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Chapter 20: Thermodynamics

Thermodynamics is the study of energy

(including heat) and chemical processes.

First Law of Thermodynamics: Energy cannot be

created nor destroyed.

Euniverse = Esystem + Esurroundings

∆Euniverse = ∆Esystem + ∆Esurroundings

If ∆Euniverse = 0, then –∆Esystem = ∆Esurroundings

The first law of thermodynamics explains the energy exchange between the system and surroundings.

Does it explain the direction a reaction proceeds toward equilibrium?

For systems under constant pressure and no work, ∆H = ∆E. Does the sign of ∆H tell us the direction a reaction proceeds toward equilibrium?

Let’s try the Second Law of Thermodynamics.

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Second Law of Thermodynamics: The total entropy (S) of a system and its surroundings increases for spontaneous reactions.

A spontaneous reaction is one that occurs naturally under certain conditions without a continuous input of outside energy.

Note: If a reaction is spontaneous in one direction, it will be nonspontaneous in the reverse direction.

Consider the following diagram:

When the valve is opened some of the gas particles move spontaneously into the other side as shown in B.

Will all of the molecules spontaneously move back as depicted in A?

In B of the diagram above, the gas particles occupy twice the volume thus the gas particles can exist in twice as many locations. Thus greater disorder.

Disorder

Sdisorder > Sorder , in other words, the entropy of

disordered matter is greater than the entropy

of ordered matter.

Examples to consider:

Ice melting � disorder ∆Ssystem increases

Water freezing � order ∆Ssystem decreases

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All processes occur spontaneously in the

direction that increases the total entropy of

the universe.

∆Suniverse > 0,

Therefore,

∆Ssystem + ∆Ssurroundings > 0

Let’s consider two rxns, one exothermic and one endothermic.

4 Fe(s) + 3 O2(g) � 2 Fe2O3(s) + Heat

Now, let’s consider another reaction.

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Ba(OH)2.8H2O(s) + 2 NH4NO3(s) + Heat �

Ba2+(aq) + 2NH3(g) + 2 NO3

-(aq) + 10H2O(l)

Therefore changes in entropy of the system and the surroundings must be important to the spontaneity of reactions.

Third Law of Themodynamics: A perfect crystalline substance at 0 K has 0 entropy.

Perfect means that all of the particles are flawlessly aligned in the crystal with no imperfections.

Standard Molar Entropies (So) at 298 K are on pg A-5 of text.

Units: J/mol-K

Note: Standard Molar Enthalpies (∆Ho) are also located on pg A-5.

Let’s Make Some Comparisons

Temperature versus Entropy

Copper metal T(K) So (J/mol-K)

273 31.0

295 32.9

298 33.1

What general statement can you make about this comparison?

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Physical State Compared with Entropy

Values

Substance So (J/mol-K)

Ba(s) 62.5

Ba(g) 170.3

Br2(l) 152.2

Br2(g) 245.4

C(s) 5.7

C(g) 158.0

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Comparison of the Solid State with

Aqueous Solution

Substance Entropy (J/mol-K)

NaCl(s) 72.1

NaCl(aq) 115.1

AlCl3(s) 167

AlCl3(aq) -148

CH3OH(l) 127

CH3OH(aq) 132

Note: When gases are dissolved in liquids the system decreases in entropy. Explain.

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Compare Molar Mass and Entropy

Substance MW (g/mol) Entropy (J/mol-K)

Cu 63.6 33.1

Au 197 51

Ag 107.9 43

What general statement can you make about

this comparison?

Compare Compound Complexity with

Entropy

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Substance Entropy (J/mol-K)

MgCO3 65.9

Na2CO3 139

C2H6 229

CH4 186

What general statement can you make about

this comparison?

Calculating the Change in Enthalpy of a

Reaction

Recall the equation:

∆Horxn = Σ n∆Ho

products – Σ n∆Horeactants

We can use a similar equation to calculate the

change in entropy of a reaction.

∆Sorxn = Σ nSo

products – Σ nSoreactants

Problem Solving

Calculate the change in entropy of the combustion

of methane.

H2O(g) CO2(g) CH4(g)O2(g)

So (J/mol-K) 188.72 213.7 186.1 205.0

Did the entropy of the system increase or

decrease?

What does this mean for the surroundings?

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Entropy of the Surroundings

In order for ∆Suniverse to be increasing if ∆Ssystem

decreases for a spontaneous reaction, then

∆Ssurroundings must increase.

Let’s consider the following diagram.

Exothermic Rxn Exothermic Rxn Endothermic Rxn

Let’s Talk Heat Exchange (∆Η)

If heat is lost to the surroundings, then the

entropy of the surroundings should increase.

If heat is gained by the surroundings, then the

KE goes up and so does the entropy.

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If heat is gained by the system (lost by the surroundings), the system becomes more disordered (entropy increases).

Also,

If the temperature of the surroundings is low, the particles are more ordered and a change in heat of the surroundings would have an even greater effect on the change in entropy.

Then ∆Ssurroundings is directly related to an opposite sign of ∆Hsystem, and inversely related to temperature at which the heat is transferred.

Mathematically Speaking

∆Ssurroundings is proportional to - qsystem

∆Ssurroundings is proportional to 1/T

∆Ssurroundings = -qsystem/T

∆Ssurroundings = -∆Hsystem/T

So…, calculate ∆Ssurroundings for the combustion of

methane at 25oC from ∆Hfo values. ∆Hf

o(CH4) = -

74.87kJ/mol, ∆Hfo(CO2) = -398.5 kJ/mol, and

∆Hfo(H2O) = -241.83kJ/mol

Using your information, calculate ∆Suniverse for

the combustion of methane. Did the entropy

of the universe increase?

