1. Dissolution of NH4NO3 What does it mean?

2. Control of Chemical Reactions

3. 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.

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

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

6. Trends in entropy

7. Which has the greatest entropy?
Cl2(g)
Cl(g)
2 Cl(g)
H2O(s)

8. Entropy Change For the System If the system gets more random,
DS is positive. (Favors the reaction)
If the system gets more ordered,
DS is negative. (Disfavors the reaction)

9. 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 oC?

10. Calculating S All other reactions
C3H8(g) O2(g)  3 CO2(g) H2O(l)
-377 J/Kmol

11. Return to: What is entropy, anyway?
Consider two atoms in a box: forced onto left side,
viewed as quadrants

12. Consider two atoms in a box: forced onto left side,
viewed as quadrants

14. Calculate entropy change for formation of rain:
H2O(g)  H2O(l)

15. What types of reactions lead to increased entropy?

17. 1
Inc
Dec
Can’t tell

18. 2
Inc
Dec
Can’t tell

19. 3
Inc
Dec
Can’t tell

20. 4
Inc
Dec
Can’t tell

21. Entropy vs. Enthalpy Control of Reactions
Second law of thermodynamics:
Suniverse = Ssystem + Ssurroundings

22. Question: How can rain form? H2O(g)  H2O(l)
S = J/K

23. Calculate Suniverse for H2O(g)  H2O(l) at:
25oC
130 oC

24. 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.

25. 2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
H = +468 kJ
S = +561 J/K G = H - TS
What is G at 25 oC and at 1000 oC?

26. Breaking bonds releases energy:
True
False

27. Bond energies and enthalpy change review:

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

29. Temperature Domains and Reaction FavorabilityH + S -

30. A:
High temperature
Low temperature
All temperatures
No temperatures

31. A:
High temperature
Low temperature
All temperatures
No temperatures

32. A:
High temperature
Low temperature
All temperatures
No temperatures

33. A:
High temperature
Low temperature
All temperatures
No temperatures

34. Temperature free energy plots:

35. Temperature free energy plots:

36. Temperature free energy plots:

37. Temperature free energy plots:

38. 2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
H = +468 kJ
S = +561 J/K
In what temperature range
will this reaction be favored?
High or low?
What temperature?

40. Free Energy of Formation: Only used at 25 oC

41. Free Energy of Formation: Only used at 25 oC