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

Thermodynamics

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)

5Entropy

6Entropy

7Entropy

8Entropy

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