Chapter · Chapter 17 Free Energy and Thermodynamics Chemistry: A Molecular Approach, 2nd Ed....

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Copyright 2011 Pearson Education, Inc.

Chapter 17

Free Energy

and

Thermodynamics

Chemistry: A Molecular Approach, 2nd Ed.

Nivaldo Tro

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MACopyright 2011 Pearson Education, Inc.

First Law of Thermodynamics

• You can’t win!

• First Law of Thermodynamics: Energy

cannot be created or destroyed

the total energy of the universe cannot change

though you can transfer it from one place to another

DEuniverse = 0 = DEsystem + DEsurroundings

2Tro: Chemistry: A Molecular Approach, 2/e


First Law of Thermodynamics• Conservation of Energy

• For an exothermic reaction, “lost” heat from the system goes into the surroundings

• Two ways energy is “lost” from a system converted to heat, q

used to do work, w

• Energy conservation requires that the energy change in the system equal the heat released + work done DE = q + w

DE = DH + PDV

• DE is a state function internal energy change independent of how done

3Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

The Energy Tax

• You can’t break even!

• To recharge a battery with 100 kJ of

useful energy will require more than

100 kJ

because of the Second Law of

Thermodynamics

• Every energy transition results in a

“loss” of energy

an “Energy Tax” demanded by nature

and conversion of energy to heat which

is “lost” by heating up the surroundings



Heat Tax

fewer steps

generally results

in a lower total

heat tax


Thermodynamics and Spontaneity

• Thermodynamics predicts whether a process will occur under the given conditions

processes that will occur are called spontaneous

nonspontaneous processes require energy input to go

• Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction

if the system after reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable.

• Spontaneity ≠ fast or slow


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Comparing Potential Energy

The direction of

spontaneity can

be determined

by comparing

the potential

energy of the

system at the

start and the

end


Reversibility of Process• Any spontaneous process is irreversible because

there is a net release of energy when it proceeds in that direction

it will proceed in only one direction

• A reversible process will proceed back and forth between the two end conditions

any reversible process is at equilibrium

results in no change in free energy

• If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction



Thermodynamics vs. Kinetics


Diamond → Graphite

Graphite is more stable than diamond, so the conversion

of diamond into graphite is spontaneous – but don’t worry,

it’s so slow that your ring won’t turn into pencil lead in

your lifetime (or through many of your generations)



Spontaneous Processes

• Spontaneous processes occur because they

release energy from the system

• Most spontaneous processes proceed from a

system of higher potential energy to a system at

lower potential energy

exothermic

• But there are some spontaneous processes that

proceed from a system of lower potential energy to

a system at higher potential energy

endothermic

• How can something absorb potential energy, yet

have a net release of energy?


Melting Ice

12

Melting is an

Endothermic process,

yet ice will

spontaneously melt

above 0 °C.

When a solid melts, the

particles have more

freedom of movement.

More freedom of motion

increases the

randomness of the

system. When systems

become more random,

energy is released. We

call this energy,

entropyTro: Chemistry: A Molecular Approach, 2/e

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Factors Affecting Whether a

Reaction Is Spontaneous

• There are two factors that determine whether a

reaction is spontaneous. They are the

enthalpy change and the entropy change of

the system

• The enthalpy change, DH, is the difference in

the sum of the internal energy and PV work

energy of the reactants to the products

• The entropy change, DS, is the difference in

randomness of the reactants compared to the

products


Enthalpy Change

DH 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, DH is negative

• A reaction is generally endothermic if the bonds in

the products are weaker than the bonds in the

reactants

endothermic = energy absorbed, DH is positive

• The enthalpy change is favorable for exothermic

reactions and unfavorable for endothermic reactions



Entropy

• Entropy is a thermodynamic function

that increases as the number of

energetically equivalent ways of

arranging the components increases, S

S generally J/mol

• S = k ln W

k = Boltzmann Constant = 1.38 x 10−23 J/K

W is the number of energetically equivalent

ways a system can exist

unitless

• Random systems require less energy

than ordered systems15Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

W

These are energetically

equivalent states for the

expansion of a gas.

