Chapter 20: ThermodynamicsEntropy, Free Energy, and the Direction of Chemical

Reactions

20.1 Predicting Spontaneous Change The Second Law of Thermodynamics:

20.2 Calculating the Change in Entropy of a Reaction

20.3 Entropy, Free Energy, and Work

20.4 Free Energy, Equilibrium, and Reaction Direction

The First Law of Thermodynamics

E = q + w = heat + work = Energy

Everything that is not part of a system (sys) is the surroundings (surr), and vice versa:

Euniv = Esys + Esurr

qsys = -qsurr wsys = -wsurr

The total mass-energy of the universe is constant. Thereforemass-energy cannot be created or destroyed.

Mass can be converted into energy in nuclear reactions, and energy can be converted into mass in nuclear accelerators.

q = heat w = work

Spontaneous ProcessesSpontaneous Processes

• Thermodynamics is concerned with the question: can a reaction occur?

• First Law of Thermodynamics: Energy is conserved.• Any process that occurs without outside intervention is

spontaneous.• When two eggs are dropped they spontaneously break.• The reverse reaction (two eggs leaping into your hand

with their shells back intact) is not spontaneous.• We can conclude that a spontaneous process has a

direction.

Spontaneous ProcessesSpontaneous Processes• A process that is spontaneous in one direction is not

spontaneous in the opposite direction.• Example: Ice turns to water spontaneously at T > 0C,

Water turns to ice spontaneously at T < 0C.

Spontaneous ProcessesSpontaneous Processes• Chemical systems in equilibrium are reversible

and are not spontaneous.

• In any spontaneous process, the path between reactants and products is irreversible.

• Thermodynamics provides the direction of a process. It cannot predict the speed (rate) at which the process will occur.

• Why can both endothermic and exothermic reactions be spontaneous?

Heat of reaction (H) & Spontaneous Change

All combustion reactions are spontaneous and exothermic:

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g) Horxn = -802 kJ

Iron rusts spontaneously and exothermically:

2 Fe(s) + O2 (g) Fe2O3 (s) Horxn = -826 kJ

Ionic compounds form spontaneously from their elements with a largerelease of heat:

Na(s) + Cl2 (g) NaCl(s)

32

12

Horxn = -411 kJ

At 1 atm, water freezes below 0°C but melts above 0°C.Both processes are spontaneous, but the first is exothermic and thesecond endothermic.

H2O(l) H2O(s) Horxn = -6.02 kJ

(exothermic; spontaneous at T < 0oC)H2O(s) H2O(l) Ho

rxn = +6.02 kJ (endothermic; spontaneous at T > 0oC)

A Spontaneous, Endothermic Chemical Reaction

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Spontaneous Expansion of a Gas

Entropy and the Second Law of Thermodynamics

Entropy (S) is a measure of disorderEntropy (S) is a measure of disorder

A system with little freedom, such as a crystalline solid or a deck of cards in a specific sequence, has relatively small disorder and low entropy.

A system with many degrees of freedom (possible arrangements), such as a gas or a shuffled deck of cards, has relatively large disorder and high entropy.

Sdisorder > Sorder

Ssys = Sfinal - Sinitial

Entropy and the Second LawEntropy and the Second Law

The Second Law of Thermodynamics The Second Law of Thermodynamics – All spontaneous processes occur in the direction that increases the total entropy of the universe (system plus surroundings).

Suniv = Ssys + Ssurr

• Entropy is not conserved: Suniv is increasing

• For a reversible process: Suniv = 0.

• For a spontaneous process (i.e. irreversible): Suniv > 0.

• Note: the second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases.

• For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process.

Temperature Changes and the Increase in Entropy from a Solid to a Liquid to a Gas

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The Increase in Entropy From Solid to Liquid to Gas

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy• As we heat a substance, the entropy must increase.• The entropy increases markedly at the solid to liquid

phase change.• Boiling corresponds to a much greater change in entropy

than melting.• Entropy will increase when:

�liquids or solutions are formed from solids,�gases are formed from solids or liquids,�the number of gas molecules increase,�the temperature is increased.

The Small Increase in Entropy when Methanol Dissolves in Water

The Large Decrease in Entropy of a Gas When it Dissolves in a Liquid

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy

There are three atomic modes of motion:�translation (the moving of a molecule

from one point in space to another),�vibration (the shortening and

lengthening of bonds, including the change in bond angles),

�rotation (the spinning of a molecule about some axis).

http://www.chem.purdue.edu/gchelp/vibmodes/normal_mode_1.htm

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropyand the Effects of Temperature on Molecular and the Effects of Temperature on Molecular

MotionMotionhttp://fy.chalmers.se/~brodin/MolecularMotions/CCl4molecule.html

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The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy

• Energy is required to get a molecule to translate, vibrate or rotate.

• The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy.

• In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order.

• The Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

Predicting Relative Entropy Values

Problem: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in(f)]:

(a) 1 mol NaCl(s) or 1 mol NaCl(aq)

(b) 1 mol SF6 or 1 mol SCl6

(c) 1 mol CO(g) or 1 mol CO2 (g)

(d) 1 mol S8 or 4 mol S2

(e) 1 mol H2O(s) or 1 mol H2O(g)

(f) Lipton’s noodle soup at 24oC or at 95oC

Plan: The less ordered a system, the greater the entropy. A higher temperature increases entropy. Assume constant temperature in a-e.

