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Thermodynamics: Free Energy and Entropy

Suggested Reading: Chapter 19

An object or collection of objects being studied.

System:

Everything outside of the system.

Surroundings:

the surroundings

the system

System and Surroundings

When E is transferred as heat between a system and its surroundings, the direction is described as endo or exothermic.

Energy is transferred from the system to its surroundings. E of the system decreases, E of the surroundings increases.

Exothermic:

Energy is transferred from the surroundings to the system. E of the system increases, E of the surroundings decreases.

Endothermic:

Direction of Energy Transfer

The energy change for a system (ΔU) is the sum of the E transferred as heat (q) and the E transferred as work (w).

Work:

Energy transfer that occurs as a mass is moved against an opposing force.

If a system does work on its surroundings, Esys will decrease. If work is done by the surroundings, Esystem will increase.

internal energy change

work energy transferred ΔU = q + w

heat energy transferred

First Law of Thermodynamics

Heat energy transferred at constant pressure.

ΔU = qp + wp

ΔU = qp - PΔV

qp = ΔU + PΔV

ΔH = ΔU + PΔV

ΔH = qp

ΔV is typically very small therefore work is small.

ΔH and ΔU differ by PΔV.

ΔV is large when gases are formed or consumed.

ΔH (-) E transfer from the system ΔH (+) E transfer to the system

Enthalpy

A state function depends on the initial and final state of the system but is independent of the path taken. (Doesn’t matter how you got to the end product!)

No matter how you go from reactants to products in a reaction, the values of ΔH and ΔU are always the same.

P, V and T are also state functions q and w are not state functions

State Functions

The state of matter effects the magnitude of the enthalpy.

H2 (g) + ½ O2 (g) H2O (g) ΔrH = - 241.8 kJ/mol

H2 (g) + ½ O2 (g) H2O (l) ΔrH = - 285.8 kJ/mol

The difference is equal to the enthalpy change for the condensation of 1 mol of H2O vapor to 1 mol of liquid H2O.

Keep in Mind: 1. ΔH is specific to each reaction; it depends on the

reactants, products and their states of matter. 2. ΔH depends on the number of moles of reactant. 3. ΔH (-) = exothermic reaction ΔH (+) = endothermic reaction

Standard Reaction Enthalpy

The enthalpy change for the formation of 1 mol of a cmpd directly from its elements in their standard state.

For an element, ΔHf = 0 If ΔHf is negative, formation of the compound is exothermic is positive, formation of the compound is endothermic

The more negative the ΔHf, the more stable the compound

Compound ΔHf (kJ/mol)

HF (g) -273.3

HCl (g) -92.31

HBr (g) -35.29

HI (g) +25.36

Which of the hydrogen halides is most stable?

ΔHf does not always describe a rxn that can be done in lab

Standard Molar Enthalpy of Formation

Which of the following chemical equations does not correspond to a standard molar enthalpy of formation?

a. Ca (s) + C (s) + 3/2 O2 (g) CaCO3 (s)

b. C (s) + O2 (g) CO2 (g)

c. NO (g) + ½ O2 (g) NO2 (g)

d. N2 (g) + 2 O2 (g) N2O4 (g)

e. H2 (g) + ½ O2 (g) H2O (l)

Have to form ONLY 1 mol of product Elements must be in standard states

(No compounds as reactants!)

Standard Molar Enthalpy of Formation

Enthalpy changes accompany all chemical reactions. The standard reaction enthalpy (ΔrH°) is used when all

reactants and products are in their standard state [ie O2 (g), graphite (s)].

H2O (g) H2 (g) + ½ O2 (g) ΔrH = + 241.8 kJ/mol

Positive value for enthalpy indicates an endothermic rxn.

2 H2 (g) + O2 (g) 2 H2O (g) ΔrH = - 483.6 kJ/mol

Reverse reaction has an enthalpy with an opposite sign; If two moles are used in the rxn, the enthalpy is doubled.

Standard Reaction Enthalpy

Is a Reaction Spontaneous?

Spontaneous: a change that occurs without intervention

The change occurs to lead to equilibrium

Has nothing to do with rate of reaction!

All chemical reactions proceed in the direction to achieve equilibrium.

