Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect.

Chapter 16

Spontaneity, Entropy and Free energy

Contents Spontaneous Process and Entropy Entropy and the second law of thermodynamics The effect of temperature on spontaneity Free energy Entropy changes in chemical reactions: The

Third Law of Thermodynamics Free energy and chemical reactions The dependence of free energy on pressure Free energy and equilibrium Free energy and work

16.1 Spontaneous Process and Entropy A process that will occur without outside

intervention. Thermodynamics can’t determine how fast the

process is (may be fast or slow). Kinetics tell us that the rate of reaction depends

on: activation energy; temperature, concentration and catalysts; i.e, it depends on the pathway of process.

Thermodynamics compares initial and final states and does not require knowledge of the pathway.

Kinetics describes pathway between reactants and products.

We need both thermodynamics and kinetics to describe a reaction completely.

Spontaneity process that will occur without outside intervention. Rolling a ball down a hill (gravity); steel rusting (?);

wood burning (exothermic process); transfer of heat from hot to cold (exothermic); freezing of water (exothermic); Melting of ice (endothermic!!); etc.

What common characteristic derives all those processes to be spontaneous?

It is an increase in a property called Entropy, S.

Thus all spontaneous processes occur as a result of an increase of the S of the universe.

What is Entropy?

Entropy, S, is measure of randomness, or disorder in the system. For example molecular randomness.

Naturally, things move from order to disorder! (from lower S to higher S.)

EntropyS defined in terms of probability.Substances take the most likely

arrangement (that have more probabilities).

The most likely arrangement is the most randomrandom.

Can we calculate the number of arrangements for a system?

Entropy and Gases Gas placed in one bulb of a

container will spontaneously expand to fill the entire vessel (vessel of two bulbs) evenly.

Probabilities of finding equal number of molecules in each bulb are huge.

Finding molecules in one bulb is highly improbable; thus the process does not occur spontaneously.

Entropy and state of matter Gases completely fill their chamber

because there are many more ways to do that than to leave half empty.

Ssolid <Sliquid <<Sgas In solids, molecules are very close and thus

they have relatively few positions available to them

Entropy also describes the number of possible positions of a molecule

There are many more ways for the molecules to be arranged as a liquid than a solid.

Gases have a huge number of positions possible.

Example

Which has higher positional S?– Solid CO2 or gaseous CO2?– N2 gas at 1 atm or N2 gas at 0.01 atm?

Predict the sign of S– Solid sugar is added to water

• + because larger volume– Iodine vapor condenses on cold surface to

form crystals

• - because gs, less volume

Positional EntropyPositional Entropy

Entropy also describes the number of

possible positions of a molecule A gas expands into a vacuum because

the expanded state has the highest

positional probability of states available

to the system; i.e., the highest positional

entropy Probability depends on the number of

configurations in space ( positional

microstates)

Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate; there is an increase in entropy Generally, in any spontaneous process, there is always an increase in the entropy of the universe This is the second law of thermodynamicssecond law of thermodynamics;

Suniv = Ssys + Ssurr

IfSuniv is positive the process is spontaneous. IfSuniv is negative the process is spontaneous in the opposite direction. Suniv > 0 for any spontaneous processFirst law: The energy of a universe is constantFirst law: The energy of a universe is constant

16.2 Entropy and the second law of thermodynamics

For exothermic processes Ssurr is positive For endothermic processes Ssurr is negative Consider this process (endothermic)

H2O(l)H2O(g) Ssys is positive (change from L to G) Ssurr is negative (random motion of atoms in the

surrounding decreases) Which one will control the process: Ssys or Ssurr? Suniv;

If the ∆SIf the ∆Ssyssys and ∆S and ∆Ssurrsurr have different signs, the have different signs, the temperature determines the ∆Stemperature determines the ∆Sunivuniv

16.3 The effect of temperature on spontaneity

Determining ∆Ssurr Sign of ∆Ssurr

– depends on direction of heat flow

– ∆Ssurr + for exothermic reactions

– ∆Ssurr - for endothermic reactions Magnitude of ∆Ssurr

– Depends on temperature– Heat flow = ∆H at constant P– Very small at high T, increases as T

decreases

T

HS reactionsurr

SsysSsurr Suniv Spontaneous?

