1. Thermochemistry Heat and Chemical Change

2. Agenda Intro Video Discussions
Notes over Exothermic and Endothermic reactions
Review of Energy Stoichiometry
Notes over Heat Capacity and Specific Heat

3. Intro Videos

4. What Triggers a Chemical Reaction Video Questions
Everything you see around you is made of chemicals: the air that you breathe, the water that you drink, and the chair you are sitting on. Entropy is a characteristic of all chemicals, and more ordered chemicals have lower entropy. List two chemicals that you think have low entropy and two that have high entropy. Explain the reason for your choices.

5. Intro Videos

6. Ice Pack Questions
The breaking of bonds on the molecular level always requires an input of energy. Therefore forming bonds always releases energy. Based on this, build an explanation for why a reaction would release heat (exothermic) and why a reaction would consume heat (endothermic).

7. Ice Pack Questions
Imagine you were going to design a heat pack that worked in a similar fashion as the cold pack described in the video. Describe how the solid dissolution would occur and how the device would work as a heat pack instead of a cold pack.

8. Energy and Heat
Thermochemistry - concerned with heat changes that occur during chemical reactions
Energy - capacity for doing work or supplying heat

9. Energy and Heat
Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.

10. Exothermic and Endothermic Processes
In studying heat changes, think of defining these two parts:
the system
the surroundings

11. Exothermic and Endothermic Processes
The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed.
All the energy is accounted for as work, stored energy, or heat.

12. Exothermic and Endothermic Processes
heat flowing into a system from it’s surroundings:
defined as positive
q has a positive value
called endothermic
system gains heat as the surroundings cool down

13. Exothermic and Endothermic Processes
heat flowing out of a system into it’s surroundings:
defined as negative
q has a negative value
called exothermic
system loses heat as the surroundings heat up

14. Exothemic and Endothermic
Every reaction has an energy change associated with it
Exothermic reactions release energy, usually in the form of heat.
Endothermic reactions absorb energy
Energy is stored in bonds between atoms

15. Chemistry Happens in MOLES
An equation that includes energy is called a thermochemical equation
CH4 + 2O2 ® CO2 + 2H2O kJ
1 mole of CH4 releases kJ of energy.
When you make kJ you also make 2 moles of water

16. Thermochemical Equations
A heat of reaction is the heat change for the equation, exactly as written
The physical state of reactants and products must also be given.
Standard conditions for the reaction is kPa (1 atm.) and 25 oC

17. CH4 + 2 O2 ® CO2 + 2 H2O kJ
If grams of CH4 are burned completely, how much heat will be produced?
1 mol CH4
802.2 kJ
10. 3 g CH4
16.05 g CH4
1 mol CH4
= 514 kJ

18. Heat Capacity and Specific Heat
A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.
Used except when referring to food
a Calorie, written with a capital C, always refers to the energy in food
1 Calorie = 1 kilocalorie = 1000 cal.

19. Heat Capacity and Specific Heat
Joule-- the SI unit of heat and energy
4.184 J = 1 cal
Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”)

20. Heat Capacity and Specific Heat
For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC)
Thus, for water:
it takes a long time to heat up, and
it takes a long time to cool off!
Water is used as a coolant!
Note Figure 11.7, page 297

21. Heat Capacity and Specific Heat
To calculate, use the formula:
q = mass (g) x T x C
heat abbreviated as “q”
T = change in temperature
C = Specific Heat
Units are either J/(g oC) or cal/(g oC)

22. Sample Problem # 1
The temperature of a a piece of copper with a mass of 95.4 g increases from 25 °C to 48 °C when the metal absorbs 849 J of heat. What is the specific heat of copper?
Solution:
Knowns:
Mass= 95.4 g
Δt =48-25=23 °C
Q= 849 J
Equation: C= q/m* Δt
Plug in numbers: 849/(95.4*23) = J/g* °C

23. Sample Problem # 2
How much heat is required to raise the temperature of g of mercury 52°C?
Solution:
Knowns:
Mass: g
Δt = 52°C
C= 0.14 J/g*°C (found on chart on pg. 296)
Equation: q= m*C* Δt
Plug in numbers: q= 250g*(0.14 J/g*°C) * 52 °C
Answer: 1.82 kJ or 1820 J

24. Practice Problem 1 (Yay!!!!)
What is the specific heat capacity of Mercury if J of heat are used to raise the temp of g of Mercury from 14.0 degrees to 34.0 degrees?

25. Practice Problem 2 (WooHoo)
How much heat energy is used to raise the temp of 56.0 g of Al from 0.0 degrees to 95.0 degrees?

26. Practice Problem 3 (Awesome)
A 3.0 kg piece of iron is cooled from 300oC to 20oC. How much heat energy was lost from the iron?

27. Homework Finish any and all missing assignments
Do specific heat practice

28. Warm-Up Problem #1
14KMnO4 + 4C3H5(OH) > 7K2CO3 + 7Mn2O3 +5CO2 + 16H2O
ΔH = -566 kJ
If I used 24.0 g of Potassium Permangante with an excess of glycerin, then how much heat was released from this reaction?

