1. Chapter 17
The Flow of Energy

2. Energy Transformations
Energy is the capacity for doing work or supplying heat.
Unlike matter, energy has neither mass nor volume
Energy is detected only because of its effects – ex: the
motion of a race car.
Thermochemistry is the study of energy changes that
occur during chemical reactions and changes in state.
Every substance has a certain amount of energy stored
inside it. The energy stored in the chemical bonds of a
substance is called chemical potential energy.

3. Energy Transformations
The kinds of atoms and their arrangement in the
substance determine the amount of energy stored in the
substance.
During a chemical reaction, a substance is transformed
into another substance with a different amount of
chemical potential energy.
When you buy gasoline, you are actually buying the
stored potential energy it contains.
The controlled explosions of the gasoline in a car’s
engine transform the potential energy into useful work,
which can be used to propel the car.

4. Energy Transformations
At the same time, heat is also produced, making the car’s
engine extremely hot.
Energy changes occur as either heat transfer or
work, or a combination of both.
Heat (q) is energy that transfer from one object to
another because of a temperature difference between
them.
Heat always flows from a warmer object to a cooler
object.
If two objects remain in contact, heat will flow from the
warmer object to the cooler object until the temperature

5. The Universe
Chemical reactions and changes in physical state
generally involve either the release or the absorption of
heat.
System is the part of the universe on which you focus
your attention.
Surroundings include everything else in the universe.
Together the system and its surroundings make up the
universe.
Thermochemistry examines the flow of heat between the
system and its surroundings.

6. Law of Conservation of Energy
The law of conservation of energy states that in any
chemical or physical process, energy is neither created
nor destroyed.
If the energy of the system decreases during a process,
the energy of the surroundings must increase by the
same amount so that the total energy of the universe
remains unchanged.
In thermochemical calculations, the direction of the heat
flow is given from the point of view of the system.

7. Endothermic Process
An endothermic process is one that absorbs heat from
the surroundings.
In an endothermic process, the system gains heat as the
surroundings cool down.
Heat flowing into a
system from its
surrounding is
defined as positive
(+ q value)

8. Exothermic Process
An exothermic process is one that releases heat into its
surroundings.
In an exothermic process, the system loses heat as the
surroundings heat up.
Heat flowing out of a system
into its surroundings is
defined as negative (- q value)

9. Units for Measuring Heat Flow
Heat flow is measured in two common units, the calorie
and the joule.
A calorie (cal) is the quantity of heat needed to raise the
temperature of 1 g of pure water 1ºC.
The word calorie is written with a small c except when
referring to the energy contained in food.
The dietary Calorie, written with a capital C, always
Calorie C
refers to the energy in food.
1 Calorie (dietary) = 1 kilocalorie = 1000 calories (heat flow)

10. Units for Measuring Heat Flow
The statement “10 g of sugar has 41 Calories” means
that 10 g of sugar releases 41 kilocalories of heat when
completely burned.
The joule (J) is the SI unit of energy. One joule of heat
raises the temperature of 1 g of pure water 0.2390ºC.
1 J = 0.2390 cal 4.164 J = 1 cal

11. Conceptual Problem
A container of melted paraffin wax is allowed to stand at room
temperature until the wax solidifies. What is the direction of
heat flow as the liquid wax solidifies. Is the process
exothermic or endothermic?
Heat flows from the system (paraffin wax) to the surroundings
(air). The process is exothermic.
When solid barium hydroxide octahydrate is mixed in a beaker
with solid ammonium thiocynanate, a reaction occurs. The
beaker quickly becomes very cold. Is the reaction exothermic
or endothermic?
Since the beaker becomes cold, heat is absorbed by the
system (chemical within the beaker) from the surroundings
(the beaker and surrounding air). The process is
endothermic.

12. Heat Capacity
The amount of heat needed to increase the temperature
of an object exactly 1ºC is the heat capacity of that
object.
The heat capacity of an object depends on both its mass
and its chemical composition.
The greater the mass of the object, the greater its heat
capacity.

13. Heat Capacity
Different substances with the same mass may have
different heat capacities.
On a sunny day, a 20kg puddle of water may be cool,
while a nearby 20 kg iron sewer cover may be too hot to
touch.

