Spontaneity of ReactionChapter 16I am a sleepless

Slowfaring eater,Maker of rust and rot

In your bastioned fastenings,I am the crumbler:

tomorrow.– Carl Sandburg

Under

I am a sleeplessSlowfaring eater,

Maker of rust and rotIn your bastioned fastenings,

I am the crumbler:tomorrow.

– Carl SandburgUnder

Key Concepts of Chapter 16

• Identifying Spontaneous Processes.

• Identifying reversible and irreversible processes.

• Entropy and its relation to randomness.

• Second Law of Thermodynamics.

• Predicting Entropy Changes of a Process.

• Third Law of Thermodynamics.

• Relate temperature change to entropy change.

• Calculating change in standard entropy.

• Free energy in terms of enthalpy and entropy.

• Relating free energy change to spontaneity.

• Calculating standard free energy change.

• Relationship between free energy and work.

• Calculating free energy Δ under nonstandard conditions.

The First law of Thermodynamics- energy cannot be created or destroyed only converted from one form to another.

Chemical thermodynamics is the study of energy relationships in chemistry.

From Chapter 8

• Enthalpy– heat transfer between the system and its

surroundings under constant pressure.

– Enthalpy is a guide to whether a reaction is likely to proceed.

– It is not the only factor that determines whether a reaction proceeds.

• Occur without outside intervention

• Have a definite direction.– The reverse process is not spontaneous.

• Temperature has an impact on spontaneity.– Ex: Ice melting or forming

– Ex: Hot metal cooling at room temp.

KI (aq) + Pb(NO3)2 (aq) PbI2(s) + KNO3 (aq)

When mixed Precipitate forms spontaneously.

*It does not reverse itself and become two clear solutions.

• Reversible:System changes state and can be restored by

reversing original process.

Ex: Water (s) Water (l)

• Irreversible:System changes state and must take a different

path to restore to original state.

Ex: CH4 + O2 CO2 + H2O

• Whenever a system is in equilibrium, the reaction can go reversibly to reactants or products (water water vapor at 100 º C).

• In a Spontaneous process, the path between reactants and products is irreversible. (Reverse of spontaneous process is not spontaneous).

*Scrambled eggs don’t unscramble*

- The entropy of the universe always increases in a spontaneous process and remains unchanged in an equilibrium process.

“But ma, it’s not my fault… the universe wants my room like this!”

• A measure of randomness or disorder• S = entropy in J/K·mole• Increasing disorder or increasing

randomness is increasing entropy.• Three types of movement can lead to an

increase in randomness.

Entropy is a state function• Change in entropy of a system

S= Sfinal- Sinitial

• Depends only on initial and final states, and not the pathway.

-S indicates a more ordered state(think: < disorder or - disorder)

Positive (+) S = less ordered state (think: > disorder or + disorder)

Entropy, S - a measure of disorder

Ssolid Sliquid Sgas

If entropy always increases, how can we account for the fact that water spontaneously freezes when placed in the freezer?

• Movement of compressor

+

• Evaporation and condensation of refrigerant

+

• Warming of air around container

Net increase in the entropy of the universe

On the AP exam, you will likely be asked to:

1) predict whether a process leads to an increase in entropy or a decrease in entropy.

2) Determine if ΔS is + or –

3) Determine substances or reactions that have the highest entropy.

Processes that lead to an Increase in Entropy

1) When a solid melts.

2) When a solid dissolves in solution.

3) When a solid or liquid becomes a gas.

4) When the temperature of a substance increases.

5) When a gaseous reaction produces more molecules.

6) If no net change in # of gas molecules, can be + or -, but small.

Predict whether the entropy change is greater than or less than zero for each

of the following processes:

a)Freezing liquid bromine

b)Evaporating a beaker of ethanol at room temperature

c)Dissolving sucrose in water

d)Cooling N2 from 80ºC to 20ºC

S<0

S>0

S>0

S<0

Predict whether the entropy change of the system in each of the following reactions is positive or negative:

1.) Ag+(aq)+ Cl-

(aq)AgCl(s)

2.) NH4Cl(s) NH3(g)+ HCl(g)

3.) H2(g) + Br2(g)2HBr(g)

1)S –

2)S+

3)S?

According to the 2nd law of thermodynamics; the entropy of the universe always increases.

