Thermodynamics Part 5 - Spontaneity. Thermodynamics Thermodynamics = the study of energy changes...
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Transcript of Thermodynamics Part 5 - Spontaneity. Thermodynamics Thermodynamics = the study of energy changes...
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