we can compare q vs ksp to predict whether a precipitate ... · pdf filepredicting whether a...

Q = Ksp

Q < Ksp

Q > Ksp

Saturated solutionUnsaturated solution No precipitate

Supersaturated solution Precipitate will form

Compare Qsp vs Ksp.

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq)

Ksp = [Ag+]2[CO32-]Q = [Ag+]2[CO3

2-] before equilbrium after equilbrium

We can compare Q vs Ksp to predict whether a precipitate forms.

Predicting Whether a Precipitate Will FormA common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

1) Write out a reaction equation and Ksp 2) Look up the Ksp values in a table (Appendix C).

3) Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp.

4) Remember to consider the final diluted solution when calculating concentrations.

Predicting Whether a Precipitate Will FormA common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11

mol Ca2+ = 0.100 L(0.30 mol/L) = 0.030 mol[Ca2+] = 0.030 mol/0.300 L = 0.10 M

mol F- = 0.200 L(0.060 mol/L) = 0.012 mol[F-] = 0.012 mol/0.300 L = 0.040 M

Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4 Q is >> Ksp and the CaF2 WILL precipitate.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Why Does This Happen With Most Mixtures?

Chapter 20Thermodynamics: Entropy, Free Energy and

the Direction of Chemical Reactions

KMnO4(s) spontaneously mixes and will not un-mix without outside intervention. WHY?

Why do homogeneous mixtures never revert spontaneously to their original form?

Spontaneity refers to any chemical process that appears to proceed “naturally” to a final state without outside intervention (does not mean “now”).

The rusting of a nail

Our goal in this Chapter is to identify those factors which determine whether a chemical reaction will proceed spontaneously.

Freezing and melting of water

T > 0˚C

T < 0˚C

In the 1870’s, it was thought that ΔH determined the “spontaneity of a chemical reaction”. This notion was proved wrong as both exothermic and endothermic reactions are spontaneous (i.e. no outside intervention is needed to make it happen).

Examples of spontaneous reactions:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ΔH0 = -890.4 kJ

H2O (s) H2O (l) ΔH0 = 6.01 kJ

H2O (l) H2O (s) ΔH0 = - 6.01 kJ

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ΔH0 = 25 kJH2O

ΔHrnx does not determine whether a reaction is spontaneous!

Entropy can be described (incorrectly) as a “positionally messy room” or “disordered cards” and described as “disorder”. This is not correct in the chemical sense but it is useful!

How many ways can you have your room or a deck of cards organized vs disorganized?

Entropy is a measure of energy dispersal in a chemical system. Energy tends to be dispersed over a larger number of “ quantum states” in a system of molecules rather than dispered over fewer states.

Increasing the volume decreases energy level spacing increasing entropy by populating more.

Review: The internal energy, E, of a system is a bulk property of matter that describes all energy within a defined system.

• Translational kinetic energy.• Molecular rotation.• Bond vibrations.• Intermolecular attractions.• Electrons.• Nuclear • Electrochemical

Internal Energy is the sum of energies that exists in atoms/molecules.

• Translational kinetic energy.

• Rotational energy

• Vibrational energy

• Electrostatic energy

Molecules translate, vibrate, rotate in quantized states. The more atoms there are, the more ways there are a molecule can vibrate or “wiggle” and store energy (degrees of freedom).

Microscopic: Internal energy is distributed over all the quantized translational, vibronic and rotational energy states in molecules and atoms.

The quantized energy levels are of different energy. Rotational gaps are on the order of the energy of microwave radiation. Vibrational spacing is higher in energy (IR).

Trans Roto Vibro

EnergySpacing

It is the distribution of quanta in these energy levels that is essential to understanding the entropy and the “dispersal of energy” that occurs during a spontaneous chemical reaction.

A state function is a special type of mathematical function that has ONE UNIQUE VALUE between an initial and a final state. That value does not depend on the path taken between the two states.

The potential energy of hiker 1 and hiker 2 is the same even though they took very different paths to get to the mountain top. This is the power of all the state functions in thermodynamics---we only need to now the start and the finish bulk parameters.

PE = mghh

ΔE = Efinal - Einitial

ΔP = Pfinal - Pinitial

ΔV = Vfinal - Vinitial

ΔT = Tfinal - Tinitial

ΔH = Hfinal - Hinitial

All are state functions!

0

Final State

Initial State

The relevance and power of a state function is that there is one unique value between any two states (which represents a change or a reaction) and this change does not depend on the path between the two states.

mgh2

h = 1

Δh

Initial State

Final State

Arbitray State

Two paths up the mountain, one hard--the other easy. Both height and potential energy are state functions in this scenario, but work required, or calories burned, time, or distance traveled to the top are not state functions!

