Chapter 19 principles of chemical reactivity entropy and free energy
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Chapter 19 Principles of Chemical Reactivity: Entropy and Free Energy. Entropy & Free Energy. Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous.

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Chapter 19 Principles of Chemical Reactivity: Entropy and Free Energy

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Chapter 19 principles of chemical reactivity entropy and free energy

Chapter 19Principles of Chemical Reactivity: Entropy and Free Energy


Entropy free energy

Entropy & Free Energy

Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous.

Chemical Kinetics provides us with information about the rate of reaction and how the reaction proceeds.


Entropy free energy1

Entropy & Free Energy

Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous.

Spontaneous changes occur only in the direction that leads to equilibrium.

Systems never change spontaneously in a direction that takes them farther from equilibrium.

Example: Heat Transfer

Heat (thermal energy) always flows spontaneously from a hot object to a cold object.

The process occurs until thermal equilibrium is achieved, that is when both objects are at the same temperature.


Spontaneous process

Spontaneous Process

  • Additional Examples:

  • Solvation of a Soluble salt: NH4NO3(s) dissolves spontaneously even the process is endothermic (H > 0)

  • Expansion of a gas and Diffusion. Gasses mix to form homogeneous mixtures spontaneously.

  • Certain Phase Changes: Ice melts spontaneously above 0oC. Water evaporates even though the enthalpy of vaporization is endothermic.

  • Certain Chemical Reactions: Na(s) reacts vigorously when dropped in water. Iron rusts when exposed to the atmosphere.


Spontaneous process1

Spontaneous Process

But many spontaneous reactions or processes are endothermic or even have ∆H  0.

NH4NO3(s) + heat  NH4NO3(aq)

∆H = 0


Thermodynamics

Thermodynamics

A Review of Concepts of Thermodynamics

  • First law of thermodynamics:The law of conservation of energy; energy cannot be created or destroyed.

  • State Function: Quantity in which its determination is path independent.

  • U = q + w:The change in internal energy of a system is a function of heat and work done on or by the system.

  • H: Heat transferred at constant pressure.

  • Exothermic Process: H < 0

  • Endothermic Process: H > 0


Thermodynamics1

Thermodynamics

Enthalpy alone does not predict spontaneity:

Some processes are energetically favored (rH < 0) but not spontaneous.

Equilibrium alone cannot determine spontaneity:

Some processes are favored based on Equilibrium (K >> 1) yet they are non-spontaneous.

There must be another factor that plays a role in determination of spontaneity!


Thermodynamics kinetics

Thermodynamics & Kinetics

Diamond is thermodynamically favored to convert to graphite, but not kinetically favored.

Paper burns once the reaction is initiates. The process is product-favored & kinetically favored.


Thermodynamics kinetics1

Thermodynamics & Kinetics

Reactants

Products

Kinetics

Thermodynamics


Thermodynamics kinetics2

Thermodynamics & Kinetics

  • Factors that Affect Spontaneity (Thermodynamic favorability):

  • Enthalpy: Comparison of bond energy (H)

  • Entropy: Randomness vs. Order of a system (S)

  • In general, enthalpy is more important than entropy.


Dispersal of energy entropy

Dispersal of Energy: Entropy

  • In a spontaneous processes the energy of the final state is more dispersed.

  • The system moves to a higher state of disorder.

  • The thermodynamic quantity associated with disorder and energy dispersal is called ENTROPY, S.

  • The 2nd law of thermodynamics states that a spontaneous process results in an increase in the entropy of the universe. S > 0

Reaction of K and water


Dispersal of energy entropy1

Dispersal of Energy: Entropy

Observation of a spontaneous process shows that it is associated with a dispersal of energy.

Energy Dispersal


Dispersal of energy entropy2

Dispersal of Energy: Entropy

The change in entropy for a spontaneous process is given by:

Where qrev is the heat gained or lost by the system during the process and T is the absolute temperature.

