chapter 17 free energy and...

Chapter 17Free Energy

and Thermodynamics

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Tro, Chemistry: A Molecular Approach 2

First Law of Thermodynamics

• you can’t win!• First Law of Thermodynamics: Energy

cannot be Created or Destroyedthe total energy of the universe cannot changethough you can transfer it from one place to another

• ∆Euniverse = 0 = ∆Esystem + ∆Esurroundings


First Law of Thermodynamics• Conservation of Energy• For an exothermic reaction, “lost” heat from the system

goes into the surroundings• two ways energy “lost” from a system,

converted to heat, qused to do work, w

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

∆E = q + w∆E = ∆H + P∆V

• ∆E is a state functioninternal energy change independent of how done


Energy Tax• you can’t break even!• to recharge a battery with 100 kJ of

useful energy will require more than 100 kJ

• every energy transition results in a “loss” of energy

conversion of energy to heat which is “lost” by heating up the surroundings


Heat Taxfewer steps

generally results in a lower total

heat tax


Thermodynamics and Spontaneity• thermodynamics predicts whether a process will

proceed under the given conditionsspontaneous process

nonspontaneous processes require energy input to go• spontaneity is determined by comparing the free

energy of the system before the reaction with the free energy of the system after reaction.

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

• spontaneity ≠ fast or slow


Comparing Potential EnergyThe 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

it will proceed in only one direction• a reversible process will proceed back and forth

between the two end conditionsequilibriumresults 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).


Factors Affecting Whether a Reaction Is Spontaneous

• The two factors that determine the thermodynamic favorability are the enthalpy and the entropy.

• The enthalpy is a comparison of the bond energy of the reactants to the products.

bond energy = amount needed to break a bond. ∆H

• The entropy factors relates to the randomness/orderliness of a system

∆S• The enthalpy factor is generally more important

than the entropy factor


Enthalpy• related to the internal energy• ∆H generally kJ/mol• stronger bonds = more stable molecules• if products more stable than reactants, energy released

exothermic∆H = negative

• if reactants more stable than products, energy absorbedendothermic∆H = positive

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

• Hess’ Law ∆H°rxn = Σ(∆H°prod) - Σ(∆H°react)

Substance ∆H° kJ/mol

Substance ∆H° kJ/mol

Al(s) 0 Al2O3 -1669.8 Br2(l) 0 Br2(g) +30.71

C(diamond) +1.88 C(graphite) 0 CO(g) -110.5 CO2(g) -393.5 Ca(s) 0 CaO(s) -635.5 Cu(s) 0 CuO(s) -156.1 Fe(s) 0 Fe2O3(s) -822.16 H2(g) 0 H2O2(l) -187.8

H2O(g) -241.82 H2O(l) -285.83 HF(g) -268.61 HCl(g) -92.30

HBr(g) -36.23 HI(g) +25.94 I2(s) 0 I2(g) +62.25 N2(g) 0 NH3(g) -46.19 NO(g) +90.37 NO2(g) +33.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -296.9


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 Wk = Boltzmann Constant = 1.38 x 10-23 J/KW is the number of energetically equivalent ways, unitless

• Random systems require less energy than ordered systems


W

Energetically Equivalent States for the Expansion of a Gas


Macrostates → Microstates

This macrostate can be achieved throughseveral different arrangements of the particles

These microstates all have the same

macrostate

So there are 6 different particle arrangements that result in the same

macrostate


Macrostates and ProbabilityThere is only one possible

arrangement that gives State A and one that gives State C

There are 6 possible arrangements that give State B

Therefore State B has higher entropy than either

State A or State B

The macrostate with the highest entropy also has the greatest dispersal of energy


Changes in Entropy, ∆S• 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 disordered state.

