# Chapter 17 Free Energy and Thermodynamics - PowerPoint PPT Presentation

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Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics: The Conservation of Energy. You can’t win! First Law of Thermodynamics : Energy cannot be created or destroyed E can transfer from one place to another. 1st Law of Thermodynamics. Conservation of E

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Chapter 17 Free Energy and Thermodynamics

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## Chapter 17Free Energy and Thermodynamics

### First Law of Thermodynamics:The Conservation of Energy

• You can’t win!

• First Law of Thermodynamics:

Energycannot be created or destroyed

E can transfer from one place to another

Tro: Chemistry: A Molecular Approach, 2/e

### 1st Law of Thermodynamics

• Conservation of E

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

• 2 ways energy is “lost” from a system

Tro: Chemistry: A Molecular Approach, 2/e

### 1st Law of Thermodynamics

• E conservation requires that

E D in system = heat released + work done

• DE is a state function

Tro: Chemistry: A Molecular Approach, 2/e

### The Energy Tax

• You can’t break even!

• To recharge a battery with 100 kJ of useful E will require

Every E transition  a “loss” of E

“Energy Tax” demanded by nature & conversion of E to heat

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Tax

Tro: Chemistry: A Molecular Approach, 2/e

• Thermodynamics predicts if a process will occur under given conditions = spontaneous

• Non-spontaneous processes require

• Can you name some non-spontaneous rxtns?

Tro: Chemistry: A Molecular Approach, 2/e

• Determine “Spontaneity” by comparing chem PE of system before rxtn with free E of system after rxtn

• if system afterrxtn has less PE than beforerxtn, then rxtn is thermodynamically favorable.

Tro: Chemistry: A Molecular Approach, 2/e

### Comparing Potential Energy

direction of spontaneity can be determined by comparing

Tro: Chemistry: A Molecular Approach, 2/e

### Reversibility of Process

• Any spontaneous process is irreversible (net release of E in that direction)

• it will proceed in

• A reversible process will proceed back & forth btwn the two end conditions

• any reversible process is at ____________

• results in

Tro: Chemistry: A Molecular Approach, 2/e

### Reversibility of Process

• If a process is spontaneous in one direction,

Tro: Chemistry: A Molecular Approach, 2/e

### Thermodynamics vs. Kinetics

Tro: Chemistry: A Molecular Approach, 2/e

### Diamond → Graphite

Graphite more stable than diamond, so conversion of diamond  graphite isspontaneous

Tro: Chemistry: A Molecular Approach, 2/e

• Spontaneous processes occur because

• Most spontaneous processes go from a system of higherPE to lowerPE

e.g.

Tro: Chemistry: A Molecular Approach, 2/e

### Spontaneous Processes

• Some spontaneous processes proceed from a system of lowerPE to a system at higherPE

e.g.

• How can something absorb PE,

yet have a net release of E?

Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

### Melting Ice

When a solid melts, the particles have more

Tro: Chemistry: A Molecular Approach, 2/e

### Melting Ice

More freedom of motion __________________________ of the system.

When systems become more random,

E is _______________

This E=

Tro: Chemistry: A Molecular Approach, 2/e

### Factors Affecting Whether a Reaction Is Spontaneous

Two factors determine if a rxtn is spontaneous.

Tro: Chemistry: A Molecular Approach, 2/e

### Factors Affecting Whether a Reaction Is Spontaneous

• enthalpy change, DH, =

difference in sum of

internal E & PV work E of reactants

compared to

internal E & PV work E of products

Tro: Chemistry: A Molecular Approach, 2/e

### Factors Affecting Whether a Reaction Is Spontaneous

• entropy change, DS,

is difference in “randomness”

of reactants compared to products

Tro: Chemistry: A Molecular Approach, 2/e

### Enthalpy Change

• DH generally measured in

• Stronger bonds =

• A rxtn is generally exothermic if bonds in products are

• exothermic = E released

Tro: Chemistry: A Molecular Approach, 2/e

### Enthalpy Change

• A rxtn is generally endothermic if bonds in products are weaker than bonds in reactants

endothermic = E absorbed

• DS is favorable for exothermic reactions and unfavorable for endothermic reactions

Tro: Chemistry: A Molecular Approach, 2/e

### Entropy

• Entropy= thermodynamic function

• Entropy, S, increases as # of energetically equivalent ways of arranging components increases

Tro: Chemistry: A Molecular Approach, 2/e

S = k lnW

k = Boltzmann Constant = R/Avogadro’s #

k =

Tro: Chemistry: A Molecular Approach, 2/e

### Entropy

• S = k lnW

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

W = (W is unitless) =

• Random systems require

Tro: Chemistry: A Molecular Approach, 2/e

These are energetically equivalent states for expansion of a gas.

