1. Energy and chemical change
Zhanna Charniauskaya
Energy and chemical change

2. A System and the surroundings
A system is the part of universe chosen for study.
The surroundings are the part of universe outside of the system with which the system interacts.

3. An Open system
An open system freely exchange energy and matter with the surroundings.

4. A Closed system
A closed system can exchange energy with the surroundings, but not matter.

5. An Isolated system
An isolated system does not interact with the surroundings.

6. Energy Energy (Greek): “work within”
Energy is the capacity to do work and produce heat.
Work (in chemistry) is done when system expands or contracts: balloon filled with gas expands or contracts
Energy = Kinetic Energy + Potential Energy (energy that is stored in an object or substance)
Kinetic energy is energy of motion.

7. Work

8. Work

9. Chemical potential energy
is stored in the chemical bonds
depends on the strength of the bonds
is released or absorbed as heat during chemical reaction when bonds break and rearrange.
Chemical potential energy of Methane CH4 results from the arrangement of the carbon and hydrogen atoms and the strength of the bonds that join them.

10. Direction of heat flow

11. Analyze the diagram

12. Heat
Heat is energy transferred between a system and its surroundings as a result of temperature difference.
At the molecular level
Molecules of the warmer body, through the collisions, loose kinetic energy to those of the colder body.
Thermal energy is transferred until the average kinetic energies of two bodies become the same, and until the temperatures become equal.

13. Heat versus temperature
Heat is a the form of energy that flows from a warmer object to a cooler object.
Temperature is a measure of heat.
Temperature is a measure of the average kinetic energy of the particles in a sample of matter.

14. Units for measuring heat flow
The SI unit for measuring energy and heat is the joule (J).
Also energy and heat are measured in calories (cal).
1 J = cal cal = J

15. Discussion Question
Is it possible to calculate quantity of heat transferred from one object to another?
What do you think quantity of heat absorbed or released will depend on?

16. quantity of heat Quantity of heat (q) depends on:
How much the temperature is to be changed/change in temperature
The quantity of substance, mass
The nature of substance/chemical composition of substance

17. Change in temperature Change in Temperature ∆𝑻
∆ is a Greek letter called delta
∆ always means change
∆𝑻 = 𝑻 𝒇 − 𝑻 𝒊
𝑻 𝒇 is a final temperature
𝑻 𝒊 is an initial temperature

18. Specific heat describes how well substance absorbs heat.

19. Specific heat Specific heat, c Unit of measurement: 𝑱 𝒈℃
describes how well substance absorbs/takes in heat
is determined by the nature of substance/chemical composition and state of matter.
is the amount of heat needed to raise temperature of one gram of substance by one degree Celsius.
Unit of measurement: 𝑱 𝒈℃

20. quantity of heat 𝒒=𝒄 ∙𝒎 ∙∆𝑻
𝒒−𝒂𝒏 𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅 𝒐𝒓 𝒓𝒆𝒍𝒆𝒂𝒔𝒆𝒅, 𝒋𝒐𝒖𝒍𝒆 (𝑱)
𝒄 −𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒉𝒆𝒂𝒕 𝒐𝒇 𝒂 𝒔𝒖𝒃𝒔𝒕𝒂𝒏𝒄𝒆, 𝑱 𝒈 ℃
𝒎−𝒎𝒂𝒔𝒔 𝒐𝒇 𝒂 𝒔𝒖𝒃𝒔𝒕𝒂𝒏𝒄𝒆, 𝒈𝒓𝒂𝒎 (𝒈)
∆𝑻 𝒊𝒔 𝒂 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒄𝒉𝒂𝒏𝒈𝒆, ℃
𝑺𝒊𝒏𝒄𝒆 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒊𝒔 ∆𝑻= 𝑻 𝟐 − 𝑻 𝟏
We can rewrite the formula for quantity of heat
𝒒=𝒄 ∙𝒎 ∙( 𝑻 𝟐 − 𝑻 𝟏 )