16

It doesn’t matter, in terms

of potential energy,

whether the molecules

are all in one flask, or

evenly distributed

But one of these states is

more probable than the

other two

Tro: Chemistry: A Molecular Approach, 2/e


This macrostate can be achieved through

several different arrangements of the particles

Macrostates → Microstates

17

These

microstates all

have the same

macrostate

So there are six

different particle

arrangements

that result in the

same macrostate

Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Macrostates and Probability

There is only one possible

arrangement that gives State

A and one that gives State B

There are six possible

arrangements that give State C

Therefore State C has

higher entropy than

either State A or State B

The macrostate with the

highest entropy also has the

greatest dispersal of energy

There is six times the

probability of having the

State C macrostate than

either State A or State B


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Changes in Entropy, DS

DS = Sfinal − Sinitial

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

DS 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

reactions that have larger numbers of product molecules than reactant molecules

increase in temperature

solids dissociating into ions upon dissolving


Increases in Entropy



DS

• For a process where the final condition is more

random than the initial condition, DSsystem is

positive and the entropy change is favorable

for the process to be spontaneous

• For a process where the final condition is more

orderly than the initial condition, DSsystem is

negative and the entropy change is

unfavorable for the process to be

spontaneous

DSsystem = DSreaction = Sn(S°products) − Sn(S°reactants)


Entropy Change in State Change

• When materials change state, the number of macrostates it can have changes as well

the more degrees of freedom the molecules have, the more macrostates are possible

solids have fewer macrostates than liquids, which have fewer macrostates than gases



Entropy Change and State Change


Practice – Predict whether DSsystem is + or −

for each of the following

• A hot beaker burning your fingers

• 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)

DS is +

DS is −

DS is −

DS is +

DS is +

DS is +

DS is −


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The 2nd Law of Thermodynamics

• The 2nd Law of Thermodynamics says that the

total entropy change of the universe must be

positive for a process to be spontaneous

for reversible process DSuniv = 0

for irreversible (spontaneous) process DSuniv > 0

• DSuniverse = DSsystem + DSsurroundings

• If the entropy of the system decreases, then the

entropy of the surroundings must increase by a

larger amount

when DSsystem is negative, DSsurroundings must be

positive and big for a spontaneous process25Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Heat Flow, Entropy, and the 2nd Law

According to the 2nd Law,

heat must flow from

water to ice because it

results in more dispersal

of heat. The entropy of

the universe increases.

26

When ice is placed in

water, heat flows from

the water into the ice



Heat Transfer and Changes in

Entropy of the Surroundings

• The 2nd Law demands that the entropy of the

universe increase for a spontaneous process

• Yet processes like water vapor condensing are

spontaneous, even though the water vapor is more

random than the liquid water

• If a process is spontaneous, yet the entropy change

of the process is unfavorable, there must have

been a large increase in the entropy of the

surroundings

• The entropy increase must come from heat released

by the system – the process must be exothermic!27Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Entropy Change in the System and

Surroundings

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When the entropy

change in system is

unfavorable (negative),

the entropy change in

the surroundings must

be favorable (positive),

and large to allow the

process to be

spontaneous



Heat Exchange and DSsurroundings

• When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings

• When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings

• The amount the entropy of the surroundings changes depends on its original temperature the higher the original

temperature, the less effect addition or removal of heat has


Temperature Dependence of DSsurroundings

• When heat is added to surroundings that are

cool it has more of an effect on the entropy than

it would have if the surroundings were already

hot

• Water freezes spontaneously below 0 °C

because the heat released on freezing

increases the entropy of the surroundings

enough to make DS positive

above 0 °C the increase in entropy of the

surroundings is insufficient to make DS positive


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Quantifying Entropy Changes in