### W

In terms of PE, doesn’t matter whether molecules are all in one flask, or evenly distributed

But one of these states is more probable than other two

Tro: Chemistry: A Molecular Approach, 2/e

These microstates all have the same macrostate

So there are six different particle arrangements that result in the same macrostate

### Macrostates → Microstates

Tro: Chemistry: A Molecular Approach, 2/e

This macrostate can be achieved through

several different arrangements of the particles

### Macrostates → Microstates

Tro: Chemistry: A Molecular Approach, 2/e

### 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

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

Tro: Chemistry: A Molecular Approach, 2/e

### Macrostates and Probability

Therefore State C has higher entropy than either State A or State B

There is 6x probability of having the State C macrostate than either State A or State B

Tro: Chemistry: A Molecular Approach, 2/e

### Changes in Entropy, DS

DS = Sfinal − Sinitial

• Entropy change is favorable when result is a more

Tro: Chemistry: A Molecular Approach, 2/e

Some D’s that increase entropy are:

• rxtns whose products are more random state

e.g.

• rxtns that have larger #’s of product molecules than reactant molecules

Tro: Chemistry: A Molecular Approach, 2/e

### Increases in Entropy

Tro: Chemistry: A Molecular Approach, 2/e

### DS

For a process where final condition is more random than initial condition,

DSsystemis ________________ and

entropy D is ___________________ for

the process to be spontaneous

Tro: Chemistry: A Molecular Approach, 2/e

### DS

For a process where final condition is more orderly than initial condition, DSsystemis ____________and

entropy change is ___________________

for the process to be spontaneous

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

Tro: Chemistry: A Molecular Approach, 2/e

### Entropy Change in State Change

• When materials D state, # of macrostates it can have D’s as well

• more degrees of freedom molecules have,

more macrostates are

possible

Tro: Chemistry: A Molecular Approach, 2/e

### Entropy Change in State Change

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

Tro: Chemistry: A Molecular Approach, 2/e

### Entropy Change and State Change

Tro: Chemistry: A Molecular Approach, 2/e

### Try this: 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)

Tro: Chemistry: A Molecular Approach, 2/e

## Chapter 17Free Energy and ThermodynamicsPART 2

### 2nd Law of Thermodynamics

• says that total entropy D of universe must be positive for a process to be spontaneous

• for irreversible (spontaneous) process

• for reversible process

• DSuniverse = DSsystem + DSsurroundings

Tro: Chemistry: A Molecular Approach, 2/e

### The 2nd Law of Thermodynamics

• DSuniverse = DSsystem + DSsurroundings

• If entropy of system ____________________, then entropy of surroundings must increase by ___________________________

when DSsystem is __________________, DSsurroundingsmust be ____________________ for a spontaneous process

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Flow, Entropy, and the 2nd Law

When ice is placed in water, heat flows from

2nd Law: heat must flow from water to ice because it results in

Entropy of universe

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Transfer &DS of Surroundings

• 2nd Law demands that Suniverseincrease for a spontaneous process

• Yet processes like water vapor condensing are spontaneous, even though water vapor is more random than liquid water.

• What’s driving that process then?

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Transfer &DS of Surroundings

• If a process is spontaneous, yet DS of process is unfavorable, there must have been

• Entropy increase must heat released by system:

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Transfer &DS of Surroundings

When the DSsystem is unfavorable (neg)

DSsurroundings must be favorable (pos), and large

to allow process to be spontaneous

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Exchange and DSsurroundings

• When a process is exothermic, heat is added to surroundings, increasing Ssurroundings

• When a process is endothermic, it takes heat from surroundings, decreasing Ssurroundings

