ThermodynamicsSpontaneity, Entropy and Free

Energy

Thermodynamics

- Thermodynamics studies how changes in energy,

entropy and temperature affect the spontaneity of

a process or chemical reaction. - Using thermodynamics we can predict the

direction a reaction will go, and also the

driving force of a reaction or system to go to

equilibrium.

Spontaneity

- A spontaneous process is one that occurs

without outside intervention. Examples include - - a ball rolling downhill
- - ice melting at temperatures above 0oC
- - gases expanding to fill their container
- - iron rusts in the presence of air and water
- - two gases mixing

Spontaneity

- Spontaneous processes can release energy (a

ball rolling downhill), require energy (ice

melting at temperatures above 0oC), or involve no

energy change at all (two gases mixing) . - Spontaneity is independent of the speed or rate

of a reaction. A spontaneous process may proceed

very slowly.

Spontaneity

- There are three factors that combine to predict

spontaneity. They are - 1. Energy Change
- 2. Temperature
- 3. Entropy Change

Entropy

- A measure of randomness or disorder

Entropy

- Entropy, S, is a measure of randomness or

disorder. The natural tendency of things is to

tend toward greater disorder. This is because

there are many ways (or positions) that lead to

disorder, but very few that lead to an ordered

state.

Entropy

- The entropy of a system is defined by the

Boltzmann equation - S k ln W
- k is the Boltzmann constant, and W is the

number of energetically equivalent ways to

arrange the components of the system.

Entropy

- Gases will spontaneously and uniformly mix

because the mixed state has more possible

arrangements (a larger value of W and higher

entropy) than the unmixed state.

Entropy

- The driving force for a spontaneous process is an

increase in the entropy of the universe.

?So and Phase Changes

Gases have more entropy than liquids or solids.

?So and Mixtures

Mixtures have more entropy than pure substances.

Entropy Values

- All substances have zero disorder, and an

entropy of zero, at 0 K. As a result,

calorimetry can be used to measure absolute

values of entropy, rather than changes in

entropy. - Entropy values are tabulated, and can be looked

up in thermodynamic tables of data.

Entropy Values of Common Substances

Predicting the sign of ?So

- For many chemical reactions or physical

changes, it is relatively easy to predict if the

entropy of the system is increasing or

decreasing. - If a substance goes from a more ordered phase

(solid) to a less ordered phase (liquid or gas),

its entropy increases.

Predicting the sign of ?So

- For chemical reactions, it is sometimes

possible to compare the randomness of products

versus reactants. - 2 KClO3(s) ? 2 KCl(s) 3 O2(g)
- The production of a gaseous product from a

solid reactant will have a positive value of ?So.

Calculating Entropy Changes

- Since entropy is a measure of randomness, it is

possible to calculate absolute entropy values.

This is in contrast to enthalpy values, where we

can only calculate changes in enthalpy. - A perfect crystal at absolute zero has an

entropy value (S) 0. All other substances have

positive values of entropy due to some degree of

disorder.

Calculating Entropy Changes

- Fortunately, the entropy values of most common

elements and compounds have been tabulated. Most

thermodynamic tables, including the appendix in

the textbook, include standard entropy values,

So.

Entropy Values of Common Substances

Entropy Values

- For comparable structures, the entropy increases

with increasing mass

Entropy Values

- For molecules with similar masses, the more

complex molecule has greater entropy. The

molecule with more bonds has additional ways to

absorb energy, and thus greater entropy.

Calculating Entropy Changes

- For any chemical reaction,
- ? Soreaction Smolprod Soproducts- Smolreact

Soreactants - The units of entropy are joules/K-mol.

The 2nd Law of Thermodynamics

- In any spontaneous process there is always
- an increase in the entropy of the universe.

The 2nd Law of Thermodynamics

- Water spontaneously freezes at a temperature

below 0oC. Therefore, the process increases the

entropy of the universe. - The water molecules become much more ordered as

they freeze, and experience a decrease in

entropy. The process also releases heat, and

this heat warms gaseous molecules in air, and

increases the entropy of the surroundings.