+ + +

- --

+ - ?

+- ?

No

Yes

Yes at Low temp.

Yes at High temp.

Entropy of Sys and Ssurr

in determining the sign of Suniv

Ssurr = -H/T

Example A process has a H of +22 kJ and a S of -13

J/K. At which temperatures is the process spontaneous?

– if there is no subscript, S = Ssys

Suniv > 0 to be spontaneous Ssys + Ssurr > 0

KTK

J

T

JT

J

K

JT

HS reactionsys

1700

13000,22

0000,22

13

0

Example

For methanol, the enthalpy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 213 J/K. What is the normal boiling point of methanol?

Ssys= 213 J/K and H = 71.8 kJ/molK = 71,800 J/molK

– at the boiling point, the vaporization begins to be spontaneous

Suniv = 0 to be at bp Ssys + Ssurr = 0

CKTK

J

T

JT

J

K

J

T

HS reactionsys

64337213800,71

0800,71

2130

16.4 Free Energy, G G: Gibbs free energy G: helps the determination of the T dependence

on spontaneity G ≡ H - TS definition of G ∆G ≡ ∆ H - T ∆ S for constant T

All quantities refer to the system. When no subscript the quantity refers to

the system

T-

S T -

T-

H

T-

G

syssurr S S

T

G

If G is negative at constant T and P, the process is spontaneous.

The process is spontaneous in the direction G decreases

- G means + Suniv

So what sign would the ∆G of reaction with a + ∆Suniv have?

– Negative

syssurruniv S S T

GS

at constant T and P

Example

Is the following reaction spontaneous at -10°C in

the forward direction? H2O(s) H2O(l)

– where ∆H°=6.03x103 J/mol and ∆S°=22.1 J/mol.K

mol

J102.2G

)molK

J22.1( 63K)2( - )

mol

J106.03( G

2

3

∆Go = ∆ Ho - T ∆ So

J/molK -0.82.12 9.22 Suniv

Another way to check for spontaneity of a reaction?

– Check to see if the ∆Suniv is positive

– How can we solve for ∆Ssurr?– use ∆H° and T

At -10°C, is it spontaneous?

molKJsurr /9.22263K

J/mol1003.6 S

3

NO!- the reverse is spontaneous

T

HS reactionsurr

syssurruniv S SS

Example 2 At what T is this reaction

spontaneous at 1 atm

Of pressure?

– Br2(l) Br2(g)– ∆Ho=31.0 kJ/mol, – ∆So=93.0 J/molK

K 333 )

molKJ

0.39(

)mol

J100.13

T

ΔS

ΔH T

ST- H ΔG

0 G point, boiling At the

3

o

o

oo

o

o

G=H-TS

HS Spontaneous?

+ - At all Temperatures

+ + Spontaneous at high temperatures

- - Spontaneous at low temperatures

+- Not spontaneous at any temperature,

Reverse is spontaneous

16.5 Entropy changes in chemical reactionsThe Third Law of Thermodynamics

16.5 Entropy changes in chemical reactionsThe Third Law of Thermodynamics

So far we dealt with physical changesIn a chemical reaction, when the number of

gaseous molecules increases, the positional

disorder will increase and consequently S would

be Positive and Visa Versa Consider the following examples:

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

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

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

The changes in enthalpy determines the exothermicity and endothermicity at constant P

The changes in entropy determines the

spontaneity at constant P and T Are there values given for S? The entropy of a pure crystal at 0 K is 0.

– Molecular motion is almost zero and only one arrangement is possible.

Thus, the entropy of a perfect crystal is zero; this statement is called the Third Law of Thermodynamics

At T >0, some changes in order will occur, consequently, S >0

This value gives us a starting point. Standard Entropies Sº ( at 298 K and 1 atm) of

substances are listed. Sº can then be determined for products and

reactants

Sreaction = npS(products) nrS(reactants)

Entropy is a state function of the system (It is not pathway dependent)

Entropy is an extensive property. It depends on the amount of substance present

Number of moles of reactants and products must be taken into account

What is the expected S (0, 0r +ve or –ve) for the following reaction?