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Entropy and Equilibrium

What is the change in entropy of the universe

for rxns at equilibrium?

∆Suniverse = 0 @ Equilibrium

∆Ssurroundings + ∆Ssystem = ∆Suniverse

Thus, ∆Ssurroundings = -∆Ssystem @ Equilibrium

Problem

Calculate the change in entropy for the following

reaction:

If the change in enthalpy is 40,700 J/mol at 373 K, what

is the change in entropy of the universe?

H2O(l)H2O(g)

∆S(J/mol-K) 86.8 195.9 @ 373 K

Entropy, Free Energy, and Work

One criterion for spontaneity is Gibbs Free Energy (∆G)

Proof:

∆Ssurroundings + ∆Ssystem = ∆Suniverse

∆Ssurroundings = -∆Hsystem/T

-∆Hsystem/T + ∆Ssystem = ∆Suniverse

∆Hsystem - T∆Ssystem = -T∆Suniverse

Given that: -T∆Suniverse = ∆Guniverse

∆Hsystem - T∆Ssystem = ∆Guniverse

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Gibbs Free Energy and Spontaneity

If ∆G < 0, The rxn is spontaneous

If ∆G > 0, the rxn is nonspontaneous

If ∆G = 0, the rxn is at equilibrium

Note: ∆Gof values are on page A-5.

These values for an element in its most stable state are 0. Also, if a rxn has a particular value for ∆G in one direction, it will be the same value with reversed sign in the opposite direction. Just like ∆Η. ;)

Problem

Determine whether or not the following reaction is spontaneous. Use TWO methods and compare: (a) use only delta G values (kJ/mol), (b) use delta Ho

f

(kJ/mol) and So (J/mol-K) values

4 KClO3(s) � 3 KClO4(s) + KCl(s)

∆Hof -397.7 -432.75 -436.7

So 143.1 151.0 82.59

∆Gof -296.3 -303.2 -409.2

Most exothermic reactions are spontaneous. However, temperature will affect the T ∆S factor of the equation:

∆Hsystem - T∆Ssystem = ∆Guniverse

Therefore, reaction spontaneity can change with temperature.

Example: At 298 K, ∆Go = -141.6kJ, ∆Ho = -198.4 kJ, and ∆So = -187.9 J/K for a reaction.

(1) Is the rxn spontaneous at 298K?

(2) If temperature is increased what happens to ∆Go ?

(3) Assuming ∆Ηo and ∆So are constant with temperature, is the rxn spontaneous at 900.oC?

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Crossover Temperature

At what temperature does a reaction become

spontaneous?

+ ∆Go == � - ∆Go

At some point ∆Go equals 0 ( @ equil)

Thus, ∆Hsystem - T∆Ssystem = ∆Guniverse

∆Hsystem - T∆Ssystem = 0

∆Hsystem = T∆Ssystem or

∆Hsystem/∆Ssystem = Tcrossover

What is the crossover temperature for the

previous problem?

Problem: Determine if the following reaction is

spontaneous at 298K. Also determine the

crossover temperature.

Cu2O(s) + C(s) � 2 Cu(s) + CO(g)

∆So = 165 J/K and ∆Ho = 58.1 kJ

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Free Energy, Equilibrium, and Rxn Direction

Recall,

If ∆G < 0, rxn is spontaneous, product favored,

and Q is < K.

If ∆G > 0, rxn is nonspontaneous, reactant

favored, and Q > K.

If ∆G = 0, rxn is at equilibrium and Q = K.

Given that ∆Gorxn is proportional to ln (Q/K) and their

signs are identical for a given reaction, we have:∆Grxn = RT ln (Q/K)

∆Grxn = RTln Q - RTln K

If we set [ ]’s = 1.00 M for sol’ns and 1.00 atm for gases (std conditions),

then, ∆Go

rxn = RTln 1 - RTln K

Or ∆Gorxn = - RTln K

Note: K can be determined for any rxn at any temp if we know ∆Go

rxn

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Problem Solving

Determine the thermodynamic equilibrium constant at

25oC for the following rxn:

N2O4(g) 2 NO2(g)

∆Gof 97.82 51.30

(kJ/mol)

However, most reactions are not with 1.0 M solutions

and/or 1.00 atm for gases, thus we also need to

remember this form of the equation:

∆Grxn = RT lnQ - RT lnK

Since ∆Gorxn = - RT lnK

∆Grxn = ∆Gorxn + RTlnQ

The above relationship is very important to our next

chapter, you will definitely see it again.

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Problem SolvingConsider the following reaction:

2 SO2(g) + O2(g) 2 SO3(g)

∆Gorxn -141.6 kJ at 298K and -12.12 kJ at 973K.

(a) Determine Keq at 298K.

(b) Determine Keq at 973K.

(c) If a vessel contains 0.500M SO2 , 0.0100M O2, and 0.100M SO3 at 298K, what is ∆Grxn ?

(d) Under the conditions in (c) which way would the reaction proceed?

More Problem SolvingConsider the reaction:

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

So (J/mol-K) 191.5 131 193.0

∆Ho (kJ/mol) 0 0 -46.3

(a) Determine the equilibrium constants at 298 K and at 984 K.

(b) What is the cross-over temperature for this reaction in Kelvin.

(c) If, at 298K, a vessel contains 0.250M H2, 0.870 M N2, and 12.9M NH3, predict the direction the reaction will proceed at this temperature.