solve for the equilibrium temperatures in the table above. Specify the

spontaneous temperature range/conditions at 1 atm, (above or below

equilibrium temperature).

Example 8:

An important reaction in the production of sulfuric acid is the oxidation of

sulfur dioxide gas into sulfur trioxide gas.

a) Write the chemical reaction.

b) Use appendix: Solve for G°rxn

c) Is the reaction spontaneous at standard conditions?

d) Solve H°f and S°

e) Will this reaction be spontaneous at 900°C?

f) Solve for the equilibrium temperature and determine the spontaneity

conditions.

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Gibbs Free Energy and Equilibrium Constant, Keq:

G = G° + RTlnQeq

and at equilibrium

G° = -RTlnKeq

For a spontaneous reactions…G< 0 is negative and K >>1 is very large.

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Example 9:

Consider the reaction 2 NO2 (g) N2O4 (g)

a) Calculate G˚ at 25˚C using Appendix values.

b) Calculate G at 25˚C if the partial pressures of NO2 and N2O4 are 0.40 atm and

1.60 atm, respectively.

Example 10:

Consider the decomposition of barium carbonate, BaCO3(s) BaO(s)+ CO2(g)

a) Calculate the equilibrium pressure of CO2 using Appendix values at 25˚C.

b) Calculate the equilibrium pressure of CO2 at 1100 K.

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Example 11:

The Ka for HNO2 = 4.5 x 10-4.

a) Write the Ka equation for HNO2.

b) Calculate G˚ at 25˚C using Ka.

c) What is G at equilibrium.

d) What is G when [NO2-1] = 6.0 x 10-4M, [HNO2] = 0.20 M, and pH = 1.30

Calculate Coupling Reactions:

Consider the direct decomposition of rust to iron…

2 Fe2O3 (s) 4 Fe (s) + 3 O2 (g) : G = +1487 kJ

This large positive value for free energy tells us the reaction is nonspontaneous.

However this does not mean you cannot change the rust back to pure iron; you just

need to do some useful work to reduce it by coupling with a reaction with a large

negative free energy.

2 CO (g) + O2 (g) 2 CO2 (g) : G = -514 kJ

The common term in both reactions is O2 (g). Notice we have 3 as product in the

rust decomposition and 1 O2 (g) in the CO (g) combustion.

2 Fe2O3 (s) 4 Fe (s) + 3 O2 (g) : G = +1487 kJ

3 x [2 CO (g) + O2 (g) 2 CO2 (g) : G = -514 kJ]

Overall: 2 Fe2O3 (s) + 6 CO (g) 4 Fe (s) + 6 CO2 (g) ;G = -56 kJ

This reaction occurs in a blast furnace.

In Biochemistry ATP provides the coupling energy necessary to accomplish many

reactions in a cell.

ATP + H2O (l) ADP + HPO4-2 (aq) ; G = -30.5 kJ

Glucose + Fructose sucrose + H2O (l) ; G = +29.3 kJ

Glucose + Fructose + ATP sucrose + ADP + HPO4-2 (aq); G = -1.2 kJ

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Example 12:

Given the two step reaction to form methane gas from CO and H2...

Step 1 CO (g) + 2 H2 (g) CH3OH (l)

Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g)

And using only the information on the table solve all parts in this problem.

compound H°f S°f G°f

CO(g) ------ ------ -137.2 kJ/mol

H2 (g) you know this! ------ you know this!

CH3OH (l) -238.7 kJ/mol 126.8 J/mol K -166.3 kJ/mol

CH4 (g) -74.87 kJ/mol 186.1 J/mol K ------

O2 (g) you know this! 205.0 J/mol K you know this!

a) What is H°f and G°f for H2 (g) and O2 (g)?

b) Calculate G°rxn in kJ/mol for step 1 (25°C)

Step 1 CO (g) + 2 H2 (g) CH3OH (l)

c) Is step 1 spontaneous under standard conditions?

d) Given that H° is negative and you have the ability to predict the sign of S,

answer under what conditions is step 1 spontaneous?

(always, at low temperatures, at high temperatures, never)

e) Calculate H°rxn in kJ/mol for step 2

Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g)

f) Calculate S°rxn in J/mol*K for step 2

Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g)

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g) Calculate G°rxn in kJ/mol for step 2 (25°C)

h) Is step 2 spontaneous under standard conditions?

i) At what temperature will step 2 have G = 0?

j) Specifically, what temperature range is step 2 spontaneous?

k) Write the overall coupled reaction of step 1 and step 2 when together and

Calculate the coupled overall G° under standard condition, 25°C.

Step 1 CO (g) + 2 H2 (g) CH3OH (l) G°

Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g) G°

Coupled reaction G°

l) Is the coupled reaction spontaneous under standard conditions, 25°C?

m) You are designing a production facility. Considering the information above,

would you carry out the reactions in separate steps under different

temperature conditions or coupled simultaneously together in the same

vessel under the same conditions.