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Chapter 19 – Chemical Thermodynamics

Using thermodynamics we can predict the direction of a reaction (whether it will be product favored or reactant

favored), and the extent to which it will take place.

We cannot predict the rate of reaction (thermodynamics has no connection to kinetics).

19.1 Spontaneous Processes

Processes that, once started, proceed without any outside intervention.

Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

A spontaneous process or reaction will go forward, given enough time (this does not mean it happens fast!).

Spontaneity can depend on temperature.

Ice melts spontaneously above 0℃. Liquid water freezes spontaneously below 0℃.

At 0℃ , 𝐻2𝑂(𝑙) ⇌ 𝐻2𝑂(𝑠) there is equilibrium; no preferred direction (reversible).

First law of Thermodynamics: Energy is conserved (not created or destroyed, just converted to different forms).

State function Path functions

∆𝐸 = 𝑞 + 𝑤

if absorbed if released

𝑞 = ℎ𝑒𝑎𝑡 𝑞 + 𝑞 −

𝑤 = 𝑤𝑜𝑟𝑘 𝑤 + 𝑤 −

System and Surroundings interact.

Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in

their original states by exactly reversing the process.

A reversible process will proceed back and forth between the two end conditions (any reversible process is at

equilibrium).

Irreversible Processes

Irreversible processes cannot be undone by exactly reversing the change to the system.

Spontaneous processes are irreversible.

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19.2 Entropy and the Second Law of Thermodynamics

There are two factors that determine whether a reaction is spontaneous or not. They are the enthalpy change

(∆𝐻) and the entropy change (∆𝑆) of the system.

The two factors must be considered to find out if the process is spontaneous.

Enthalpy (𝐻)

1.- ∆𝐻 generally measured in kJ/mol

Stronger bonds = more stable molecules

A reaction is generally exothermic if the bonds in the products are stronger than the bonds in the reactants

– exothermic = energy released, ∆𝐻 is negative

A reaction is generally endothermic if the bonds in the products are weaker than the bonds in the reactants

– endothermic = energy absorbed, ∆𝐻 is positive

The enthalpy change is favorable for exothermic reactions and unfavorable for endothermic reactions.

Enthalpy – Another way to see it

∆𝐻.- Most (not all) exothermic reactions are product-favored (spontaneous).

In an exothermic reaction , chemical potential energy stored in the bonds of the reactants gets converted to heat

(it increases the KE of the molecules in the surroundings).

Result: the energy is dispersed over more molecules, which is a more probable situation than lots of energy in

few molecules.

Entropy (𝑆)

2. ∆𝑆.- Entropy can be thought of as a measure of the randomness of a system.

S generally has units of: J/mol•K

Things tend to become more disordered because is a more probable state.

What would be more probable once the stopcock is opened?

Gases do not un-mix (it is too improbable!).

If a system contains N molecules, the probability that any given molecule is in flask "𝐴" =1

2 (one side or the

other). The probability that all N are in flask “A” is: (1

2)

𝑁

.

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Doing some math….

If 𝑁 = 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 , (1

2)

10

= 10−3𝑜𝑟1

1000.

All 10 would be in flask “A” 1 out of 1000 times.

If 𝑁 = 100 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠, (1

2)

100

= 10−30.

If you looked at the system every second, it would take 2.3 × 1012 times the estimated age of the universe too

see all in flask “A”. (Imagine the calculation for 1 mol!!!)

Conclusion (the point):

in macroscopic systems, wild fluctuations are not observed (or if you wait long enough, it will happen!).

So… There are two “driving forces” for reactions and processes in general:

1) Dispersal of energy (exothermic ∆𝐻 ⊖)

2) Dispersal of matter (∆𝑆 ⊕, more randomness)

Entropy - Another way to see it

∆𝑆 can be linked to heat flows.

For a process that occurs at constant 𝑇 (isothermal process):

∆𝑆 =𝑞𝑇

𝑇

Where 𝑞𝑇 is the heat flow at constant 𝑇.

𝑇 is the temperature in 𝐾.

(examples: phase changes at mp or bp.)

