1. THERMOCHEMISTRY

2. Thermochemistry
Thermochemistry is the study of energy changes (HEAT) that occur during chemical reactions and changes in state.

3. Heat (q)
Heat is the energy that transfers from one object to another because of a temperature difference between them.
Heat ALWAYS flows from a warmer object to a cooler one.

4. Heat movement
Heat moves between the system (reaction) and the surroundings
*** must obey the law of conservation of energy (heat (energy) is never created nor destroyed, just transferred)
Thermochemical equations tell you the direction of heat flow by the “sign”, + or -

5. Endo vs. Exo-
Endothermic reactions: absorbs heat from surroundings (+).
If you touch an endothermic reaction it feels COLD
Exothermic reactions: release heat to the surroundings (-)
If you touch an exothermic reaction it feels HOT

6. HEAT energy UNIT of energy = JOULE (J)
or Calorie (cal) 1 cal = J
Heat capacity is how much heat (energy) is needed to increase the temperature of an object by 1 C.
*** heat capacity of an object depends on both its mass and its chemical composition
The greater the mass, the greater its heat capacity

7. Example
On a sunny day, a 20 kg puddle of water may be cool, while a nearby 20 kg iron sewer cover may be too hot to touch.
Both have the same mass, BUT the are made of different materials (chemical composition) BECAUSE they have different SPECIFIC HEAT CAPACITIES

8. SPECIFIC HEAT (C) WATER has a specific heat of 4.184 J/gC
SPECIFIC Heat Capacity (specific heat, C) = specific to a substance
Amount of heat it takes to raise the temp. of 1 g of a substance by 1 C
Units = (J/gC)
WATER has a specific heat of J/gC

9. q = mCT CALCULATING HEAT Where: q = heat (Joules) m = mass (grams)
C = specific heat (J/gC)
T = change in temperature (C) Tf-Ti

10. Calculations
How much heat is absorbed when a 95.4 g piece of copper increases from 25.0 C to 48.0 C. The specific heat of copper is J/gC.

11. 849 J m= 95.4 g C= 0.387 J/gC T= 48.0 C - 25.0 C= 23 C
q= (95.4) (.387) (23)
849 J

12. Ex. 2
How much heat is absorbed when 3.4 g of olive oil is heated from 21.0 C to 85.0 C. The specific heat of olive oil is 2.0 J/g C.

13. m= 3.4 g C= 2.0 J/gC
T= 85.0 C C= 64 C
q= (3.4) (2.0) (64)
435 J

14. Ex. 3
How much heat is required to raise the temperature of g of mercury 52.0 C? The specific heat of mercury is 0.14 J/g C.

15. m= 250 g C= 0.14 J/gC
T= 52 C
q= (250) (0.14) (52)
1820 J = 1.8 kJ

16. Ex. 4
How many joules (J) of heat are absorbed when 1000.g of water is heated from 18.0 C to 85.0 C? C for water = J/g C.

17. 280000 J =2.80 x 105 J q= (1000) (4.184) (67) m= 1000 g C= 4.184 J/gC
T= 85.0 C C= 67 C
q= (1000) (4.184) (67)
J =2.80 x 105 J

18. Practice Problems
1. What is the specific heat of a substance that has a mass of 25.0 g and requires 2197 J of energy to raise its temperature by 15.0 C?
2. Suppose g of ice absorbs J of heat. What is the corresponding temperature change? The specific heat of ice is 2.1 J/g C.
3. How many Joules of heat energy are required to raise the temperature of g of aluminum by C. The specific heat of aluminum is 0.90 J/g C. 

19. ANSWERS
1) 5.86 J/g°C
2) 6.0 °C
3) J = 11,000 J or 1.1 x 104 J

20. STOP
Work on Specific Heat Calcs WS

21. ENTHALPY
Enthalpy = a type of chemical energy, sometimes referred to as “heat content”, ΔH (the heat of reaction for a chemical reaction)
exothermic reactions (feels hot):
q (heat) = ΔH (enthaply, heat of rxn) < 0 (negative values)
endothermic reactions (feels cold):
q = ΔH > 0 (positive values)

22. Thermochemical Equations
A chemical equation that shows the enthalpy (H) is a thermochemical equation.

23. Rule #1
The magnitude (value) of H is directly proportional to the amount of reactant or product.
H2 + Cl2  2HCl H = kJ
* meaning there are 185 kJ of energy RELEASED for every:
1 mol H2
1 mol Cl2
2 moles HCl

24. Rules of Thermochemistry
Example 1:
H2 + Cl2  2HCl H = kJ
Calculate H when 2.00 moles of Cl2 reacts.

25. Rules of Thermochemistry
Example 2: Methanol burns to produce carbon dioxide and water:
2CH3OH + 3O2  2CO2 + 4H2O H = kJ
What mass of methanol is needed to produce 1820 kJ?

