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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
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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. ΔHrxn= nΔHproducts- nΔHreactants
Exothermic vs. Endothermic • Exothermic reactions/changes: release energy in the form of heat; have negative ΔH values. H2O(g) H2O(l)ΔH = -2870 kJ • Endothermic 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 Ea Ea energy energy P R R P time time Ea= activation energy; P = products; R = reactants
Entropy • Entropy (S): the measure of the degree of disorder in a system; in nature, things tend to increase in entropy, or disorder. ΔSrxn= nΔ Sproducts– nΔSreactants • All physical & chemical changes involve a change in entropy, or ΔS. (Remember that a high entropy is favorable)
Entropy gases solutions solids liquids
Entropy pure substances mixtures
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
Example #1 • For the decomposition of O3(g) to O2(g): 2O3(g) 3O2(g) ΔH = -285.4 kJ ΔS = 137.55 J/·K @25 °C a) Calculate ΔG for the reaction. ΔG = (-285.4 kJ) – (298K)(0.13755kJ/K) ΔG = -326.4 kJ
Example #1 • For the decomposition of O3(g) to O2(g): 2O3(g) 3O2(g) ΔH = -285.4 kJ ΔS = 137.55 J/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 ΔS = 137.55 J/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; ΔS = +24.3 J/K (Hint: assume ΔG = -0.100 kJ) ΔG = ΔH – TΔS -0.100 = (144.5) – (T)(0.0243) T ≈ 5950 K T = 5677 °C