1. Thermodynamics Review

2. Energy Ability to do work Units– Joules (J), we will use “kJ”
Can be converted to different types
Energy change results from forming and breaking chemical bonds in reactions

3. System vs. Surroundings

4. Heat (q)
Energy transfer between a system and the surroundings due to a temperature change
Transfer is instantaneous from high----low temperature until thermal equilibrium
Temperature—
Measure of heat, “hot/cold”

5. Heat (q) continued Kinetic theory of heat Increase in heat results in
Heat increase resulting in temperature change causes an increase in the average motion of particles within the system.
Increase in heat results in
Energy transfer
Increase in both potential and kinetic energies

6. 1st Law of Thermodynamics (Conservation of Energy)
Energy cannot be created or destroyed.
With physical and chemical changes, energy can be transferred or converted.
Total energy = Σenergy of its components
ΔU = q + w , ΔEtotal = ΔEsys + ΔEsurr = 0

7. Enthalpy

8. Thermodynamics 101 First Law of Thermodynamics
Energy is conserved in a reaction (it cannot be created or destroyed)---sound familiar???
Math representation: ΔEtotal = ΔEsys + ΔEsurr = 0
Δ= “change in”
ΔΕ= positive (+), energy gained by system
ΔΕ= negative (-), energy lost by system
Total energy = sum of the energy of each part in a chemical reaction

9. Enthalpy (H) Measures 2 things in a chemical reaction:
Energy change
Amount of work done to or by chemical reaction
2 types of chemical reactions:
Exothermic—heat released to the surroundings, getting rid of heat, -ΔΗ
Endothermic—heat absorbed from surroundings, bringing heat in, +ΔΗ
**Enthalpy of reaction—heat from a chemical reaction which is given off or absorbed, units = kJ/mol
Enthalpy of reaction
Heat from a chemical reaction which is given off or absorbed
At constant pressure
Units = kJ/mol

10. Exothermic Temperature increase (--isolated system)
Heat is released to surroundings (--open/closed system)
q = - value
Chemical  Thermal Energy

11. Endothermic Temperature decrease (--isolated system)
All energy going into reaction, not into surroundings
Heat absorbed by system, surroundings have to put energy into reaction
q = + value
Thermal  Chemical Energy

13. Methods for Calculating Enthalpy--
Stoichiometric Calculations using Balanced Chemical Equation
Calorimetry (lab based method)
Hess’s Law
Enthalpy of Formation
Bond Enthalpies
**Which method is the LEAST accurate?

14. Calorimetry
How do we find the change in energy/heat transfer that occurs in chemical reactions???

15. Calorimetry
Experimentally “measuring” heat transfer for a chemical reaction or chemical compound
Calorimeter
Instrument used to determine the heat transfer of a chemical reaction
Determines how much energy is in food
Observing temperature change within water around a reaction container
** assume a closed system, isolated container
No matter, no heat/energy lost
Constant volume

16. Specific Heat Capacity
Amount of heat required to increase the temperature of 1g of a chemical substance by 1°C
Units--- J/g°K
Unique to each chemical substance
Al(s) = 0.901J/g°K
H2O(l) = 4.18 J/g°K

17. q = smΔT

18. “Coffee Cup” calorimeter (cont.)
qchemical = -qwater

19. Δqrxn ΔHrxn Heat gained/lost in experiment with calorimeter
Heat gained/lost in terms of the balanced chemical equation

20. Example 2: Using the following data, determine the metal’s specific heat.
Metal mass = 25.0g Water mass = 20.0g
Temperature of large water sample = 95°C
Initial temperature in calorimeter = 24.5°C
Final temperature in calorimeter = 47.2°C
Specific heat of water = 1.00 cal/g°C OR J/g°K (KNOW!!!!)
0.380 cal/g°C

21. Bond Energy Energy required to make/break a chemical bond
Endothermic reactions
Products have more energy than reactants
More energy to BREAK bonds
Exothermic reactions
Reactants have more energy than products
More energy to FORM bonds