26. Rule #2
H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction.
(If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

27. Rules of Thermochemistry
Example 1:
Given:
H2 + ½O2  H2O H = kJ
Reverse:
H2O  H2 + ½O2 H = kJ

28. Example 2 CaCO3 (s)  CaO (s) + CO2 (g) H = 178 kJ
What is the H for the REVERSE RXN?
CaO (s) + CO2 (g)  CaCO3 (s) H = ?

29. Practice Problem: (rule 1 + rule 2)
Given:
H2 + ½O2  H2O H = kJ
Calculate H for the equation:
2H2O  2H2 + O2

30. Alternate form of thermochem. eq.
Putting the heat content of a reaction INTO the actual thermochemical eq.
EX:
H2 + ½O2  H2O H = kJ
EXOTHERMIC:
Heat is ___________ as a ____________.

31. ALTERNATE FORM EX: H2 + ½O2  H2O H = -285.8 kJ
The alternate form is this: H2 + ½O2  H2O kJ

32. EX:
2 NaHCO kJ  Na2CO3 + H2O + CO2
Put in the alternate form

33. Ex:
2 NaHCO kJ  Na2CO3 + H2O + CO2
The alternate form is this:
2 NaHCO3  Na2CO3 + H2O + CO2 H = kJ

34. Put the following in alternate form
1. H2 + Cl2  2 HCl H = -185 kJ
2. 2 Mg + O2  2 MgO kJ
3. 2 HgO  2 Hg + O2 H = kJ

35. CALORIMETRY
The enthalpy change associated with a chemical reaction or process can be determined experimentally.
Measure the heat gained or lost during a reaction at CONSTANT pressure

36. Calorimeter
Device used to measure the heat absorbed or released during a chemical or physical process

37. Example
If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter?
Why is there a difference in temperature between the two objects?

38. Because…
Different substances have different specific heats (amount of energy needed to raise the temperature of 1 g of a substance by 1 degree Celsius).

39. Need to know how to calc. heat (review)
Heat (q) = mCT
If the specific heat of Al is 0.90 J/gC, how much heat is required to raise the temperature of 10,000 g of Al from 25.0 C to 30.0 C?

40. Calculator says……..
= 45,000 J

41. For example (review):
If 418 J is required to increase the temperature of 50.0 g of water by 2.0 C, what is the specific heat of water?

42. What happens in a calorimeter
One object will LOSE heat, and the other will ABSORB the heat
System loses heat to surroundings = EXO = -q
System absorbs heat from surroundings = ENDO = +q

43. EXAMPLE
A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble?
The specific heat of water is J/g C.
Which was warmer? The pebble or the water? 

44. The pebble because the water heated up from 25.0 C to 26.4 C.
Pebble loses heat (-q, exothermic) while water gains heat (+q, endothermic)
Do the calc… 

45. All values are for WATER
A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble?
Calc the heat absorbed by the water (+q).
The heat (J) released by the warm pebble = - of the heat absorbed by the water.
q water = - q pebble

46. -150 J

47. PRACTICE 1 (LAB type of CALC)
Suppose that g of water at 22.4 °C is placed in a calorimeter. A g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal.
The specific heat of water is J/g C. 

48. Heat gained by water = Heat lost by the Al q of water = - q Al
Make a chart
Water Al
Mass (g)
Specific heat (J/g °C )
Initial temp (°C )
Final temp (°C )

49. We don’t know specific heat of Al, but we know all the values for water
So, calc q for WATER
The q for water is the same for q of Al
(the value of q is the same)

50. 0.879 J/g °C

51. Practice 2
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 g of water at 17.0C. The water reaches a temperature of 20.0 C.
How many joules of heat were released by the lead?
The specific heat of water is J/g C.