The 1st Law of Thermodynamics says that total energy of a system + surroundings remains constant—i.e. energy is neither created nor destroyed just transformed between the two boundries.

ΔEsystem + ΔEsurroundings = 0

ΔEsystem = -ΔEsurroundings

“what the system gives out.....the surroundings takes and vis versa”!

An Exothermic reaction releases heat to the surroundings, ΔH < 0.

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

H2O (g) H2O (l) + heat

heat + 2HgO (s) 2Hg (l) + O2 (g)

heat + H2O (s) H2O (l)

An Endothermic reaction absorbs heat from the surroundings into the system, ΔH > 0.

ΔH = <0

ΔH = >0

heat written as a product (given off)

heat written as a reactant (consumed)

A B

Our goal is to identify those factors that will allow us to predict whether a chemical reaction will proceed spontaneously (no outside intervention).Suppose a chemical reaction:

The answer is yes and the details are explained in thermodynamic theory.

Can we develop a parameter or a function that can predict whether A ==>B will occur “spontaneously”?

What determines whether a reaction will occur spontaneously? It was thought that ΔH determined the “spontaneity of a chemical reaction”. This notion was proved wrong as both exothermic and endothermic reactions are spontaneous .

Examples of spontaneous reactions:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ΔH0 = -890.4 kJ

H2O (s) H2O (l) ΔH0 = 6.01 kJ

H2O (l) H2O (s) ΔH0 = - 6.01 kJ

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ΔH0 = 25 kJH2O

ΔHrnx does not determine whether a reaction is spontaneous!

Boltzmann founded a thermodynamic state function called entropy, S, that is a measure of the “dispersion or spread of energy” that occurs when all spontaneous reactions occur.

When Wf > Wi then ΔS > 0

When Wf < Wi then ΔS < 0

!S = k ln Wf - k ln Wi

Change in Entropy

S = k ln WEntropy

# of microstates

Boltzman’s constant1.38 × 10−23 J/K.

Boltzmann’s Tomb In Vienna, Austria

There is a thermodynamic state function called entropy, S, that is a measure of the energy dispersal that occurs when a change of state occurs.

-qsys reversible

TΔS =

1. Macroscopic or bulk defintion (it needs no molecules). Represents heat flow.

!S = k (ln Wf - ln Wi)

Change in Entropy

2. Microscopic or Molecular Description

A system can be described by bulk or macroscopic that we can observe and measure, and the microscopic or molecular that we can’t see but can model statistically. You must see both!

Macrostate is the observed state of the system that represents on a molecular level that microstate with the highest probability or number.

Microstate is particular distrubution that corresponds to some macrostate.

Multiplicity is the number of microstates that give a specific macrostate.

Consider 4 labeled molecules A,B,C,D

5-observable macrostates

A,B,CA,B,DA,C,DB,C,D

DCBA

A,B,C,D

Left Side

Right Side

Multiplicity

-

4

A,B C,DA,C B,DA,D B,CB,C A,DB,D A,CC,D A,B

6

1

A,B,C,D- 1

MicrostateBulk PropertyProbability

16

1/16

4/16

6/16

1/16

Macrostate

The number of microstates scales as 2N in a two state system.

1 atom

2 atom

3 atom

NA 2NA

23 = 8

22 = 4

21 = 2

The second law of thermodynamics says that all processes that are spontaneous produce an increase in the entropy of the universe.

ΔSuniv = ΔSsys + ΔSsurr > 0

Note: Entropy is not energy! It is an index or measure of the dispersal of energy that occurs in all spontaneous processes passing from State 1 to State 2.

ΔSuniv = ΔSsys + ΔSsurr = 0

ΔSuniv = ΔSsys + ΔSsurr < 0 2. No spontaneous change

3. Equilibrium condition

1. Criteria for Spontaneous change!

While it appears that entropy focuses on the most statistically favored position distribution entropy is more concerned with the fact that energy levels become more closely spaced and more occupied. Reactions tend towards dispersing energy!

There are many different ways we can

Trans Roto Vibro

EnergySpacing

Matter being dispersed into a larger number of statistical microstates disperses energy into a larger number of thermal energy levels increasing entropy.

0.5 atmevacuated1.0 atm

Gas expandsspontaneously

into larger volume

Energy level Energy level

Quantum mechancis dictates closer energy

level spacing as V increases.

Same amount of energy is dispersed or spread among more

energy levels.