14. Specific Heat
Assuming that both the water and the sewer cover
absorb the same amount of radiant energy from the sun,
the temperature of the water changes less than the
temperature of the cover because the specific heat
capacity of water is larger.
The specific heat (C) of a substance is the amount of
heat it takes to raise the temperature of 1 g of the
substance 1ºC.
Water has a very high specific heat (it takes more energy
to raise the temperature)
Metals have low specific heats (it takes less energy to
raise the temperature)

15. Specific Heat
Heat affects the temperature of objects with a high
specific heat much less than the temperature of those
with a low specific heat.
It takes a lot of heat to raise the temperature of water,
water also releases a lot of heat as it cools.
Water in lakes and oceans absorbs heat from the air on
hot days and releases it back into the air on cool days.
This property of water is
responsible for moderate
climates in coastal areas.

16. High Specific Heat of Water
Two other common effects associated with the high
specific heat of water.
1.Spraying oranges with water to protect the fruit from
frost damage during icy weather.
 As the water freezes, it releases heat, which
helps prevent the fruit from freezing.
2.The filling of an apple pie stays hotter than the crust.
 The filling that is mostly water, has a higher
specific heat than the crust. In order to cool down,
the filling must give off a lot of heat.

17. Calculating Specific Heat
C= q = heat (joules or calories)
m x ∆T = mass (g) x change in temperature (ºC)
∆T = Tfinal - Tinitial
Specific heat may be expressed in terms of joules or
calories.
Therefore, the units of specific heat are either J / g ⋅ ºC
or cal / g ⋅ ºC
What factors do you think affect the specific heat of a
substance?
Amount of heat and the change in temperature.

18. Questions
When 435 J of heat is added to 3.4g of olive oil at 21ºC,
the temperature increases to 85ºC. What is the specific
heat of the olive oil?
C= q = 435 J = 2.0 J / g ⋅ ºC
m x ∆T (3.4g) (64ºC)
How much heat is required to raise the temperature of
250.0g of mercury 52ºC?
C x m x ∆T = q (0.14J/g⋅ºC)(250.0g)(52ºC) = 1.8kJ

19. Questions
In what direction does heat flow between two objects?
From the object of higher temperature to the object of
lower temperature.
How do endothermic processes differ from exothermic
processes?
Endothermic absorbs heat from surroundings; exothermic
releases heat to the surroundings.
What units are used to measure heat flow?
Calories and joules

20. Questions
On what factors does the heat capacity of an object
depend?
Mass and chemical composition
Using calories, calculate how much heat 32.0g of water
absorbs when it is heated from 25.0ºC to 80.0ºC. How
many joules is this?
q=C x m x ∆T (1.00cal/g⋅ºC)(55ºC)(32.0g) = 1.76kcal
4.164 J = 1 cal (1760cal)(4.164J/cal) = 7.33 kJ

21. Questions
A chunk of silver has a heat capacity of 42.8 J/g ⋅ºC and
a mass of 181g. Calculate the specific heat of silver.
C = 42.8 J/ºC = 2.36 x 10-1 J/g⋅ºC
181g
How many kilojoules of heat are absorbed when 1.00L
(1L = 1kg) of water is heated from 18ºC to 85ºC?
q=C x m x ∆T (4.18J/g⋅ºC)(67ºC)(1kg) = 2.8 x 102 kJ

22. End of Section 17.1

23. Calorimetry
Heat that is released or absorbed during many chemical
reactions can be measured by a technique called
calorimetry.
Calorimetry is the precise measurement of the heat flow
into or out of a system for chemical and physical
processes.
In calorimetry, the heat released by the system is equal
to the heat absorbed by its surroundings
Conversely, the heat absorbed by a system is equal to
the heat released by its surroundings.
The insulated device used to measure the absorption or
release of heat is called a calorimeter.

24. Enthalpy
Heat flows for many chemical reactions can be measured
in a constant pressure calorimeter .
Because most chemical reactions and physical changes
carried out in the laboratory are open to the
atmosphere, these changes occur at constant pressure.
The heat content of a system at constant pressure is the
same as a property called enthalpy (H) of the system.
The heat released or absorbed by a reaction at constant
pressure is the same as the change in enthalpy (∆H)
The terms heat and enthalpy change are used
interchangeably when reaction occur under constant
pressure. ( q = ∆H)

25. Measuring Enthalpy
(heat absorbed by surroundings) qsurr = ∆H = m x C x ∆T
Because the heat absorbed by the surroundings is equal
to (but has the opposite sign of) the heat released by
the system, the enthalpy change (∆H) for the reaction
can be written as follows.
(heat released by the system) qsys = ∆H = -qsurr = - (m x C x ∆T)
The sign of ∆H is negative for an exothermic reaction and
positive for an endothermic reaction.