?What if the entire senior class assembles in the

auditorium? Aren’t we decreasing disorder, and therefore

decreasing entropy? If so, how can the second law of thermodynamics

be true?

If we consider the senior class as the system, the ΔS of the system would indeed decrease.

ΔS of the system is –

In order for the students to gather, they would:1) Metabolize food (entropy increase of surroundings)

2) Generate heat (entropy increase of surroundings)

The magnitude of the entropy increase of the surroundings will

always be greater than the entropy decrease of the system.

Theoretical values

Suniverse = Ssystem + Ssurroundings

Suniverse = (-10) + (+20)

Suniverse = +10

+ means entropy increases+ means entropy increases

The same can be considered in a chemical process.

When a piece of metal rusts:

4Fe(s) + O2(g) 2Fe2O3(s)

The entropy of the solid slowly decreases.

Although this is a slow process, it is exothermic, and heat is released into the surroundings causing an overall

increase in entropy of the universe!

Spontaneous Physical and Chemical Processes

• A waterfall runs downhill

• A lump of sugar dissolves in a cup of coffee

• At 1 atm, water freezes below 0 0C and ice melts above 0 0C

• Heat flows from a hotter object to a colder object

• A gas expands in an evacuated bulb

• Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

spontaneous

nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H0 = -890.4 kJ

H+ (aq) + OH- (aq) H2O (l) H0 = -56.2 kJ

H2O (s) H2O (l) H0 = 6.01 kJ

NH4NO3 (s) NH4+(aq) + NO3

- (aq) H0 = 25 kJH2O

Spontaneous reactions at 25 °C

TWO Trends in Nature

• Order Disorder

• High energy Low energy

The Driving Forces

• Energy Factor (ΔH)

• Randomness Factor (ΔS) or more lately the “dispersal” factor.

Entropy (S) is a measure of the randomness or disorder of a system…the tendency to spread energy out – disperse energy.

order SdisorderS

S = Sf - Si

If the change from initial to final results in an increase in randomness

Sf > Si S > 0

For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state

Ssolid < Sliquid << Sgas

H2O (s) H2O (l) S > 0

First Law of Thermodynamics

Energy can be converted from one form to another but energy cannot be created or destroyed.

Second Law of Thermodynamics

The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

Suniv = Ssys + Ssurr > 0Spontaneous process:

Suniv = Ssys + Ssurr = 0Equilibrium process:

The Second Law of Thermodynamics

Suniverse = Ssystem + Ssurroundings

Suniverse > 0 for spontaneous rxn

Suniverse = 0 at equilibrium

- The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

In terms of temperature, how would you describe an object that has an entropy value of 0?

0 KPerfect solid crystal with no motion

Only Theoretical

It is not possible to reach absolute 0!

Entropy of universe is always increasing!

3rd Law of Thermodynamicsthe entropy of a perfect crystalline substance is zero at absolute zero

*Based on 0 entropy as a reference point, and calculations involving calculus beyond the scope of

this course, data has been tabulated for

Standard Molar Entropies

ΔSº

Pure substances, 1 atm pressure, 298 K

Standard Molar Entropies

ΔSº1) Standard molar entropies of elements are not 0 (unlike ΔHºf).

(0 entropy is only theoretical; not really possible)

2) S.M.E of gases > S.M.E of liquids and solids.

(gases move faster than liquids)

3) S.M.E. increase with increasing molar mass.

(more potential vibrational freedom with more mass)

4) S.M.E. increase as the number of atoms in a formula increase.

(same as above)

Calculating the Entropy Change

Sorxn = n So

(products) - m So(reactants)

Units for S

S=J/mol•K

Since we are considering ΔS°

J/K are often used because moles are assumed and cancel in the calculations

when considering standard states.

Calculate the standard entropy change (Sº) for the following reaction at 298K

Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)

Substance Sº(J/mol-K) at 298K

Al 28.32

Al2O3 51.00

H2O(g) 188.8

H2(g) 130.58

So = [2Sº(Al) + 3Sº(H2O)] - [Sº(Al2O3) + 3Sº(H2)]

Sorxn = n So

(products) - m So(reactants)

Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)

Kmol

J

Kmol

J

Kmol

J

Kmol

JS

)58.130(300.51)8.188(3)32.28(2

= 180.4 J/K

Predict the sign of ΔSº of the following reaction.