21. The sign of the quantity of heat
Endothermic process:
Heat (q) is absorbed/taken in by a system.
q >𝟎 q is positive
Exothermic process:
Heat (q) is released/lost to the surroundings.
q<𝟎 q is negative

22. Keep in mind….
When analyzing a problem, keep in mind that heat can be negative or positive quantity.
If the problem states:
A sample of metal of 22.2 g lost 3536 J of heat…
You should assign negative sign to the amount of heat, because heat was lost by the system and released to the surroundings.
q = J

23. Keep in mind… If the problem states:
The sample with the initial temperature of 20 ℃ was heated to 35℃. The amount of heat that was absorbed by a sample was 7865 J.
You should assign a positive sign to the quantity of heat, as the heat was absorbed by a system.
q = 7865 J.

24. Example problem
A 38.8 g piece of metal alloy absorbs 181 J as its temperature increases from 25.0℃ to ℃. What is the alloy’s specific heat?
m = 38.8 g
q = 181 J
T1 = 25.0℃
T2 = 36.0 ℃
c - ?

25. 𝒒=𝒄∙𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 )
We need to isolate c, specific heat.
Divide both sides of the equation by 𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 )
𝒒 𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 ) = 𝒄 ∙𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 ) 𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 ) 𝒄= 𝒒 𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 )
𝒄= 𝟏𝟖𝟏 𝑱 𝟑𝟖.𝟖 𝒈∙(𝟑𝟔.𝟎℃ −𝟐𝟓.𝟎℃ ) = 𝑱 𝒈℃

26. Example Problem
A cup of hot tea is left on the kitchen table. After few hour its temperature is 22 ℃. The mass of tea in the cup is 220 g, and quantity of heat released was J. What was the initial temperature of the tea? Specific heat of H2O(l) is 𝑱 𝒈℃ .

27. Example problem
A cup of hot water was left on the counter. The temperature of water decreased from 82℃ to 23℃. It was found that J of heat was released to the environment. The specific heat of water is 4.18 𝑱 𝒈℃ . What was the mass of water?

28. Equal masses of aluminum, gold, iron, and silver were left in the Sun at the same time and for the same length of time. Use the table to arrange four metals according to the increase in their temperatures from largest increase to smallest.
Substance
Specific heat, J/(g℃)
Aluminum
0.897
Iron
0.449
Silver
0.235
Gold
0.129

29. Calorimetry/measuring heat

30. calorimeter measures heat of chemical reaction/physical process.
An isolated system
Well insulated container
Has thermometer and stirrer
Water inside is to absorb or release heat from/to an object placed into calorimeter.
Heat released and heat absorbed between object (system) and surroundings (water) in a calorimeter are equal but opposite in sign.
𝒒 𝒓𝒆𝒍𝒆𝒂𝒔𝒆𝒅 = −𝒒 𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅

31. Discussion question Tf object = Tf water
When you place an object in a calorimeter, how can we describe the final temperature of the object and the final temperature of water?
Final temperature of the object and the final temperature of water are equal.
Tf object = Tf water

32. Example Problem
A 2.50 g sample of zinc is heated, then placed in a calorimeter containing 65.0 g of water. Temperature of water increases from oC to oC. The specific heat of zinc is J/goC. What was the initial temperature of the zinc metal sample?
Specific heat of water is 4.18 𝑱 𝒈℃

33. Calorimetry problem Water (surroundings) mwater = 65.0 g T1 = 20.00℃
Zinc sample (system/object)
mwater = 65.0 g
T1 = 20.00℃
T2 = 22.50℃
Cwater = 𝑱 𝒈℃
We can calculate heat absorbed by water
𝒒=𝒄 ∙𝒎 ∙( 𝑻 𝟐 − 𝑻 𝟏 )
𝒒=4.18 𝑱 𝒈℃ ∙65.0 g∙(22.50℃−20.00℃)
𝒒=𝟔𝟕𝟗 𝑱
mZn= 2.50 g
cZn = 𝑱 𝒈℃
T1 - ?
Tf object = Tf water T2 = 22.50℃
𝒒 𝒁𝒏 = −𝒒 𝒘𝒂𝒕𝒆𝒓 𝒒 𝒁𝒏 = −𝟔𝟕𝟗 𝑱
𝒒=𝒄 ∙𝒎 ∙( 𝑻 𝟐 − 𝑻 𝟏 )
𝑻 𝟐 − 𝑻 𝟏 = 𝒒 𝒄 ∙𝒎
22.50℃− 𝑻 𝟏 = −𝟔𝟕𝟗 𝑱 𝑱 𝒈℃ ∙2.50 g
22.50℃− 𝑻 𝟏 =− 696.4℃
− 𝑻 𝟏 =− 696.4℃−22.50℃
𝑻 𝟏 =718.9℃
Calorimetry problem