A reversible process can be returned to its original state. (chemical equilibrium), an irreversible process cannot.

Example: The breaking of a coffee mug into many pieces is an irreversible process.


Dispersal of energy entropy3

Dispersal of Energy: Entropy

To begin, particle 1 has 2 units of energy and 2-4 have none.


Dispersal of energy entropy4

Dispersal of Energy: Entropy

Particle 1 can transfer one unit of energy to particle 2, then the other to 3 or 4.


Dispersal of energy entropy5

Dispersal of Energy: Entropy

Particle 2 can initially have two units of energy.


Dispersal of energy entropy6

Dispersal of Energy: Entropy

Particle 2 transfer one unit to particles 4 or 3. (2 to 1 has already been counted.)


Dispersal of energy entropy7

Dispersal of Energy: Entropy

Particle 4 can initially have two units of energy.


Dispersal of energy entropy8

Dispersal of Energy: Entropy

Particle 4 transfer one unit to particle 3. (4 to 1 & 4 to 2 have already been counted.)


Dispersal of energy entropy9

Dispersal of Energy: Entropy

Particle 3 can start with two units of energy. Energy transfers between particle 3 were previously counted.


Dispersal of energy entropy10

Dispersal of Energy: Entropy

Each unique combination that results in a dispersion of energy is called a microstate. There are 10 microstates in this system. The greater the number of microstates, the greater the entropy of the system.


Dispersal of energy entropy11

Dispersal of Energy: Entropy


Dispersal of energy entropy12

Dispersal of Energy: Entropy

As the size of the container increases, the number of microstates accessible to the system increases. Therefore the entropy of the system increases.


Dispersal of energy entropy13

Dispersal of Energy: Entropy

  • The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C.

  • Energy is more dispersed in liquid water than in solid water due to the lack of an ordered network as in the solid state.


Entropy states of matter

Entropy & States of Matter

So (J/K•mol)

H2O(liq)69.95

H2O(gas)188.8

S (solids) < S (liquids) < S (gases)

Energy dispersal


Entropy states of matter1

Entropy & States of Matter

S˚(Br2 liq) < S˚(Br2 gas)

S˚(H2O sol) < S˚(H2O liq)


Dispersal of energy entropy14

Dispersal of Energy: Entropy

Entropy and Microstates:

As the number of microstates increases, so does the entropy of the system.

S = klnW

k = Boltzman’s constant (1.381 1023 J/K)

W = the number of microstates


Entropy entropy change energy dispersal a summary

Entropy, Entropy Change, & Energy Dispersal: A Summary

Entropy and Microstates:

The change in entropy associated with a process is a function of the number of final and initial microstates of the system.

If Wfinal > Winital, S > 0

If Wfinal < Winital, S < 0


Dispersal of energy entropy15

Dispersal of Energy: Entropy

When a solute dissolves in a solvent the process is spontaneous owing to the increase in entropy.

Matter (and energy) are more dispersed. The number of microstates is increased.


Entropy measurements values

Entropy Measurements & Values

The entropy of a substance increases with temperature.

Molecular motions of heptane at different temps.

Molecular motions of heptane, C7H16


Entropy measurements values1

Entropy Measurements & Values

An Increase in molecular complexity generally leads to increase in S.


Entropy measurements values2

Entropy Measurements & Values

Entropies of ionic solids depend on coulombic attractions.

S° (J/K•mol)

MgO26.9

NaF51.5

Mg2+ & O2-

Na+ & F-


Standard molar entropies

Standard Molar Entropies

Defined by Ludwig Boltzmann, the third law states that a perfect crystal at 0 K has zero entropy; that is, S =0.

The entropy of an element or compound under any other temperature and pressure is the entropy gained by converting the substance from 0 K to those conditions.

To determine the value of S, it is necessary to measure the energy transferred as heat under reversible conditions for the conversion from 0 K to the defined conditions and then to use Equation 19.1

Because it is necessary to add energy as heat to raise the temperature, all substances have positive entropy values at temperatures above 0 K.