Tro: Chemistry: A Molecular Approach, 2/e

### Heat Exchange and DSsurroundings

• Amount of Dssurroundingsdepends on original temp

• higher original temp,

Tro: Chemistry: A Molecular Approach, 2/e

### Temp Dependence of DSsurroundings

• When heat is added to surroundings that are cool greater effect on DS than it would have if the surroundings were already hot

Tro: Chemistry: A Molecular Approach, 2/e

### Temp Dependence of DSsurroundings

• Water freezes spontaneously below 0 °C because heat released on freezing increases Ssurroundings enough to make

• above 0 °C increase in Ssurroundings is

Tro: Chemistry: A Molecular Approach, 2/e

### Quantifying DSsurroundings

• DSsurroundings is a amount of heat gained or lost

qsurroundings = −qsystem

• DSsurroundings is also

• At constant P and T, overall relationship is

Tro: Chemistry: A Molecular Approach, 2/e

T, DH

DS

### Calculate DSsurroundings at 25oC for rxtn below C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)DHrxn= −2044 kJ

Given:

Find:

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

DSsurroundings, J/K

Conceptual Plan:

Relationships:

Solution:

Check:

Tro: Chemistry: A Molecular Approach, 2/e

Try this: The rxtn below has DHrxn = +66.4 kJ at 25 °C. (a) Determine Dssurroundings, (b) sign of Dssystem (careful! Check # particles)(c) whether process is spontaneous2 O2(g) + N2(g) 2 NO2(g)

Tro: Chemistry: A Molecular Approach, 2/e

2 O2(g) + N2(g)  2 NO2(g) has DHrxn = +66.4 kJ at 25 °C. Calculate DSsurroundings. Determine sign of DSsystem and whether reaction is spontaneous.

Given:

Find:

DHsys = +66.4 kJ, T = 25 ºC = 298 K

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

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### Gibbs Free Energy and Spontaneity

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

• Gibbs Free Energy, G,= max amount of work energy that can be released to surroundings by a system

• for a

• Gibbs Free Energy = Chem Potential because it is analogous to storing of energy in a mechanical system

Tro: Chemistry: A Molecular Approach, 2/e

### Gibbs Free Energy and Spontaneity

−TDSuniv = DHsys−TDSsys

Because DSuniv determines if a process is spontaneous,

DGalso determines spontaneity

DSuniv is (+) when spontaneous,

Tro: Chemistry: A Molecular Approach, 2/e

### Gibbs Free Energy, DG

• A process is spontaneous whenDG is neg

• DG will be negative when:

• DH is neg and DS is pos

• DH is neg and large and DS is neg but small

• DH is pos but small and DS is pos and large

Tro: Chemistry: A Molecular Approach, 2/e

### Gibbs Free Energy, DG

• DG is positive (non-spontaneous) when

DH is ________ andDS is _____________

• never spontaneous at any temp

• When DG = 0 reaction is at ______________

Tro: Chemistry: A Molecular Approach, 2/e

### DG, DH, and DS

Tro: Chemistry: A Molecular Approach, 2/e

### Free Energy Change and Spontaneity

Tro: Chemistry: A Molecular Approach, 2/e

T, DH, DS

DG

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.

Given:

Find:

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

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

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. Calculate DGand determine if it is spontaneous.

Tro: Chemistry: A Molecular Approach, 2/e

Al(s) + Fe2O3(s) Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Calculate DG and determine if it is spontaneous.

T, DH, DS

DG

Given:

Find:

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

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

DG, DH, DS

T

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.

Given:

Find:

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

T, K

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

Practice – The reaction Al(s) + Fe2O3(s) Fe(s) + Al2O3(s)has DH = −847.6 kJ and DS = −41.3 J/K. Calculate maximum temperature it will be spontaneous.

Tro: Chemistry: A Molecular Approach, 2/e

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. Determine the maximum temperature it is spontaneous.

DG,DH, DS

T

Given:

Find:

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

T, C

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

## Chapter 17Free Energy and ThermodynamicsPART 3

### Standard Conditions

Standard state= state of a material at a defined set of conditions:

• Gas = pure gas at exactly ______________

• Solid or Liquid = pure solid or liquid in its ________________________ at exactly 1 atm pressure and temp of interest (usually 25 °C or 298 K)

• Solution = substance in a solution with concentration ____________

Tro: Chemistry: A Molecular Approach, 2/e

### 3rd Law of Thermodynamics:Absolute Entropy

• absolute entropy= amount of E it has due to dispersion of energy through its particles

Tro: Chemistry: A Molecular Approach, 2/e

### 3rd Law of Thermodynamics:Absolute Entropy

• 3rd Law states: a perfect crystal at absolute zero: absolute entropy = 0 J/mol∙K

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

• So, absolute entropy of substances is

Tro: Chemistry: A Molecular Approach, 2/e

### Standard Absolute Entropies

• Extensive (what does that mean?)