Q. ∆𝐻𝑣𝑎𝑝benzene (𝐶6𝐻6) = 30.9 𝑘𝐽/𝑚𝑜𝑙 at 𝑏𝑝 = 80.1℃. Calculate ∆𝑆 for 𝐶6𝐻6 (𝑔) → 𝐶6𝐻6 (𝑙) at 80.1℃.

Something else about Entropy

Entropy – a measure of disorder (S) is a state function (i.e. it depends on the current state, not the history of the

system); higher entropy, more disorder. ∆𝑆 = 𝑆𝑓𝑖𝑛𝑎𝑙 − 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙

Second Law of Thermodynamics: In any spontaneous process, the total entropy of the universe increases (gets more disordered). It does not

change for reversible processes.

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19.3 The Molecular Interpretation of Entropy and the Third Law of Thermodynamics

Entropy, S, is a thermodynamic function that increases as the number of energetically equivalent ways of

arranging the components increases.

Random systems require less energy than ordered systems

Ludwig Boltzmann described the concept of entropy on the molecular level.

Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.

– This would be akin to taking a snapshot of all the molecules.

He referred to this sampling as a microstate of the thermodynamic system.

These are energetically equivalent states for the expansion of a gas. Which state has higher entropy?

In other words, which one is more probable? (for which of the three states there is a larger number of

energetically equivalent ways of arranging the components?)

Entropy on the Molecular Scale

Each thermodynamic state has a specific number of microstates, W, associated with it.

Entropy is

𝑆 = 𝑘 ln 𝑊

𝑘 = 1.38 × 10−23 𝐽/𝐾 (Boltzmann constant).

𝑊 is the # of ways (microstates) the system can be described (number of “degrees of freedom”).

If 2 marbles in 1 box, 𝑊 = 9. If 2 marbles in different boxes, 𝑊 = 72.

Higher 𝑾 more disorder.

The change in entropy for a process, then, is:

∆𝑆 = 𝑘 ln 𝑊𝑓𝑖𝑛𝑎𝑙 − 𝑘 ln 𝑊𝑖𝑛𝑖𝑡𝑖𝑎𝑙

∆𝑆 = 𝑘 ln𝑊𝑓𝑖𝑛𝑎𝑙

𝑊𝑖𝑛𝑖𝑡𝑖𝑎𝑙

Entropy increases with the number of microstates, 𝑊, in the system.

(We won’t use these equations because it is too difficult to apply them to macroscopic systems.)

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Entropy

Entropy change is favorable when the result is a more random system

– ∆𝑆 is positive

Some changes that increase the entropy are

– reactions whose products are in a more random state

• solid more ordered than liquid more ordered than gas

Some changes that increase the entropy are

– reactions that have larger numbers of product molecules than reactant molecules

Some changes that increase the entropy are

– increase in temperature

– solids dissociating into ions upon dissolving

Q. Predict whether ∆𝑆system is + or − for each of the following

Water vapor condensing

Separation of oil and vinegar salad dressing

Dissolving sugar in tea

2 PbO2(s) 2 PbO(s) + O2(g)

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

Ag+(aq) + Cl−(aq) AgCl(s)

Molecules exhibit several types of motion:

Translational: Movement of the entire molecule from one place to another.

Vibrational: Periodic motion of atoms within a molecule.

Rotational: Rotation of the molecule about an axis or rotation about bonds.

The Third Law of Thermodynamics

The entropy of a pure perfect crystalline solid substance at absolute zero is 0 (it has no disorder, no molecular

motion, no vibration).

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19.4 Entropy Changes in Chemical Reactions

There is no way to measure absolute enthalpy values, but we can determine absolute entropy.

The entropy change of a substance from absolute zero to a given temperature.

Determining Absolute Entropy

∆𝑆 =𝑞𝑇

𝑇

𝑆 = 0 +𝑞1

1 𝐾+

𝑞2

2 𝐾+

𝑞3

3 𝐾+ ⋯ = ∑

𝑞𝑇

𝑇

298

𝑇=0

𝑆 = ∫𝑑𝑞

𝑇

298

𝑇=0

= ∫𝐶𝑃(𝑇)𝑑𝑇

𝑇

298

𝑇=0

We won’t do this! Values are known…

Standard Molar Entropies

𝑆°: absolute entropy of a substance, standard state pure substance, at 1 atm pressure. If it is dissolved, 1 M

concentration.