Standard molar entropies1

Standard Molar Entropies

The standard molar entropy, S°, of a substance is the entropy gained by converting 1 mol of it from a perfect crystal at 0 K to standard state conditions (1 bar, 1m for a solution) at the specified temperature.


Entropy changes for phase changes

Entropy Changes for Phase Changes

For a phase change,

where q = heat transferred in the phase change


Entropy temperature

Entropy & Temperature

S increases slightly with T

S increases a large amount with phase changes


Determining entropy changes in physical chemical processes

Determining Entropy Changes in Physical & Chemical Processes

Standard molar entropy values can be used to calculate the change in entropy that occurs in various processes under standard conditions.

The standard entropy change for a reaction (rS°) can be found in the same manner as rH° were:

Where n & m are the stoichiometric balancing coefficients.

This calculation is valid only under reversible conditions.


Chapter 19 principles of chemical reactivity entropy and free energy

Determining Entropy Changes in Physical & Chemical Processes

Problem:

Consider the reaction of hydrogen and oxygen to form liquid water.

What is the standard molar entropy for the reaction?


Chapter 19 principles of chemical reactivity entropy and free energy

Determining Entropy Changes in Physical & Chemical Processes

Problem:

Consider the reaction of hydrogen and oxygen to form liquid water.

What is the standard molar entropy for the reaction?


Chapter 19 principles of chemical reactivity entropy and free energy

Determining Entropy Changes in Physical & Chemical Processes

Problem:

Consider the reaction of hydrogen and oxygen to form liquid water.

What is the standard molar entropy for the reaction?


Chapter 19 principles of chemical reactivity entropy and free energy

Determining Entropy Changes in Physical & Chemical Processes

Problem:

Consider the reaction of hydrogen and oxygen to form liquid water.

What is the standard molar entropy for the reaction?

The enthalpy change is negative (net decrease in dispersion) due to the change in number of moles: 3 reduced to 2


2 nd law of thermodynamics

2nd Law of Thermodynamics

The second law of thermodynamics states:..

S°universe = ∆S°system + ∆S°surroundings

Any change in entropy for the system plus the entropy change for the surroundings must equal the overall change in entropy for the universe.

A process is considered to be spontaneous under standard conditions if S°(universe) is greater than zero.


2 nd law of thermodynamics1

2nd Law of Thermodynamics

The solution process for NH4NO3 (s) in water is an entropy driven process.

∆S°universe =

∆S°system + ∆S°surroundings


2 nd law of thermodynamics2

2nd Law of Thermodynamics

Calculating S°Surroundings:


2 nd law of thermodynamics3

2nd Law of Thermodynamics

Calculating S°Surroundings:


2 nd law of thermodynamics4

2nd Law of Thermodynamics

Calculating S°Surroundings:


2 nd law of thermodynamics5

2nd Law of Thermodynamics

Calculating S°Surroundings:

at constant pressure, qsystem = rH°system


2 nd law of thermodynamics6

2nd Law of Thermodynamics

Calculating S°Surroundings:

at constant pressure, qsystem = rH°system


2 nd law of thermodynamics7

2nd Law of Thermodynamics

Problem:

Calculate the entropy change for the surroundings for the reaction: 2H2(g) + O2(g)  2H2O(l)


2 nd law of thermodynamics8

2nd Law of Thermodynamics

Problem:

Calculate the entropy change for the surroundings for the reaction: 2H2(g) + O2(g)  2H2O(l)

The enthalpy change for the reaction is calculated using standard molar enthalpies of formation:


2 nd law of thermodynamics9

2nd Law of Thermodynamics

Problem:

Calculate the entropy change for the surroundings for the reaction: 2H2(g) + O2(g)  2H2O(l)

The enthalpy change for the reaction is calculated using standard molar enthalpies of formation:


2 nd law of thermodynamics10

2nd Law of Thermodynamics

Problem:

Calculate the entropy change for the surroundings for the reaction: 2H2(g) + O2(g)  2H2O(l)

As calculated previously:

S°universe = 327 J/K + 1917 J/K =1590. J/K


2 nd law of thermodynamics11

2nd Law of Thermodynamics

Problem:

Calculate the entropy change for the surroundings for the reaction: 2H2(g) + O2(g)  2H2O(l)

S°universe = 327 J/K + 1917 J/K = 1590. J/K

The entropy of the universe increases ( > 0) therefore the process is spontaneous at standard state conditions.