Spontaneous Reactions

Many spontaneous reactions are exothermic, such as combustion, formation of NaCl, strong acid/base reactions

Exothermicity is not a requirement of spontaneity.

•  NH4NO3 dissolves spontaneously in water and ΔHrxn = +25.7 kJ/mol. •  Expansion of gas into a vacuum is energy neutral (or endothermic for real gases). •  Phase changes (s à l à g) require E input (and depend on Temperature!)

These examples illustrate spontaneity cannot be equated to whether a reaction is endo/exothermic.

Entropy (S)

In a spontaneous process, energy goes from being concentrated to being more disperse.

Example: a book falling from a table

How do we predict directionality and quantify a process?

Entropy (S): state function

Second Law of Thermodynamics: A spontaneous process results in an increase in the

entropy of the universe (i.e ΔSuniverse > 0)

Energy is conserved, so it is not a good indicator of spontaneity.

The book will fall spontaneously, but will not jump back up!

Calculating Entropy

Thermal energy is caused by the random motion of particles, so potential energy is dispersed when it is

converted to thermal energy (i.e. heat). Therefore heat (q) is related to entropy.

Energy dispersal due to heat is inversely proportional to T.

T = Temperature (in Kelvin)

qrev = heat transferred under reversible conditions (adding energy by heating an object slowly)

ΔS = qrev

T ΔS units = J/K

Entropy on the Microscopic Level

Why does entropy increase in a spontaneous process?

Why must energy be more dispersed?

E transferred as heat between hot and cold gaseous atoms

Initial Equilibrium

T

Why does energy disperse? Consider a statistical approach.

Imagine 4 atoms, 1 which has 2 “units” of energy

Overall energy of all 4 atoms = 2 units

Over time, 10 different ways of distributing the energy between particles can be observed


T

Each of the different configurations is called a microstate.

In only one state do both energy units remain on atom 1

4 ways to distribute 2 units on a single atom 6 ways to distribute the units between different atoms


There is a preference that energy by dispersed over a greater number of atoms (60 % chance).

T

Statistics dictate a better chance of having the energy dispersed between more atoms (71.4 % chance).

For a 6 atom system:

6 microstates with the energy concentrated on one atom 15 microstates with the energy present on two atoms

As number of atoms increases, probability of energy dispersing increases dramatically.



Instead of increasing # atoms, what about more E units?

Consider 6 energy units, dispersed over four atoms

Initially two atoms have 3 quanta of E and two atoms have no E.

Through collisions, E is transferred to achieve different distributions of E.

Now there are 84 possible microstates.


Increasing # energy units also increases the # microstates


There are 4 different ways to have four particles such that one particle has 3 quanta of energy and the other three

each have one quantum of energy.

Boltzmann’s Equation

Entropy of a system is determined by # of microstates available

Entropy is proportional to the number of microstates

S = k lnW

W = # of microstates

k = Boltzmann’s constant = 1.381 x 10-23 J/K

Dispersal of Energy: Gas Diffusion

Why does a gas diffuse throughout a container?

Think of each energy as a microstate

As the gas disperses across the two flasks, there are more spaces for the molecules, and therefore

more microstates available!

Energy is quantized

Molecules have a distribution of energies

Why does a gas diffuse throughout a container?

The energy is conserved, so the microstates are closer.

Dispersal of Energy: Gas Diffusion

A Chemist’s Perspective on Entropy

Chemists are interested in the change in entropy (ΔS)

ΔS = Sfinal – Sinitial = k ln(Wfinal/Winitial)

ΔS depends on the number of microstates available before and after (a state function).

We are interested in determining a numerical value for ΔS

Third Law of Thermodynamics: A perfect crystal at 0 K has S = 0.

Entropy (S) is always positive (but ΔS can be negative).

This is used as a reference point.

The entropy for any other set of conditions is measured relative to the absolute zero point.

Entropy increases with increase in T

Standard Entropy (S°)

Changes of state accompany large entropy increases

Standard entropy (S°) of a substance: Entropy gained by converting 1 mol of a substance from a perfect crystal at 0 K to standard conditions (1 bar, 1 m).

Units = J/K mol

Generally find in tables (Appendix L)

General Trends:

Larger molecules generally have larger S°

Big molecules have more ways to orient therefore more microstates

S° gas > S° liquid > S° solid

Compound S° (J/K mol) I2 (s) 116 Br2 (l) 152 Cl2 (g) 223

Large increase in S° accompanies a phase change

Standard Entropy (S°)

Which substance has the higher entropy?