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

More complex molecules possess higher Sº

Example

Find the ∆S° at 25°C for:

– 2NiS(s) + 3O2(g) 2SO2(g) + 2NiO(s)

53 205 248 38

K

J

molK

Jmol

molK

Jmol

molK

Jmol

molK

JmolSreaction

149)205(3)53(2

)38(2)248(2

16.6 Free Energy and Chemical Reactions

Gº is needed when dealing with chemical reactions

Gº = standard free energy change. Free energy change that will occur if

reactants in their standard state turn to products in their standard state.

N2(g) + 3H2(g) 2NH3(g) Gº= -33.3 kJ

Gº can’t be measured directly, but can be calculated from other measurements.

Gº=Hº-TSº

Free Energy in Reactions

There are tables of Gºf . The standard free energy of formation for

any element in its standard state is 0. Why Gº is useful? To compare relative tendencies of

reactions to occur. The more –ve the value of Gº, the further

a reaction will go to the right to reach equilibrium

How do we calculate Gº?There are three Ways to Calculate ∆G°

Use equation: ∆G° ≡ ∆ H ° - T ∆ S ° Use Hess’s Law

– Rearrange equations to get the given equation

– Add ∆G° values for the equations Use ∆Gf° : standard free energy of formation

)tan()( tsreacfrproductsfpreaction GnGnG

1. Use the equation: Gº=Hº-TSº This equation is applicable for

reactions taking place at constant temperature

2. Claculation of Gº using Hess’s law

Free energy is a state function just like enethalpy, i.e., it depends upon the pathway of the reaction

It can be found same as H using Hess’s law

3. Calculation of Gº using the standard free energies of formation Gºf

Gºf of a substance is – the change in free energy that

accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states

Gº for a specific reaction is obtained from the equation:

G = npGf(products) nrGf(reactants)

There are tables of Gºf . The standard free energy of

formation for any element in its standard state is 0.

Remember also that the number of moles of each reactant and product must be used in calculating Gº for a reaction

Exercise Is the following reaction spontaneous under the

standard conditions? :

C2H4(g) + H2O(l) C2H5OH(l)– Find Gº for the reaction using Gºf values

from the table– If Gº is –Ve, the process is spontaneous and

the formation of ethanol is favorable– Of course to judge whether the reaction is

feasible or not, we have to study its kinetics to know whether the reaction is fast or slow.

– If it is slow adding a catalyst may be considered

– Also, remember that Gº depends on temperature: Go=Ho-TSo

16.7 The Dependence of Free Energy on Pressure

At a large volume the gas has many more positions available for its molecules than at low volumes

S large V > S small V S low P > S high P Since S depends on P then G will depend on P.

It can be shown that: G = Go + RTln(P) (Go = free energy at 1 atm)

By proper derivation we got: G = Gº +RTln(Q)

– where Q is the reaction quotient (P of the products /P of the reactants).

16.8 Free energy and equilibrium

According to thermodynamics the equilibrium occurs at the lowest value of free energy available

At equilibrium G = 0, Q = K G = Gº +RT ln(Q) Thus, Gº = -RT lnK

Gº K

=0 =1

<0 >0

>0 <0

Gº = -RT lnK

Temperature dependence of K

Gº= -RT lnK = Hº - TSº ln(K) = - Hº/RT + Sº/RPlot ln(K) VS 1/TA straight line gives a:

Slope = Hº/R

Intercept = Sº/R

Example

16.9 Free energy and Work

Free energy is that energy free to do work. The maximum amount of work possible at a

given temperature and pressure. Never really achieved because some of the

free energy is changed to heat during a change, so it can’t be used to do work.

Reversible v. Irreversible Processes

Reversible: The universe is exactly the same as it was before the cyclic process.

Irreversible: The universe is different after the cyclic process.

All real processes are irreversible -- (some work is changed to heat)