Surroundings

• The entropy change in the surroundings is

proportional to the amount of heat gained or lost

qsurroundings = −qsystem

• The entropy change in the surroundings is also

inversely proportional to its temperature

• At constant pressure and temperature, the

overall relationship is


Example 17.2a: Calculate the entropy change of the

surroundings at 25ºC for the reaction below

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ

32

combustion is largely exothermic, so the entropy of

the surroundings should increase significantly

DHsystem = −2044 kJ, T = 25 ºC = 298 K

DSsurroundings, J/K

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DST, DH



Practice – The reaction below has DHrxn = +66.4 kJ at

25 °C. (a) Determine the Dssurroundings, (b) the sign of

DSsystem, and (c) whether the process is spontaneous

2 O2(g) + N2(g) 2 NO2(g)


Practice – The reaction 2 O2(g) + N2(g) 2 NO2(g) has DHrxn

= +66.4 kJ at 25 °C. Calculate the entropy change of the

surroundings. Determine the sign of the entropy change in

the system and whether the reaction is spontaneous.

DHsys = +66.4 kJ, T = 25 ºC = 298 K

DSsurr, J/K, DSreact + or −, DSuniverse + or −

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DST, DH

the major difference is that there are

fewer product molecules than

reactant molecules, so the DSreaction

is unfavorable and (−)

because DSsurroundings is

(−) it is unfavorable

both DSsurroundings and DSreaction are (−),

DSuniverse is (−) and the process is

nonspontaneous



Gibbs Free Energy and Spontaneity

• It can be shown that −TDSuniv = DHsys−TDSsys

• The Gibbs Free Energy, G, is the maximum amount of work energy that can be released to the surroundings by a system for a constant temperature and pressure system

the Gibbs Free Energy is often called the Chemical Potential because it is analogous to the storing of energy in a mechanical system

DGsys = DHsys−TDSsys

• Because DSuniv determines if a process is spontaneous, DG also determines spontaneity

DSuniv is + when spontaneous, so DG is −


Gibbs Free Energy, DG

• A process will be spontaneous when DG is

negative

DG will be negative when

DH is negative and DS is positive

exothermic and more random

DH is negative and large and DS is negative but small

DH is positive but small and DS is positive and large

or high temperature

• DG will be positive when DH is + and DS is −

never spontaneous at any temperature

• When DG = 0 the reaction is at equilibrium36Tro: Chemistry: A Molecular Approach, 2/e

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DG, DH, and DS


Free Energy Change and Spontaneity



Example 17.3a: The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has

DH = +95.7 kJ and DS = +142.2 J/K at 25 °C.

Calculate DG and determine if it is spontaneous.

39

Because DG is +, the reaction is not spontaneous

at this temperature. To make it spontaneous, we

need to increase the temperature.

DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K

DG, kJ

Answer:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGT, DH, DS


Practice – The reaction Al(s) + Fe2O3(s) Fe(s) + Al2O3(s) has

DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.





DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.


Because DG is −, the reaction is spontaneous at

this temperature. To make it nonspontaneous, we

need to increase the temperature.

DH = −847.6 kJ and DS = −41.3 J/K, T = 298 K

DG, kJ

Answer:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGT, DH, DS


Example 17.3b: The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has

DH = +95.7 kJ and DS = +142.2 J/K.

Calculate the minimum temperature it will be spontaneous.

42

the temperature must be higher than 673K for

the reaction to be spontaneous

DH = +95.7 kJ, DS = +142.2 J/K, DG < 0

T, K

Answer:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

TDG, DH, DS


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DH = −847.6 kJ and DS = −41.3 J/K.

Calculate the maximum temperature it will be spontaneous.



DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.