26. Bomb Calorimeter
Calorimetry experiments can also be performed at
constant volume using a device called a bomb
calorimeter.
In a bomb calorimeter, a sample of a compound is
burned in a constant-volume chamber in the presence
of oxygen at high pressure.
The heat that is released warms the water surrounding
the chamber. By measuring the temperature increase of
the water, it is possible to calculate the quantity of heat
released.

27. Foam Cup Calorimeter
(Constant pressure)

28. Bomb Calorimeter
(Constant volume)

29. Questions
When 25.0 mL of water containing 0.025 mol HCl at
25.0ºC is added to 25.0 mL of water containing 0.025
mol NaOH at 25ºC in a foam cup calorimeter, a reaction
occurs. Calculate the enthalpy change in kJ during this
reaction if the highest temperature observed is 32.0ºC.
Assume the densities of the solutions are 1.00g/mL
∆H = - (m x C x ∆T) this is an exothermic reaction
The total volume is 25.0 mL + 25.0 mL = 50.0mL
You need the mass of water, so use the densities given
to calculate. 50.0mL (1.00 g/mL) = 50.0g

30. Questions
∆H = - (m x C x ∆T)
You know the specific heat of water is 4.18 J/g⋅ºC
∆T = Tf – Ti = 32.0ºC – 25.0ºC = 7.0ºC
∆H = - (50.0g) (4.18 J/g ⋅ºC) (7.0ºC) = -1463J
= 1.46 x 103 kJ

31. Questions
When 50.0 mL of water containing 0.050 mol HCl at
22.5ºC is added to 50.0 mL of water containing 0.50 mol
NaOH at 22.5ºC in calorimeter the temperature of the
solution increases to 26.0ºC. How much heat in kJ was
released by this reaction?
q = (m x C x ∆T)
The total volume is 50.0 mL + 50.0 mL = 100.0mL
You need the mass of water, so use the densities given
to calculate. 100.0mL (1.00 g/mL) = 100.0g
You know the specific heat of water is 4.18 J/g⋅ºC
∆T = Tf – Ti = 26.0ºC – 22.5ºC = 3.5ºC
q = (100.0g) (4.18 J/g ⋅ºC) (3.5ºC) = 1463J = 1.5 x 103 kJ

32. Questions
A small pebble is heated and placed in a foam cup
calorimeter containing 25.0 mL of water at 25.0ºC. The
water reaches a maximum temperature of 26.4ºC. How
many joules of heat were released by the pebble?
q = m x C x ∆T
You need the mass of water, so use known 1L = 1kg to
calculate. .0250L (1000 g/L) = 25.0g
You know the specific heat of water is 4.18 J/g⋅ºC
∆T = Tf – Ti = 26.4ºC – 25.0ºC = 1.4ºC
q = (25.0g) (4.18 J/g ⋅ºC) (1.4ºC) = 146J

33. Thermochemical Equations
When you mix calcium oxide with water, 1 mole of
calcium hydroxide forms and 65.2 kJ of heat is released.
In a chemical equation, the enthalpy change for the
reaction can be written as either a reactant or a product.
In the equation describing the exothermic reaction of
CaO and H2O, the enthalpy change can be considered a
product.
CaO (s) + H2O (l) → Ca(OH)2 (s) + 65.2 kJ
A chemical equation that includes the enthalpy change is
called a thermochemical equation.
equation
CaO (s) + H2O (l) → Ca(OH)2 (s) ∆H= -65.2 kJ

34. Thermochemical Equations
The heat of reaction is the enthalpy change for the
chemical equation exactly as it is written. You will see
heats of reaction reported as ∆H, which is equal to the
heat flow at constant pressure.
The physical state of the reactants and products must
also be given.
The standard conditions are that the reaction is carried
out at 101.3 kPa (1atm) and that the reactants and
products are in their usual physical states at 25ºC.
The heat or reaction, or ∆H, in the CaO reaction example
is -65.2kJ.
Each mole of CaO and H2O that react to form Ca(OH)2
produces 65.2 kJ of heat.