29. Warm-Up Problem #2
Ba(OH)2 . 8H2O + 2NH4SCN > Ba(SCN)2 + 2NH3 + 10H2O
ΔH = 330 kJ
If I used 50.0 g of Ammonium Thiocyanate in excess Barium Hydroxide how much heat would the reaction absorb?

30. Warm-Up Problem # 3
Calculate the specific heat capacity of copper given that J of energy raises the temperature of 15g of copper from 25o to 60o.

31. Warm-Up Problem # 4
How much heat energy is required to raise the temperature of 55.0g of water from 25.0°C to 28.6°C? The specific heat of water is 4.18J/g°C.

32. Section 11.2 Measuring and Expressing Heat Changes
OBJECTIVES:
Construct equations that show the heat changes for chemical and physical processes.

33. Section 11.2 Measuring and Expressing Heat Changes
OBJECTIVES:
Calculate heat changes in chemical and physical processes.

34. Calorimetry
Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.

35. Calorimetry
For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system

36. Calorimetry Changes in enthalpy = H
q = H These terms will be used interchangeably in this textbook
Thus, q = H = m x C x T
H is negative for an exothermic reaction
H is positive for an endothermic reaction (Note Table 11.3, p.301)

37. C + O2 ® CO2
+ 395 kJ
Energy
Reactants
Products
C + O2
395kJ
C O2

38. In terms of bonds C O O C O Breaking this bond will require energy. C
Making these bonds gives you energy.
In this case making the bonds gives you more energy than breaking them.

39. Exothermic The products are lower in energy than the reactants
Releases energy

40. CaCO3 + 176 kJ ® CaO + CO2 CaCO3 ® CaO + CO2 Energy Reactants Products

41. Summary, so far...

42. Enthalpy
The heat content a substance has at a given temperature and pressure
Can’t be measured directly because there is no set starting point
The reactants start with a heat content
The products end up with a heat content
So we can measure how much enthalpy changes

43. Enthalpy Symbol is H Change in enthalpy is DH (delta H)
If heat is released, the heat content of the products is lower
DH is negative (exothermic)
If heat is absorbed, the heat content of the products is higher
DH is positive (endothermic)

44. Energy
Change is down
DH is <0
Reactants
Products

45. Energy
Change is up
DH is > 0
Reactants
Products

46. More Fun with Thermo Practice Problem 5
25.0 mL of water containing moles HCl is added to 25.0 mL of water containing mols NaOH in a foam cup calorimeter. At the start, the solutions and the calorimeter are all at 25.0oC. During the reaction, the highest temperature observed is 32.0oC. Calculate the heat (in kJ) released during this reaction. Assume the densities of the solutions are 1.00 g/mL.

47. Heat of Reaction
The heat that is released or absorbed in a chemical reaction
Equivalent to DH
C + O2(g) ® CO2(g) kJ
C + O2(g) ® CO2(g) DH = kJ
In thermochemical equation, it is important to indicate the physical state
H2(g) + 1/2O2 (g)® H2O(g) DH = kJ
H2(g) + 1/2O2 (g)® H2O(l) DH = kJ

48. Heat of Combustion
The heat from the reaction that completely burns 1 mole of a substance
Note Table 11.4, page 305

49. Heats of Fusion and Solidification
Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid
Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies

50. Heats of Vaporization and Condensation
Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.

51. Heats of Vaporization and Condensation
Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses
Hvap = - Hcond

52. Heats of Vaporization and Condensation
The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous
You can receive a scalding burn from steam when the heat of condensation is released!

53. Heat of Solution
Heat changes can also occur when a solute dissolves in a solvent.
Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substance

54. Hess’s Law
If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.
Called Hess’s law of heat summation
Example shown on page 314 for graphite and diamonds

55. Why Does It Work? If you turn an equation around, you change the sign:
If H2(g) + 1/2 O2(g)® H2O(g) DH= kJ
then, H2O(g) ® H2(g) + 1/2 O2(g) DH = kJ
also,
If you multiply the equation by a number, you multiply the heat by that number:
2 H2O(g) ® 2 H2(g) + O2(g) DH = kJ

56. Standard Heats of Formation
The DH for a reaction that produces 1 mol of a compound from its elements at standard conditions
Standard conditions: 25°C and 1 atm.
Symbol is
The standard heat of formation of an element = 0
This includes the diatomics

57. What good are they? DHo = ( Products) - ( Reactants)
Table 11.6, page 316 has standard heats of formation
The heat of a reaction can be calculated by:
subtracting the heats of formation of the reactants from the products
DHo
= (
Products) - (
Reactants)

58. Examples CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) CH4 (g) = - 74.86 kJ/mol
O2(g) = 0 kJ/mol
CO2(g) = kJ/mol
H2O(g) = kJ/mol
DH= [ (-241.8)] - [ (0)]
DH= kJ