34. Example problem
Suppose you put 125 g of water into a calorimeter and find that the initial temperature is 25.60℃. Then you heat a 50.0 g sample of the unknown metal to 115.0℃ and put the metal sample into the water. As the resulted the water is heated to ℃. What is the specific heat of an unknown metal?

35. Thermochemistry
Thermochemistry studies heat changes during chemical reactions or phase changes.

36. Heat and enthalpy change
When chemical bonds and/or intermolecular forces break and rearrange during chemical reactions or physical process heat is released or absorbed.
Heat released or absorbed in a chemical reaction, phase change, or solution formation is called an enthalpy change ∆H (Joules).
The enthalpy (H) is related to chemical potential energy stored in the bonds.
Enthalpein (from Greek): heat from within
En - within
Thalpein – to heat

37. Enthalpy (heat) of chemical reaction

38. Enthalpy change of chemical reaction

39. Enthalpy (heat) of chemical reaction
A + B → C + D
Enthalpy change (heat) produced or absorbed in a reaction depends only on enthalpies of reactants and products.
∆ 𝑯 𝒓𝒙𝒏 =∆ 𝑯 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 −∆ 𝑯 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔

40. Thermochemical equations
Thermochemical equation includes
balanced chemical equation
the energy change/enthalpy change
Example:
4Fe(s) + 3O2(g) → 2Fe2O3(s) kJ

41. Analyze diagram
4Fe(s) + 3O2(g) → 2Fe2O3(s)

42. Exothermic process 4Fe(s) + 3O2(g) → 2Fe2O3(s) ∆𝑯=−𝟏𝟔𝟐𝟓 𝒌𝑱
Heat is released.
Heat is produced during reaction, it is considered as a product, and is included with the products with a positive sign.
4Fe(s) + 3O2(g) → 2Fe2O3(s) kJ

43. Analyze diagram
NH4NO3(s) → NH4+(aq) + NO3−(aq)

44. Endothermic process NH4NO3(s) → NH4+(aq) + NO3−(aq) ∆𝑯=𝟐𝟕 𝒌𝑱
Heat is consumed during the reaction and is needed for the reaction to happen.
Heat is considered to be part of the reactants and included on the reactants side.
27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq)

45. Heat involved in phase changes
Heat involved in: Melting/Fusion is called Heat of fusion ∆ 𝑯 𝒇𝒖𝒔𝒊𝒐𝒏 , J/mol Freezing : Heat of solidification ∆ 𝑯 𝒔𝒐𝒍𝒊𝒅 , J/mol Vaporization/Boiling: Heat of vaporization ∆ 𝑯 𝒗𝒂𝒑. Condensation: Heat of condensation ∆ 𝑯 𝒄𝒐𝒏𝒅.

47. Calorimeter and enthalpy change of chemical reaction
A + B → C + D
∆ 𝑯 𝒓𝒙𝒏 =∆ 𝑯 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 −∆ 𝑯 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔
When ∆ 𝑯 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝒂𝒏𝒅 ∆ 𝑯 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 are unknown, calorimeter could be used to find ∆ 𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 .

