Chapter 17 free energy and thermodynamics
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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 17 free energy and thermodynamics

Chapter 17Free Energy and Thermodynamics


First law of thermodynamics the conservation of energy

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

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 thermodynamics1

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

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

Heat Tax

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

  • 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


Chapter 17 free energy and thermodynamics

  • 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

Comparing Potential Energy

direction of spontaneity can be determined by comparing

Tro: Chemistry: A Molecular Approach, 2/e


Reversibility of process

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 process1

Reversibility of Process

  • If a process is spontaneous in one direction,

Tro: Chemistry: A Molecular Approach, 2/e


Thermodynamics vs kinetics

Thermodynamics vs. Kinetics

Tro: Chemistry: A Molecular Approach, 2/e


Diamond graphite

Diamond → Graphite

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

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

  • 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

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


Chapter 17 free energy and thermodynamics

Tro: Chemistry: A Molecular Approach, 2/e


Melting ice

Melting Ice

When a solid melts, the particles have more

Tro: Chemistry: A Molecular Approach, 2/e


Melting ice1

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

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 spontaneous1

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 spontaneous2

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

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 change1

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

  • Entropy= thermodynamic function

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

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

S = k lnW

k = Boltzmann Constant = R/Avogadro’s #

k =

Tro: Chemistry: A Molecular Approach, 2/e


Entropy1

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


Chapter 17 free energy and thermodynamics

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


Macrostates microstates

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


Macrostates microstates1

This macrostate can be achieved through

several different arrangements of the particles

Macrostates → Microstates

Tro: Chemistry: A Molecular Approach, 2/e


Macrostates and probability

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 probability1

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 d s

Changes in Entropy, DS

DS = Sfinal − Sinitial

  • Entropy change is favorable when result is a more

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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

Increases in Entropy

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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

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 change1

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

Entropy Change and State Change

Tro: Chemistry: A Molecular Approach, 2/e


Try this predict whether d s system is or for each of the following

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 17 free energy and thermodynamics part 2

Chapter 17Free Energy and ThermodynamicsPART 2


2 nd law of thermodynamics

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 2 nd law of thermodynamics

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 2 nd law

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 d s of surroundings

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 d s of surroundings1

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 d s of surroundings2

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 d s surroundings

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 d s surroundings1

Heat Exchange and DSsurroundings

  • Amount of Dssurroundingsdepends on original temp

    • higher original temp,

Tro: Chemistry: A Molecular Approach, 2/e


Temp dependence of d s surroundings

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 d s surroundings1

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 d s surroundings

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


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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

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 spontaneity1

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 d g

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 d g1

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


D g d h and d s

DG, DH, and DS

Tro: Chemistry: A Molecular Approach, 2/e


Free energy change and spontaneity

Free Energy Change and Spontaneity

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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:

Answer:

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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:

Answer:

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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:

Answer:

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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:

Answer:

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics part 3

Chapter 17Free Energy and ThermodynamicsPART 3


Standard conditions

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


3 rd law of thermodynamics absolute entropy

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


3 rd law of thermodynamics absolute entropy1

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

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 entropies1

Standard Absolute Entropies

Tro: Chemistry: A Molecular Approach, 2/e


Relative standard entropies states

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

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

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

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 entropies dissolution

Relative Standard EntropiesDissolution

  • Dissolved solids generally have largerentropy

  • Distributing particles throughout mixture

Tro: Chemistry: A Molecular Approach, 2/e


Standard entropy change d s

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


Calculate d s for the reaction 4 nh 3 g 5 o 2 g 4 no g 6 h 2 o g

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 d s for the reaction 2 h 2 g o 2 g 2 h 2 o g

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

Tro: Chemistry: A Molecular Approach, 2/e


Calculate d s for the reaction 2 h 2 g o 2 g 2 h 2 o g

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 d g

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


Chapter 17 free energy and thermodynamics

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:

Answer:

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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

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


Calculate d g at 25 c for the reaction ch 4 g 8 o 2 g co 2 g 2 h 2 o g 4 o 3 g

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


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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


D g relationships

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 d g rxn for the following reaction using the given equations 3 c s 4 h 2 g c 3 h 8 g

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 d g rxn for the following reaction using the given equations 3 c s 4 h 2 g c 3 h 8 g1

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 k 2 h 2 o g o 2 g 2 h 2 o 2 g

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 k 2 h 2 o g o 2 g 2 h 2 o 2 g

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

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

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

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

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


D g under nonstandard conditions

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


Chapter 17 free energy and thermodynamics

Tro: Chemistry: A Molecular Approach, 2/e


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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


D g o and k

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


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

Tro: Chemistry: A Molecular Approach, 2/e


Calculate k at 298 k for the reaction n 2 o 4 g 2 no 2 g

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


Chapter 17 free energy and thermodynamics

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


Chapter 17 free energy and thermodynamics

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

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 dependent1

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


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