Spontaneous reactions always occur in that direction that maximizes energy dispersal over more microstates. We call this an increase an increase in entropy from some initial state to some final state .

Atoms and molecules with widely-spaced energy levels have fewer ways to distribute or disperse energy.

Wide and few states Close and many states

Energy

The second law of thermodynamics provides the criteria for spontaneous chemical reactions.

The rusting of a nail

Freezing and melting of water

T > 0˚C

T < 0˚C

1. Temperature increases

2. Solid => Liquid => Gas phase changes

3. Dissolution of a solid or liquid

5. Atomic size or molecular complexity

4. Dissolution of a gas decreases entropy.

ΔS increases as temperature rises as more energy states are filled

ΔS increases as phase changes to a more dispersed phase.

Dissolving a solid or liquid into a solvent increases entropy.

A gas becomes less dispersed when it dissolves in a liquid or solid.

In similar substances, increases in mass relate directly to entropy.In allotropic substances, increases in complexity (e.g., bond flexibility) relate directly to entropy.

There are many events that result in a higher number of microstates or higher entropy.

There are many chemical

reactions that lead to an

increase in dispersal of

energy spread over a larger

number microstates

entropy (ΔS > 0)

ΔS > 0

ΔS > 0

ΔS > 0

Dissolution

Dissolution

Mixing

Increasing T

Entropy increases when a substance is heated as thermal energy is distributed over more energy states. Phase changes lead to large increases.

Phase transitions: solid to liquid to gas have very

large increases in entropy!

High TLow T

Temp increases populates more energy levels increasing entropy

TemperatureEn

tropy

(S)

pure solid

pure liquid solution

MIX

When a solid crystalline salt dissociates into a greater number of ions entropy increases as the number of microstates that energy can be dispersed is larger.

ΔSsys > 0

When two miscible liquids combine the most probable microstate (the observed macrostate) is a homogeneous mixture.

The dispersal of matter leads to the dispersal of energy among more energy states in a mixture just as it does when gases expand.

Benzene (C6H6) Toluene (C6H5CH3)

Impossible Result

Solution

ΔSsys > 0

O2 gas

O2 dissolved

Entropy decreases significantly as a gas is dissolved in a liquid, or as a gas condenses to liquid, or as a liquid freezes to a solid.

ΔSsys < 0

1. Substances down a group (increasing mass) have higher absolute entropies.

2. As the number of atoms increase in a molecule absolute entropy tends to increase (more degrees of freedom => more microstate available)

3. As the number of atoms increase in a molecule absolute entropy tends to increase (more degrees of freedom => more microstate available)

Other trends in absolute entropies (see text p.657-660)

How does the entropy of a system change for each of the following processes?

(a) Condensing water vapor to a liquid

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

How does the entropy of a system change for each of the following processes?

(a) Condensing water vapor to a liquiddispersal of energy decreases Entropy decreases (ΔS < 0)

(b) Forming sucrose crystals from a supersaturated solutionEntropy decreases (ΔS < 0)

(c) Heating hydrogen gas from 600C to 800CEntropy increases (ΔS > 0)

(d) Subliming dry ice

Entropy increases (ΔS > 0)

dispersal of energy decreases

dispersal of energy increases

dispersal of energy increases

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Why Does This Happen With Most Mixtures?

Onto Chapter 20 Thermodynamics in Silberberg. P. 864-877 and sample problems 20.1, 20.2. Skip 878 and Sample 20.3 up to p881.Start again Section 20.3 to table p.887. Skip problem 20.6

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Make-Up Exam Comprehensive50 Multiple Choice Questions Coverage: Chapter 12/13/16/17/18/19 OnlyDate: March 17, 2009Time: 6-7:30PMPlace: Berchmans 102 and 103 (same as Exam 3)Why: This will replace your lowest Hour Exam score.

Final Exam Comprehensive Coverage: Chapter 12/13/16/17/18/19/20Date: March 24, 2009

Choose the member with the higher entropy in each of the following pairs, and justify your choice. Assume constant temperature, except in part (e).

(a) 1 mol of SO2(g) or 1 mol of SO3(g)(b) 1 mol of CO2(s) or 1 mol of CO2(g)(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)

(d) 1 mol of KBr(s) or 1 mol of KBr(aq)(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC

(f) 1 mol of CF4(g) or 1 mol of CCl4(g)

SOLUTION:

Choose the member with the higher entropy in each of the following pairs, and justify your choice. Assume constant temperature, except in part (e).

