Thermodynamics Part 5 - Spontaneity

ΔHrxn = nΔHproducts - nΔHreactants 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 = nΔHproducts - nΔHreactants

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 Endothermic Exothermic 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

ΔSrxn = nΔ Sproducts – nΔSreactants Entropy Entropy (S): the measure of the degree of disorder in a system; in nature, things tend to increase in entropy, or disorder. ΔSrxn = nΔ Sproducts – nΔSreactants All physical & chemical changes involve a change in entropy, or ΔS. (Remember that a high entropy is favorable)

Entropy

Entropy

Entropy gases solids liquids solutions

Entropy pure substances mixtures

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+18H2O2C8H18+25O2 never spon.

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 YES 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 Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) ΔG = ΔH – TΔS 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