Standard Molar Entropies: States

Entropy increases with the freedom of motion of molecules.

Therefore:

S(g) > S(l) > S(s)

Standard Molar Entropies: Molar Mass

The larger the molar mass, the larger the entropy. It is a quantum effect: the available energy states are more

closely spaced, allowing more dispersal of energy through the states. (Or, more 𝑒−, more disorder.)

NaClheatq1

S = 0 0 K

NaClheatq2

1 K

NaClheatq3

2 K

NaCletc.

3 K

NaCl

298 K

∆𝑆1 =𝑞1

1 𝐾 ∆𝑆2 =

𝑞2

2 𝐾 ∆𝑆3 =

𝑞3

3 𝐾

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Standard Molar Entropies: Allotropes

The less constrained the structure of an allotrope is, the larger its entropy. (More degrees of freedom.)

𝑆ℎ𝑎𝑟𝑑 < 𝑆𝑠𝑜𝑓𝑡

Standard Molar Entropies: Molecular Complexity

Larger and more complex molecules have greater entropies (more degrees of freedom – more types of bending

and vibrations possible – and more microstates).

Standard Molar Entropies: Ionic Compounds

Weaker attractions, higher entropy (weaker: large atoms, low charges).

Standard Molar Entropies: Dissolusion

Dissolved solids generally have larger entropy due to the distribution of particles throughout the mixture.

Standard Molar Entropies: Gas

For a dissolved gas coming out of the solution, 𝑆 increases.

Entropy Changes

Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:

∆𝑆° = ∑ 𝑛∆𝑆°(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑚∆𝑆°(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

n and m are the coefficients in the balanced eq.

Look up 𝑆° values from appendix. The ° means standard states: s or l at 1 atm, g at 1 atm partial P. Solutes at

1M.

Q. Calculate ∆𝑆 for the reaction 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

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19.5 Gibbs Free Energy

Heat that flows into or out of the system changes the entropy of the surroundings.

For an isothermal process:

∆𝑆𝑠𝑢𝑟𝑟 =−𝑞𝑠𝑦𝑠

𝑇

At constant pressure, 𝑞𝑠𝑦𝑠 is simply ∆𝐻 for the system.

Entropy Change in the Universe

The universe is composed of the system and the surroundings. Therefore:

∆𝑆𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠

For spontaneous processes:

∆𝑆𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑒 > 0 Since:

∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 =−𝑞𝑠𝑦𝑠𝑡𝑒𝑚

𝑇=

−∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚

𝑇

Then:

∆𝑆𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 +−∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚

𝑇

Gibbs Free Energy

Multiplying both sides by T, we get:

−𝑇∆𝑆𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = ∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚 − 𝑇∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚

−𝑇∆𝑆𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑒 is defined as the Gibbs free energy, ∆𝐺.

When ∆𝑆𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑒 is positive, ∆𝐺 is negative (constant P and T are implied).

Sign of the Gibbs Free Energy

- If ∆𝐺 < 0, the rxn/process is spontaneous in the forward direction.

- If ∆𝐺 = 0, the rxn/process is at equilibrium.

- If ∆𝐺 > 0, the rxn/process is nonspontaneous in the forward direction (work

must be done to make it happen), but the reverse rxn/process is spontaneous.

Potential Energy Analogy

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Interpretation of the Gibbs Free Energy terms

∆𝐺 = ∆𝐻 − 𝑇∆𝑆

∆𝐻 is the available energy of the process at P ct.

𝑇∆𝑆 is the minimum energy that must be “wasted” as heat. (?)

∆𝐺 is the maximum amount of work that can be done (“free energy”- energy available to do work).

The Energy Tax

We can’t break even! Every energy transition results in a “loss” of energy…

The 2nd Law of Thermodynamics says that entropy (unusable energy) tends to increase. No machine is ever

100% efficient.