(a) 1 mol of SO2(g) or 1 mol of SO3(g)(b) 1 mol of CO2(s) or 1 mol of CO2(g)(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)

(d) 1 mol of KBr(s) or 1 mol of KBr(aq)(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC

(f) 1 mol of CF4(g) or 1 mol of CCl4(g)

(a) 1 mol of SO3(g) - more atoms

(b) 1 mol of CO2(g) - gas > solid

(c) 3 mol of O2(g) - larger # mols

(d) 1 mol of KBr(aq) - solution > solid

(e) 23 oC - higher temperature

(f) CCl4 - larger mass

A “perfectly” ordered crystal at 0K gives way to a less-structured liquid with more available microstates for internal energy to be dispersed.

3rd Law of Thermodynamics: The entropy, S of a perfect crystal is zero at the 0 Kelvin.

T = 0KS = 0

T > 0 KS > 0

Increasing Temp

Perfect crystal with 1 microstate

The reference point for entropy is called the “zero point energy” at 0K. The entropy of a crystal is 0.

T = 0K

T = 100K

T = 298K

S = 0

S > 0

S > 0

A Mountain Peak

Sea-level

Height > 0

Height = 0

∆H!f = 0

1M concentration

Standand State Reaction Conditions

A “standard state” is a defined set of experimental conditions T, P, [x], phase, denoted symbolically by a superscript degree sign above the thermodynamic functions: ΔH˚, ΔG˚, ΔS˚

ΔHrxn

Non-standard

ΔGrxn

ΔHorxn

ΔGorxn

Standard state

ΔS0rxn= [ dS˚(D)cS˚(C) + ] - bS˚(B)]aS˚(A)[ +

∆Ssys° = ∑ni Si°(products) – ∑niSi°(reactants)

• Standard Molar Entropy, S°rxn: The standard molar entropy of a chemical reaction can be determined from the standard molar entropies of reactants and products found in tables in Handbooks.

S° for each component is found in a table. Also note that S° = 0 for elements (unlike H).

aA + bB cC + dD ΔS0rxn = ?

• standard conditions are 1 atm pressure, 1M and at 25°C (298K).

• Standard entropies tend to increase with increasing molar mass as solids==>liquids==> gases.

• Standard molar entropies of compounds are tabulated just like ∆H˚

What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)

Calculating the standard entropy of reaction, ΔSorxn

What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)

ΔS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

ΔS0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

S0(CO) = 197.9 J/K•molS0(O2) = 205.0 J/K•mol

S0(CO2) = 213.6 J/K•mol

Calculating the standard entropy of reaction, ΔSorxn

From a table of standard entropies we find:

∆Ssys° = ∑ni Si°(products) – ∑niSi°(reactants)

Calculate ΔSorxn for the combustion of 1 mol of

propane at 25 oC.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

SOLUTION: Find standard entropy values in an appropriate table.

ΔSorxn = [(3 mol)(So CO2) + (4 mol)(So H2O)] - [(1 mol)(So C3H8) + (5 mol)(So O2)]

ΔSorxn = [(3 mol)(213.7 J/mol.K) + (4 mol)(69.9 J/mol.K)] - [(1 mol)(269.9

J/mol.K) + (5 mol)(205.0 J/mol.K)]

ΔSorxn = - 374 J/K

∆Ssys° = ∑ni Si°(products) – ∑niSi°(reactants)

The second law of thermodynamics says that all “spontaneous processes” are accompanied by an increase in the entropy of the universe.

ΔSuniv = ΔSsys + ΔSsurr > 0

Entropy is a measure of the dispersal of energy that occurs as a system passes from State 1 to State 2.

ΔSuniv = ΔSsys + ΔSsurr = 0

ΔSuniv = ΔSsys + ΔSsurr < 0 2. No spontaneous change

3. Equilibrium condition

1. Criteria for Spontaneous change!

In a reversible process, the surroundings can add heat to the system, or it can accept it from the system.

ΔSsurr =−Δ sys

Tq −ΔHsys

TΔ sur

T

q= =

-qp

-qsurr

+

_+qp

+qsurr

For an endothermic process: qsys > 0, qsurr < 0, ΔSsurr < 0

For an exothermic process: qsys < 0, qsurr > 0, ΔSsurr > 0

ΔSuniv = ΔSsys + ΔSsurr > 0

Calculate from Standard Entropies of compounds from a table

ΔSsurr = T-qsys −ΔHsys

T=

Substituting -ΔHsys/T for ΔSsurr we get an expression not so useful.