22. Bond Enthalpy
Focuses on the energy/heat between products and reactants as it relates to chemical bonding
Amount of energy absorbed to break a chemical bond--- amount of energy released to form a bond.
Multiple chemical bonds take more energy to break and release more energy at formation
Amount of energy absorbed = amount of energy released
to break chemical bond to form a chemical bond

23. Calculating ΔHrxn. by bond enthalpies (4th method)
Least accurate method
ΔH = ΣBE (bonds broken) ΣBE (bonds formed)

24. Example 1:
Using average bond enthalpy data, calculate ΔH for the following reaction.
CH4 + 2O2  CO2 + 2H2O ΔH = ?
Bond
Average Bond Enthalpy
C-H
413 kJ/mol
O=O
495 kJ/mol
C-O
358 kJ/mol
C=O
799 kJ/mol
O-H
467 kJ/mol

25. Hess’ Law
Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step
ΔHrxn = ΣΔHA+B+C (sum of ΔH for each step)
Allows us to break a chemical reaction down into multiple steps to calculate ΔH
Add the enthalpies of the steps for the enthalpy for the overall chemical reaction

26. Example 1:
H2O(l)  H2O (g) ΔH° = ? Based on the following: H2 + ½ O2  H2O(l) ΔH° = kJ/mol H2 + ½ O2  H2O(g) ΔH° = kJ/mol

27. Enthalpy of Formation (ΔHf°)
Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state.
Elements present in “most thermodynamically stable state”
25°C°, 1atm

28. Example 5
Isopropyl alcohol (rubbing alcohol) undergoes a combustion reaction
2(CH3)2CHOH + 9O2  6CO H2O
ΔH° = kJ/mol
Calculate the standard enthalpy of formation for isopropyl alcohol.

29. Example 2:
Calculate the ΔH for the following reaction when 12.8 grams of hydrogen gas combine with excess chlorine gas to produce hydrochloric acid.
H2 + Cl2  2HCl ΔH = kJ

30. Entropy

31. Spontaneous vs. Nonspontaneous
Spontaneous Process
Occurs WITHOUT help outside of the system, natural
Many are exothermic—favors energy release to create an energy reduction after a chemical reaction
Ex. Rusting iron with O2 and H2O, cold coffee in a mug
Some are endothermic
Ex. Evaporation of water/boiling, NaCl dissolving in water

32. Spontaneous vs. Nonspontaneous
2) Nonspontaneous Process
REQUIRES help outside system to perform chemical reaction, gets aid from environment
Ex. Water cannot freeze at standard conditions (25°C, 1atm), cannot boil at 25°C
**Chemical processes that are spontaneous have a nonspontaneous process in reverse **

33. Entropy (S) Measure of a system’s disorder
Disorder is more favorable than order
ΔS = S(products) - S(reactants)
ΔS is (+) with increased disorder
State function
Only dependent on initial and final states of a reaction
Ex. Evaporation, dissolving, dirty house

34. Thermodynamic Laws 1st Law of Thermodynamics 2nd Law of Thermodynamics
Energy cannot be created or destroyed
2nd Law of Thermodynamics
The entropy of the universe is always increasing.
Naturally favors a disordered state

35. When does a system become MORE disordered from a chemical reaction
When does a system become MORE disordered from a chemical reaction? (ΔS > 0)
Melting
Vaporization
More particles present in the products than the reactants
4C3H5N3O9 (l)  6N2 (g) + 12CO2 (g) + 10H2O (g) + O2 (g)
Solution formation with liquids and solids
Addition of heat

36. 3rd Law of Thermodynamics
The entropy (ΔS) of a perfect crystal is 0 at a temperature of absolute zero (0°K).
No particle motion at all in crystal structure
All motion stops