52. 502 J or 5.0 x 102 J

53. Work on WS: Calorimetry
STOP

54. Review Heat in Changes of STATE
Heating and Cooling Curve
LABEL: *physical states
*boiling point
*melting point
*heat of fusion
*heat of vaporization

55. Need to calculate heat for the WHOLE process of changing physical state
Review
Specific heat is the amount of heat that must be added to a stated mass of a substance to raise its temperature by 1°C, with NO change in state.
CHANGING STATE requires “heat of vaporization” and “heat of fusion”
Heat of vaporization = amount of heat that must be added to 1 g of a liquid at its boiling point to convert it to vapor with NO temp. change
Heat of fusion = amount of heat needed to melt 1 g of a solid at its melting point

56. MAKE YOUR OWN HEATING/COOLING CURVE
Specific heat of water = J/g°C
Specific heat of ice =2.09 J/g°C
Specific heat of steam = 2.03 J/g°C
Heat of fusion = 334 J/g
Heat of vaporization = 2260 J/g

57. Now let’s take it a step further….
How can you use a heating curve to explain specific heat?

58. Heating curves and DH
temperature
added energy

59. Heating curves and DH
gas
liquid
temperature
solid
added energy

60. Heating curves and DH gas temperature liquid melting/ freezing pt
solid
added energy

61. Heating curves and DH boiling/ cond. pt gas temperature liquid
melting/
freezing pt
temperature
solid
added energy

62. Heating curves and DH boiling/condensing occurring here
melting/freezing
occurring here
boiling/
cond. pt
gas
liquid
melting/
freezing pt
temperature
solid
added energy

63. We can use DH in our specific heat equation in place of “q”
How is the total enthalpy change (DH) calculated for a substance whose temperature change includes a change in state?
We can use DH in our specific heat equation in place of “q”
DH = m x C x DT

64. temperature
Dt of solid
absorbing energy
added energy

65. temperature
DH = m x Csolid x Dt
added energy

66. as a solid melts becomes potential energy, so no Dt
the energy absorbed
as a solid melts becomes
potential energy,
so no Dt
temperature
DH = m x Csolid x Dt
added energy

67. DH = DHfus x m
temperature
DH = m x Csolid x Dt
added energy

68. Dt of liquid absorbing energy DH = DHfus x m DH = m x Csolid x Dt
temperature
DH = m x Csolid x Dt
added energy

69. DH = m x Cliquid x Dt DH = DHfus x m DH = m x Csolid x Dt temperature
added energy

70. as a liquid boils becomes potential energy, so no Dt
the energy absorbed
as a liquid boils becomes
potential energy,
so no Dt
DH = DHfus x m
DH = m x Cliquid x Dt
temperature
DH = m x Csolid x Dt
added energy

71. DH = DHvap x m temperature DH = DHfus x m DH = m x Cliquid x Dt
DH = m x Csolid x Dt
added energy

72. Dt of gas absorbing energy DH = DHvap x m temperature DH = DHfus x m
DH = m x Cliquid x Dt
temperature
DH = m x Csolid x Dt
added energy

73. DH = m x Cgas x Dt DH = DHvap x m temperature DH = DHfus x m
DH = m x Cliquid x Dt
temperature
DH = m x Csolid x Dt
added energy

74. DH = m x Cgas x Dt
DH = DHvap x m
DH = DHfus x m
DH = m x Cliquid x Dt
temperature
DH = m x Csolid x Dt
added energy

75. being heated will be the sum of the DH of any Dt occurring plus
The DH of any substance
being heated will be the sum of the
DH of any Dt occurring plus
DH of any phase change occurring
DH = m x Cgas x Dt
DH = DHvap x m
DH = DHfus x m
DH = m x Cliquid x Dt
temperature
DH = m x Csolid x Dt
added energy

76. EX: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
DH = m x Cgas x Dt
DH = DHvap x m
DH = DHfus x m
DH = m x Cliquid x Dt
temperature
DH = m x Csolid x Dt
added energy

77. EX: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
temperature
0 oC
-20 oC
added energy

78. EX: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 334J/g, CH2O liq = J/goC
50 oC
temperature
0 oC
-20 oC
added energy

79. EX: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = J/goC
50 oC
DH3 = m x Cliquid x Dt
DH2 = DHfus x m
temperature
0 oC
DH1 = m x Csolid x Dt
-20 oC
added energy

80. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = J/goC
50 oC
DH3 = m x Cliquid x Dt
DH2 = DHfus x m
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20oC
-20 oC
added energy

81. DH2=334J/g x 10g DH1 = 10g x 2.1 J/goC x 20oC
DH2=10 g x 1mol/18g x 6.01kJ/mol
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 334J/g, CH2O liq = J/goC
50 oC
DH2=334J/g x 10g
DH3 = m x Cliquid x Dt
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20oC
-20 oC
added energy