• Entropies for 1 mole at 298 K for one state,

• a particular allotrope,

• particular molecular complexity,

• a particular molar mass,

• and a particular degree of dissolution

Tro: Chemistry: A Molecular Approach, 2/e

### Standard Absolute Entropies

Tro: Chemistry: A Molecular Approach, 2/e

### Relative Standard Entropies:States

• gas state has a larger S than liquid state at a particular temp

• liquid state has a larger S than solid state at a particular temp

Tro: Chemistry: A Molecular Approach, 2/e

### Relative Standard Entropies:Molar Mass

• larger the molar mass, the larger the entropy

• Available E states more closely spaced, allowing more dispersal of E through the states

Tro: Chemistry: A Molecular Approach, 2/e

### Relative Standard Entropies: Allotropes

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

• layers in graphite are not bonded together, makes it less constrained

Tro: Chemistry: A Molecular Approach, 2/e

### Relative Standard Entropies:Molecular Complexity

• Larger, more complex molecules generally have larger entropy

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

Tro: Chemistry: A Molecular Approach, 2/e

### Relative Standard EntropiesDissolution

• Dissolved solids generally have largerentropy

• Distributing particles throughout mixture

Tro: Chemistry: A Molecular Approach, 2/e

### Standard Entropy Change, DS

• is the diff in absolute entropy btwn reactants & products under standard conditions

DSºreaction = (∑npSºproducts) − (∑nrSºreactants)

remember:

though DHf° of an element is 0 kJ/mol,

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

Tro: Chemistry: A Molecular Approach, 2/e

SNH3, SO2, SNO, SH2O

DS

### Calculate DS for the reaction 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Given:

Find:

standard entropies from Appendix IIB

DS, J/K

Conceptual Plan:

Relationships:

Sol’n:

Check:

Tro: Chemistry: A Molecular Approach, 2/e

### Practice – Calculate the DS for the reaction2 H2(g) + O2(g)  2 H2O(g)

Tro: Chemistry: A Molecular Approach, 2/e

SH2, SO2, SH2O

DS

### Calculate DS for the reaction2 H2(g) + O2(g) 2 H2O(g)

Given:

Find:

standard entropies from Appendix IIB

DS, J/K

Conceptual Plan:

Relationships:

Solution:

Check:

Tro: Chemistry: A Molecular Approach, 2/e

### Calculating DG

• At 25 C

DGoreaction = SnDGof(products) - SnDGof(reactants)

• At temps other than 25 C

assuming DHoreaction & DSoreaction is negligible

Tro: Chemistry: A Molecular Approach, 2/e

T, DH, DS

DG

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.

Given:

Find:

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

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### Practice – Determine free energy change in the following reaction at 298 K2 H2O(g) + O2(g) 2 H2O2(g)

Tro: Chemistry: A Molecular Approach, 2/e

### Standard Free Energies of Formation

• Free energy of formation (DGf°) = change in G 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 _____

Tro: Chemistry: A Molecular Approach, 2/e

DGf° of prod & react

DG

### Calculate DG at 25 C for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)

Given:

Find:

standard free energies of formation from Appendix IIB

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### Practice – Determine the free energy change in the following reaction at 298 K2 H2O(g) + O2(g) 2 H2O2(g)

Tro: Chemistry: A Molecular Approach, 2/e

DGf° of prod & react

DG

### Determine the free energy change in the following reaction at 298 K (standard temp)2 H2O(g) + O2(g) 2 H2O2(g)

Given:

Find:

standard free energies of formation from Appendix IIB

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### DG Relationships

• If a rxtn can be expressed as a series of rxtns, the sum of DG values of the individual rxtn is the DG of the total rxtn

• DG is a state function

• If a rxtn is reversed, ______________ of its DG value reverses

• If amount of materials is multiplied by a factor, value of DG is multiplied by _______________

