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Spontaneity of Reaction. Chapter 16. 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.

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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.


From Chapter 8

Chemical thermodynamicsis the study of energy relationships in chemistry.

The First law of Thermodynamics

- energy cannot be created or destroyed only converted from one form to another.


  • 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.


Spontaneous Processes

  • 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 & Irreversible

  • 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 reversiblyto reactants or products (water  water vapor at 100 º C).

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

    *Scrambled eggs don’t unscramble*


The Second Law

of Thermodynamics

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



Entropy (S) 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
Entropy is a state function like this!”

  • 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 like this!”

- a measure of disorder

Ssolid

Sliquid

Sgas


Increasing Entropy like this!”


Increasing Entropy like this!”


Increasing Entropy like this!”


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:

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

  • Determine if ΔS is + or –

  • Determine substances or reactions that have the highest entropy.


  • Processes that lead to an Increase in Entropy

  • When a solid melts.

  • When a solid dissolves in solution.

  • When a solid or liquid becomes a gas.

  • When the temperature of a substance increases.

  • When a gaseous reaction produces more molecules.

  • 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:

  • Freezing liquid bromine

  • Evaporating a beaker of ethanol at room temperature

  • Dissolving sucrose in water

  • 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)S –

2)S+

3)S?

1.) Ag+(aq)+ Cl-(aq)AgCl(s)

2.) NH4Cl(s) NH3(g)+ HCl(g)

3.) H2(g) + Br2(g)2HBr(g)


According to the 2 the following reactions is positive or negative:nd 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 the following reactions is positive or negative:

  • ΔS of the system would indeed decrease.

  • ΔS of the system is –

  • In order for the students to gather, they would:

  • Metabolize food (entropy increase of surroundings)

  • 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 the following reactions is positive or negative:

Suniverse = Ssystem+Ssurroundings

Suniverse = (-10) + (+20)

Suniverse = +10

+ means entropy increases


The same can be considered in a chemical process. the following reactions is positive or negative:

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!


nonspontaneous the following reactions is positive or negative:

spontaneous

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 the following reactions is positive or negative:

nonspontaneous


CH the following reactions is positive or negative:4(g) + 2O2(g) CO2(g) + 2H2O (l)DH0 = -890.4 kJ

H+(aq) + OH-(aq) H2O (l)DH0 = -56.2 kJ

H2O (s) H2O (l)DH0 = 6.01 kJ

H2O

NH4NO3(s) NH4+(aq) + NO3-(aq)DH0 = 25 kJ

Does a decrease in enthalpy mean a reaction proceeds spontaneously?

Spontaneous reactions at 25 °C


Two trends in nature
TWO Trends in Nature the following reactions is positive or negative:

  • Order  Disorder

     

  • High energy  Low energy


The driving forces
The Driving Forces the following reactions is positive or negative:

  • Energy Factor (ΔH)

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


S the following reactions is positive or negative:

order

disorder

S

H2O (s) H2O (l)

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

DS = Sf - Si

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

DS > 0

Sf > Si

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

DS > 0


First Law of Thermodynamics the following reactions is positive or negative:

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.

DSuniv = DSsys + DSsurr > 0

Spontaneous process:

DSuniv = DSsys + DSsurr = 0

Equilibrium process:


The Second Law of Thermodynamics the following reactions is positive or negative:

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

Suniverse = Ssystem +Ssurroundings

Suniverse > 0 for spontaneous rxn

Suniverse = 0 at equilibrium


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

0 K

Perfect solid crystal with no motion

Only Theoretical

It is not possible to reach absolute 0!

Entropy of universe is always increasing!


3rd Law of Thermodynamics that has an entropy value of 0?

the 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 that has an entropy value of 0?

  • ΔSº

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

  • (0 entropy is only theoretical; not really possible)

  • 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 that has an entropy value of 0?

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 s for the following reaction at 298k
Calculate the standard entropy change ( that has an entropy value of 0?Sº) for the following reaction at 298K

Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)


S o rxn n s o products m s o reactants
that has an entropy value of 0?Sorxn =n So(products) - m So(reactants)

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

Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)

= 180.4 J/K


Predict the sign of that has an entropy value of 0?ΔSº of the following reaction.

2SO2(g) + O2(g) 2SO3(g)

Entropy decreases, -

Lets’ Calculate


Calculate the standard entropy change s for the following reaction at 298k1
Calculate the standard entropy change ( that has an entropy value of 0?Sº) for the following reaction at 298K

2SO2(g) + O2(g) 2SO3(g)

ΔSº = -187.8 J K-1


Predicting spontaneous reactions that has an entropy value of 0?

  • 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.


Gibbs Free Energy that has an entropy value of 0?

(G)

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
If Both T and P are constant, the relationship between that has an entropy value of 0?G and spontaneity is:

  • G is (-), forward rxn is spontaneous.

  • G is 0, rxn is at equilibrium.

  • G is (+) forward rxn is not spontaneous (requires work) reverse rxn is spontaneous.


(sum of standard free energies of formation of products) that has an entropy value of 0?

minus

(sum of standard free energies of formation of reactants)

Coefficients from equation

the sum of


The values of that has an entropy value of 0?ΔGºf of elements in their most stable form is 0, just as with enthalpy of formations.


Calculate the that has an entropy value of 0?ΔG°rxnfor the combustion of methane at 298K and determine if the reaction is spontaneous.