ΔSuniv = ΔSsys + -ΔHsys/T > 0

ΔSosys

= Σ miSofi(products) - Σ niSo

fi(reactants)

This is related to heat lost by system

It is very difficult to measure ΔSuniv, ΔSsys and ΔSsurr so we recast the 2nd Law equation into something more useful called the Gibb’s Free Energy, ΔG.

ΔGuniverse = ΔHsys - TΔSsys

ΔSuniv = ΔSsys + ΔSsurr 1. substitute ΔSsurr = -qsys/T = -ΔHsys/T

ΔSuniv = ΔSsys - ΔHsys/T

-TΔSuniv = - TΔSsys + ΔHsys

3. Define ΔGuniverse = -TΔSuniv

2. multiply by (-T)

Criteria for spontaneity!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Why Does This Happen With Most Mixtures?

Onto Chapter 20 Thermodynamics in Silberberg. P. 864-877 and sample problems 20.1, 20.2. Skip 878 and Sample 20.3 up to p881. Start again Section 20.3 to table p.887. Skip problem 20.6

2,3,12,19,20,25,34,36,46,47,51,55,58,62,64,67,69,83,86,90

Final Exam Comprehensive Date: March 24, 2009Time: 7:30-9:30AM C-109 (watch blog for changes)Coverage: Chapter 12/13/16/17/18/19/2060-multiple choice items

Make-up Exam

Student ID63118 50.00 70352 54.00 71497 78.00 7193680136 64.00 80157 46.00 80167 58.00 80194 78.00 80276 72.00 80301 58.00 80445 74.00 80520 56.00 80534 80.00 80604 42.00 80605 74.00 80724 88.00 80816 72.00 80870 84.00 80949 68.00 81001 68.00 81111 82.00 81192 64.00 81242 80.00 81292 76.00 81466 46.00 81476 56.00 81514 66.00 81554 90.00 81582 68.00 81592 62.00 81697 46.00 81757 86.00 81766 66.00 81793 76.00 81796 62.00 81803 80.00 82043 50.00 82155 84.00 82162 48.00 8221182213 70.00 82368 76.00 82426 80.00 82640 52.00 82680 42.00 82694 80.00 82790 54.00 82804 80.00 82829 52.00 82851 76.00 82865 52.00 82961 82.00 83246 70.00 83297 66.00 83380 66.00 83410 74.00 83506 64.00 83563 62.00 83595 82.00 83608 70.00 83667 72.00 83705 92.00 83765 82.00 83811 74.00 83919 88.00 83927 70.00 84030 42.00 84044 48.00 84089 70.00 84243 56.00 84286 42.00 84370 84.00 84450 52.00 84455 70.00 84514 50.00

Make-up Exam

Student ID63118 50.00 70352 54.00 71497 78.00 7193680136 64.00 80157 46.00 80167 58.00 80194 78.00 80276 72.00 80301 58.00 80445 74.00 80520 56.00 80534 80.00 80604 42.00 80605 74.00 80724 88.00 80816 72.00 80870 84.00 80949 68.00 81001 68.00 81111 82.00 81192 64.00 81242 80.00 81292 76.00 81466 46.00 81476 56.00 81514 66.00 81554 90.00 81582 68.00 81592 62.00 81697 46.00 81757 86.00 81766 66.00 81793 76.00 81796 62.00 81803 80.00 82043 50.00 82155 84.00 82162 48.00 8221182213 70.00 82368 76.00 82426 80.00 82640 52.00 82680 42.00 82694 80.00 82790 54.00 82804 80.00 82829 52.00 82851 76.00 82865 52.00 82961 82.00 83246 70.00 83297 66.00 83380 66.00 83410 74.00 83506 64.00 83563 62.00 83595 82.00 83608 70.00 83667 72.00 83705 92.00 83765 82.00 83811 74.00 83919 88.00 83927 70.00 84030 42.00 84044 48.00 84089 70.00 84243 56.00 84286 42.00 84370 84.00 84450 52.00 84455 70.00 84514 50.00