Solutions: (a) 1 mol NaCl(aq). The two samples have the same number of ions, but in the solid they are highly ordered, and in the solution they are randomly dispersed.(b) 1 mol SCl6. For similar compounds, entropy increases with molar mass. (c) 1 mol CO2 (g) . For equal numbers of moles of substances with the same types of atoms in the same physical state, the more atoms per molecule, the more types of motion available to it and, thus, the higher its entropy. (d) 4 mol S2. The two samples contain the same number of sulfur atoms, but different numbers of molecules. Despite the greater complexity of S8 , the greater number of molecules dominates in this case because there are many more ways to arrange 4 moles of particles than one mole. (e) 1 mol H2O(g). For a given substance entropy increases in the sequence: solid < liquid < gas. (f) Soup at 95oC. Entropy increases with increasing temperature.

Calculations of Entropy ChangesCalculations of Entropy Changes

• Absolute entropy can be determined from complicated measurements.

• Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H.

• Units: J/mol-K. Note units of H: kJ/mol.• Standard molar entropies of elements are not zero.• For a chemical reaction which produces n products from

m reactants:

( ) ( )∑ ∑ °−°=° reactantsproducts mSnSS

The Standard Entropy of Reaction, Sorxn

For many chemical reactions: So = Soproducts - So

reactants > 0

Standard entropy of reaction, Sorxn:

∑mSoproducts - ∑nSo

reactantsSorxn =

But for reactions in which the total # of moles of products decrease, particularly gases which have very high entropy, we predict that the entropy of the products is less than that of the reactants. Therefore, the entropy will decrease during the reaction:

N2 (g) + 3 H2 (g) 2 NH3 (g) So = Soproducts - So

reactants< 0

Sorxn = (2 mol NH3 x So of NH3) - [(1 mol x 191.5 J/mol K) +

(3 mol x 130.6 J/mol K)] So

rxn = -197 J/K

..

As predicted, So < 0

Calculating the Standard Entropy of Reaction, Sorxn

Problem: Calculate the Sorxn for the oxidation of

one mole of S8 to form either SO2 (g), or SO3 (g) at 25oC:

S8 (s) + 8 O2 (g) 8 SO2 (g)

versus

S8 (s) + 12 O2 (g) 8 SO3 (g)

Plan: To determine Sorxn, apply Equation 20.3. We

predict the sign of Sorxn from the change in the number of

moles of gas: 8 = 8 or 12 = 8, so the entropy will decrease ( So

rxn< 0).

Calculating the Standard Entropy of Reaction, Sorxn

Solution: Calculate Sorxn using Appendix B values,

Rxn #1 So

rxn = ( 8 mol SO2 x So of SO2) - [(1 mol S8 x So of S8) + ( 8 mol O2 x So of O2)] = ( 8 mol x 248.2 J/mol K) - [(1 mol x 430.211 J/mol K) + (8 mol x 205.0 J/mol K)] = (1,985.6 J/K) - [(430.211 J/K) + (1,640.0 J/K)] = 1985.6 J/K - 2,070.211 J/K = - 84.611 J/K

Rxn #2 So

rxn = ( 8 mol SO3 x So of SO3) - [(1 mol S8 x So of S8) + ( 12 mol O2 x So of O2)] = ( 8 mol x 256.66 J/mol K) - [(1 mol x 430.211 J/mol K) + (12 mol x 205.0 J/mol K)] = (2,053.28 J/K) - [(430.211 J/K) + (2,460.0 J/K)] = 2,053.28 J/K - 2,890.211 J/K = - 836.931 J/K

Calculating the Standard Entropy of Reaction, Sorxn

Summary: For Reaction #1 (SO2 is product) So

rxn = - 84.611 J/K

For Reaction #2 (SO3 is the product) Sorxn = - 836.931 J/K

As predicted they are both negative, but Rxn #1 is close to zero, and we would also predict that it would be close to zero, since the number of moles of gaseous molecules did not change from reactant to product.

Calculating the Standard Entropy of Reaction, Sorxn

Entropy Changes in the Surroundings

The second law of thermodynamics restated says:

Decreases in the entropy of the system can occur Decreases in the entropy of the system can occur only if only if increasesincreases in the entropy of the surroundings in the entropy of the surroundings outweigh them.outweigh them.

•The surroundings either add energy to the system or remove energy from it.

•Therefore there are two possible types of enthalpy changes in chemical reactions.

1) Can you name the two types of processes: (a)? & (b)?

2) If (b) is a spontaneous process, what is happening to Suniv?

Entropy Changes in the Surroundings

Those two possible types of enthalpy changes are:

1. Exothermic. Heat lost by the system is gained by the surroundings.

2. Endothermic. Heat gained by the system is lost by the surroundings.

qsys < 0, qsurr > 0 and Ssurr > 0

qsys > 0, qsurr < 0 and Ssurr < 0

The change in entropy of the surroundingsis directly related to an opposite change inthe energy of the system and inversely relatedto the temperature of the surroundings beforeheat is transferred.

Ssurr = -Hsys

T

Components of So

univ for Spontaneous Reactions