35. Thermochemical Equations
Other reactions absorb heat from the surroundings.
Baking soda decomposes when heated. The carbon
dioxide released in the reaction causes a cake to rise
while baking. This process in endothermic.
2NaHCO3 (s) + 129kJ → Na2CO3 (s) + H2O (g) + CO2 (g)
Remember that ∆H is positive for endothermic reactions.
Therefore, you can write the reactions as follows:
2NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g) ∆H=129kJ

36. Thermochemical Equations
Chemistry problems involving enthalpy changes are
similar to stoichiometry problems.
The amount of heat released or absorbed during a
reaction depends on the number of moles of the
reactants involved.
The decomposition of 2 mol of sodium bicarbonate
requires 129kJ of heat.
2NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g) ∆ H=129kJ
Therefore, the decomposition of 4 mol of the same
substance would require twice as much heat or 258 kJ.

37. Thermochemical Equations
In endothermic processes, the potential energy of the
products(s) is higher than the potential energy of
the reactants.
The physical state of the reactants and products must
also be given.
H2O (l) → H2 (g) + 1/2O2 (g) ∆H = 285.8 kJ
H2O (g) → H2 (g) + 1/2O2 (g) ∆H = 241.8 kJ
Although the two equations are very similar, the different
physical states of H2O result in different ∆H values.

38. Questions
Calculate the amount of heat (in kJ) required to
decompose 2.24 mol NaHCO3 (s)
2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H=129kJ
The thermochemical equation indicates that 129 kJ of
heat are needed to decompose 2 mole of NaHCO3 (s)
∆H = 2.24 mole NaHCO3 (s) | 129 kJ = 144kJ
| 2 mol NaHCO3 (s)
Think Logically: Because the decomposition of 2 mol of
NaHCO3 requires 129kJ, then the decomposition of 2.24
mol should absorb about 10% more heat than 129kJ.

39. Questions
When carbon disulfide is formed from its elements, heat
is absorbed. Calculate the amount of heat (in kJ)
absorbed when 5.66 g of carbon disulfide is formed.
C(s) + 2S(s) → CS2(l) ∆H= 89.3kJ
The thermochemical equation indicates that 89.3 kJ of
heat are needed to form 1 mole of CS2 (l)
∆H = 5.66 g CS2 (l) | 1 mol CS2 (l) | 89.3 kJ = 6.64kJ
| 76.1g CS2 (l) | 1mol CS2 (l)

40. Questions
The production of iron and carbon dioxide from Iron(III)
oxide and carbon monoxide is an exothermic reaction.
How many kJ of heat are produced when 3.40 mol
Fe2O3 reacts with an excess of CO?
Fe2O3(s) + 3CO(g) → 2Fe(s) +3CO2(g) + 26.3kJ
∆H = 3.40 mol Fe2O3(s) | 26.3 kJ = -89.4kJ
| mol Fe2O3(s)

41. Questions
Explain the difference between H and ∆H.
H is enthalpy or heat content; ∆H represents a change in
heat content.
Calorimetry is based on what basic concepts?
The heat released by the system is equal to the heat
absorbed by the surroundings. Vice versa.
How are enthalpy changes treated in chemical equations
The enthalpy change in a chemical reaction can be
written as either a reactant or a product.

42. Questions
When 2 mol of solid magnesium combines with 1 mole of
oxygen gas, 2 mol of solid magnesium oxide is formed
and 1204kJ of heat is released. Write the
thermochemical equation for this combustion reaction.
2Mg (s) + O2 (g) → 2MgO + 1204kJ
2Mg (s) + O2 (g) → 2MgO ∆H = -1204kJ

43. Questions
Gasohol contains ethanol (C2H5OH) (l) which when
burned reacts with oxygen to produce CO2(g) and
H2O (g). How much heat is released when 12.5 g of
ethanol burns?
C2H5OH (l) + 3O2 (g) → 2CO3 (g) + 3H2O (l) + 1368kJ
∆H = 12.5g C2H5OH(l) | 1 mol C2H5OH(l) |1368kJ
| 46g C2H5OH(l) | mol
= -372 kJ
Explain the term heat of combustion
The heat of reaction for the complete burning of one mole

44. End of Section 17.2

45. Heats of Fusion and Solidification
All solids absorb heat as they melt to become liquids.
The gain of heat causes a change of state instead of a
change in temperature.
Whenever a change of state occurs by a gain or loss of
heat, the temperature of the substance remains
constant.
The heat absorbed by one mole of a solid substance as it
melts to a liquid at constant temperature is the molar
heat of fusion. (∆Hfus)
The molar heat of solidification (∆Hsolid) is the heat lost
when one mole of a liquid solidifies at constant
temperature.

46. Heats of Fusion
The quantity of heat absorbed by a melting solid is
exactly the same as the quantity of heat released when
a liquid solidifies.
∆Hfus = -∆Hsolid
When a liquid solidifies, it loses heat, thus the negative sign.