Entropy Practice

NO2 (g) or N2O4 (g)

N2O4 is a larger, more complex molecule. In the same state of matter (g), it will have a greater S°.

I2 (g) or I2 (l)

For the same substance, the gas phase will have the higher S°.

range the substances on the left in order of increasing entropy. Assume 1 mole of each at standard conditions.

CH3COOH (l) CO2 (g) HCOOH (l) Al (s)

Arrange in order of increasing entropy, assuming 1 mol.

(1) lowest entropy

1 2 4 3

Predict whether ΔS for each reaction is greater than zero, less than zero or too close to determine.

Entropy Practice range the substances on the left in order of increasing entropy. Assume 1 mole of each at standard conditions.

H2 (g) + F2 (g) 2 HF (g) Too close

4 HCl (g) + O2 (g) 2 H2O (g) + 2 Cl2 (g) ΔS < 0

2 H2O (l) 2 H2 (g) + O2 (g) ΔS > 0

2 NOBr (g) 2 NO (g) + Br2 (g) ΔS > 0

2 HBr (g) + Cl2 (g) 2 HCl (g) + Br2 (g) Too close

Consider which side of the reaction has more disorder (count moles!)

Calculating ΔSrxn°

ΔSrxn° = ΣS°(products) - ΣS°(reactants)

Calculate entropy change for a system in which reactants are completely converted to products.

What is ΔSrxn° for the formation of NO2?

2 NO (g) + O2 (g) 2 NO2 (g)

Compound S° (J/K mol)

NO 210.8

O2 205.1

NO2 240.0

ΔSrxn° = [2 mol NO2 x 240.0] – [2 mol NO x 210.8 + 1 mol O2 x 205.1] = - 146.7 J/K mol

Calculate ΔSrxn° for the formation of ammonia.

Predict the sign of ΔSrxn° Less gas moles in product therefore entropy must decrease

ΔSrxn° for a system can be positive or negative

Calculating ΔSrxn°

N2 (g) + 3 H2 (g) 2 NH3 (g)


Compound S° (J/K mol)

N2 191.56

H2 130.7

NH3 192.77

ΔSrxn° = [2 mol NH3 x 192.77] – [1 mol N2 x 191.56 + 3 mol H2 x 130.7] = - 198.12 J/K mol

Entropy Changes & Spontaneous Rxns

ΔS for the system can be positive or negative

ΔS°(universe) must increase in a spontaneous process

ΔS°(universe) = ΔS°(system) + ΔS°(surroundings)

ΔS°(system) can be negative if the magnitude of ΔS°(surrounding) is larger so that ΔS°(universe) is positive.

ΔS°(universe) > 0 if the reaction is spontaneous

ΔS°(universe) = 0 at equilibrium

ΔS°(universe) < 0 if the reaction is not spontaneous

Calculating ΔS°(universe)

CO (g) + 2 H2 (g) CH3OH (g)

Calculate ΔS°(universe) for the following reaction at 25 °C



ΔSrxn° = [1 mol CH3OH x 127.2] – [1 mol CO x 197.7 + 2 mol H2 x 130.7] = -331.9 J/K mol

ΔS°(surroundings) = qrev

T

ΔH(surroundings)

T =

ΔH(system)

T = -

ΔHsys° = ΣΔHf°(products) - ΣΔHf°(reactants) ΔHsys° = [1 mol CH3OH x -238.4] – [1 mol CO x -110.5 + 2 mol H2 x 0] = -127.9 kJ/mol

ΔS° (universe) > 0 therefore the reaction is spontaneous


ΔS°(surroundings) = - ΔH(system)

T =

127,900 J/mol 298 K

= 429.2 J/K mol


ΔS°(universe) = -331.9 J/K mol + 429.2 J/K mol

= 97.3 J/K mol


Calculate ΔS°(universe) when 1.966 moles of NO (g) react under standard conditions at 298.15 K.

ΔH° = -752.2 kJ and ΔS° = -351.6 J/K

2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (l)


ΔS°(surroundings) = - ΔH(system)

T =

752,200 J

298.15 K

= 2522.9 J/K

ΔS°(universe) = -351.6 J/K + 2522.9 J/K = 2171.3 J/K

1.966 moles NO

2 moles NO x 2171.3 J/K = 2134 J/K

The reaction is product favored and enthalpy favored.