82. DH3 = 10g x 4.186 J/goC x 50 oC DH1 = 10g x 2.1 J/goC x 20oC
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 334J/g CH2O liq = J/goC
DH3 = 10g x J/goC x 50 oC
50 oC
DH2=334J/g x 10g
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20oC
-20 oC
added energy

83. total DH = DH1 + DH2 + DH3 DH1 = 10g x 2.1 J/goC x 20oC
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = J/goC
total DH = DH1 + DH2 + DH3
DH3 = 10g x J/goC x 50 oC
50 oC
DH2=334J/g x 10g
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20oC
-20 oC
added energy

84. total DH = DH1 + DH2 + DH3 = 420 J + 3340 J + 2093 J DH2=334J/g x 10g
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = J/goC
total DH = DH1 + DH2 + DH3
= 420 J J J
DH3 = 10g x J/goC x 50 oC
50 oC
DH2=334J/g x 10g
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20oC
-20 oC
added energy

85. total DH = DH1 + DH2 + DH3 5853 J = 420 J + 3340 J + 2093 J
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
use the following values:
Cice = 2.1 J/goc, DHfus H2O = 6.01 kJ/mol, CH2O liq = J/goC
total DH = DH1 + DH2 + DH3
5853 J = 420 J J J
DH3 = 10g x J/goC x 50 oC
50 oC
DH2=334J/g x 10g
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20oC
-20 oC
added energy

86. It takes 5853 joules to heat up 10 grams
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
It takes 5853 joules to heat up 10 grams
of water from -20 oC to +50 oC.
DH3 = 10g x J/goC x 50 oC
50 oC
DH2=334J/g x 10g
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20 oC
-20 oC
added energy

87. It takes 5853 joules to heat up 10 grams
EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
It takes 5853 joules to heat up 10 grams
of water from -20 oC to +50 oC.
DH3 = 10g x J/goC x 50 oC
50 oC
DH2=334J/g x 10g
temperature
0 oC
DH1 = 10g x 2.1 J/goC x 20 oC
-20 oC
added energy

88. EXAMPLE: What is DH for 10 g water with a total Dt from -20 oC to +50 oC?
It takes 5853 joules to heat up 10 grams of water from -20 oC to +50 oC.
50 oC
temperature
0 oC
-20 oC
5853 J
added energy

89. USE YOUR HEATING/COOLING CURVE to help guide your calculations
1. How much heat is released by g of water as it cools from 85.0 °C to 40.0 °C?

90. ANSWER
1. How much heat is released by g of water as it cools from 85.0 °C to 40.0 °C?
Calculation is within the liquid phase of water (no phase changes)
q = mcΔT = (250.0 g)(4.184 J/g°C)(40.0°C °C)
= J

91. Example 2
How much heat energy is required to bring g of water at 55.0 °C to its boiling point and then vaporize it?
THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE? Use you heating and cooling curve to help you understand the steps!

92. ANSWER (2 steps) Step 1 = raise temp of water from 55.0 °C to 100.0 °C
How much heat energy is required to bring g of water at 55.0 °C to its boiling point and then vaporize it?
ANSWER (2 steps)
Step 1 = raise temp of water from 55.0 °C to °C
Step 2 = vaporize water using heat of vaporization
q = mcΔT
q = (135.5g)(4.184 J/g°C)(100.0°C-55.0°C) = to (3 sig figs) = J
q = mass x heat of vaporization
q = (135.5g)(2260 J/g) =
= to (3 sig figs) = J
***ADD UP THE 2 NUMBERS TO GET THE OVERALL HEAT REQUIRED FOR THIS PROCESS 

93. Example 3
How much heat energy is required to convert 15.0 g of ice at °C to steam at °C?
THINK!!! HOW MANY STEPS SHOULD THIS CALCULATION BE??

94. J

95. How much heat energy is required to convert 15. 0 g of ice at -12
How much heat energy is required to convert 15.0 g of ice at °C to steam at °C? = 5 steps
q = (15.0 g)(2.09 J/g°C)( °C) = J
q = (15.0 g)(334 J/g) = J
q = (15.0 g)(4.184 J/g°C)( °C) = J
q = (15.0 g)(2260 J/g) = J
q = (15.0 g)(2.03 J/g°C)( °C) = J
46282 J

96. WS: PROBLEM SET (heat transfer)
Complete the problems