The process is spontaneous due to the entropy change in the surroundings, not the system.


Spontaneous or not

Spontaneous or Not?


Gibbs free energy g

Gibbs Free Energy, G

The method used so far to determine whether a process is spontaneous required evaluation of two quantities, S°(system) and S°(surroundings).

J. Willard Gibbs asserted that that the maximum non-PV work available to the system must be a function of Enthalpy and Entropy:

G = H – TS

G = Gibbs Free energy of the system, H = system enthalpy and S = the entropy of the system.

J. Willard Gibbs1839-1903


Gibbs free energy g1

Gibbs Free Energy, G

Since it is impossible to measure individual values of enthalpy, we often express free energy in terms of the changes of thermodynamic quantities.

(G = H – TS)

At constant temperature:

G° = Gibbs Free energy change of the system, H° = enthalpy change and S° = the entropy change at SS conditions.


Gibbs free energy g2

Gibbs Free Energy, G

If the reaction is exothermic (H < 0)

And the change in entropy is positive (H > 0) at a given temperature, then S < 0.

We then can assert that if G < 0 that the reaction is spontaneous as well as product favored!

Spontaneity is a function of energy and dispersion!


Gibbs free energy g3

Gibbs Free Energy, G


R g o equilibrium

rGo & Equilibrium

Since rG° is related to S°universe, it follows that:

  • If rG° < 0: The process is spontaneous in the direction written under standard conditions.

  • If rG° = 0: The process is at equilibrium under standard conditions and K=1

  • If rG° > 0: The process is non-spontaneous in the direction written under standard conditions.

  • Conclusion: A reaction proceeds spontaneously toward the minimum in free energy, which corresponds to equilibrium.


G g q k

∆G, ∆G°, Q, & K

Product Favored Reactions, ∆G° negative, K > 1

Q < K: Heading to equilibrium G < 0

Q = K: At equilibrium G = 0

Q > K: Heading away from equilibrium G > 0


G g q k1

∆G, ∆G°, Q, & K

  • Product-favored

  • 2 NO2 N2O4

  • ∆rG° = – 4.8 kJ

  • State with both reactants and products present is more stable than complete conversion.

  • K > 1, more products than reactants.


G g q k2

∆G, ∆G°, Q, & K

Reactant Favored Reactions, ∆G° positive, K < 1

Q < K: Heading to equilibrium G < 0

Q = K: At equilibrium G = 0

Q > K: Heading away from equilibrium G > 0


G g q k3

∆G, ∆G°, Q, & K

  • Reactant-favored

  • N2O4 2 NO2 ∆rG° = +4.8 kJ

  • State with both reactants and products present is more stable than complete conversion.

  • K < 1, more reactants than products


G g q k4

∆G, ∆G°, Q, & K

rG° represents the free energy change for a process at standard state conditions. (Equilibrium)

What if this is not the case?

Under nonstandard conditions:

Where R is the

gas law constant

and Q =


G g q k5

∆G, ∆G°, Q, & K

At equilibrium, we know that rG = 0 and Q = K

Therefore:

So knowing one quantity yields the other.