Make-up Exam

Student ID63118 50.00 70352 54.00 71497 78.00 7193680136 64.00 80157 46.00 80167 58.00 80194 78.00 80276 72.00 80301 58.00 80445 74.00 80520 56.00 80534 80.00 80604 42.00 80605 74.00 80724 88.00 80816 72.00 80870 84.00 80949 68.00 81001 68.00 81111 82.00 81192 64.00 81242 80.00 81292 76.00 81466 46.00 81476 56.00 81514 66.00 81554 90.00 81582 68.00 81592 62.00 81697 46.00 81757 86.00 81766 66.00 81793 76.00 81796 62.00 81803 80.00 82043 50.00 82155 84.00 82162 48.00 8221182213 70.00 82368 76.00 82426 80.00 82640 52.00 82680 42.00 82694 80.00 82790 54.00 82804 80.00 82829 52.00 82851 76.00 82865 52.00 82961 82.00 83246 70.00 83297 66.00 83380 66.00 83410 74.00 83506 64.00 83563 62.00 83595 82.00 83608 70.00 83667 72.00 83705 92.00 83765 82.00 83811 74.00 83919 88.00 83927 70.00 84030 42.00 84044 48.00 84089 70.00 84243 56.00 84286 42.00 84370 84.00 84450 52.00 84455 70.00 84514 50.00

What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)

ΔS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

ΔS0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

S0(CO) = 197.9 J/K•molS0(O2) = 205.0 J/K•mol

S0(CO2) = 213.6 J/K•mol

Calculating the standard entropy of reaction, ΔSorxn

From a table of standard entropies we find:

∆Ssys° = ∑ni Si°(products) – ∑niSi°(reactants)

The Gibb’s Free Energy, ΔGuniverse is the function that indicates whether a reaction will occur spontaneously. The sign of ΔG determines the reaction direction, the magnitude determines the maximum amount of work the system can do (if negative).

ΔGuniverse = ΔHsys - T ΔSsys

Temperature in Kelvin

Gibb’s Free Energy Change

Useless or dispersed energy

Enthalpyterm

ΔGsys = ΔHsys - T ΔSsys

ΔGrxn = ΔHrxn - T ΔSrxn

Don’t be confused by symbols. These all mean the same thing and pertain to the system!

ΔGuniverse = ΔHsys - T ΔSsys}

1. ΔGsys < 0 (ΔSuniv > 0) or for a spontaneous process 2. ΔGsys > 0 (ΔSuniv < 0) for a non-spontaneous process 3. ΔGsys = 0 (ΔSuniv = 0) or for a process at equilibrium

There are 3 Cases for the Sign of ΔGsys:

-ΔG < 0 is the maximum “useful” work that can be produced by a chemical reaction. ΔG = workmax

- For ΔG > 0 (non-spontaneous process) ΔG is the minimum work that must be done on the system to make the process take place.

Use Table of ΔHof and ΔSof

ΔGsys˚

Use Table of ΔGof

ΔHorxn

= Σ miΔHofi (products) - Σ niΔHo

fi(reactants)

ΔGorxn = ΔHo

rxn - TΔSorxn

Using Hess’s Law and thermodynamic data, we can compute ΔGorxn under standard state conditions in two different ways as well.

dΔG˚ (D)fcΔG˚ (C)fΔG˚rxn = [ + ] - bΔG˚ (B)faΔG˚ (A)f[ + ]

ΔSorxn

= Σ miSofi (products) - Σ niSo

fi(reactants)

1 2

Thermodynamic Tables

Substance ΔG°fΔH°

f S° ΔH°f and ΔG°

f have units of kJ/mol (energy/mol)

S° has units of J/mol K

ΔGrxn˚, the standard free energy of a reaction can be computed directly from its formation data, or from ΔHo

f and Sof formation data, ΔGf˚

ΔG°rxn

= Σ ni ΔGif°(products) - Σ mi ΔGjf

°(reactants)= [cΔG°f (C) + dΔG°f (D)] – [aΔG°f (A) + bΔG°f (B)]

aA + bB cC + dD ΔG°rxn = ?

OR

ΔHorxn

= Σ miΔHofi (products) - Σ niΔHo

fi(reactants)

ΔGorxn = ΔHo

rxn - TΔSorxn

ΔSorxn

= Σ miSofi (products) - Σ niSo

fi(reactants)

= [cΔH°f (C) + dΔH°f (D)] – [aΔH°f (A) + bΔH°f (B)]

= [cS°f (C) + dS°f (D)] – [aS°f (A) + bS°f (B)]

Recall: The Standard Enthalpy of Formation, ΔHf°, is

the enthalpy change associated with the formation of 1 mole of compound from its naturally occurring elements under standard state conditions for both.

H2(g) + O2(g) → H2O(l ) ΔHf° = -285.8 kJ/mol

Ag(s) + 1/2Cl2(g) → AgCl(s) ΔHf° = -127.0 kJ/mol

•Oxygen exists as O2 gas at 25 °C ΔHf° = 0

•Carbon exists as solid graphite (C) at 25 °C. ΔHf° = 0

•Sulfur exits as S8 as a solid at 25˚C•Water is H2O(l ) in its standard state (not ice or water vapor).