Spontaneous Reactions: A Summary

Four possible outcomes of ΔH°(system) and ΔS°(system)

2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (g)

ΔH = -5315.1 kJ/mol, ΔS = 312.4 J/K mol N2 (g) + 2 H2 (g) N2H4 (l)

ΔH = 50.63 kJ/mol, ΔS = -331.4 J/K mol

Gibbs Free Energy (G)

A single equation to predict reaction spontaneity.

Considers only the system, not the surroundings.

Gibbs free energy is a state function.

ΔG° = ΔH° - TΔS°

Where Does ΔG° Come From?

-TΔS°(universe) = ΔH(system) – TΔS(system)

ΔG = -TΔS°(universe)

Under standard conditions:


ΔS°(universe) = ΔS°(surroundings) + ΔS°(system)

ΔS°(universe) = - ΔH(system)

T + ΔS°(system)

ΔG° = 0 reaction at equilibrium

ΔG° < 0 reaction spontaneous in direction written

ΔG° > 0 reaction not spontaneous

Interpreting ΔG°

Lowest possible free energy at equilibrium.

Reactant Favored Product Favored

T

Converting between ΔGrxn° and ΔGrxn

For a reaction under nonstandard conditions

ΔGrxn = ΔGrxn° + RT lnQ

R = ideal gas constant Q = reaction quotient

At equilibrium: (i.e. ΔGrxn = 0)

0 = ΔGrxn° + RT lnK

ΔGrxn° = -RT lnK

If K >1 (products favored) then ΔGrxn° must be negative (spontaneous reaction)

If K <1 (reactants favored) the ΔGrxn° must be positive (not spontaneous reaction)

Gibbs Free Energy: A Summary

At equilibrium, Gibbs free energy is at a minimum

Reactions always proceed toward equilibrium

When ΔGrxn is negative (Q < K), the rxn is spontaneous

When ΔGrxn is positive (K > Q), the reaction is

not spontaneous in that direction

When ΔGrxn = 0, the reaction is at equilibrium (Q = K)

For ΔGrxn° (under standard conditions):

ΔGrxn° = 0 rxn is at equilibrium (K = 1)

ΔGrxn° < 0 rxn spontaneous in direction written (K > 1)

ΔGrxn° > 0 rxn not spontaneous (K < 1)

Calculating Free Energy for a Reaction

Standard free energy of formation (ΔGf°):

The free energy change when one mole of compound is formed from elements in their standard states

ΔGf° = 0 for an element in its standard state

ΔGf° can be used to calculate ΔGrxn°

ΔGrxn° = ΣΔGf°(products) - ΣΔGf°(reactants)

free energy products < free energy reactants: spontaneous reaction (ΔGrxn° < 0)

free energy products > free energy reactants not a spontaneous reaction (ΔGrxn° > 0)

Problems Involving Gibbs Free Energy

Three types of problems:

1. Using ΔHf° and ΔS° values, calculate ΔGrxn°

2. Using ΔGf° values, calculate ΔGrxn°

3. Knowing K or Q, calculate ΔGrxn° or ΔGrxn




ΔGrxn° = -RT lnK

Calculating Free Energy

Calculate ΔGrxn° for the following reaction at 298 K, using values of ΔHf° and ΔS°. Is the reaction spontaneous?

Does the reaction favor products or reactants?

C (graphite) H2 (g) CH4 (g)

ΔHf° (kJ/mol) 0 0 -74.9

S° (J/Kmol) 5.6 130.7 186.3

Remember units of ΔHf° and ΔS° and are different! (kJ versus J)

C (graphite) + 2 H2 (g) CH4 (g)

ΔHrxn° = ΣΔHf°(products) - ΣΔHf°(reactants) ΔHrxn° = [1 mol CH4 x -74.9] – [1 mol C x 0 + 2 mol H2 x 0] = -74.9 kJ/mol



ΔSrxn° = [1 mol CH4 x 186.3] – [1 mol C x 5.6 + 2 mol H2 x 130.7] = -80.7 J/K mol

ΔGrxn° = -74.9 kJ/mol – [(298 K)(-0.0807 kJ/K mol)]

ΔGrxn° is negative therefore the reaction is spontaneous

ΔGrxn° = -RT lnK

ΔGrxn° = ΔHrxn° - TΔSrxn°

= -50.9 kJ/mol

Product favored (K > 1)

We can actually calculate K!