48. Enthalpy change (∆𝐻) versus heat transfer (𝓆)
Quantity of heat (heat transfer, or just heat) 𝒒
𝒒 =𝒄∙𝒎∙( 𝑻 𝟐 − 𝑻 𝟏 )
measured in J, joules
is calculated for any amount of substance.
Enthalpy change of chemical reaction ∆𝑯
∆𝑯= 𝒒 𝒎𝒐𝒍
Measured in 𝑱 𝒎𝒐𝒍
∆𝑯 is calculated per 1 mole of substance

49. Heat of solution NaOH(s) → Na+(aq) + OH-(aq) ∆𝑯=−𝟒𝟒.𝟓 𝒌𝑱/𝒎𝒐𝒍
When a solute dissolves in a solvent heat could be released or absorbed.
The enthalpy change (heat of solution) ∆𝑯 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 is calculated for one mole of a substance dissolved.
NaOH(s) → Na+(aq) + OH-(aq) ∆𝑯=−𝟒𝟒.𝟓 𝒌𝑱/𝒎𝒐𝒍
NH4NO3(s) → NH4+(aq) + NO3−(aq) ∆𝑯=𝟐𝟕 𝒌𝑱/𝒎𝒐𝒍

50. Heat of solution
To find ∆𝑯 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 calorimeter is used.
𝒒 𝒓𝒆𝒍𝒆𝒂𝒔𝒆𝒅 = −𝒒 𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅
∆𝑯= 𝒒 𝒎𝒐𝒍

51. Enthalpy of a solution problem
A calorimeter is filled with 75 grams of water at an initial temperature of 19.8 oC. A 2.0 g sample of NaOH is added, and the temperature increases to 27.6 oC. What is the enthalpy change for dissolution of NaOH?
NaOH(s)  Na+(aq) + OH-(aq)
H = ?

52. Heat of solution problem
Water (surroundings)
NaOH (system/object)
mwater = 75.0 g
T1 = 19.8℃
T2 = 27.6℃
Cwater = 𝑱 𝒈℃
We can calculate heat absorbed by water
𝒒=𝒄 ∙𝒎 ∙( 𝑻 𝟐 − 𝑻 𝟏 )
𝒒=4.18 𝑱 𝒈℃ ∙75.0 g×(27.6℃−19.8℃)
𝒒=𝟐𝟒𝟒𝟓 𝑱
𝒎 𝑵𝒂𝑶𝑯 = 2.0 g
H = ?
∆𝑯= 𝒒 𝒎𝒐𝒍
𝒒 𝒘𝒂𝒕𝒆𝒓 = −𝒒 𝑵𝒂𝑶𝑯
𝒒 𝑵𝒂𝑶𝑯 =−𝟐𝟒𝟒𝟓 𝑱
𝒎𝒐𝒍 𝑵𝒂𝑶𝑯 = 𝟐.𝟎𝒈 𝑵𝒂𝑶𝑯 𝟏 ∙ 𝟏 𝒎𝒐𝒍 𝑵𝒂𝑶𝑯 𝟒𝟎.𝟎 𝒈𝑵𝒂𝑶𝑯
𝒎𝒐𝒍 𝑵𝒂𝑶𝑯 = 0.05 mol
∆𝑯= −𝟐𝟒𝟒𝟓 𝑱 𝟎.𝟎𝟓𝒎𝒐𝒍 =−𝟒𝟖𝟗𝟎𝟎 𝑱 𝒎𝒐𝒍
Solvation/dissolution of NaOH is an exothermic process.
Heat of solution problem

53. Heat of solution problem
A sample of 4.50 g of ammonium chloride is dissolved in 53.o g of water in a calorimeter. The temperature of solution decreases from 20.40℃ to ℃. Specific heat of water is 4.18 𝑱 𝒈℃ . Find the heat of solution.

54. Heat of solution problem
A 12.8 g sample of KCl is added to 75 g of water and the temperature drops from 31.0 ℃ to ℃. Specific heat of H2O(l) is J/(g oC).
KCl (s) ® K+(aq) + Cl-(aq)
1. Calculate the heat change.
2. Calculate H for the dissolution of KOH.

55. Hess Law: Analyze the diagram

56. Hess law

57. Enthalpy (heat) of chemical reaction
A B
This equation summarizes the initial and final steps (reactants and products).
But few reactions only proceed in one step.
Usually reaction goes through intermediate steps to form final products.
∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 does not depend on the path the reaction takes.

