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Control of Chemical Reactions. Thermodynamic Control of Reactions Enthalpy Bond Energies – Forming stronger bonds favors reactions. – Molecules with strong.

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Presentation on theme: "Control of Chemical Reactions. Thermodynamic Control of Reactions Enthalpy Bond Energies – Forming stronger bonds favors reactions. – Molecules with strong."— Presentation transcript:

1 Control of Chemical Reactions

2 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.

3 Remember Enthalpy? Enthalpy of formation for elements in their natural states = 0.

4 H 2 O(g) Gaseous water -242

5 The Laws of Thermodynamics 1st Law: Energy is Conserved 2 nd Law: Any “spontaneous” process leads to an increase in entropy of the universe.

6 Entropy A measure of randomness. Units of J/K.

7 Trends in entropy

8 Entropy Change First define: SystemSurroundings

9 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)

10 Calculating  S Chemical Reactions C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)

11 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?

12 Calculate entropy change for formation of rain: H 2 O(g)  H 2 O(l)

13 What types of reactions lead to increased entropy?

14

15 Entropy vs. Enthalpy Control of Reactions Second law of thermodynamics:  S universe =  S system +  S surroundings

16 Question: How can rain form? H 2 O(g)  H 2 O(l)

17 Calculate  S universe for H 2 O(l)  H 2 O(g) at: 90 o C 110 o C

18 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.

19 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?

20 Enthalpy vs. Entropy Control of Reactions  G =  H - T  S At high temperatures: At low temperatures:

21 Temperature Domains and Reaction Favorability  H  S -

22 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?

23 Free Energy vs. Temperature Curves

24 Catalytic Converters Nitrogen oxides cause smog. N 2 (g) + O 2 (g)  2 NO(g)  H = +180 kJ  S = +25 J/K

25 Free Energy of Formation: Only used at 25 o C 2 BaO(s) + C(s)  2 Ba(s) + CO 2 (g)


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