2SO2(g) + O2(g) 2SO3(g)

Entropy decreases, -

Lets’ Calculate

Calculate the standard entropy change (Sº) for the following reaction at 298K

2SO2(g) + O2(g) 2SO3(g)

Substance Sº(J/mol-K) at 298K

SO2(g) 248.1

SO3(g) 256.7

O2(g) 205.0

ΔSº = -187.8 J K-1

• Spontaneous reactions result in an increase in entropy in the universe.

• Rx’s that have a large and negative tend to occur spontaneously.

• Spontaneity depends on enthalpy, entropy, and temperature.

Provides a way to predict the spontaneity of a reaction using a combination of enthalpy and entropy of a reaction.

If Both T and P are constant, the relationship between G and spontaneity is:

1) G is (-), forward rxn is spontaneous.

2) G is 0, rxn is at equilibrium.3) G is (+) forward rxn is not

spontaneous (requires work) reverse rxn is spontaneous.

)reactants()products( ffrxn GmGnG

rxnG(sum of standard free energies of formation of products)

minus(sum of standard free energies of formation of reactants)

the sum of Coefficients from equation

The values of ΔGºf of elements in their most stable form is 0, just as with enthalpy of formations.

Calculate the ΔG°rxn for the combustion of methane at 298K and determine if the reaction is spontaneous.

Substance Gº(KJ/mol) at 298K

CH4 -50.8

CO2 -394.4

H2O(l) -237.2

-818.0 KJ Spontaneous

Calculate the (Gº) for the thermite reaction (aluminum with iron(III) oxide).

Substance Gº(KJ/mol) at 298K

Al2O3 (s) -1576.5

Fe2O3 (s) -740.98

-835.5 KJ spontaneous

• This equation allows us to determine if a process provides energy to do work. A spontaneous reaction in the forward direction provides energy for work.

• If not spontaneous, ΔG equals the amount of energy needed to initiate the reaction.• Allows us to calculate the value of ΔG as temperature changes.• Gibbs free energy (G°) is a state function defined as:

ΔG° = ΔH° – TΔS° (Given on AP Exam)

T is the absolute temperature ΔG° = ΔH° – TΔS° ΔG = ΔH – TΔS (when nonstandard) If given a temperature change and asked to determine spontaneity or value

of ΔG, this is the equation you would use.Value of ΔG tells us if a reaction is spontaneous.

G = H – T(S)

If we know the conditions of ΔH and ΔS, we can predict the sign of G.

We will see that:Two conditions always produce the same result, andtwo conditions depend on temperature.

Predicting Sign of ΔG in Relation to Enthalpy and Entropy

ΔH ΔS ΔG

- +

+ -

- -

+ +

Always negative (spontaneous)

Always positive (nonspontaneous)

Neg. (spontaneous) at low temp

Pos. (nonspontaneous) at high temp.

Pos. (nonspontaneous) at low temp

Neg. (spontaneous) at high temp.

ΔH ΔS ΔG

- +

+ -

- -

+ +

Different sign, not temperature dependent.

Same sign, temperature dependent.

Freezing

Melting

Some reactions are spontaneous because they give off energy in the form of heat (ΔH < 0).

Others are spontaneous because they lead to an increase in the disorder of the system (ΔS > 0).

Calculations of ΔH and ΔS can be used to probe the driving force behind a particular reaction.

G = H – T(S)

Example – The entropy change of the system is negative for the precipitation reaction:

Ag+(aq) + Cl-

(aq) AgCl(s)

Ho = -65 kJSince S decreases rather than increases in this reaction, why is this reaction spontaneous?

Yes, entropy increases, which goes against the second law. However, in this case, the entropy decrease is minimal compared to the magnitude of change in enthalpy. Therefore, the release of heat drives the reaction to stability, which is why it is spontaneous.

Theoretical Values

G = H – TSG = -65 – 298(-.030) = -73.94

Problem

For a certain reaction, ΔHº = -13.65 KJ and ΔSº = -75.8 J K-1.

A) What is ΔGº at 298 K?

B) Will increasing or decreasing the temperature make the reaction become spontaneous? If so, at what temperature will it become spontaneous?

At 298 K the free energy is

ΔG° = ΔH° – TΔS°

ΔG° = -13.65 KJ – 298(-.0758 KJ K-1) = +8.94 KJ

(Reaction is not spontaneous at 298 K)

Given: ΔHº = -13.65 KJ ΔSº = -75.8 J /K T = 298 K

A) What is ΔGº at 298 K?