•Examples:Note: 1 mol product!

Note: fractional coefficients allowed

Elements as they exist in elemental form

We can measure ΔH˚f for all compounds and put them into a table.

ΔHrxn° for any reaction can be calculated using

ΔH˚f data from thermodynamic tables.

aA + bB cC + dD ΔH°rxn = ?

When expanded the equation above translates to this:

!H!rxn =

all products!

i=1

ni!H!fi !

all reactants!

i=1

ni!H!fi

!H!rxn = [c!H!

fC + d!H!fD]! [b!H!

fB + a!H!fA]

Recall ΔHfo and ΔGfo of any element in its most stable form is defined as zero.

Both ΔHfo and ΔGfo of many compounds are tabulated in handbooks. This gives us predictive capability and is someone useful in the real world.

ΔH0 (O2) = 0f

ΔH0 (O3) = 142 kJ/molf

ΔH0 (C, graphite) = 0f

ΔH0 (C, diamond) = 1.90 kJ/molf

ΔH0 (S8, rhombic) = 0 kJ/molfΔH0 (N2) = 0f

ΔH0 (Na(s) = 0f

Calculating ΔGorxn from ΔGo

f valuesUse ΔGo

f values to calculate ΔGorxn for the following

reaction:

PLAN:Look up the values of products and reactants in a thermodynamic table (appendix of your book!) Use the ΔGo

rxn summation equation!

4KClO3(s) 3KClO4(s) + KCl(s)Δ

-303.2 kJ/mol -409.2 kJ/mol-296.3 kJ/molΔGof

KClO3(s) KClO4(s) + KCl(s)

Calculating ΔGorxn from ΔGo

f valuesUse ΔGo

f values to calculate ΔGorxn for the following

reaction:4KClO3(s) 3KClO4(s) + KCl(s)Δ

PLAN:

Use the ΔGorxn summation equation and look up the values of

products and reactants in a thermodynamic table.

ΔGorxn

= Σ miΔGofi (products) - Σ niΔGo

fi (reactants)

ΔGorxn

= [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] - [(4 mol)(-296.3 kJ/mol)]

ΔGorxn

= -134 kJ (reaction is spontaneous as written)

-303.2 kJ/mol -409.2 kJ/mol-296.3 kJ/mol4KClO3(s) 3KClO4(s) + KCl(s)Δ

ΔGof

]

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

What is the standard free-energy change for the following reaction at 25 0C?

Is the reaction spontaneous at 25 0C and what does the free energy tell us?ΔG0 = -6405 kJ < 0 and therefore a spontaneous reaction

ΔG0rxn 6ΔG0 (H2O)f12ΔG0 (CO2)f= [ + ] - 2ΔG0 (C6H6)f[

ΔG0rxn = [ 12(-394.4) + 6(-237.2) ] – [ 2 x 124.5 ] = -6405 kJ

ΔGorxn

= Σ miΔGofi (products) - Σ niΔGo

fi (reactants)

It tells us that under standard state conditions the combustion of propane is product favored and that -6405kJ

is the maxium work that could be done.

Please place your books on the floor!

Quiz 7

Quiz 7

1) Which of the following has the highest entropy?

(a) H2O(l) (b) LiCl(s) (c) H2(g) (d) Ni(s)

2) What possible values of ΔSuniverse and ΔGuniverse does a spontaneous reaction take?

Use Table of ΔHof and ΔSof

ΔGsys˚

Use Table of ΔGof

ΔHorxn

= Σ miΔHofi (products) - Σ niΔHo

fi(reactants)

ΔGorxn = ΔHo

rxn - TΔSorxn

Using thermodynamic tables of data we can easily compute ΔGorxn under standard state conditions in two different ways.

dΔG˚ (D)fcΔG˚ (C)fΔG˚rxn = [ + ] - bΔG˚ (B)faΔG˚ (A)f[ + ]

ΔSorxn

= Σ miSofi (products) - Σ niSo

fi(reactants)

1 2 ✓

The standard free-energy (ΔG0 ) is the free-energy change for the complete reaction of reactants to products with both reactants and products under standard-state conditions.

rxn

!H!rxn = [d!H!

fD + c!H!fC ]! [b!H!

fB + a!H!fA]

!G!rxn = !H!

rxn ! T!S!rxn

!S!rxn = [dS!

fD + cS!fC ]! [bS!

fB + aS!fA]

ΔG0 values are a special hypothetical case of 100% conversion. We are usually more interested in real world non-standard conditions and equilibrium values for ΔG.

aA + bB cC + dD ΔG°rxn = ?

