Presentation on theme: "Control of Chemical Reactions. Thermodynamic Control of Reactions Enthalpy Bond Energies – Forming stronger bonds favors reactions. – Molecules with strong."— Presentation transcript:
Control of Chemical Reactions
Thermodynamic Control of Reactions Enthalpy Bond Energies – Forming stronger bonds favors reactions. – Molecules with strong bonds are more stable. Entropy Randomness – Reactions that increase random- ness are favored. – Forming gases favors reactions.
Remember Enthalpy? Enthalpy of formation for elements in their natural states = 0.
H 2 O(g) Gaseous water -242
The Laws of Thermodynamics 1st Law: Energy is Conserved 2 nd Law: Any “spontaneous” process leads to an increase in entropy of the universe.
Entropy A measure of randomness. Units of J/K.
Trends in entropy
Entropy Change First define: SystemSurroundings
Entropy Change For the System If the system gets more random, S is positive. (Favors the reaction) If the system gets more ordered, S is negative. (Disfavors the reaction)
Calculating S Chemical Reactions C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l)
Calculating S Special Case: Phase Changes Heat of fusion (melting) of ice is 6000 J/mol. What is the entropy change for melting ice at 0 o C?
Calculate entropy change for formation of rain: H 2 O(g) H 2 O(l)
What types of reactions lead to increased entropy?
Entropy vs. Enthalpy Control of Reactions Second law of thermodynamics: S universe = S system + S surroundings
Question: How can rain form? H 2 O(g) H 2 O(l)
Calculate S universe for H 2 O(l) H 2 O(g) at: 90 o C 110 o C
Putting S, H and Temperature Together Gibb’s Free Energy: G = H - T S When G is negative, reaction is favored. When G is positive, reaction is disfavored.
2 Fe 2 O 3 (s) + 3 C(s) 4 Fe(s) + 3 CO 2 (g) H = +468 kJ S = +561 J/K G = H - T S What is G at 25 o C and at 1000 o C?
Enthalpy vs. Entropy Control of Reactions G = H - T S At high temperatures: At low temperatures:
Temperature Domains and Reaction Favorability H S -
2 Fe 2 O 3 (s) + 3 C(s) 4 Fe(s) + 3 CO 2 (g) H = +468 kJ S = +561 J/K In what temperature range will this reaction be favored? High or low? What temperature?
Free Energy vs. Temperature Curves
Catalytic Converters Nitrogen oxides cause smog. N 2 (g) + O 2 (g) 2 NO(g) H = +180 kJ S = +25 J/K
Free Energy of Formation: Only used at 25 o C 2 BaO(s) + C(s) 2 Ba(s) + CO 2 (g)