Standard Conditions

Solids and liquids: pure, at 1 atm of P.

Gases: 1 atm of partial pressure.

Solutions: 1 M concentration.

T: temperature of interest (not necessarily 298 K).

∆𝐺° = ∆𝐻° − 𝑇∆𝑆°

Standard Molar Free Energies of Formation (∆𝐺𝑓° )

∆𝐺 when 1 mole of substance is formed from elements. E.g.;

∆𝐺𝑓° 𝐶2𝐻5𝑂𝐻(𝑙) = −174.8

𝑘𝐽

𝑚𝑜𝑙 @ 25℃

2 𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) + 3 𝐻2 (𝑔) +1

2𝑂2 (𝑔) → 𝐶2𝐻5𝑂𝐻(𝑙)

∆𝐺° = −174.8 𝑘𝐽

∆𝐺𝑟𝑥𝑛° = ∑ 𝑛∆𝐺𝑓

° (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑚∆𝐺𝑓° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

Note: Only valid @ 25℃. ∆𝐺°strongly depends on T.

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Another way to calculate: ∆𝐺𝑟𝑥𝑛°

We can also use: ∆𝐺° = ∆𝐻° − 𝑇∆𝑆°.

The calculation is valid even for 𝑇 ≠ 25℃ (any!).

∆𝐻° and ∆𝑆° don’t change much with 𝑇.

1. Calculate ∆𝐻° & ∆𝑆° using values in appendix.

2. Insert values in ∆𝐺° eq., using the proper 𝑇.

This will be an approximation, but not too far off.

Q. Calculate ∆𝐺𝑟𝑥𝑛° for:

2 𝐶3𝐻6 (𝑔) + 2 𝑁𝐻3 (𝑔) + 3 𝑂2 (𝑔) → 2 𝐶3𝐻3𝑁(𝑙) + 6 𝐻2𝑂(𝑙)

at 25℃ and 85℃.

∆𝐻𝑓° (

𝑘𝐽

𝑚𝑜𝑙) 20.0 -45.9 0 172.9 -285.8

∆𝐺𝑓° (

𝑘𝐽

𝑚𝑜𝑙) 74.62 -16.4 0 208.6 -237.1

𝑆° (𝐽

𝑚𝑜𝑙 𝐾) 226.9 192.8 205.15 188 69.95

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19.6 Free Energy and Temperature

∆𝐺 = ∆𝐻 − 𝑇∆𝑆

The sign of ∆𝐺, which tells us whether a process is spontaneous, depends on the signs and magnitudes of ∆𝐻

and −𝑇∆𝑆.

Keep in mind that 𝑇 is always +.

Q. State whether each reaction will be spontaneous at all T’s, no T’s, high T’s, or low T’s.

∆𝐻°(𝑘𝐽) ∆𝑆°(𝐽/𝐾)

a) 2𝐹𝑒2𝑂3 (𝑠) + 3𝐶(𝑔𝑟) → 4𝐹𝑒(𝑠) + 3𝐶𝑂2 (𝑔) +468 +560

b) 𝐶(𝑔𝑟) + 𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) – 393.5 +2.9

Finding 𝑇 at which a reaction becomes spontaneous (std. conditions).

This is where ∆𝐺° = 0, so ∆𝐻° − 𝑇∆𝑆° = 0.

∆𝐻° = 𝑇∆𝑆° ; 𝑇 =∆𝐻°

∆𝑆°

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Q. At what 𝑇 does the following reaction become spontaneous under standard conditions?

𝑀𝑔𝑂(𝑠) + 𝐶(𝑔𝑟) → 𝑀𝑔(𝑠) + 𝐶𝑂(𝑔)

∆𝐻𝑓° (

𝑘𝐽

𝑚𝑜𝑙) -601.70 0 0 -110.525

𝑆° (𝐽

𝑚𝑜𝑙 𝐾) 26.94 5.74 32.68 197.674

19.7 Free Energy and the Equilibrium Constant

Under any conditions, standard or nonstandard, the free energy change can be found this way:

∆𝐺 = ∆𝐺° + 𝑅𝑇 ln 𝑄

Under standard conditions, all concentrations are 1 M (P = 1 atm for gases), so 𝑄 = 1 and ln 𝑄 = 0; the last

term drops out.