The 2nd Law of Thermodynamics

- Since the process is spontaneous below 0oC,

?Ssurr, which is positive, must be greater in

magnitude than ?S of the water molecules.

Entropy

- Entropy can be viewed as the dispersal or

randomization of energy. The freezing of water

(an exothermic process) releases heat to the

surroundings, and thus increases the entropy of

the surroundings. The process is spontaneous at

or below 0oC because the increase in entropy of

the surroundings is greater than the decrease in

entropy of the water as it freezes.

? S and Spontaneity

Spontaneity

- Entropy, temperature and heat flow all play a

role in spontaneity. A thermodynamic quantity,

the Gibbs Free Energy (G), combines these factors

to predict the spontaneity of a process. - ?G ?H - T?S

Spontaneity

- ?G ?H - T?S
- If a process releases heat (?H is negative) and

has an increase in entropy (?S is positive), it

will always be spontaneous. - The value of ?G for spontaneous processes is

negative.

Spontaneity

?G ?H - T?S

Spontaneity and ?G

- If ?G is negative, the process is spontaneous

(and the reverse process is non-spontaneous). - If ?G is positive, the process is

non-spontaneous, and the reverse process is

spontaneous. - If ?G 0, the system is at equilibrium.

?G

- Although ?G can be used to predict in which

direction a reaction will proceed, it does not

predict the rate of the reaction. - For example, the conversion of diamond to

graphite has a ?Go -3 kJ, so diamonds should

spontaneously change to graphite at standard

conditions. However, kinetics shows that the

reaction is extremely slow.

The Significance of ?G

- ?G represents the driving force for the

reaction to proceed to equilibrium.

The Significance of ?G

- If negative, the value of ?G in KJ is the

maximum possible useful work that can be obtained

from a process or reaction at constant

temperature and pressure. - In practice, some energy is always lost, so the

actual work produced will be less than the

calculated value.

The Significance of ?G

- If positive, the value of ?G in KJ is the

minimum work that must be done to make the

non-spontaneous process or reaction proceed. - In practice, some additional work is required

to make the non-spontaneous process or reaction

proceed.

Calculation of ?Go

- ?Go, the standard free energy change, can be

calculated in several ways. - ?Go ?Ho - T ?So
- It can be calculated directly, using the

standard enthalpy change and entropy change for

the process.

Calculation of ?Go

- ?Go ?Ho - T ?So
- ?Ho is usually calculated by using standard

enthalpies of formation, ?Hfo. - ?Horxn Snprod ?Hoproducts- Snreact

?Horeactants

Calculation of ?Go

- ?Go ?Ho - T ?So
- Once ?Ho and ?So have been calculated, the

value of ?Go can be calculated, using the

temperature in Kelvins.

Calculation of ?Go

- ?Go can also be calculated by combining the

free energy changes of related reactions. This

is the same method used in Hess Law to calculate

enthalpy changes. If the sum of the reactions

gives the reaction of interest, the sum of the

?Go values gives ?Go for the reaction.

Calculation of ?Go

- Lastly, ?Go can be calculated using standard

free energies of formation, ?Gfo. Some tables of

thermodynamic data, including the appendix of

your textbook, include values of ?Gfo. - ?Gorxn Smolprod ?Gfo prod - Smolreact ?Gfo

react

Calculation of ?Go

- When calculating ?Go from standard free

energies of formation, keep in mind that ?Gfo for

any element in its standard state is zero. As

with enthalpies of formation, the formation

reaction is the reaction of elements in their

standard states to make compounds (or

allotropes).

Calculation of ?Go

Calculation of ?Go

Note the values of zero for nitrogen, hydrogen

and graphite.

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Calculate ?Go using the tables in the appendix

of your textbook. Is the process spontaneous at

this temperature? Is it spontaneous at all

temperatures? If not, at what temperature does

it become spontaneous?

