1. Thermochemistry
Unit 10 Lesson 2

2. Thermochemistry  Study of energy changes that occur during chemical reactions and changes of state

3. Endothermic Process that absorbs/gains heat
Heat flows into the system (+q)
Heat removed from the surroundings (-q)
Surroundings feel cold

4. Exothermic Process that releases/produces heat
Heat flows out of the system (-q)
Heat added to the surroundings (+q)
Surroundings may feel warm/hot

5. assuming pressure is constant, ΔH = q
Enthalpy (H)  the heat content of a system
units for enthalpy are kJ
assuming pressure is constant, ΔH = q
ΔH  change in enthalpy
Exothermic = - ΔH
Endothermic = + ΔH

6. ΔH & Changes of State
Change in state always involves a change in heat energy

7. Melting (fusion) Vaporization Sublimation
Endothermic changes of state absorb heat energy,
increasing in temperature
Melting (fusion)
Vaporization
Sublimation

8. As more heat is added the temperature increases, meaning the particles are moving faster.
To make a substance melt or vaporize you continuously add heat to:
*increase the speed of the
molecules
*break the (IMFs) attractions
between molecules causing
them to separate

9. q = m·Hf Melting (fusion)
The Heat of Fusion is the amount of heat needed to melt one gram of a substance at standard pressure.
Heat of Fusion (Hf) of water is 334 J/g.
LOOK in your ref. packet!
q = m·Hf

10. How much heat is added when 15.7 moles of water melts?
q = m·Hf
15.7 mol
1 mol g
= g
q = ( g)(334 J/g)
q = 94,471 J
q = 94.5 kJ

11. In the diagram, how much heat is needed for the ice to melt?
q = m · Hf
Hf (water) = 334 J/g
in ref. packet!
q = 2000 g · 334 J/g
q = J
= 670 kJ

12. q = m·Hv Vaporization/Boiling
The Heat of Vaporization is the amount of heat needed for one gram of the substance to change from liquid to gas at standard pressure.
Heat of Vaporization (Hv) of water is 2260 J/g.
LOOK in your ref. packet!
q = m·Hv

13. You NEED to recognize that 100°C is the boiling point for water
How much heat is absorbed when 24.8 g H2O at oC is converted to steam at oC?
You NEED to recognize that 100°C is the boiling point for water
q = m·Hv
q = (24.8 g)(2260 J/g)
q = 56,048 J
q = 56.0 kJ

14. Freezing Condensation Deposition
Exothermic changes of state release heat energy,
decreasing in temperature
Freezing
Condensation
Deposition

15. As heat is removed the temperature decreases, meaning the particles are moving slower.
To make a substance freeze or condense you continuously remove heat to:
*decrease the speed of the
molecules
*increase the (IMFs) attractions
between molecules causing them
to clump

17. Heating/Cooling Curves

18. Heat can be transferred even if there is no change in state
What happens when you add heat to a solid?
HEATING curve for water
100
Time
Temp
Heat can be transferred even if there is no change in state
Ice at –30.0°C absorbs heat but stays SOLID while temperature rises to 0°C.

19. What happens as a solid melts?
HEATING curve for water
During a phase change, the temperature remains CONSTANT. This is because the heat being added is used to break IMFs holding the molecules together not to raise the temperature.
100
Time
Temp
At 0°C ice begins to melt

20. What happens when you add heat to a liquid?
HEATING curve for water
100
Time
Temp
Water remains a LIQUID as it absorbs heat and goes from 0°C to 100°C.

21. What happens a liquid begins to change to a gas?
HEATING curve for water
100
Time
Temp
At 100°C water begins to boil. The temperature will be constant until phase change is complete.

22. What happens you add heat to gas?
HEATING curve for water
100
Time
Temp
After 100°C, the temperature of steam will continue to increase as heat is added.

23. Heating curves are endothermic as they absorb/gain heat.
100
Time
Temp
boiling
gas
liquid/gas
Heating curves are endothermic as they absorb/gain heat.
liquid
melting
solid/liquid
solid

24. During a phase, KE is increasing (temp is increasing).
What happens to Kinetic energy (KE) & potential energy (PE) at each stage of the curve?
HEATING curve
During a phase, KE is increasing (temp is increasing).
100
Time
Temp
KE constant PE KE KE
During a phase change, KE is constant and PE is increasing. PE
KE constant KE

25. Calculating heat for each segment of the heating curve for water.
100
Time
Temp
The temperature of the ice is increasing.
The specific heat for ice is 2.05 J/g°C.
LOOK in your ref. packet!
q1 = mcpΔT

26. LOOK in your ref. packet for Cp for iron
Calculate the amount of energy it takes to heat iron from 0°C to 557°C. (remains a solid the entire time)
LOOK in your ref. packet for Cp for iron
q = mcpΔT
q = (2000 g) (0.449 J/g°C) (557°C)
q = 500,000 J
= 5.0 x 10 5J

27. Calculating heat for each segment of the heating curve for water.
100
Time
Temp
A phase change occurs at a constant temperature. Use the heat of fusion since ice is melting Hf (water) = 334 J/g
q2 = m·Hf
LOOK in your ref. packet!
q1 = mcpΔT