Determine the maximum temperature it is spontaneous.

any temperature above 1775 C will make the

reaction nonspontaneous

DH = −847.6 kJ and DS = −41.3 J/K, DG > 0

T, C

Answer:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

TDG, DH, DS

2048 − 273 = 1775 C



Standard Conditions

• The standard state is the state of a

material at a defined set of conditions

• Gas = pure gas at exactly 1 atm pressure

• Solid or Liquid = pure solid or liquid in its

most stable form at exactly 1 atm pressure

and temperature of interestusually 25 °C

• Solution = substance in a solution with

concentration 1 M


The 3rd Law of Thermodynamics:

Absolute Entropy

• The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles

• The 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K

therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy

therefore, the absolute entropy of substances is always +



Standard Absolute Entropies

• S°

• Extensive

• Entropies for 1 mole of a substance at 298 K

for a particular state, a particular allotrope,

particular molecular complexity, a particular

molar mass, and a particular degree of

dissolution


Standard Absolute Entropies


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Relative Standard Entropies:

States

• The gas state has a larger entropy than the

liquid state at a particular temperature

• The liquid state has a larger entropy than the

solid state at a particular temperature

SubstanceS°,

(J/mol∙K)

H2O (l) 70.0

H2O (g) 188.8



Molar Mass

• The larger the molar

mass, the larger the

entropy

• Available energy states

more closely spaced,

allowing more dispersal

of energy through the

states




Allotropes

• The less constrained

the structure of an

allotrope is, the

larger its entropy

• The fact that the

layers in graphite

are not bonded

together makes it

less constrained



Molecular Complexity

• Larger, more complex

molecules generally

have larger entropy

• More available energy

states, allowing more

dispersal of energy

through the states

SubstanceMolar

Mass

S°,

(J/mol∙K)

Ar (g) 39.948 154.8

NO (g) 30.006 210.8

52

SubstanceMolar

Mass

S°,

(J/mol∙K)

CO (g) 28.01 197.7

C2H4 (g) 28.05 219.3



Relative Standard Entropies

Dissolution

• Dissolved solids

generally have larger

entropy

• Distributing particles

throughout the mixture

SubstanceS°,

(J/mol∙K)

KClO3(s) 143.1

KClO3(aq) 265.7


The Standard Entropy Change, DS

• The standard entropy change is the difference

in absolute entropy between the reactants and

products under standard conditions

DSºreaction = (∑npSºproducts) − (∑nrSºreactants)

remember: though the standard enthalpy of

formation, DHf°, of an element is 0 kJ/mol, the

absolute entropy at 25 °C, S°, is always positive


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Example 17.4: Calculate DS for the

reaction

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

55

DS is +, as you would expect for a reaction with more gas

product molecules than reactant molecules

standard entropies from Appendix IIB

DS, J/K

Check:

Sol’n:

Conceptual

Plan:

Relationships:

Given:

Find:

DSSNH3, SO2, SNO, SH2O

Substance S, J/molK

NH3(g) 192.8

O2(g) 205.2

NO(g) 210.8

H2O(g) 188.8


Practice – Calculate the DS for the reaction

2 H2(g) + O2(g) 2 H2O(g)

Substance S, J/molK

H2(g) 130.7

O2(g) 205.2

H2O(g) 188.8



Example 17.4: Calculate DS for the

reaction

2 H2(g) + O2(g) 2 H2O(g)

57

DS is −, as you would expect for a reaction with more

gas reactant molecules than product molecules

standard entropies from Appendix IIB

DS, J/K

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DSSH2, SO2, SH2O

Substance S, J/molK

H2(g) 130.6

O2(g) 205.2

H2O(g) 188.8


Calculating DG

• At 25 C

DGoreaction = SnDGo

f(products) - SnDGof(reactants)

• At temperatures other than 25 C

assuming the change in DHoreaction and DSo

reaction is negligible

DGreaction = DHreaction – TDSreaction

• or

DGtotal = DGreaction 1 + DGreaction 2 + ...



Example 17.6: The reaction SO2(g) + ½ O2(g) SO3(g) has

DH = −98.9 kJ and DS = −94.0 J/K at 25 °C.

Calculate DG at 125 C and determine if it is more or less

spontaneous than at 25 °C with DG° = −70.9 kJ/mol SO3.

