CHAPTER 12: Thermodynamics Why Chemical Reactions …postonp/ch223/pdf/Ch12-s15.pdf4/3/2015 6 Thermodynamics: Entropy Second Law of Thermodynamics: • The entropy of the universe

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CHAPTER 12: Thermodynamics

Why Chemical Reactions Happen

Useful energy is being "degraded" in

the form of unusable heat, light, etc.

A tiny fraction of the sun's

energy is used to produce

complicated, ordered, high-

energy systems such as life

• 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

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

Chapter Outline

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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):

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Ho = -2200 kJ

1. propane combustion:

2. water melting:

H2O(s) H2O(l) Ho = 6.01 kJ

3. gas expansion:


12.2 Entropy




12.6 Free Energy



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

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12.2 Entropy




12.6 Free Energy



Chapter Outline

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Standard Molar Entropy, So

(absolute entropy content, J/mol K

The entropy of one mole of

a substance in its standard

state at 1 atm and 298 K.

Sosolid < So

liquid < Sogas

Trends in Entropies of Phase Changes Ssolid < Sliquid < Sgas

S = Sfinal - Sinitial

Units: J/molK

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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)


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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12.2 Entropy




12.6 Free Energy



Chapter Outline

Changes in Entropy

S = Sf - Si

S > 0

S < 0

Suniverse

Spontaneous Nonspontaneous

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

Suniv = Ssys + Ssurr > 0 Spontaneous process:

sys = system

surr = surroundings One can be negative but the

other will be even more positive

Entropy Changes in the Surroundings (Ssurr)

Exothermic Process

Ssurr > 0

Endothermic Process

Ssurr < 0

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The change in entropy of the surroundings

can be calculated:

Ssurr -Hsys

if Hsys < 0 (exothermic), then

Ssurr > 0 (entropy of the

surroundings increases)

if Hsys > 0 (endothermic), then

Ssurr < 0 (entropy of the

surroundings decreases)

Ssurr 1

Tsurr

If the temperature of the

surroundings is already high,

then pumping heat in or out

causes less change in disorder

than at lower temperatures

Combining the two:

and

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.

e.g. N2(g) + 3H2(g) 2NH3(g) Ssys = -198.3 J/K

Hsys = -92.6 kJ*

*from Ch 9: H0 rxn nH0 (products) f = S mH0 (reactants) f S -

Ssurr -Hsys Ssurr 1

Tsurr

so Ssurr = -Hsys

T

(Tsurr usually = Tsys)

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Suniv = Ssys + Ssurr

Is the reaction spontaneous at 25 oC?

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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12.2 Entropy




12.6 Free Energy



Chapter Outline

Standard Entropy of Reaction (Srxn)

The standard entropy of reaction (S0 ) is the entropy

change for a reaction carried out at 1 atm and 250C. rxn

aA + bB cC + dD

S0 rxn nS0(products) = S mS0(reactants) S -

S0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) [ + ]

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Calculating So Values

Given the following standard molar entropy values at 298 K (found in

Appendix 4, Table A4.3), what is the of Sorxn for the dissolution of

ammonium nitrate under standard conditions?

NH4NO3(s) NH4+(aq) + NO3

-(aq)

So [J/molK] 151.1 113.4 146.4


12.2 Entropy




12.6 Free Energy



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:

G = Hsys -TSsys Gibbs free

energy (G)

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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aA + bB cC + dD

Calculating Free-Energy Changes Using

G = Hsys -TSsys

∆𝐻𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑 ∙ ∆𝐻𝑓

𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∙ ∆𝐻𝑓𝑜 𝑟𝑒𝑎𝑐𝑡

∆𝑆𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑 ∙ 𝑆𝑓

𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∙ 𝑆𝑓𝑜 𝑟𝑒𝑎𝑐𝑡

G = Hsys -TSsys

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)

(a)Before doing any calculations, predict the sign of ∆𝑆𝑟𝑥𝑛𝑜

(b)What is the actual value of ∆𝑆𝑟𝑥𝑛𝑜 ?

(c) What is the value of ∆𝐻𝑟𝑥𝑛𝑜 ?

(d)What is the value of ∆𝐺𝑟𝑥𝑛𝑜 ?

(e) Is the reaction spontaneous at 298 K and 1 atm?

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Standard free energy of

formation (G0) is the free-energy

change that occurs when 1 mole

of the compound is formed from its

elements in their standard states.

f

Calculating Free-Energy Changes Using

∆𝐺𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑∆𝐺𝑓

𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡∆𝐺𝑓𝑜 𝑟𝑒𝑎𝑐𝑡

G0 of any element in its most

stable allotropic form is zero, e.g.

graphite and not diamond

f

Sample Exercise 12.5: Calculating ∆𝐺𝑟𝑥𝑛𝑜 Using

Appropriate ∆𝐺𝑓𝑜 Values

∆𝐺𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑∆𝐺𝑓

𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡∆𝐺𝑓𝑜 𝑟𝑒𝑎𝑐𝑡

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)

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12.2 Entropy




12.6 Free Energy



Chapter Outline

G = H –TS rearranging terms to match

the general formula for a

straight line -

G = –TS + H

G = –ST + H

y = m x + b

m = –S

b = H

At what temperature

does the reaction

spontaneity change?

Temperature and Spontaneity

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At what temperature does the reaction

spontaneity change?

G = H –TS

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 –TS and

H = TS and therefore

T =

Always

Spontaneous:

H < 0 and S > 0

G

T

0.0

b = H (negative)

m = - S (overall

negative)

G < 0 spontaneous at

all temperatures

b = H

m = - S

G = H - T S

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Never

Spontaneous:

H > 0 and S < 0

G

T

0.0

b = H (positive)

m = - S (overall positive)

G > 0 not spontaneous

at all temperatures

b = H

m = - S

G = H - T S

Enthalpy-driven:

spontaneous at low T

H < 0 and S < 0

G

T

0.0

b = H (negative)

m = - S (overall positive)

G < 0 spontaneous only

at low T

b = H

m = - S

T = H

S

G = H - T S

G < 0

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Entropy-driven:

spontaneous at high T

H > 0 and S > 0

G

T

0.0

b = H (positive)

m = - S (overall negative)

G < 0 spontaneous only

at high T b = H

m = - S

T = H

S

G = H - T S

G < 0

G = H - T S

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G = 0

T

G

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)

Documents

CHAPTER 12: Thermodynamics Why Chemical Reactions …postonp/ch223/pdf/Ch12-s15.pdf4/3/2015 6 Thermodynamics: Entropy Second Law of Thermodynamics: • The entropy of the universe