47. Heats of Fusion and Solidification
Melting 1 mol of ice at 0ºC to 1 mol of water at 0ºC
requires the absorption of 6.01kJ of heat. (this quantity
of heat is the molar fusion of water)
The conversion of 1 mol of water at 0ºC to 1 mol of ice at
0ºC releases 6.01kJ of heat. (this quantity of heat is the
molar heat of solidification of water)
H2O (s) → H2O (l) (∆Hfus)= 6.01 kJ/mol
H2O (l) → H2O (s) (∆Hsolid)= 6.01 kJ/mol

48. Sample Problem
How many grams of ice at 0ºC will melt is 2.25kJ of heat
are added?
2.25 kJ 1 mol ice 18.0 g ice = 6.74 g ice
6.01 kJ 1 mol ice
Use your common sense to check. 6.01 kJ of heat is
required to melt 1 mol of ice. You are only adding about
1/3 of that heat, so only about 1/3 of the ice should melt.

49. Sample Problem
How many kJ of heat are required to melt a 10.0g
popsicle at 0ºC? Assume the popsicle has the same
molar mass and heat of fusion as water.
10.0 g pop 1 mol pop 6.01 kJ = 3.34 kJ
18 g pop 1 mol pop
How many grams of ice at 0ºC could be melted by the
addition of 0.400 kJ of heat?
0.400 kJ 1 mol ice 18 g ice = 1.20 g ice
6.01 kJ 1 mol ice

50. Heats of Vaporization and Condensation
When liquids absorb heat at their boiling points, they
become vapors. The amount of heat necessary to
vaporize one mole of a given liquid is called its molar
heat of vaporization (∆Hvap)
The molar heat of vaporization of water is 40.7 kJ /mol
It takes 40.7 of energy to convert 1 mol of water to 1
mole of water vapor at the normal boiling point of water.
H2O (l) → H2O (g) ∆Hvap = 40.7 kJ/mol
Condensation is the exact opposite of vaporization

51. Heats of Vaporization and Condensation
When a vapor condenses, heat is released. The amount
of heat released when 1 mol of vapor condenses at the
normal boiling point is called its molar heat of
condensation. (∆Hcond)
The value is numerically the same as the molar heat of
vaporization, however, the value has the opposite sign.
∆Hvap = -∆Hcond
Heat is released during condensation, thus the negative
sign.
Condensation is the exact opposite of vaporization

52. (∆Hcond)
(∆Hvap)
∆Hfus
∆Hsolid

53. Changes of State

54. Sample Problem
How much heat (in kJ) is absorbed when 24.8 g H20 (l) at
100ºC and 101.3 kPa is coverted to steam at 100ºC?
24.8g H2O 1 mol H2O 40.7 kJ = 56.1 kJ
18 g H2O 1 mol H2O
How much heat is absorbed when 63.7 g H2O at 100ºC
and 101.3 kPa is converted to steam at 100ºC?
63.78g H2O 1 mol H2O 40.7 kJ = 144 kJ
18 g H2O 1 mol H2O

55. Heat of Solution
During the formation of a solution, heat is either released
or absorbed.
The enthalpy change caused by dissolution of one mole
of substance is the molar heat of solution (∆Hsoln)
Hot packs are an example. When CaCl2 and H2O are
mixed, heat is produced. (solution releases heat and the
reaction is exothermic)
A cold pack is an example of an endothermic reaction,
where the solution absorbs heat.

56. Sample Problem
How much heat is released when 0.677 mol NaOH is
dissolved in water.
0.677 mol NaOH -445.1 kJ = -301 kJ
1 mol NaOH
How many moles of NH4NO3 must be dissolved in water
so that 88.0 kJ of heat is absorbed from the water?
(∆Hsoln for NH4NO3 = 25.4 kJ/mol)
88.0 kJ 1 mol NH4NO3 = 3.42 mol NH4NO3
25.4 kJ

57. Calculating Heats of Reaction
Hess’s law of heat summation states that 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.
Use Hess’s law to find the enthalpy change for the
conversion of diamond to graphite as follows:
C(s,graphite) + O2 → CO2(g) ∆H = -393.5 kJ
C(s, diamond) + O2 → CO2(g) ∆H = -395.4 kJ
Write the first equation in reverse because you want
graphite on the product side. When you reverse the
equation, the sign of ∆H is also reversed.