Calculating ΔGo from standard enthalpy and entropy valuesPotassium chlorate, one of the common oxidizing agents in explosives, fireworks and match heads, undergoes a solid-state redox reaction when heated. In this reaction, the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (a disproportionation reaction).

4KClO3(s) 3KClO4(s) + KCl(s)Δ +7 -1+5

Use ΔHof and So values to calculate ΔGo

sys (ΔGorxn) at 25 oC for this reaction.

PLAN: Obtain appropriate thermodynamic data from a data table; insert them into the Gibbs free energy equation and solve.

ΔHorxn

= [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)] - [(4 mol)(-397.7 kJ/mol)]

ΔHorxn

= -144 kJ

ΔH°rxn

= Σ ni ΔHif°(products) - Σ mi ΔHjf

°(reactants)ΔH° = [cΔH°f (C) + dΔH°f (D)] – [aΔH°f (A) + bΔH°f (B)]

ΔSorxn

= [(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] - [(4 mol)(143.1 J/mol.K)]

ΔSorxn

= -36.8 J/K

ΔGorxn

= ΔHorxn - TΔSo

rxn

= -144 kJ - (298 K)(-36.8 J/K)(kJ/103J) = -133 kJ

The reaction is spontaneous or the reaction goes to proceeds to the right.

ΔHorxn

= -144 kJ

ΔSorxn

= Σ miSofi (products) - Σ niSo

fi(reactants)= [cS°f (C) + dS°f (D)] – [aS°f (A) + bS°f (B)]

ΔHorxn TΔSo ΔGo Process Description

- + + -+ - - ++ +- -

spontaneous at all T

nonspontaneous at all T

spontaneous at high T; nonspontaneous at low T

spontaneous at low T; nonspontaneous at high T

Case

1

2

3

4

ΔGosys = ΔHo

sys - TΔSosys

Temperature plays an important role in determining whether a reaction is spontaneous. It determines the spontaneity in 2 of 4 cases.

ΔSosys

+ +-

+--

ΔGo and Kc at 25 oC are inversely related....big K means smaller G and vis versa.

FORW

ARD REACTION

REVERSE REA

CTIO

N

ΔGo (kJ) K Significance

200100

5010

10

-1-10-50-100-200

9 x 10-36

3 x 10-18

2 x 10-9

2 x 10-2

7 x 10-1

11.55 x 101

6 x 108

3 x 1017

1 x 1035

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to the same extent

Forward reaction goes to completion; essentially no reverse reaction

1. ΔGo tells us the maximum amount of work a spontaneous reaction can do (when negative)

2. Q and ΔGrxn are linked Q = Kc, ΔGrxn = 0 and ΔSuniverse = 0 ====> Equilibrium. Q < Kc, ΔGrxn < 0 and ΔSuniverse > 0 ====> Product-Favored Q > Kc, ΔGrxn > 0 and ΔSuniverse < 0 ====> Reactant-Favored

Summary of Key Points:

3. ΔGorxn

gives the position of equilibrium and can be calculated three ways: ΔGo

rxn = Σ miΔGo

fi (products) - Σ niΔGofi (reactants)

ΔGorxn

= ΔHorxn - TΔSorxn

ΔGorxn = -RT ln K 4. ΔGrxn

describes the direction in which a non-standard reaction proceeds to reach equilibrium calculated by: ΔGrxn = ΔGo + RT ln Q

ΔGosys = ΔHo

sys - TΔSosys

1. By inspection only of the chemical equation determine if entropy of the system increases or decreases?2. Calculate ΔH˚rxn 3. Calculate ΔS˚rxn 4. Calculate ΔG˚rxn, using ΔG˚rxn = ΔH˚rxn – TΔS˚rxn5. Caculate ΔG˚rxn using the thermodynamic ΔG˚f data6. Determine if the reaction is spontaneous under standard state conditions (298K, 1M, 1atm)?7. What impact would temperature have on the spontaneity of the reaction assuming constant enthalpy and entropy vs T?8. Calculate the “cross-over” temperature which must be exceeded to make this reaction spontaneous.9. Determine Kp for this reaction.

Consider the reaction of copper(I) oxide with carbon with the following thermodynamic data. .

Cu2O(s) + C(s) ===> 2Cu(s) + CO(g)

ΔG˚fS˚

ΔH˚f -168.6-146.0

93.1

00

5.68

00

33.1

-110.5-137.2197.5J/mol K

kJ/molkJ/mol