(solid > liquid > gas)reactions which have larger numbers of product molecules than reactant molecules.increase in temperaturesolids dissociating into ions upon dissolving


Increases in Entropy


The 2nd Law of Thermodynamics• the total entropy change of the universe must be

positive for a process to be spontaneous for reversible process ∆Suniv = 0, for irreversible (spontaneous) process ∆Suniv > 0

• ∆Suniverse = ∆Ssystem + ∆Ssurroundings• if the entropy of the system decreases, then the

entropy of the surroundings must increase by a larger amount

when ∆Ssystem is negative, ∆Ssurroundings is positive• the increase in ∆Ssurroundings often comes from the

heat released in an exothermic reaction


Entropy Change in State Change• when materials change state, the number of

macrostates it can have changes as wellfor entropy: solid < liquid < gasbecause the degrees of freedom of motion increases solid → liquid → gas


Entropy Change and State Change


Heat Flow, Entropy, and the 2nd Law

Heat must flow from water to ice in order for the entropy of the universe to increase


Temperature Dependence of ∆Ssurroundings

• 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 the temperature it is at originally

the higher the original temperature, the less effect addition orremoval of heat has

TH

S systemgssurroundin

∆−=∆


Gibbs Free Energy, ∆G• maximum amount of energy from the system

available to do work on the surroundingsG = H – T·S

∆Gsys = ∆Hsys – T∆Ssys∆Gsys = – T∆Suniverse

∆Greaction = Σ n∆Gprod – Σ n∆Greact• when ∆G < 0, there is a decrease in free

energy of the system that is released into the surroundings; therefore a process will be spontaneous when ∆G is negative

( )

KJ3

KkJ

surr

syssurr

106.86 86.6SK 298

kJ 2044TH

S

×==∆

−−=

∆−=∆

Ex. 17.2a – The reaction C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) has ∆Hrxn = -2044 kJ at 25°C.

Calculate the entropy change of the surroundings.

combustion is largely exothermic, so the entropy of the surrounding should increase significantly

∆Hsystem = -2044 kJ, T = 298 K

∆Ssurroundings, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

∆ST, ∆H

TH

S syssurr

∆−=∆


Free Energy Change and Spontaneity


Gibbs Free Energy, ∆G• process will be spontaneous when ∆G is negative• ∆G will be negative when

∆H is negative and ∆S is positiveexothermic and more random

∆H is negative and large and ∆S is negative but small∆H is positive but small and ∆S is positive and

largeor high temperature

• ∆G will be positive when ∆H is + and ∆S is −never spontaneous at any temperature

• when ∆G = 0 the reaction is at equilibrium


∆G, ∆H, and ∆S

Ex. 17.3a – The reaction CCl4(g) → C(s, graphite) + 2 Cl2(g) has ∆H = +95.7 kJ and ∆S = +142.2 J/K at 25°C.

Calculate ∆G and determine if it is spontaneous.

Since ∆G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.

∆H = +95.7 kJ, ∆S = 142.2 J/K, T = 298 K

∆G, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

∆GT, ∆H, ∆S

STHG ∆−∆=∆

( ) ( )( )J1033.5

2.142K 298J 1095.7

STHG

4KJ3

×+=

+−×+=

∆−∆=∆

Ex. 17.3a – The reaction CCl4(g) → C(s, graphite) + 2 Cl2(g) has ∆H = +95.7 kJ and ∆S = +142.2 J/K.

Calculate the minimum temperature it will be spontaneous.