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Calculate ?Go using the tables in the appendix

of your textbook. Is the process spontaneous at

this temperature? - Calculation of ?Grxno will indicate spontaneity

at 25oC. It can be calculated using - ?Gfo values or from ?Hfo and ?So values.

Calculation of ?Go

- CaCO3(s) ?CaO(s) CO2(g)
- ?Grxno Snprod ?Gfo prod - Snreact ?Gfo react

Calculation of ?Go

- CaCO3(s) ?CaO(s) CO2(g)
- ?Grxno (1 mol) (-604.0 kJ/mol) (1

mol)(-394.4 kJ/mol) 1 mol(-1128.8 kJ/mol)

Calculation of ?Go

- CaCO3(s) ?CaO(s) CO2(g)
- ?Grxno (1 mol) (-604.0 kJ/mol) (1

mol)(-394.4 kJ/mol) 1 mol(-1128.8 kJ/mol)

130.4 kJ

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Calculate ?Go using the tables in the appendix

of your textbook. Is the process spontaneous at

this temperature? - Since ?Grxno 130.4 kJ, the reaction is not

spontaneous at 25oC.

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Is it spontaneous at all temperatures? If

not, at what temperature does it become

spontaneous?

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Is it spontaneous at all temperatures? If

not, at what temperature does it become

spontaneous? - At 25oC, ?Grxno is positive, and the reaction

is not spontaneous in the forward direction.

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Is it spontaneous at all temperatures? If

not, at what temperature does it become

spontaneous? - Inspection of the reaction shows that it

involves an increase in entropy due to production

of a gas from a solid.

Spontaneity Problem

- Consider the reaction
- CaCO3(s) ?CaO(s) CO2(g) at 25oC.
- Is it spontaneous at all temperatures? If

not, at what temperature does it become

spontaneous? - We can calculate the entropy change and the

enthalpy change, and then determine the

temperature at which spontaneity will occur.

CaCO3(s) ?CaO(s) CO2(g)

- Since ?Go ?Ho - T?So, and there is an

increase in entropy, the reaction will become

spontaneous at higher temperatures. - To calculate ?So, use the thermodynamic tables

in the appendix.

CaCO3(s) ?CaO(s) CO2(g)

?Srxno 1mol(213.6J/K-mol)1mol(39.7J/K-mol) -

1mol(92.9J/K-mol) 160.4 J/K

CaCO3(s) ?CaO(s) CO2(g)

- ?Go ?Ho - T?So
- Since we know the value of ?Go (130.4 kJ) and

?So (160.4 J/K), we can calculate the value of

?Ho at 25oC. - 130.4 kJ ?Ho (298K) (160.4 J/K)
- ?Ho 130.4 kJ (298K) (.1604 kJ/K)
- ?Ho 178.2 kJ

CaCO3(s) ?CaO(s) CO2(g)

- ?Go ?Ho - T?So
- If we assume that the values of ?Ho and ?So

dont change much with temperature, we can

estimate the temperature at which the reaction

will become spontaneous.

CaCO3(s) ?CaO(s) CO2(g)

- ?Go ?Ho - T?So
- ?Go is positive at lower temperatures, and will

be negative at higher temperatures. Set ?Go

equal to zero, and solve for temperature. - 0 ?Ho - T?So
- T ?Ho
- ?So

CaCO3(s) ?CaO(s) CO2(g)

- ?Go ?Ho - T?So
- 0 ?Ho - T?So
- T ?Ho
- ?So
- T (178.2 kJ)/(160.4 J/K)(10-3kJ/J)
- 1111K or 838oC
- The reaction will be spontaneous in the forward

direction at temperatures above 838oC.

Coupled Biological Processes

- Many biochemical reactions have positive values

of ?Go, and would not occur spontaneously. These

processes, such as the synthesis of ATP

(adenosine triphosphate), are driven by favorable

reactions such as the conversion of glucose to

carbon dioxide and water. - ATP is essentially a way for the body to store

free energy until it is needed for cellular

processes.

Coupled Biological Processes