58. End of Section 17.3

59. Calculating Heats of Reaction
CO2(g) → C(s,graphite) + O2(g) ∆H = 393.5 kJ (in reverse)
Add both equations to get:
CO2(g) → C(s,graphite) + O2 ∆H = 393.5 kJ
C(s, diamond) + O2 → CO2(g) ∆H = -395.4 kJ
C(s, diamond) → C(s,graphite) ∆H = -1.9 kJ
The conversion of diamond to graphite is an exothermic
process, so its heat of reaction has a negative sign.
Conversely, the change of graphite to diamond is an
endothermic process.

60. Calculating Heats of Reaction
Find the enthalpy change of the change of graphite to CO
C(s,graphite) + O2(g) → CO2(g) ∆H = -393.5 kJ
CO(g) + 1/2O2(g) → CO2(g) ∆H = -283.0 kJ
Write the second equation in reverse to get CO on the
product side. (don’t forget to change the sign)
C(s,graphite) + O2(g) → CO2(g) ∆H = -393.5 kJ
CO2(g) → CO(g) + 1/2O2(g) ∆H = 283.0 kJ
C(s,graphite) + 1/2O2(g) → CO(g) ∆H = -110.5 kJ

61. Standard Heats of Formation
Enthalpy changes generally depend on conditions of the
process. In order to compare enthalpy changes,
scientists specify a common set of conditions as a
reference point.
These conditions, called the standard state, refer to the
stable form of a substance at 25ºC and 101.3 kPa.
The standard heat of formation (∆Hf0) of a compound is
the change in enthalpy that accompanies the formation
of one mole of a compound from its elements with all
substances in their standard states.
The ∆Hf0 of a free element is arbitrarily set at zero.

62. Standard Heats of Formation
Standard heats of formation provide an alternative to
Hess’s law in determining heats of reaction indirectly.
For a reaction that occurs at standard conditions, you can
calculate the heat of reaction by suing standard heats of
formation.
This enthalpy change is called the standard heats of
reaction (∆H0)
The standard heat of reaction is the difference between
the standard heats of formation of all the reactants and
products.
∆H0 =
∆Hf0 (products) -
∆Hf0 (reactants)

63. Sample Problem
What is the standard heat of reaction for the reaction of
CO(g) with O2 (g) to form CO2 (g)
∆Hf0 O2 = 0kJ/mol (free element)
∆Hf0 CO2 = -393.5kJ/mol
∆Hf0 CO = -110.5kJ/mol
First write a balanced equation:
2CO (g) + O2 (g) → 2CO2 (g)
Next find and add the ∆Hf0 of all of the reactants, taking
into account the number of moles of each.
∆Hf0 (reactants) = (2 mol CO)(-110.5kJ/mol) + 0kJ = -221.0kJ

64. Sample Problem
∆Hf0 (products) = (2 mol CO2) (-393.5kJ/mol) = -787 kJ
Lastly, plug your values calculated for ∆Hf0 (products) and
∆Hf0 (reactants) into the equation.
∆H0 =
∆Hf0 (products) -
∆Hf0 (reactants)
∆H0 = (-787.0kJ) – (-221.0kJ) = -566.0 kJ

65. Sample Problem
What is the standard heat of reaction for the reaction
Br2(g) → Br2 (l)
∆Hf0 Br2 (g) = 0kJ/mol (free element)
∆Hf0 Br2 (l) = -393.5kJ/mol ∆
The equation is already balanced
Next find and add the ∆Hf0 of all of the reactants, taking
into account the number of moles of each.
∆Hf0 (reactants) = (1 mol Br2 (g))(30.91kJ/mol) = 30.91 kJ
∆Hf0 (products) = 0
∆H0 = (0kJ) – (30.91kJ) = -30.91kJ

66. Sample Problem
What is the standard heat of reaction for the reaction
CaCO3(s) → CaO(s) + CO2(g)
∆Hf0 CaCO3(s) ) = -1207.0 kJ/mol
∆Hf0 CaO(s) = -635.1 kJ/mol ∆Hf0
CO2(g) = -393.5 kJ/mol
The equation is already balanced
Next find and add the ∆Hf0 of all of the reactants, taking
into account the number of moles of each.
∆Hf0 (products) = (1 mol CaO(s) )(-635.1 kJ/mol ) +
(1 mol CO2(g) )(-393.5 kJ/mol ) = -1028.6 kJ
∆Hf0 (reactants) = -1207.0kJ
∆H0 = (-1028.6 kJ) – (-1207.9 kJ) = 179.3 kJ

67. End of Chapter 17