58. Enthalpy (heat) of reaction
Hess Law:
The overall enthalpy change of a chemical reaction is independent of the path taken.
The enthalpy change can be calculated by adding the enthalpy changes for the intermediate reactions A to X, X to Y, and Y to B.
∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = ∆𝑯 𝟏 + ∆𝑯 𝟐 + ∆𝑯 𝟑 + ….

59. What do you notice? Forward reaction:
C(s, graphite) + O2(g) → CO2(g) ∆𝑯=−𝟑𝟗𝟑.𝟓 𝒌𝑱
Reverse reaction:
CO2(g) → C(s, graphite) + O2(g) ∆𝑯=𝟑𝟗𝟑.𝟓 𝒌𝑱

60. Reversing the process changes the sign of ∆𝑯 According to Hess law
A → B forward reaction
B → A reverse reaction
∆𝑯 𝒓𝒆𝒗𝒆𝒓𝒔𝒆 𝒓𝒙𝒏 = −∆𝑯 𝒇𝒐𝒓𝒘𝒂𝒓𝒅 𝒓𝒙𝒏

61. Hess Law According to Hess law
∆𝑯 𝒓𝒆𝒗𝒆𝒓𝒔𝒆 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = −∆𝑯 𝒇𝒐𝒓𝒘𝒂𝒓𝒅 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏
A(s) → A(l) ∆𝑯 = 25 kJ
A(l) → A(s) ∆𝑯 = ?
B (l) → B(g) ∆𝑯=𝟑𝟎𝟎 𝑱
B(g) → B(l) ∆𝑯= ?

62. Hess Law
A → B
What is enthalpy change for the above reaction if we know enthalpy changes for the following reactions?
A → X ∆𝑯=−𝟐𝟎𝟎 𝑱
B → X ∆𝑯=𝟏𝟐𝟎 𝑱
How can we use given reactions with known enthalpies to calculate ∆𝑯 for reaction A → B?

63. Hess law problem 1. Consider the path the reaction has taken.
Substance A did not turn straight to substance B.
Substance A turned into substance X first, and then substance X turned into substance B.
A → X
X → B
2. ∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = ∆𝑯 𝟏 + ∆𝑯 𝟐
As we can add enthalpies we can add reactions.

64. Hess Law problem
2. Add two reactions together: reactants with reactants and products with products.
A → X
X → B
A + X → X + B, simplify to get A → B

65. Hess Law problem 3. Calculate the ∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = ∆𝑯 𝟏 + ∆𝑯 𝟐
We know that for reaction A → X, ∆𝑯=−𝟐𝟎𝟎 𝑱
Do we know ∆𝑯 for the reaction X → B?
We know that for B → X, ∆𝑯=𝟏𝟐𝟎 𝑱
∆𝑯 𝒓𝒆𝒗𝒆𝒓𝒔𝒆 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = −∆𝑯 𝒇𝒐𝒓𝒘𝒂𝒓𝒅 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏
X → B ∆𝑯=−𝟏𝟐𝟎 𝑱
∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = ∆𝑯 𝟏 + ∆𝑯 𝟐
∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = −𝟐𝟎𝟎 𝑱 + (−𝟏𝟐𝟎 𝑱) = −𝟑𝟐𝟎 𝑱

66. Hess law problem 2N(g) → M(l) ∆𝑯=?
What is enthalpy change for the above reaction if we know enthalpy changes for the following reactions?
2N(g) → M(g) ∆𝑯=−𝟏𝟎𝟎 𝑱
M(l) → M(g) ∆𝑯=𝟏𝟐𝟎 𝑱
2N(g) → M(g)
M(g) → M(l)

