1. The study of the heat flow of a chemical reaction or physical change
Thermochemistry
The study of the heat flow of a chemical reaction or physical change

2. 1st Law of Thermodynamics
Energy cannot be created or destroyed but only converted from one form to another
The energy of the universe is constant

3. Ex - Elevation vs distance
State Function
A state function is independent of the pathway (does not rely on the past or future of the system
A state function depends only on the present state of the system
Ex - Elevation vs distance

4. Energy The capacity to do work or provide heat
Potential Energy (PE) = mhv
Kinetic Energy (KE) = 1/2mv2
The total energy of the system is equal to the sum of the potential and kinetic energy
Can be changed by a flow of heat/work
ΔE = q + w

5. Chemical Energy
Energy will flow either from the system to its surroundings (exothermic) or from the surroundings to the system (endothermic)
The system will be the reactants and products
ΔE < 0 energy flows out of the system
ΔE > 0 energy flows into the system

6. From the system’s point of view…q +
Heat flows into the system -
Heat flows out of the system w
Work done on the system
Work done by the system

7. Internal Energy
Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system
ΔE = q + w
ΔE = 15.6 kJ kJ = 17.0 kJ
Work vs Energy flow

8. Work Done by or to Gases

9. Work is done by the system so w is negative
PV Work
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm
w = -PΔV
ΔV = 64 L – 46 L = 18 L
w = -(15 atm x 18 L)
w = -270 L∙atm
Work is done by the system so w is negative

10. Internal Energy, Heat & Work
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. Use 1 L∙atm = J

11. w = -(1.0 atm x (5.0 x 105 L)) = -5.0 x 105 L∙atm
ΔE = q + w
q = +1.3 x 108 J
w = -PΔV
ΔV = 5.0 x 105 L
w = -(1.0 atm x (5.0 x 105 L)) = -5.0 x 105 L∙atm
ΔE = (+1.3 x 108 J) + (-5.1 x 107 J)
ΔE = 8 x 107 J

12. Enthalpy (H) H = E + PV E = internal energy P = pressure of the system
V = volume of the system
Enthalpy is a state function

13. ΔH = ΔE + PΔV (pressure is constant)
What is Enthalpy?
Consider a process at constant pressure where the only work is PV work (w = -PΔV)
ΔE = qp + w
ΔE = qp – PΔV
qp = ΔE + PΔV
H = E + PV or ΔH = ΔE + Δ(PV)
ΔH = ΔE + PΔV (pressure is constant)
ΔH = qp

14. ΔH = Hproducts - Hreactants
In other words…
the terms heat of reaction and change in enthalpy are the same so
ΔH = Hproducts - Hreactants

15. 2H2(g) + O2(g)  2H2O(l) ΔH = -572 kJ
Enthalpy
How much heat is released when 4.03 g of hydrogen is reacted with excess oxygen?
2H2(g) + O2(g)  2H2O(l) ΔH = -572 kJ

16. Enthalpy #2
When 1 mole of CH4 is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which 5.8 g of CH4 is burned at constant pressure.
qp = ΔH = -890 kJ/mole CH4

17. Calorimetry
Calorimeter – used to determine the heat energy change during a reaction
Carried out under constant pressure measures enthalpy (ΔH)
Carried out under constant volume measures energy (ΔE)

18. Heat Capacity
Specific heat capacity – the amount of heat needed to raise one gram on a substance by 1oC (J/Co∙g or J/K∙g)
Molar heat capacity – the amount of heat needed to raise one mole of a substance by 1oC (J/Co∙mol or J/K∙mol)

19. Heat Gained or Lost Depends on…
The change in temperature during the reaction
The amount of substance present
The heat capacity of the substance

20. Calorimetry
A 110. g sample of copper (specific heat capacity = 0.20 J/Co∙g) is heated to 82.4oC and then placed in a container of water at 22.3oC. The final temperature of the water and the copper is 24.9oC. What was the mass of the water in the original container, assuming complete transfer of heat from the copper to the water?