The temperature must be higher than 673K for the reaction to be spontaneous

∆H = +95.7 kJ, ∆S = 142.2 J/K, ∆G < 0

Τ, Κ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

T∆G, ∆H, ∆S

STHG ∆−∆=∆

( )( ) ( )( )( ) ( )( )K

J3KJ3

2.142TJ 1095.7

02.142TJ 1095.7

0 STHG

+<×+

<+−×+

<∆−∆=∆ ( )( )

TK 673

T 2.142

J 1095.7

KJ

3

<

<+

×+


The 3rd Law of ThermodynamicsAbsolute 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 entropytherefore, the absolute entropy of substances is always +


Standard Entropies

• S°• extensive• entropies for 1 mole at 298 K for a particular

state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution

Substance S° J/mol-K

Substance S° J/mol-K

Al(s) 28.3 Al2O3(s) 51.00 Br2(l) 152.3 Br2(g) 245.3

C(diamond) 2.43 C(graphite) 5.69 CO(g) 197.9 CO2(g) 213.6 Ca(s) 41.4 CaO(s) 39.75 Cu(s) 33.30 CuO(s) 42.59 Fe(s) 27.15 Fe2O3(s) 89.96 H2(g) 130.58 H2O2(l) 109.6

H2O(g) 188.83 H2O(l) 69.91 HF(g) 173.51 HCl(g) 186.69

HBr(g) 198.49 HI(g) 206.3 I2(s) 116.73 I2(g) 260.57 N2(g) 191.50 NH3(g) 192.5 NO(g) 210.62 NO2(g) 240.45 Na(s) 51.45 O2(g) 205.0 S(s) 31.88 SO2(g) 248.5


Relative Standard EntropiesStates

• 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

188.8H2O (l)70.0H2O (g)

S°, (J/mol·K)

Substance


Relative Standard EntropiesMolar Mass

• the larger the molar mass, the larger the entropy

• available energy states more closely spaced, allowing more dispersal of energy through the states


Relative Standard EntropiesAllotropes

• the less constrained the structure of an allotrope is, the larger its entropy


Relative Standard EntropiesMolecular Complexity

• larger, more complex molecules generally have larger entropy

• more available energy states, allowing more dispersal of energy through the states

30.00639.948

MolarMass

210.8NO (g)154.8Ar (g)

S°, (J/mol·K)

Substance


Relative Standard EntropiesDissolution

• dissolved solids generally have larger entropy

• distributing particles throughout the mixture

265.7KClO3(aq)143.1KClO3(s)

S°, (J/mol·K)

Substance

Ex. 17.4 –Calculate ∆S° for the reaction4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l)

∆S is +, as you would expect for a reaction with more gas product molecules than reactant molecules

standard entropies from Appendix IIB∆S, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

∆SS°NH3, S°O2, S°NO, S°H2O,

( ) ( )reactantsproducts SSS ooorp nn Σ−Σ=∆

( ) ( )

KJ

KJ

KJ

KJ

KJ

)(O)(NH)O(H)NO(

reactantsproducts

8.178

)]2.205(5)8.192(4[)]8.188(6) 8.210(4[

)]S(5)S(4[)]S(6)S(4[

SSS

232

=

+−+=

+−+=

Σ−Σ=∆

gggg

rp nnoooo

ooo

188.8H2O(g)210.8NO(g)205.2O2(g)192.8NH3(g)

S°, J/mol⋅KSubstance


Calculating ∆G°

• at 25°C:∆Go

reaction = ΣnGof(products) - ΣnGo

f(reactants)

• at temperatures other than 25°C:assuming the change in ∆Ho

reaction and ∆Soreaction is

negligible∆G°reaction = ∆H°reaction – T∆S°reaction

42

Substance ∆G°f kJ/mol

Substance ∆G°f kJ/mol

Al(s) 0 Al2O3 -1576.5 Br2(l) 0 Br2(g) +3.14

C(diamond) +2.84 C(graphite) 0 CO(g) -137.2 CO2(g) -394.4 Ca(s) 0 CaO(s) -604.17 Cu(s) 0 CuO(s) -128.3 Fe(s) 0 Fe2O3(s) -740.98 H2(g) 0 H2O2(l) -120.4

H2O(g) -228.57 H2O(l) -237.13 HF(g) -270.70 HCl(g) -95.27

HBr(g) -53.22 HI(g) +1.30 I2(s) 0 I2(g) +19.37 N2(g) 0 NH3(g) -16.66 NO(g) +86.71 NO2(g) +51.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -300.4

