Thermodynamics

Part 5 - Spontaneity

Thermodynamics

Thermodynamics = the study of energy changes that accompany physical and chemical changes.

Enthalpy (H): the total energy “stored” within a substance

Enthalpy Change (ΔH): a comparison of the total enthalpies of the product & reactants.

ΔH = Hproducts - Hreactants

Exothermic vs. Endothermic

Exothermic reactions/changes: release energy in the form of heat; have negative ΔH values.

H2O(g) H2O(l) ΔH = -2870 kJEndothermic reactions/changes: absorb

energy in the form of heat; have positive ΔH values.

H2O(l) H2O(g) ΔH = +2870 kJ

Reaction Pathways

Changes that involve a decrease in enthalpy are favored!

Endothermic Exothermic

time time

EaEa

P

P

R R

Entropy

Entropy (S): the measure of the degree of disorder in a system; in nature, things tend to increase in entropy, or disorder.

ΔS = Sproducts – Sreactants

All physical & chemical changes involve a change in entropy, or ΔS. (Remember that a high entropy is favorable)

Entropy

Entropy

Entropy

Entropy

Driving Forces in Reactions

Enthalpy and entropy are DRIVING FORCES for spontaneous reactions (rxns that happen at normal conditions)

It is the interplay of these 2 driving forces that determines whether or not a physical or chemical change will actually happen.

Free Energy

Free Energy (G): relates enthalpy and entropy in a way that indicates which predominates; the quantity of energy that is available or stored to do work or cause change.

Free Energy

ΔG = ΔH – TΔS

Where: ΔG = change in free energy (kJ)

ΔH = change in enthalpy (kJ)

T = absolute temp (K)

ΔS = change in entropy (kJ/K)

Free Energy

ΔG: positive (+) value means change is NOT spontaneous

ΔG: negative (-) value means change IS spontaneous

Relating Enthalpy and Entropy to Spontaneity

Example ΔH ΔS Spontaneity

2K + 2H2O 2KOH + H2 - + always spon.

H2O(g) H2O(l) - - spon. @ lower temp.

H2O(s) H2O(l) + + spon. @ higher temp.

16CO2+18H2O2C8H18+25O2 + - never spon.

Example #1

For the decomposition of O3(g) to O2(g):

2O3(g) 3O2(g)

ΔH = -285.4 kJ/mol

ΔS = 137.55 J/mol·K @25 °C

a) Calculate ΔG for the reaction.

ΔG = (-285.4 kJ/mol) – (298K)(0.13755KJ/mol·K)

ΔG = -326.4 kJ

Example #1

For the decomposition of O3(g) to O2(g):

2O3(g) 3O2(g)

ΔH = -285.4 kJ/mol

ΔS = 137.55 J/mol·K @25 °C

b) Is the reaction spontaneous?

YES

Example #1

For the decomposition of O3(g) to O2(g):

2O3(g) 3O2(g)

ΔH = -285.4 kJ/mol

ΔS = 137.55 J/mol·K @25 °C

c) Is ΔH or ΔS (or both) favorable for the reaction?

Both ΔS and ΔH are favorable (both are driving forces)

Example #2

What is the minimum temperature (in °C) necessary for the following reaction to occur spontaneously?

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)ΔH = +144.5 kJ/mol; ΔS = +24.3 J/K·mol

(Hint: assume ΔG = -0.100 kJ/mol)

ΔG = ΔH – TΔS

-0.100 = (144.5) – (T)(0.0243)

T ≈ 5950 K

T = 5677 °C