59

because DG is −, the reaction is spontaneous at

this temperature, though less so than at 25 C

DH = −98.9 kJ, DS = −94.0 J/K, T = 398 K

DG, kJ

Answer:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGT, DH, DS


Practice – Determine the free energy change in the

following reaction at 298 K

2 H2O(g) + O2(g) 2 H2O2(g)

Substance DH, kJ/mol S, J/mol

H2O2(g) −136.3 232.7

O2(g) 0 205.2

H2O(g) −241.8 188.8


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2 H2O(g) + O2(g) 2 H2O2(g)

61

standard energies from Appendix IIB, T = 298 K

DG, kJ

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGDHf° Sº of prod & react DH, DSº

DH = 211.0 kJ, DS = −117.4 J/K, T = 298 K

DG, kJ


Standard Free Energies of Formation

• The free energy of formation (DGf°) is the change

in free energy when 1 mol of a compound forms

from its constituent elements in their standard

states

• The free energy

of formation of

pure elements in

their standard

states is zero



Example 17.7: Calculate DG at 25 C

for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)

63

standard free energies of formation from Appendix IIB

DG, kJ

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGDGf° of prod & react

Substance DGf°, kJ/mol

CH4(g) −50.5

O2(g) 0.0

CO2(g) −394.4

H2O(g) −228.6

O3(g) +163.2




2 H2O(g) + O2(g) 2 H2O2(g)

Substance DG, kJ/mol

H2O2(g) −105.6

O2(g) 0

H2O(g) −228.6





2 H2O(g) + O2(g) 2 H2O2(g)

65


DG, kJ

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGDGf° of prod & react


DG Relationships

• If a reaction can be expressed as a series of

reactions, the sum of the DG values of the

individual reaction is the DG of the total reaction

DG is a state function

• If a reaction is reversed, the sign of its DG value

reverses

• If the amount of materials is multiplied by a factor,

the value of the DG is multiplied by the same

factor

the value of DG of a reaction is extensive


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Example 17.8: Find DGºrxn for the following reaction

using the given equations

3 C(s) + 4 H2(g) C3H8(g)

Given:

Find:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DGº = −2074 kJ

C(s) + O2(g) CO2(g) DGº = −394.4 kJ

2 H2(g) + O2(g) 2 H2O(g) DGº = −457.1 kJ

DGº of 3 C(s) + 4 H2(g) C3H8(g)

Rel: Manipulate the given reactions so they add up to the

reaction you wish to find. The sum of the DGº’s is the

DGº of the reaction you want to find.

Solution: 3 CO2(g) + 4 H2O(g) C3H8(g) + 5 O2(g) DGº = +2074 kJ

3 x [C(s) + O2(g) CO2(g) ] DGº = 3(−394.4 kJ)

2 x [2 H2(g) + O2(g) 2 H2O(g)] DGº = 2(−457.1 kJ)

Given:

Find:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DGº = −2074 kJ

C(s) + O2(g) CO2(g) DGº = −394.4 kJ

2 H2(g) + O2(g) 2 H2O(g) DGº = −457.1 kJ

DGº of 3 C(s) + 4 H2(g) C3H8(g)




Solution: 3 CO2(g) + 4 H2O(g) C3H8(g) + 5 O2(g) DGº = +2074 kJ

3 C(s) + 3 O2(g) 3 CO2(g) DGº = −1183.2 kJ

4 H2(g) + 2 O2(g) 4 H2O(g) DGº = −914.2 kJ

3 C(s) + 4 H2(g) C3H8(g) DGº = −23 kJ




2 H2O(g) + O2(g) 2 H2O2(g)

Substance DG, kJ/mol

H2 (g) + O2(g) H2O2(g) −105.6

2 H2 (g) + O2(g) 2 H2O(g) −457.2





2 H2O(g) + O2(g) 2 H2O2(g)

Given:

Find:

H2(g) + O2(g) H2O2(g) DGº = −105.6 kJ

2 H2(g) + O2(g) 2 H2O(g) DGº = −457.2 kJ

DGº of 2 H2O(g) + O2(g) 2 H2O2(g)




Solution: 2 H2(s) + 2 O2(g) 2 H2O2(g) DGº = −211.2 kJ

2 H2O(g) 2 H2(g) + O2(g) DGº = +457.2 kJ

2 H2O(g) + O2(g) 2 H2O2(g) DGº = +246.0 kJ


What’s “Free” About Free Energy?