Ex. 17.7 –Calculate ∆G° at 25°C for the reaction

CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g)

standard free energies of formation from Appendix IIB∆G°, kJ

Solution:

Concept Plan:

Relationships:

Given:Find:

∆G°∆G°f of prod & react

( ) ( )reactantsproducts GGG ooofrfp nn ∆Σ−∆Σ=∆

( ) ( )

kJ 3.148)]kJ 0.0(8)kJ 5.50[(]kJ) 2.163()kJ 6.228(2)kJ 4.394[(

)]G(8)G[()]G()G(2)G[(

GGG

24322 OCHOOHCO

reactantsproducts

−=+−−++−+−=

∆+∆−∆+∆+∆=

∆Σ−∆Σ=∆ooooo

ooo

fffff

frfp nn

163.2O3(g)-228.6H2O(g)-394.4CO2(g)

0.0O2(g)-50.5CH4(g)

∆G°f, kJ/molSubstance

Ex. 17.6 – The reaction SO2(g) + ½ O2(g) → SO3(g) has ∆H° = -98.9 kJ and ∆S° = -94.0 J/K at 25°C.

Calculate ∆G° at 125°C and determine if it is spontaneous.

Since ∆G is -, the reaction is spontaneous at this temperature, though less so than at 25°C

∆H° = -98.9 kJ, ∆S° = -94.0 J/K, T = 398 K

∆G°, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

∆G°T, ∆H°, ∆S°ooo STHG ∆−∆=∆

( ) ( )( )kJ5.61J105.61

0.94K 398J 1098.9

STHG

3KJ3

−=×−=

−−×−=

∆−∆=∆ ooo


∆G Relationships• if a reaction can be expressed as a series of reactions,

the sum of the ∆G values of the individual reaction is the ∆G of the total reaction

∆G is a state function

• if a reaction is reversed, the sign of its ∆G value reverses

• if the amounts of materials is multiplied by a factor, the value of the ∆G is multiplied by the same factor

the value of ∆G of a reaction is extensive


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


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

“lost” as heatif not most

• therefore, real reactions are irreversible


∆G under Nonstandard Conditions⋅ ∆G = ∆G° only when the reactants and products

are in their standard states⋅ there normal state at that temperature⋅ partial pressure of gas = 1 atm⋅ concentration = 1 M

⋅ under nonstandard conditions, ∆G = ∆G° + RTlnQ⋅ Q is the reaction quotient

⋅ at equilibrium ∆G = 0⋅ ∆G° = ─RTlnK

50

Example - ∆G• Calculate ∆G at 427°C for the reaction below if the

PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atmN2(g) + 3 H2(g) → 2 NH3(g)

Q = PNH3

2

PN21 x PH2

3

(2.0 atm)2

(33.0 atm)1 (99.0)3= = 1.2 x 10-7

∆G = ∆G° + RTlnQ∆G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)

∆G = -46300 J = -46 kJ

∆H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J

∆S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K∆G° = -92380 J - (700 K)(-198.2 J/K)

∆G° = +46400 J

52

Example - K• Estimate the equilibrium constant and position of

equilibrium for the following reaction at 427°CN2(g) + 3 H2(g) ⇔ 2 NH3(g)

∆G° = -RT lnK+46400 J = -(8.314 J/K)(700 K) lnK

lnK = -7.97K = e-7.97 = 3.45 x 10-4

since K is << 1, the position of equilibrium favors reactants

∆H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J∆S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K

∆G° = -92380 J - (700 K)(-198.2 J/K)∆G° = +46400 J


Temperature Dependence of K• for an exothermic reaction, increasing the

temperature decreases the value of the equilibrium constant

• for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant

RS

T1

RHln rxnrxn

oo ∆+

∆

−=K

chapter 17 free energy and...

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