ThermodynamicsPart 5 - Spontaneity1ThermodynamicsThermodynamics = the study of energy changes that accompany physical and chemical changes.Enthalpy (H): the total energy stored within a substanceEnthalpy Change (H): a comparison of the total enthalpies of the product & reactants.H = Hproducts - Hreactants2Exothermic vs. Endothermic Exothermic reactions/changes: release energy in the form of heat; have negative H values.H2O(g) H2O(l)H = -2870 kJEndothermic reactions/changes: absorb energy in the form of heat; have positive H values.H2O(l) H2O(g)H = +2870 kJ
3Reaction PathwaysChanges that involve a decrease in enthalpy are favored!EndothermicExothermictimetimeEaEaPPRR4EntropyEntropy (S): the measure of the degree of disorder in a system; in nature, things tend to increase in entropy, or disorder.S = Sproducts Sreactants
All physical & chemical changes involve a change in entropy, or S. (Remember that a high entropy is favorable)
9Driving Forces in ReactionsEnthalpy 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.10Free EnergyFree 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.11Free EnergyG = H TS
Where: G = change in free energy (kJ)H = change in enthalpy (kJ) T = absolute temp (K)S = change in entropy (kJ/K)
12Free EnergyG: positive (+) value means change is NOT spontaneous
G: negative (-) value means change IS spontaneous
13Relating Enthalpy and Entropy to SpontaneityExampleHSSpontaneity2K + 2H2O 2KOH + H2-+always spon.H2O(g) H2O(l)--spon. @ lower temp.H2O(s) H2O(l)++spon. @ higher temp.16CO2+18H2O2C8H18+25O2+-never spon.14Example #1For the decomposition of O3(g) to O2(g):2O3(g) 3O2(g)H = -285.4 kJ/mol S = 137.55 J/molK @25 Ca) Calculate G for the reaction.
G = (-285.4 kJ/mol) (298K)(0.13755KJ/molK)
G = -326.4 kJ
15Example #1For the decomposition of O3(g) to O2(g):2O3(g) 3O2(g)H = -285.4 kJ/mol S = 137.55 J/molK @25 Cb) Is the reaction spontaneous?
YES16Example #1For the decomposition of O3(g) to O2(g):2O3(g) 3O2(g)H = -285.4 kJ/mol S = 137.55 J/molK @25 Cc) Is H or S (or both) favorable for the reaction?
Both S and H are favorable (both are driving forces)
17Example #2What is the minimum temperature (in C) necessary for the following reaction to occur spontaneously?Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)H = +144.5 kJ/mol; S = +24.3 J/Kmol(Hint: assume G = -0.100 kJ/mol)
G = H TS-0.100 = (144.5) (T)(0.0243) T 5950 KT = 5677 C18