CHAPTER 12: Thermodynamics Why Chemical Reactions Happen

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A tiny fraction of the sun's energy is used to produce complicated, ordered, high- Useful energy is being "degraded" in energy systems such as life the form of unusable heat, light, etc.

• Our observation is that natural processes proceed from ordered, high-energy systems to disordered, lower energy states. • In addition, once the energy has been "degraded", it is no longer available to perform useful work. • It may not appear to be so locally (earth), but globally it is true (sun, universe as a whole).

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Thermodynamics - quantitative description of the factors that drive chemical reactions, i.e. temperature, enthalpy, entropy, free energy.

Answers questions such as-

. will two or more substances react when they are mixed under specified conditions? . if a reaction occurs, what energy changes are associated with it? . to what extent does a reaction occur to?

Thermodynamics does NOT tell us the RATE of a reaction

Chapter Outline

. 12.1 Spontaneous Processes . 12.2 Entropy . 12.3 Absolute Entropy and Molecular Structure . 12.4 Applications of the Second Law . 12.5 Calculating Entropy Changes . 12.6 Free Energy . 12.7 Temperature and Spontaneity . 12.8 Coupled Reactions

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Spontaneous Processes A spontaneous process is one that is capable of proceeding in a given direction without an external driving force

• A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm, water freezes below 0 0C and ice melts above 0 0C • Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb • Iron exposed to oxygen and water forms rust

Spontaneous chemical and physical changes are frequently accompanied by a release of heat (exothermic H < 0) -

C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)

Ho = -2200 kJ

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But sometimes a spontaneous process can be endothermic H > 0 -

Some processes are accompanied by no change in enthalpy at all (Ho = 0), as is the case for an ideal gas spontaneously expanding:

spontaneous

nonspontaneous

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There's another factor promoting spontaneity in these processes, and that's the increasing randomness or disorder of the system (this is a qualitative description only – quantitative coming up):

1. propane combustion:

o C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) H = -2200 kJ

2. water melting: o H2O(s)  H2O(l) H = 6.01 kJ

3. gas expansion:

Chapter Outline

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Thermodynamics: Entropy

. Second Law of Thermodynamics: • The entropy of the universe increases in any spontaneous process

• Suniv = Ssys + Ssurr > 0 . Entropy (S): • A measure of the amount of disorder (qualitative), or unusable energy in a system at a specific temperature (quantitative). • Entropy is affected by molecular motion, or disorder from volume changes (e.g. the previous gas expansion example).

Types of Molecular Motion

• Three types of motion: – Translational: Movement through space – Rotational: Spinning motion around axis perpendicular to bond – Vibrational: Movement of atoms toward/away from each other • As temperature increases, the amount of motion increases.

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Third Law of Thermodynamics

. The entropy of a perfect crystal is zero at absolute zero . Provides a point of reference or baseline for quantitating entropy (placing a numerical value on it) . Heat plays a role in the amount of entropy a system has

18 - 13

Chapter Outline

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Standard Molar Entropy, So (absolute entropy content, J/mol K The entropy of one mole of o o o a substance in its standard S solid < S liquid < S gas state at 1 atm and 298 K.

Trends in Entropies of Phase Changes

Ssolid <  Sliquid <  Sgas

S = Sfinal - Sinitial

Units: J/molK

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The Effect of Molecular Structure on Entropy

Summary:

S is expected to INCREASE for these types of processes (S > 0) :

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Sample Exercise 12.1: Predicting the Sign of an Entropy Change

Predict whether or not an increase or decrease in entropy accompanies each of these processes when they occur at constant temperature:

(a)H2O(l)  H2O(g) (b)NH3(g) + HCl(g)  NH4Cl(s) (c)C12H22O11(s)  C12H22O11(aq)

Sample Exercise 12.2: Comparing Standard Molar Entropy Changes

Without consulting any standard reference sources, select the component in each of the following pairs that has the greatest standard molar entropy at 298 K. Assume that there is one mole of each component in its standard state (the pressure of each gas is 1 bar and the concentration of each solution is 1 M).

(a)HCl(g), HCl(aq)

(b)CH3OH(l), CH3CH2OH(l)

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Chapter Outline

Spontaneous Nonspontaneous Changes in Entropy

S = Sf - Si

S > 0

S < 0

Suniverse

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The Second Law of Thermodynamics: The total entropy of the universe increases in any spontaneous process

Spontaneous process: Suniv = Ssys + Ssurr > 0

sys = system surr = surroundings One can be negative but the other will be even more positive

Entropy Changes in the Surroundings (Ssurr)

Exothermic Process Endothermic Process

Ssurr > 0 Ssurr < 0

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The change in entropy of the surroundings can be calculated:

if Hsys < 0 (exothermic), then Ssurr > 0 (entropy of the surroundings increases) Ssurr  -Hsys if Hsys > 0 (endothermic), then Ssurr < 0 (entropy of the surroundings decreases)

If the temperature of the surroundings is already high, Ssurr  1 then pumping heat in or out Tsurr causes less change in disorder than at lower temperatures

Combining the two:

Ssurr  -Hsys and Ssurr  1 so Ssurr = -Hsys Tsurr T

(Tsurr usually = Tsys)

e.g. N2(g) + 3H2(g)  2NH3(g) Ssys = -198.3 J/K * Hsys = -92.6 kJ The two main driving forces are in opposition to each other - the release of heat favors a spontaneous reaction while the

decrease in entropy does not. Calculating Suniv will decide the issue (next slide). Remember: for a spontaneous reaction the entropy of the universe increases.

