Unit 5 – Part 1: Thermodynamics

Entropy and the Second Law of Thermodynamics

Gibbs Free Energy

Free Energy and Equilibrium Constants

Entropy and the 2nd Law of Thermodynamics

Industrial chemists are responsible for designing cost-effective manufacturing processes.

2 NH3 (g) + CO2 (g) NH2CONH2 (aq) + H2O (l)

urea

Commercial uses of urea: nitrogen fertilizer for plants used to manufacture certain

plastics and adhesives

Entropy and the 2nd Law of Thermodynamics

Some of the questions a chemist must consider: Does the reaction need to be heated?

How much product will be present at equilibrium?

Does the reaction naturally proceed in this direction?

Entropy and the 2nd Law of Thermodynamics

In order to determine if a reaction proceeds naturally in the direction written, we need to know if it is spontaneous.

Capable of proceeding in the direction written without needing to be driven by an outside source of energy

Entropy and the 2nd Law of Thermodynamics

Examples of spontaneous processes: an egg breaking when dropped a ball rolling down a hill ice water at room temperature

Examples of non-spontaneous processes: a ball rolling up a hill water ice at room temperature

Entropy and the 2nd Law of Thermodynamics

Examples of spontaneous reactions: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) 2 N (g) N2 (g)

Examples of non-spontaneous reactions: 2 H2O (g) 2 H2 (g) + O2 (g)

O2 (g) 2 O (g)

Entropy and the 2nd Law of Thermodynamics

Many but not all spontaneous processes are exothermic.

Enthalpy (H) alone cannot be used to predict whether or not a reaction is spontaneous.

To predict whether a reaction is spontaneous, we need to use the second law of thermodynamics and a thermodynamic quantity called entropy.

Entropy and the 2nd Law of Thermodynamics

Entropy (S): a thermodynamic quantity related to

the disorder or randomness of a system.

The more disordered or random the system is, the larger its entropy is.

A state function not path dependent

Entropy and the 2nd Law of Thermodynamics

Every chemical has an entropy associated with it that depends on its physical state, temperature, and pressure:

H2O (l) 69.91 J/mol.K at 25oC/1 atm

H2O (g) 188.83 J/mol.K at 100oC/1 atm

Appendix C: table of thermodynamic properties including

S

Entropy and the 2nd Law of Thermodynamics

The entropy change (S) can be calculated for any process:

S = Sfinal - Sinitial

S = Sproducts - Sreactants

Sign conventions for S: S = positive more disordered S = negative less disordered

Entropy and the 2nd Law of Thermodynamics

Example: The following process occurs at 0oC and 1 atm pressure. Does the system become more ordered or more disordered?

H2O (s) H2O (l) S = 22 J /K

Entropy and the 2nd Law of Thermodynamics

The sign of S can be predicted: In general, any change that increases

the overall disorder or randomness will result in a positive value for S.

In general, the overall entropy increases when:

a molecule (or anything else) is broken into two or more smaller molecules

there is an increase in the number of moles of a gas

a solid changes to a liquid or gas a liquid changes to a gas

Entropy and the 2nd Law of Thermodynamics

Example: Without doing any calculations, predict whether S will be positive or negative.

Breaking an egg

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

2NH3(g) + CO2(g) NH2CONH2 (aq) + H2O (l)

Entropy and the 2nd Law of Thermodynamics

The entropy changeS) for a reaction or process can be calculated using the following equation:

So = n Soproducts - m So

reactants

where So = the standard molar entropy

Note: This is similar to the method used to calculate Ho for a reaction!

Entropy and the 2nd Law of Thermodynamics

Standard molar entropy (So) : the entropy value for one mole of a

chemical species in its standard state1 atm pressure1 M (for those in solution)

NOTE: Unlike Hof, the standard molar

entropy of a pure element is NOT zero.

Entropy and the 2nd Law of Thermodynamics

Example: Predict whether the entropy will increase or decrease for the following reaction. Calculate So.

C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g)

Entropy and the 2nd Law of Thermodynamics

The Second Law of Thermodynamics can be used to predict whether a reaction will occur spontaneously. The total entropy of a system and

its surroundings always increases for a spontaneous process.

How does the change in entropy relate to the spontaneity of a chemical reaction or process?

Gibbs Free Energy

Simply looking at the sign of S for a chemical reaction or process does not tell you if the reaction is spontaneous. Spontaneous reactions involve an

overall increase in the entropy of the universe.

Reactions that have a large, negative H tend to be spontaneous:

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) H = - 802 kJ

Gibbs Free Energy

The Gibbs free energy (G) is used to relate both the enthalpy change and the entropy change of a reaction to its spontaneity.

G = H - TS

where G = Gibbs free energy (“free energy”)

H = enthalpyT = temperature (K)S = entropy

Gibbs Free Energy

Free energy is a state function.

The change in free energy (G) of a system can be used to determine the spontaneity of a process or reaction.

For a process occurring at constant temperature:

G = H - TS

Gibbs Free Energy

For a reaction occurring at constant temperature and pressure, the sign of G can be used to determine if a reaction is spontaneous in the direction written: G = negative

reaction is spontaneous in the forward direction

G = zero reaction is at equilibrium

Gibbs Free Energy

The sign of G can be used to determine if a reaction is spontaneous in the direction written (cont): G = positive

reaction is not spontaneous in the direction written

work must be supplied by the surroundings to make the reaction occur in the direction written

reaction is spontaneous in the reverse direction

Gibbs Free Energy

Example: Using the definition of G, calculate the G for the following reaction at 35oC:

2 H+ (aq) + S2- (aq) H2S (g)

H = -61.9 kJS = + 183.6 J/K

Gibbs Free Energy

The standard free energy of formation (Go

f) has been tabulated for many different substances. (see Appendix C) the change in free energy associated

with the formation of 1 mole of a substance from its elements under standard conditions pure solid pure liquid gas at 1 atm pressure solution with 1 M concentration

Gibbs Free Energy

There is not a standard temperature for determining Go

f. 25oC is often used for tables of data values can be calculated at other

temperatures as well

Gof for an element in its standard state

is zero.

Gibbs Free Energy

The standard free energy change for a chemical process can be calculated using the following expression:

Go = n Gof (products) - m Go

f (reactants)

Note: This is similar to the way we calculated Ho and So

Gibbs Free Energy

Example: Calculate Go for the following reaction using the standard free energies of formation.

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)