28. Calculating heat for each segment of the heating curve for water.
100
Time
Temp
q3 = mcpΔT
The temperature of the water is increasing.
The specific heat for water is 4.18 J/g°C.
q2 = m·Hf
LOOK in your ref. packet!
q1 = mcpΔT

29. Calculating heat for each segment of the heating curve for water.
100
Time
Temp
q4 = m·Hv
q3 = mcpΔT
A phase change occurs at a constant temperature. Use the heat of vaporization Hv (water) = 2260 J/g
q2 = m·Hf
LOOK in your ref. packet!
q1 = mcpΔT

30. Calculating heat for each segment of the heating curve for water.
q5 = mcpΔT
100
Time
Temp
q4 = m·Hv
q3 = mcpΔT
The temperature of the steam is increasing.
The specific heat for steam is 2.02 J/g°C.
q2 = m·Hf
LOOK in your ref. packet!
q1 = mcpΔT

31. Calculating heat for each segment of the heating curve for water.
q5 = mcpΔT
100
Time
Temp
q4 = m·Hv
q3 = mcpΔT
Use q = mcpΔT when the temperature is changing. Use q = m·Hf or q = m·Hv when there is a phase change.
q2 = m·Hf
q1 = mcpΔT

32. The total amount of heat absorbed is the sum of all five equations:
Calculating heat for each segment of the heating curve for water.
q5 = mcpΔT
100
Time
Temp
q2 = m·Hf
q3 = mcpΔT
The total amount of heat absorbed is the sum of all five equations:
qtot= q1+q2+q3+q4+q5
q2 = m·Hf
q1 = mcpΔT

34. q4 = m·Hv Hf (water) = 334 J/g Hv (water) = 2260 J/g q2 = m·Hf
NOTICE! It takes more time (longer line) & energy for vaporization of water to occur than melting. WHY? Look at the pictures to the right. 
100
Time
Temp
q4 = m·Hv
Hf (water) = 334 J/g
Hv (water) = 2260 J/g
q2 = m·Hf

35. Cooling curves are exothermic as they release/lose heat.
gas
100
Time
Temp
condensation
liquid/gas
liquid
Hv may be used to determine amount of heat released during condensation
freezing
solid/liquid
solid
Hf may be used to determine amount of heat released during freezing

36. How much heat is released when 275 grams of water freezes?
q = m·Hf
q = (275 g)(334 J/g)
q = 91,850 J
q = 91.9 kJ

37. During a phase, KE is decreasing (temp is decreasing).
What happens to Kinetic energy (KE) & potential energy (PE) at each stage of the curve?
COOLING curve KE
During a phase, KE is decreasing (temp is decreasing).
100
Time
Temp PE
KE constant KE
During a phase change, KE is constant and PE is decreasing. PE
KE constant KE

38. ΔH & Chemical Reactions
Chemical reactions always involves a change in heat energy

39. Endothermic Reactions
C(s) + H2O(g) kJ  CO2(g) + H2(g)
Heat energy is written in the equation as a reactant since it is coming in/being used.
Surrounding area feels cool
DHrxn = +113 kJ
meaning 113kJ are absorbed

40. Potential energy diagram
Endothermic because….
1. Products have more energy than the reactants
Activation Energy DH
2. ΔH is positive

41. Exothermic Reactions -ΔH C3H8 + 5O2  3CO2 + 4H2O + 2043 kJ
Heat energy is written in the equation as a product since it is released/produced.
Surrounding area feels warm
ΔHrxn = kJ . . .
meaning 2043 kJ of heat is released

42. Potential energy diagram
Exothermic because….
1. Products have less energy than the reactants
Activation Energy
2. ΔH is negative DH

43. Thermochemistry & Stoichiometry
If you know the ΔH for a balanced equation, you may determine the amount of energy used or released by a reaction.

44. ΔH (enthalpy) is proportional to the coefficients for a balanced equation, therefore they may be used to write conversion factors.
2 H2O2(l)  2 H2O(l) + O2(g)
Hrxn = -190 kJ
2 mole H2O2 = -190 kJ
1 mole O2 = -190 kJ Or Or

45. Example
How much heat will be released if 1.0 g of hydrogen peroxide (H2O2) decomposes in a bombardier beetle to produce a steam spray?
2 H2O2(l)  2 H2O(l) + O2(g)
Hrxn = -190 kJ
1.0 g H2O2 g
1 mol
2 mol H2O2
-190 kJ
= kJ

46. C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
Example
How much heat is transferred when 9.22 g of glucose (C6H12O6) in your body reacts with O2 according to the following equation?
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
Hrxn = kJ
9.22 g C6H12O6 g
1 mol
1 mol C6H12O6
-2803 kJ
= kJ

47. Example
How much energy will be required to extract 59.5 grams of tin?
SnO2(s) + 4NO2(g) + 2H2O(l) kJ  Sn(s) + 4HNO3(aq)
Hrxn = +192 kJ g
1 mol
59.5 g Sn
1 mol Sn
192 kJ
= kJ