67. The path the reaction takes:
2N(g) → M(g)
M(g) → M(l)
2N(g) + M(g) → M(g) + M(l) N(g) → M(l)
2N(g) → M(g) ∆𝑯=−𝟏𝟎𝟎 𝑱
M(g) → M(l) ∆𝑯=−𝟏𝟐𝟎 𝑱 (reverse from M(l) → M(g))
∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = ∆𝑯 𝟏 + ∆𝑯 𝟐
∆𝑯 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = −𝟏𝟎𝟎 𝑱 + (−𝟏𝟐𝟎 𝑱) = −𝟑𝟐𝟎 𝑱

68. Hess Law example C(s, diamond) → C(s, graphite)
The enthalpy change for this reaction cannot be measured directly.
We need to go through the series of known reactions to make this calculation.
Reaction 1:
C(s, graphite) + O2(g) → CO2(g) ∆𝑯=−𝟑𝟗𝟑.𝟓 𝒌𝑱
Reaction 2:
C(s, diamond) + O2(g) → CO2(g) ∆𝑯= −𝟑𝟗𝟓.𝟒 𝒌𝑱
Reverse reaction 1
CO2(g) → C(s, graphite) + O2(g) ∆𝑯=𝟑𝟗𝟑.𝟓 𝒌𝑱

69. Hess law example Add reaction 2 and reversed reaction 1
C(s, diamond) + O2(g) → CO2(g) ∆𝑯= −𝟑𝟗𝟓.𝟒 𝒌𝑱
CO2(g) → C(s, graphite) + O2(g) ∆𝑯= 𝟑𝟗𝟑.𝟓 𝒌𝑱
C(s, diamond) → C(s, graphite)
Add enthalpies of reactions
∆𝑯= −𝟑𝟗𝟓.𝟒 𝒌𝑱
∆𝑯= 𝟑𝟗𝟑.𝟓 𝒌𝑱
∆𝑯= −𝟏.𝟗 𝒌𝑱

70. Hess Law example
C(s, diamond) → C(s, graphite)
∆𝑯=−𝟏.𝟗 𝒌𝑱

71. Heat of Fusion and solidification
H2O(s) → H2O(l) ∆ 𝑯 𝒇𝒖𝒔𝒊𝒐𝒏 =𝟔.𝟎𝟏 𝐤𝐉/𝐦𝐨𝐥
Molar heat of fusion (∆ 𝑯 𝒇𝒖𝒔𝒊𝒐𝒏 ) is the heat absorbed by 1 mol of a solid substance as it melts at 𝑻 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 .
H2O(l) → H2O(s) ∆ 𝑯 𝒔𝒐𝒍𝒊𝒅 =− 𝟔.𝟎𝟏 𝐤𝐉/𝐦𝐨𝐥
Molar heat of solidification (∆ 𝑯 𝒔𝒐𝒍𝒊𝒅 ) is the heat released when 1 mol of liquid substance solidifies at 𝑻 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 .
∆ 𝑯 𝒇𝒖𝒔𝒊𝒐𝒏 = −∆ 𝑯 𝒔𝒐𝒍𝒊𝒅

72. Vaporization/condensation
H2O(l) → H2O(g) ∆ 𝑯 𝒗𝒂𝒑 =𝟒𝟎.𝟕 𝐤𝐉/𝐦𝐨𝐥
The amount of heat required to vaporize 1 mol of a liquid at 𝑻 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 is molar heat of vaporization (∆ 𝑯 𝒗𝒂𝒑 ).
H2O(g) → H2O(l) ∆ 𝑯 𝒄𝒐𝒏𝒅 =−𝟒𝟎.𝟕 𝐤𝐉/𝐦𝐨𝐥
The amount of heat released when 1 mol of a vapor condenses at 𝑻 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 is molar heat of condensation (∆ 𝑯 𝒄𝒐𝒏𝒅 ).
∆ 𝑯 𝒗𝒂𝒑 = −∆ 𝑯 𝒄𝒐𝒏𝒅

73. heat of combustion
The heat of combustion is the heat of reaction of complete burning of 1 mol of a substance.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
∆𝑯=−𝟖𝟗𝟎 𝒌𝑱 per 1 mol of CO2

74. Example problem
If grams of CH4 are burned completely, how much heat will be produced?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) kJ