Q. Write 𝑄 expressions for each reaction:

a) 𝐶𝑎𝐶𝑂3 (𝑠) ⇌ 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2 (𝑔)

b) 𝐻3𝑂+ + 𝐻𝐶𝑂3− ⇌ 2𝐻2𝑂(𝑙) + 𝐶𝑂2 (𝑔)

Actual Conditions

Standard

Conditions 8.314

𝐽

𝑚𝑜𝑙 𝐾 Kelvin

Thermodynamic reaction quotient (solutes in M, gases in atm)

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Q. For the reaction:

2𝑁𝑂2 (𝑔) ⇌ 𝑁2𝑂4 (𝑔), ∆𝐺° = −407 𝐽 @ 50℃.

a) Which way will the reaction go if 𝑃𝑁𝑂2= 𝑃𝑁2𝑂4

= 1.0 𝑎𝑡𝑚, 50℃?

b) Which way will the reaction go if 𝑃𝑁𝑂2= 𝑃𝑁2𝑂4

= 0.10 𝑎𝑡𝑚, 50℃?

c) If 𝑃𝑁2𝑂4= 0.10 𝑎𝑡𝑚, find 𝑃𝑁𝑂2

at which the reaction becomes spontaneous at 50℃.

If a reaction is at equilibrium, ∆𝐺 = 0 (won’t go forward or reverse; this isn’t ∆𝐺°).

At equilibrium, 𝑄 = 𝐾.

∆𝐺 = ∆𝐺° + 𝑅𝑇 ln 𝐾

0 = ∆𝐺° + 𝑅𝑇 ln 𝐾

∆𝐺° = −𝑅𝑇 ln 𝐾

Not equilibrium. Equilibrium.

∆𝐺° and 𝐾 must be for the same temperature.

When ∆𝐺°is ⊝, 𝐾 > 1 (product favored).

When ∆𝐺°is ⊕, 𝐾 < 1 (reactant favored).

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Minimum Free Energy = Equilibrium

Most Stable!

∆G° or 𝐾 tell you: the position of equilibrium, composition of equilibrium mixture.

∆𝐺 or 𝑄 tell you: which way rxn go to get to equilibrium.

Q. Calculate 𝐾𝑠𝑝 for 𝑀𝑔(𝑂𝐻)2 (𝑠) at 25℃, given the following ∆𝐺° values in kJ/mol.

𝑀𝑔 (𝑎𝑞)2+ = −456.0 ; 𝑂𝐻

(𝑎𝑞)− = −157.3 ; 𝑀𝑔(𝑂𝐻)2 (𝑠) = −993.9

Q. What is ∆𝐺°for the reaction:

𝐶2𝐻3𝑂2− + 𝐻3𝑂+ → 𝐻𝐶2𝐻3𝑂2 + 𝐻2𝑂 @ 25℃

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Q. Use thermodynamic data to estimate the vapor pressure of 𝐻2𝑂 (𝑙) at 45℃.

𝐻2𝑂 (𝑙) ⇌ 𝐻2𝑂 (𝑔) ; 𝐾 = 𝑃𝐻2𝑂 (𝑔)(in atm).

Q. Estimate the normal boiling point of 𝐶𝐶𝑙4 using thermodynamic data.

𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛: 𝐶𝐶𝑙4 (𝑙) ⇌ 𝐶𝐶𝑙4 (𝑔) 𝐾 = 𝑃𝐶𝐶𝑙4

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Q. Estimate the solubility of 𝑁𝐻4𝐶𝑙 @ 90℃ using thermodynamic data.

Q. 𝐾𝑊 = 1.00 × 10−14 @ 25℃. 𝐾𝑊 = 5.62 × 10−14 @ 50℃

Estimate ∆𝐻° and ∆𝑆° for 2 𝐻2𝑂2 (𝑙) ⇌ 𝐻3𝑂+ + 𝑂𝐻−.