Because enthalpy and entropy have the same signs, spontaneity is indeed temperature dependent.

*Since going from not spontaneous to spontaneous crosses the point of equilibrium, and ΔG° = 0 at equilibrium, we can make ΔG° = to 0 to find the temperature at which equilibrium is crossed.

B) Will increasing or decreasing the temperature make the reaction become spontaneous? If so, at what temperature will it become spontaneous?

0 = ΔH° – TΔS°

0 = -13.65 KJ – T(-.0758 KJ K-1)

T = 13.65 KJ 0.0758 KJ K-1

T = 180 K

Reaction is spontaneous below 180 K, not spontaneous above 180 K

Entropy Changes in the System (Ssys)

aA + bB cC + dD

S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

S0rxn nS0(products)= mS0(reactants)-

The standard entropy of reaction (S0rxn) is the entropy change for

a reaction carried out at 1 atm and 250C.

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

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

S0(CO2) = 213.6 J/K•mol

S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

S0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

Entropy Changes in the System (Ssys)

When gases are produced (or consumed)

• If a reaction produces more gas molecules than it consumes, S0 > 0.

• If the total number of gas molecules diminishes, S0 < 0.

• If there is no net change in the total number of gas molecules, then S0 may be positive or negative BUT S0 will be a small number.

What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s)

The total number of gas molecules goes down, S is negative.

Entropy Changes in the Surroundings (Ssurr)

Exothermic ProcessSsurr > 0

Endothermic ProcessSsurr < 0

Third Law of Thermodynamics

The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

Gibb’s Free Energy --The energy of a system that is available to do work.

• ΔGorxn = ΔHo

rxn – TΔSorxn .

• Hess’ Law also applies:

ΔGrxn = ΣΔGfo(products) – ΣGf

o (reactants)

• ΔGf

o has a similar definition as ΔHfo

• If ΔGorxn is positive, the rxn is nonspontaneous at that

temperature.• If ΔGo

rxn is negative, the rxn is spontaneous at that temperature.

• If ΔGrxn is zero, the rxn is at equillibrium.

Suniv = Ssys + Ssurr > 0Spontaneous process:

Suniv = Ssys + Ssurr = 0Equilibrium process:

Gibbs Free Energy

For a constant-temperature process:

G = Hrxn -TSrxn

Gibbs free energy (G)

G < 0 The reaction is spontaneous in the forward direction.

G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.

G = 0 The reaction is at equilibrium.

aA + bB cC + dD

G0rxn dG0 (D)fcG0 (C)f= [ + ] - bG0 (B)faG0 (A)f[ + ]

G0rxn nG0 (products)f= mG0 (reactants)f-

The standard free-energy of reaction (Gorxn ) is the free-energy

change for a reaction when it occurs under standard-state conditions.

Standard free energy of formation (Gf

o) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

Go of any element in its stable form is zero.

f

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

G0rxn nG0 (products)f= mG0 (reactants)f-

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

G0rxn 6G0 (H2O)f12G0 (CO2)f= [ + ] - 2G0 (C6H6)f[ ]

G0rxn = [ 12(–394.4) + 6(–237.2) ] – [ 2(124.5) ] = -6405 kJ

Is the reaction spontaneous at 25 0C?

G0 = -6405 kJ < 0

spontaneous

ΔGfo = ΔHf

o - TΔSfo

ΔGfo ΔHf

o TΔSfo

+ +

++

- -

--

+@ all T

-@ high T

-@ low T

-@ all T

??

??

G = H - TS

CaCO3 (s) CaO (s) + CO2 (g)

H0 = 177.8 kJ

S0 = 160.5 J/K

G0 = H0 – TS0

At 25 0C, G0 = 130.0 kJ

G0 = 0 at 835 0C

Temperature and Spontaneity of Chemical Reactions

Gibbs Free Energy and Phase Transitions

H2O (l) H2O (g)

G0 = 0 = H0 – TS0

S = TH

= 40.79 kJ373 K

= 109 J/K

T = ΔH/ ΔS

Gibbs Free Energy and Chemical EquilibriumNon-standard Conditions:

G = Go + RT lnQR is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = Go + RT lnK

Go = RT lnK

Go = RT lnK

High Entropy Low Entropy

TS = H - G

Chemistry In Action: The Thermodynamics of a Rubber Band