• value of DG of a rxtn is ________________

Tro: Chemistry: A Molecular Approach, 2/e

### Find DGºrxn for the following reaction using the given equations3 C(s) + 4 H2(g) C3H8(g)

Tro: Chemistry: A Molecular Approach, 2/e

### Find DGºrxn for the following reaction using the given equations3 C(s) + 4 H2(g) C3H8(g)

• Solution: 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

Tro: Chemistry: A Molecular Approach, 2/e

### Determine free energy change in the following rxtn at 298 K2 H2O(g) + O2(g) 2 H2O2(g)

Tro: Chemistry: A Molecular Approach, 2/e

### Determine the free energy change in the following reaction at 298 K2 H2O(g) + O2(g) 2 H2O2(g)

Tro: Chemistry: A Molecular Approach, 2/e

### What’s “Free” About Free Energy?

• free E is max amount of E released from a system that is available to do ______ on surroundings

• For many exothermic reactions, some heat released (from DH) goes into increasing

• Even some G is generally lost to heating up surroundings

Tro: Chemistry: A Molecular Approach, 2/e

### Free Energy of an Exothermic Reaction

• C(s, graphite) +2 H2(g) → CH4(g)

• DH°rxn = −74.6 kJ = ______________

• DS°rxn = −80.8 J/K = _______________

• DG°rxn = −50.5 kJ = _____________

Tro: Chemistry: A Molecular Approach, 2/e

### Free Energy and Reversible Reactions

• DG is a theoretical limit of amount of work that can be done

• If rxtn achieves its theoretical limit, it is a _______________________________

Tro: Chemistry: A Molecular Approach, 2/e

### Real Reactions

• In a real reaction, some of the free energy is “lost” as heat (if not most)

• Therefore, real reactions are irreversible

Tro: Chemistry: A Molecular Approach, 2/e

### DG under Nonstandard Conditions

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

• normal state at that temp

• partial pressure of gas = 1 atm

• concentration = 1 M

• Under non-standard conditions, DG = DG + RTlnQ

• At equilibrium

Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

DG, PNO, PO2, PNO2

DG

### Calculate DG at 298 Kfor the reaction under the given conditions2 NO(g) + O2(g) 2 NO2(g)DGº = −71.2 kJ

Given:

Find:

non-standard conditions, DGº, T

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

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

Tro: Chemistry: A Molecular Approach, 2/e

DG, PNO, PO2, PNO2

DG

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

Given:

Find:

non-standard conditions, DGº, T

DG, kJ

Conceptual Plan:

Relationships:

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### DGo and K

• Because DGrxn = 0 at equilibrium, then

DGº = −RTln(K)

• When K < 1, DGo is (+) and rxtnisspontaneous in the reverse direction under standard conditions

Tro: Chemistry: A Molecular Approach, 2/e

DGº = −RTln(K)

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

• When K = 1, DGo is 0 and rxtn is at ______________ under standard conditions

Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

DGf of prod & react

DG

K

### Calculate K at 298 Kfor the reaction N2O4(g) 2 NO2(g)

Given:

Find:

standard free energies of formation from Appendix IIB

K

Conceptual Plan:

Relationships:

DGº = −RTln(K)

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### Practice – Estimate the equilibrium constant for the given reaction at 700 K N2(g) + 3 H2(g)® 2 NH3(g)DGo = +46.4 kJ

Tro: Chemistry: A Molecular Approach, 2/e

DG

K

### Estimate the Equilibrium Constant for the given reaction at 700 K N2(g) + 3 H2(g)® 2 NH3(g)DGº = +46.4 kJ

Given:

Find:

DGº

K

Conceptual Plan:

Relationships:

DGº = −RTln(K)

Solution:

Tro: Chemistry: A Molecular Approach, 2/e

### Why Is the Equilibrium Constant Temperature Dependent?

• Combining these two equations

• DG° = DH° − TDS°

• DG° = −RTln(K)

• It can be shown that

Tro: Chemistry: A Molecular Approach, 2/e

### Why Is the Equilibrium Constant Temperature Dependent?

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

• The graph of ln(K)vsinverse T is a straight line with slope and y-intercept

Tro: Chemistry: A Molecular Approach, 2/e