37. How do we determine if a chemical reaction is spontaneous?
Change in entropy (ΔS)
Gibbs Free Energy (ΔG)

38. Change in entropy (ΔS)
For a chemical reaction to be spontaneous (ΔST > 0), there MUST be an increase in system’s entropy (Δssys> 0) and the reaction MUST be exothermic (Δssurr > 0).
Exothermic reactions are favored, NOT endothermic reactions.
Exothermic (ΔH < 0, ΔS > 0)
Endothermic (ΔH > 0, ΔS < 0)
ΔST = Δssys + Δssurr
If ΔST > 0, then the chemical reaction is spontaneous

39. Example 1: Will entropy increase or decrease for the following?
N2 (g) + 3H2 (g)  2NH3 (g)
2KClO3 (s)  2KCl (s) + 3O2 (g)
CO(g) + H2O(g)  CO2 (g) H2 (g)
C12H22O11 (s)  C12H22O11
Decrease in entropy
Increase in entropy
Cannot tell

40. How do we calculate the entropy change (ΔS) in a chemical reaction?
Same method as using the enthalpies of formation to calculate ΔH and use the same table.
aA + bB  cC dD
ΔS° =[c (ΔS°C) + d(ΔS°D)] - [a (ΔS°A) + b (ΔS°B)]

41. Example 2: Calculate ΔS° for the following reaction at 25°C….
4HCl(g) O2 (g)  2Cl2 (g) H2O (g)
ΔS°= J/K, entropy decreases

42. Free Energy and Equilibrium

43. ΔG = ΔG° + RTlnQ
At equilibrium, ΔG = 0, so reaction quotient (Q) = equilibrium constant (K)
At equilibrium
ΔG° = - Rtlnk
Enables the reaction’s equilibrium constant (K) to be calculated from the change in free energy (ΔG°)

44. What is the relationship between free energy(ΔG) and K?
The magnitude of ΔG° indicates how far the chemical reaction in its standard state is from equilibrium.
ΔG° = 0 , equilibrium
ΔG°= large value, far from equilibrium
ΔG° = small value, close to equilibrium
The sign (+, - ) indicates which direction the reaction needs to shift to achieve equilibrium
Positive (+) shift to left, no reaction
Negative (-) shift to right, reaction goes to completion

45. Gibbs Free Energy

46. Change in Gibbs Free Energy (ΔG)
ΔG = ΔH – TΔS
Relates enthalpy and entropy to determine which has more importance in determining whether a reaction is spontaneous
Combines energy transfer as heat (ΔH) and energy released to contribute to disorder (ΔS)

47. Change in Gibbs Free Energy (ΔG)
ΔG = ΔH – TΔS
ΔG < 0 , spontaneous reaction
Energy available to do work
ΔG > 0, nonspontaneous reaction
Energy deficiency, no leftover energy and not enough energy for reaction

48. How can we apply the Gibbs equation to determine spontaneity of reaction?
ΔG = ΔH – TΔS ΔH ΔS ΔG
Result - +
Spontaneous
(all temperatures)
Nonspontaneous
(low temperatures)

49. Two Paths to Calculating ΔG
ΔG = ΔH – TΔS
Determine ΔH. What methods can we use?
Determine ΔS.
Then calculate ΔG

50. Two Paths to Calculating ΔG
2) Use Standard Free Energy of Formation (ΔGf °) values to determine ΔG
Standard Free Energy of Formation (ΔGf °) --- ΔG° for the formation of 1 mole of a chemical compound in its standard state.
ΔGf ° for element formation in their most stable state = 0.
aA + bB  cC dD
ΔG° =[c (ΔGf°)C + d(ΔGf°)D] - [a (ΔGf°)A + b (ΔGf°)B ]

51. Example 2:
Find ΔG for a chemical reaction given ΔH = -218 kJ and ΔS = -765 J/K at 32°C.
B) At what temperature does this reaction become spontaneous? Assume only temperature changes.

52. Example 3:
Calculate ΔG°rxn under standard conditions for the following reaction using ΔGf° values. Fe2O3 (s) + 2Al(s)  2Fe(s) + Al2O3 (s)

53. A spontaneous reaction is NOT necessarily fast!!!!
Reaction rate involves kinetics ! !