0 0 0 *from Ch 9: Hrxn = S nH f (products) - S mHf (reactants)

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Is the reaction spontaneous at 25 oC?

Suniv = Ssys + Ssurr

The previous example with ammonia illustrated that maybe entropy will decrease in the system, but this will always be accompanied by a greater increase in the entropy of the surroundings such that Suniv > 0.

Suniv = Ssys + Ssurr > 0

Another way of stating the 2nd Law is that "You Can't Win!"

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Chapter Outline

Standard Entropy of Reaction (Srxn)

0 The standard entropy of reaction (Srxn ) is the entropy change for a reaction carried out at 1 atm and 250C.

aA + bB cC + dD

0 0 0 0 0 Srxn = [c S (C) + dS (D) ] - [a S (A) + bS (B) ]

0 0 0 Srxn = S nS (products) - S mS (reactants)

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Sample Exercise 12.3: Calculating So Values

Given the following standard molar entropy values at 298 K (found in o Appendix 4, Table A4.3), what is the of S rxn for the dissolution of ammonium nitrate under standard conditions?

+ - NH4NO3(s)  NH4 (aq) + NO3 (aq)

So [J/molK] 151.1 113.4 146.4

Chapter Outline

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Gibbs Free Energy

Suniv = Ssys + Ssurr Could use this relation to calculate reaction spontaneity, but

not always easy to calculate Ssurr - so the expression is rearranged to only include terms relating to the system:

Under constant temperature and pressure:

Gibbs free energy (G) G = Hsys -TSsys

G < 0 The reaction is spontaneous in the forward direction. G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. G = 0 The reaction is at equilibrium.

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Calculating Free-Energy Changes Using

G = Hsys -TSsys

aA + bB cC + dD

G = Hsys -TSsys

표 표 표 ∆퐻푟푥푛 = Σ푛푝푟표푑 ∙ ∆퐻푓 푝푟표푑 − Σ푛푟푒푎푐푡 ∙ ∆퐻푓 푟푒푎푐푡

표 표 표 ∆푆푟푥푛 = Σ푛푝푟표푑 ∙ 푆푓 푝푟표푑 − Σ푛푟푒푎푐푡 ∙ 푆푓 푟푒푎푐푡

Sample Exercise 12.4: Predicting Reaction Spontaneity under Standard Conditions

Consider the reaction of nitrogen gas and hydrogen gas at 298 K to make ammonia at the same temperature:

N2(g) + 3 H2(g)  2 NH3(g)

표 (a) Before doing any calculations, predict the sign of ∆푆푟푥푛 표 (b) What is the actual value of ∆푆푟푥푛? 표 (c) What is the value of ∆퐻푟푥푛? 표 (d) What is the value of ∆퐺푟푥푛? (e) Is the reaction spontaneous at 298 K and 1 atm?

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Calculating Free-Energy Changes Using 표 표 표 ∆퐺푟푥푛 = Σ푛푝푟표푑∆퐺푓 푝푟표푑 − Σ푛푟푒푎푐푡∆퐺푓 푟푒푎푐푡

Standard free energy of 0 formation (Gf ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

0 G f of any element in its most stable allotropic form is zero, e.g. graphite and not diamond

표 Sample Exercise 12.5: Calculating ∆퐺푟푥푛 Using 표 Appropriate ∆퐺푓 Values

Use the appropriate standard free energy of formation values in App. 4 to calculate the change in free energy as ethanol burns under standard conditions. Assume the reaction proceeds as described by the following chemical equation:

CH3CH2OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)

표 표 표 ∆퐺푟푥푛 = Σ푛푝푟표푑∆퐺푓 푝푟표푑 − Σ푛푟푒푎푐푡∆퐺푓 푟푒푎푐푡

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Chapter Outline

Temperature and Spontaneity

G = H –TS rearranging terms to match the general formula for a straight line - G = –TS + H

G = –ST + H

y = m x + b At what temperature m = –S does the reaction b = H spontaneity change?

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At what temperature does the reaction spontaneity change?

G = H –TS

When the reaction spontaneity changes, the sign of G changes from

______to ______, passing through ______on the way. Therefore

G = 0 at the temperature that spontaneity changes, so -

 = H –TS and

H = TS and therefore

T =

G = H - T S Always b = H (negative) Spontaneous: H < 0 and S > 0 m = - S (overall negative)

G < 0 spontaneous at

 all temperatures G G

 0.0 b = H  m = - S

T 

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G = H - T S

Never b = H (positive) Spontaneous: H > 0 and S < 0 m = - S (overall positive)

G > 0 not spontaneous

at all temperatures 

G G m = - S  b = H 

0.0

T 

G = H - T S Enthalpy-driven: b = H (negative) spontaneous at low T H < 0 and S < 0 m = - S (overall positive)

G < 0 spontaneous only

at low T

 m = - S

G G 

0.0 G < 0

b = H  T = H S

T 

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G = H - T S

Entropy-driven: b = H (positive) spontaneous at high T H > 0 and S > 0 m = - S (overall negative)

G < 0 spontaneous only b = H  at high T

m = - S

 0.0

G G G < 0 

T = H S T 

G = H - T S

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Homework Problem #80

The element H2 is not abundant in nature, but it is a useful reagent in, for example, the potential synthesis of the liquid fuel methanol from gaseous carbon monoxide. Under what temperature conditions if this reaction spontaneous?

2 H2(g) + CO(g)  CH3OH(l)



G G = 0 

T 