G g q k6

∆G, ∆G°, Q, & K

Problem:

rG° for the formation of ammonia at SS conditions is 16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm)


G g q k7

∆G, ∆G°, Q, & K

Problem:

rG° for the formation of ammonia at SS conditions is 16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm)


G g q k8

∆G, ∆G°, Q, & K

Problem:

rG° for the formation of ammonia at SS conditions is 16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm)


G g q k9

∆G, ∆G°, Q, & K

Problem:

rG° for the formation of ammonia at SS conditions is 16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm)

K >> 1, product favored as predicted by G


Summary

Summary

The relation of ∆rG, ∆rG°, Q, K, reaction spontaneity, and product- or reactant favorability.


Calculating using free energy standard free energy of formation

Calculating & Using Free Energy Standard Free Energy of Formation


Calculating using free energy standard free energy of formation1

Calculating & Using Free Energy Standard Free Energy of Formation


Calculating using free energy standard free energy of formation2

Calculating & Using Free Energy Standard Free Energy of Formation


Calculating using free energy standard free energy of formation3

Calculating & Using Free Energy Standard Free Energy of Formation

rG° is negative at 298 K, so the reaction is predicted to be spontaneous under standard conditions at this temperature. It is also predicted to be product-favored at equilibrium.


Gibbs free energy g4

Gibbs Free Energy, G

Under reversible conditions, both enthalpy and entropy are state functions. It follows that the Gibbs free energy must also be.

Therefore we can write that:


Free energies of formation

Free Energies of Formation

Note that ∆fG° for an element = 0


Chapter 19 principles of chemical reactivity entropy and free energy

Free Energies of Formation

Using the fG° found in appendix L of your text, calculate rG° for the following reaction:

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)


Chapter 19 principles of chemical reactivity entropy and free energy

Free Energies of Formation

Using the fG° found in appendix L of your text, calculate rG° for the following reaction:

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)


Chapter 19 principles of chemical reactivity entropy and free energy

Free Energies of Formation

Using the fG° found in appendix L of your text, calculate rG° for the following reaction:

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)


Chapter 19 principles of chemical reactivity entropy and free energy

Free Energies of Formation

Using the fG° found in appendix L of your text, calculate rG° for the following reaction:

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)

rG° = –1101.14 kJ/mol (product-favored


Calculating r g

Calculating ∆rG°

Is the dissolution of ammonium nitrate product-favored?

If so, is it enthalpy- or entropy-driven?

NH4NO3(s) + heat  NH4NO3(aq)


Free energy temperature

Free Energy & Temperature

  • By definition:

    G = H − TS

  • Indicating that free energy is a function of temperature.

  • rG° will therefore change with temperature.

  • A consequence of this temperature dependence is that, in certain instances, reactions can be product-favored at equilibrium at one temperature and reactant-favored at another.


Free energy temperature1

Free Energy & Temperature

When a reaction has

rH° < 0

&

rS° > 0

at alltemperatures rG° is negative.

(Product favored)


Free energy temperature2

Free Energy & Temperature

When a reaction has

rH° > 0

&

rS° > 0

at high temperatures rG° is negative.

(Product favored)

When a reaction has

rH° < 0

&

rS° < 0

at low temperatures rG° is negative.

(Product favored)


Free energy temperature3

Free Energy & Temperature


Free energy temperature4

Free Energy & Temperature

At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (rG changes from + to )


Free energy temperature5

Free Energy & Temperature

At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (rG changes from + to )

Example: The reaction,

Has the following thermodynamic values.

∆rH°= +470.5 kJ

∆rS° = +560.3 J/K

∆rG° = +301.3 kJ

Reaction is reactant-favored at 298 K


Free energy temperature6

Free Energy & Temperature

At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (rG changes from + to )

Example: The reaction,

When rG  0, the reaction begins to become spontaneous.


Free energy temperature7

Free Energy & Temperature

At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (rG changes from + to )

Example: The reaction,

When rG  0, the reaction begins to become spontaneous.


Free energy temperature8

Free Energy & Temperature

At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (rG changes from + to )

Example: The reaction,

∆rH°= +467.9 kJ

∆rS° = +560.3 J/K


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