• The free energy is the maximum amount of

energy released from a system that is available

to do work on the surroundings

• For many exothermic reactions, some of the

heat released due to the enthalpy change goes

into increasing the entropy of the surroundings,

so it is not available to do work

• And even some of this free energy is generally

lost to heating up the surroundings



Free Energy of an Exothermic Reaction

• C(s, graphite) + 2 H2(g) → CH4(g)

• DH°rxn = −74.6 kJ = exothermic

• DS°rxn = −80.8 J/K = unfavorable

• DG°rxn = −50.5 kJ = spontaneous

71

DG° is less than DH°

because some of the

released heat energy

is lost to increase the

entropy of the

surroundings


Free Energy and Reversible Reactions

• The change in free energy is a theoretical limit

as to the amount of work that can be done

• If the reaction achieves its theoretical limit, it is a

reversible reaction


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Real Reactions• In a real reaction, some of the free energy is

“lost” as heat

if not most

• Therefore, real reactions are irreversible


DG under Nonstandard Conditions

DG = DG only when the reactants and products are in their standard states

their normal state at that temperature

partial pressure of gas = 1 atm

concentration = 1 M

Under nonstandard conditions, DG = DG + RTlnQ

Q is the reaction quotient

At equilibrium DG = 0

DG = −RTlnK


Copyright 2011 Pearson Education, Inc.75Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.

Example 17.9: Calculate DG at 298 K for

the reaction under the given conditions2 NO(g) + O2(g) 2 NO2(g) DGº = −71.2 kJ

76

non-standard conditions, DGº, T

DG, kJ

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGDG, PNO, PO2, PNO2

Substance P, atm

NO(g) 0.100

O2(g) 0.100

NO2(g) 2.00



Practice – Calculate DGrxn for the given

reaction at 700 K under the given conditions

N2(g) + 3 H2(g) 2 NH3(g) DGº = +46.4 kJ

Substance P, atm

N2(g) 33

H2(g) 99

NH3(g) 2.0


Practice – Calculate DGrxn for the given

reaction at 700 K under the given

conditions

N2(g) + 3 H2(g) 2 NH3(g) DGº = +46.4 kJ

78

non-standard conditions, DGº, T

DG, kJ

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DGDG, PNO, PO2, PNO2

Substance P, atm

N2(g) 33

H2(g) 99

NH3(g) 2.0


4/19/2016

14


DGº and K

• Because DGrxn = 0 at equilibrium, then

DGº = −RTln(K)

• When K < 1, DGº is + and the reaction is spontaneous in the reverse direction under standard conditions

nothing will happen if there are no products yet!

• When K > 1, DGº is − and the reaction is spontaneous in the forward direction under standard conditions

• When K = 1, DGº is 0 and the reaction is at equilibrium under standard conditions

79Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.80Tro: Chemistry: A Molecular Approach, 2/e


Example 17.10: Calculate K at 298 K

for the reactionN2O4(g) 2 NO2(g)

81


K

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

Substance DGf°, kJ/mol

N2O4(g) +99.8

NO2(g) +51.3

DGDGf of prod & react K

DGº = −RTln(K)


Practice – Estimate the equilibrium constant

for the given reaction at 700 K

N2(g) + 3 H2(g) 2 NH3(g) DGº = +46.4 kJ



Practice – Estimate the Equilibrium Constant

for the given reaction at 700 K

N2(g) + 3 H2(g) 2 NH3(g) DGº = +46.4 kJ

83

DGº

K

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DG K

DGº = −RTln(K)


• Combining these two equations

DG° = DH° − TDS°

DG° = −RTln(K)

• It can be shown that

• This equation is in the form y = mx + b

• The graph of ln(K) versus inverse T is a

straight line with slope and y-intercept

Why Is the Equilibrium Constant

Temperature Dependent?


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