K = 8.36 x 108

Calculate ΔGrxn° for the following reaction, using ΔGf° for reactants and products. Is the reaction spontaneous?

Does the reaction favor products or reactants?

CH4 (g) + 2 O2 (g) 2 H2O (g) + CO2 (g)

CH4 (g) O2 (g) H2O (g) CO2 (g)

ΔGf° (kJ/mol) -50.8 0 -228.6 -394.4

Large negative ΔGrxn° means spontaneous reaction Product favored



ΔGrxn° = [2 mol H2O x -228.6 + 1 mol CO2 x -394.4] – [1 mol CH4 x -50.8 + 2 mol O2 x 0] = -800.8 kJ/mol

Relating ΔGrxn° and K

Calculate ΔGrxn° for the following reaction and then determine the equilibrium constant at 25 °C.

½ N2 (g) + 3/2 H2 (g) NH3 (g)

ΔGf°(NH3 (g)) = -16.37 kJ/mol (from table or calculate from reaction equation)

ΔGrxn° = -RT lnKp

Convert R and ΔGrxn° to same units!

lnKp =6.64

Kp = 7.65 x 102

-16.37 kJ/mol = -(0.008314 kJ/K mol)(298 K)lnKp

-16.37 kJ/mol = (-2.48 kJ/mol)lnKp

Calculating K

What is the equilibrium constant for the following reaction at 276 K? ΔH° = 41.2 kJ, ΔS° = 42.1 J

CO2 (g) + H2 (g) CO (g) + H2O (g)


ΔGrxn° = 41.2 kJ – (276 K)(0.0421 kJ/K)

= 29.58 kJ

ΔGrxn° = -RT lnK

29.58 kJ = -(0.008314 kJ/K mol)(276 K) lnK

-12.89 = lnK

K = 2.52 x 10-6

The value of Ksp for AgCl (s) at 25 °C is 1.8 x 10-10. Calculate ΔGrxn° for the following reaction at 25 °C

Ag+ (aq) + Cl- (aq) AgCl (s)

Ksp is the equilibrium constant for the reverse reaction, so the K for the given reaction is:

K = 1/Ksp = 5.6 x 109

ΔGrxn° = -(8.3145 J/K mol)(298.15 K) ln(5.6 x 109)

ΔGrxn° = -55.6 kJ/mol

The reaction is product favored at equilibrium (i.e. favors AgCl precipitation)

Relating ΔGrxn° and K

ΔGrxn° = -RT lnK

Can a reactant favored reaction turn to product favored by a change of temperature?

Yes. This is vital to chemical industry.

Where is the line between reactant and product favored?

Solve for when ΔGrxn° = 0 and find T

ΔGrxn° = 0 marks the line between product or reactant favored

Free Energy and Temperature

Free Energy and Temperature

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

Free energy is a function of temperature. Is the following reaction spontaneous at 298 K?

Entropically favored (more moles of product than reactant)

Enthalpy disfavored (endothermic)

Reaction spontaneity depends on temperature!


ΔGf° (kJ/mol) -1129.2 -603.4 -394.4

ΔHf° (kJ/mol) -1207.6 -635.1 -393.5

S° (J/K mol) 91.7 38.2 213.7

ΔHrxn° = 179 kJ ΔSrxn° = 160.2 J/K

at 298 K

ΔGrxn° > 0 until the temperature goes above 1118 K

Qualitative Fun!

Without doing any calculations, match the following thermodynamic properties with their appropriate

numerical sign for the following exothermic reaction.

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

ΔHrxn

ΔSrxn

ΔGrxn

ΔSuniverse

1. > 0

2. < 0

3. = 0

4. > 0 low T, < 0 high T

5. < 0 low T, > 0 high T

2

1

2

1

Count the number of moles.

Summary

ΔS = qrev

T

S = k lnW


ΔGrxn° = -RT lnKp


ΔHrxn° = ΣΔHf°(products) - ΣΔHf°(reactants)




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