-818.0 KJ Spontaneous


Calculate the g for the thermite reaction aluminum with iron iii oxide
Calculate the ( that has an entropy value of 0?Gº) for the thermite reaction (aluminum with iron(III) oxide).

-835.5 KJ spontaneous


Gibbs Free Energy: Equation 2 that has an entropy value of 0?

  • 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) that has an entropy value of 0?

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, and

two conditions depend on temperature.


Predicting Sign of that has an entropy value of 0?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.


Δ that has an entropy value of 0?H

ΔS

ΔG

- +

Different sign, not temperature dependent.

+ -

- -

Freezing

Same sign, temperature dependent.

+ +

Melting


G = H – T(S) that has an entropy value of 0?

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.


Example that has an entropy value of 0? – The entropy change of the system is negative for the precipitation reaction:

Ag+(aq) + Cl-(aq) AgCl(s)

Ho = -65 kJ

Since 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 – TS

G = -65 – 298(-.030)

= -73.94


  • Problem 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.

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

  • 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?


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

A) What is ΔGº at 298 K?

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)


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

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.

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


The reaction become spontaneous? If so, at what temperature will it become spontaneous?standard entropy of reaction (DS0rxn) is the entropy change for a reaction carried out at 1 atm and 250C.

aS0(A)

bS0(B)

-

[

+

]

cS0(C)

dS0(D)

[

+

]

=

aA + bB cC + dD

-

S

mS0(reactants)

S

nS0(products)

=

DS0

DS0

DS0

DS0

rxn

rxn

rxn

rxn

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

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

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

Entropy Changes in the System (DSsys)

S0(CO) = 197.9 J/K•mol

S0(CO2) = 213.6 J/K•mol

S0(O2) = 205.0 J/K•mol


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

Entropy Changes in the System (DSsys)

When gases are produced (or consumed)

  • If a reaction produces more gas molecules than it consumes, DS0 > 0.

  • If the total number of gas molecules diminishes, DS0 < 0.

  • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number.

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


Entropy Changes in the Surroundings ( reaction? 2Zn DSsurr)

Endothermic Process

DSsurr < 0

Exothermic Process

DSsurr > 0


Third Law of Thermodynamics reaction? 2Zn

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
Gibb’s Free Energy reaction? 2Zn --The energy of a system that is available to do work.

  • ΔGorxn = ΔHorxn – TΔSorxn .

  • Hess’ Law also applies:

    ΔGrxn = ΣΔGfo(products) – ΣDGfo(reactants)

  • ΔGfo has a similar definition as ΔHfo

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

  • If ΔGorxn is negative, the rxn is spontaneous at that temperature.

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


Gibbs Free Energy reaction? 2Zn

DSuniv = DSsys + DSsurr > 0

Spontaneous process:

DSuniv = DSsys + DSsurr = 0

Equilibrium process:

For a constant-temperature process:

Gibbs free energy (G)

DG = DHrxn -TDSrxn

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

DG > 0 The reaction is nonspontaneous as written. The

reaction is spontaneous in the reverse direction.

DG = 0 The reaction is at equilibrium.


a reaction? 2Zn A + bB cC + dD

-

[

+

]

[

+

]

=

-

mDG0 (reactants)

S

S

=

f

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

DG0

DG0

rxn

rxn

DGo of any element in its stable form is zero.

f

nDG0 (products)

aDG0 (A)

dDG0 (D)

bDG0 (B)

cDG0 (C)

f

f

f

f

f

The standard free-energy of reaction (DGorxn ) is the free-energy change for a reaction when it occurs under standard-state conditions.


- reaction? 2Zn

mDG0 (reactants)

S

S

=

f

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

DG0

DG0

DG0

-

[

]

[

+

]

=

rxn

rxn

rxn

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

=

Is the reaction spontaneous at 25 0C?

12DG0 (CO2)

2DG0 (C6H6)

f

f

6DG0 (H2O)

f

nDG0 (products)

f

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

DG0 = -6405 kJ

< 0

spontaneous


G f o h f o t s f o
ΔG reaction? 2Zn fo = ΔHfo - TΔSfo

ΔGfo

ΔHfo

TΔSfo

[email protected] high T

+

+

?

-

-

[email protected] low T

?

-

[email protected] all T

+

?

+

-

[email protected] all T

?


D reaction? 2Zn G = DH - TDS


CaCO reaction? 2Zn 3(s) CaO (s) + CO2(g)

Temperature and Spontaneity of Chemical Reactions

DH0 = 177.8 kJ

DS0 = 160.5 J/K

DG0 = DH0 – TDS0

At 25 0C, DG0 = 130.0 kJ

DG0 = 0 at 835 0C


H reaction? 2Zn 2O (l) H2O (g)

40.79 kJ

=

DS =

373 K

DH

T

Gibbs Free Energy and Phase Transitions

T = ΔH/ ΔS

DG0 = 0

= DH0 – TDS0

= 109 J/K


Gibbs Free Energy and Chemical Equilibrium reaction? 2Zn

Non-standard Conditions:

DG = DGo + RT lnQ

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

DG = 0

Q = K

0 = DGo + RT lnK

DGo = -RT lnK


D reaction? 2Zn Go = -RT lnK


Chemistry In Action: reaction? 2Zn The Thermodynamics of a Rubber Band

TDS = DH - DG

Low Entropy

High Entropy


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