21. -(heat lost by copper) = (heat gained by water)

22. Calorimetry #2
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0oC is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0oC in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1oC. Assuming that the calorimeter absorbs only a negligible amount of heat, that the specific heat capacity of the solution is 4.18 J/oC∙g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

23. Solution (no pun intended)
Species present: Ba2+, NO3-, Na+, and SO42-
Spectator ions: Na+ and NO3-
Net Ionic equation:
Ba2+(aq) + SO42-(aq)  BaSO4(s)

24. q = qp = (2.0 x 103 g)(4.18J/oCg)(3.1oC)
Heat evolved by reaction
= heat absorbed by the solution
= m x c x ΔT
Since 1.00 L of each solution is used, total volume of the mixture is 2.00 L (2.0 x 103 g)
Temperature increase =
28.1oC – 25.0oC = 3.1oC
q = qp = (2.0 x 103 g)(4.18J/oCg)(3.1oC)
2.6 x 104 J = ΔH

25. Hess’s Law

26. Based on… State function
Enthalpy change is same for a reaction whether the reaction takes place in one or many steps

27. How to use Hess’s Law
Manipulate equations to reach the desired reaction
If the reaction given is reversed, so is ΔH
If multiplying the equation to balance the coefficients also multiply ΔH by the same number

28. Calculate the enthapy for the following reaction:
N2(g) + 2O2(g) ---> 2NO2(g) ΔH° = ??? kJ
N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ
2NO2(g) ---> 2NO(g) + O2(g) ΔH° = +112 kJ

29. Solution N2(g) + O2(g) ---> 2NO(g) ΔH° = +180 kJ
2NO(g) + O2(g) ---> 2NO2(g) ΔH° = -112 kJ
+ 68 kJ
The second equation has been reversed including ΔH

30. Calculate ΔH° for this reaction:
2N2(g) + 5O2(g) ---> 2N2O5(g)
H2(g) + 1/2O2(g)  H2O(l) ΔH° = kJ
N2O5(g) + H2O(l)  2HNO3(l) ΔH° = kJ
1/2N2(g) + 3/2O2(g) + 1/2H2(g)  HNO3(l) ΔH° = kJ

31. Solution

32. Standard Enthalpies of Formation
The change in enthalpy of the formation of one mole of a compound from it elements in their standard states

33. ΔH° 25oC at 1 atm and 1 M
By definition, the standard heat of formation for elements in their standard states equals zero.
Example: Which of the following will have standard heats of formation equal to zero?
H2(g), Hg(s), CO2(g), H2O(l), Br2(l)

34. Example
Write the balanced molecular equation representing the ΔHf° for ethanol.
Answer:
2C(s) + 3H2(g) + 1/2O2(g)  CH5OH(l)

35. ΔH°reaction = Σ ΔH°f(products) - Σ ΔH°f(reactants)
For any reaction…
ΔH°reaction = Σ ΔH°f(products) - Σ ΔH°f(reactants)

36. Calculate the Standard Enthalpy Change for the combustion of Methane
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
CH4(g)  C(s) + 2H2(g)
2H2(g) + C(s)  CH4(g) kJ
O2(g) kJ
C(s) + O2(g)  CO2(g) kJ
H2(g) + ½ O2(g)  H2O(l) kJ

37. Example #1
Calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water
Use table 6.2 on page on Page 262

38. Heat of Fusion & Heat of Vaporization
Heating Curves
Heat of Fusion & Heat of Vaporization

39. Heating Curves

40. Heat of Fusion
ΔHfus = enthalpy change that occurs in melting a solid at its melting point
Example: What quantity of heat is needed to melt 1.0 kg of ice at its melting point?
ΔHfus =6.0 kJ/mol

41. For liquid water ΔHvap = 43.9 kJ/mol
Heat of Vaporization
ΔHvap = the energy needed to vaporize one mole of a liquid at a pressure of 1 atm
Example: What quantity of heat is required to vaporize 130. g of water?
For liquid water ΔHvap = 43.9 kJ/mol

42. Example #2 Substance X has the following properties:
ΔHvap = 20. kJ/mol
ΔHfus = 5.0 kJ/mol
Boiling point = 75oC
Melting point = -15oC
Specific heat
Solid = 3.0 J/goC
Liquid = 2.5 J/goC
Gas = 1.0 J/goC

43. Example #2 continued…
Calculate the energy required to convert 250. g of substance X from a solid at -50oC to a gas at 100oC. Assume that X has a molar mass of g/mol.
5 Step Process
Heating solid
Melting solid
Heating liquid
Boiling liquid
Heating gas

44. Example #2 continued… q = m x c x ΔT
= 250.g x (3.0 J/goC) x 35oC = 26 kJ
mol x ΔHfus = 3.33 mol x 5.0 kJ/mol = 17 kJ
= 250.g x (2.5 J/goC) x 90oC = 56 kJ
mol x ΔHvap = 3.33 mol x 20. kJ/mol =67 kJ
= 250.g x (1.0 J/goC) x 25oC = 6.2 kJ
172 kJ