This document discusses thermochemistry and calorimetry. It defines key terms like heat, temperature, enthalpy, and specific heat. Calorimetry experiments measure heat changes from chemical or physical processes. Thermochemical equations include the quantity of heat released or absorbed during chemical reactions. Standard heats of formation and combustion are provided for reference. Hess's law allows calculation of heats of reaction from known thermochemical data using principles of reversing and combining equations.

1. Chapter 17
Reaction Energy and
Reaction Kinetics

2. THERMOCHEMISTRY
• Virtually every chemical reaction
is accomplished by a change in
energy.
• Chemical reactions usually absorb
or release energy in the form of
heat.

3. THERMOCHEMISTRY
• Thermochemistry is the study of
the changes in heat energy that
accompany chemical reactions
and physical changes.
• The heat absorbed or released in
a chemical or physical change is
measured in a calorimeter.

4. THERMOCHEMISTRY
• In a calorimeter, known quantities
of reactants are sealed in a
reaction chamber, which is
immersed in a known quantity of
water in an insulated vessel.
• The heat given off or absorbed
during the reaction is equal to the
heat absorbed or given off by the
known quantity of water.

5. THERMOCHEMISTRY
• The amount of heat is determined
from the temperature change of
the known mass of the
surrounding water.
• The data collected from
calorimetry experiments are
temperature changes because
heat cannot be measured
directly, but temperature can be.

6. THERMOCHEMISTRY
What is the difference between
heat and temperature?
• Temperature is the measure of
the average kinetic energy of the
particles in a sample of matter.
• To measure temperature we use
the Celsius and Kelvin scales.
• Kelvin = °Celsius + 273.15

7. THERMOCHEMISTRY
• The ability to measure
temperature is based on heat
transfer.
• The amount of heat transfer is
usually measured in joules (J).
• A joule is the SI unit of heat
energy as well as all other forms
of energy.

8. THERMOCHEMISTRY
• Heat can be thought of as the
sum total of the kinetic energies
of the particles in a sample of
matter.
• Heat always flows spontaneously
from matter at a higher
temperature to matter at a lower
temperature.

9. THERMOCHEMISTRY
• The quantity of heat transferred
during a temperature change
depends on:
–The nature of the material
changing temperature.
–The mass of the material
changing temperature.
–The size of the temperature
change.

10. THERMOCHEMISTRY
• A quantity called specific heat can
be used to compare heat
absorption capacities for different
materials.
• Specific Heat is the amount of
heat energy required to raise the
temperature of one gram of a
substance by one Celsius degree
or one Kelvin.

11. THERMOCHEMISTRY
• Specific Heat is measured in units
of Joules/gram•°Celsius (J/g•°C)
or calories/gram•Kelvin (cal/g•K)
or calories/gram•°Celsius
(cal/g•°C).
• Table 17-1, page 513.

12. THERMOCHEMISTRY
• Specific Heat is usually measured
under constant pressure
conditions, so its
symbol, cp, contains a subscripted
p as a reminder.
• Specific heat can be found using
the following equation:
cp = q/(m × T)

13. THERMOCHEMISTRY
• cp = Specific Heat
• q = Heat lost or gained
• m = mass of the sample
• T = Difference between the
initial and final temperatures.

14. THERMOCHEMISTRY
EXAMPLE:
A 4.0 g sample of glass was heated
from 274 K to 314 K, a temperature
increase of 40 K, and was found to
have absorbed 32 J of heat. What is
the specific heat of this type of glass?
How much heat will the same glass
sample gain when it is heated from
314 K to 344 K?

15. THERMOCHEMISTRY
EXAMPLE:
• Determine the specific heat of a
material is a 35 g sample
absorbed 48 J as it was heated
from 293 K to 313 K.

16. THERMOCHEMISTRY
EXAMPLE:
• A piece of copper alloy with a
mass of 85.0 g is heated from 30
°C to 45 °C. In the process, it
absorbs 523 J of heat. What is the
specific heat of this copper alloy?
How much heat will the sample
lose if it is cooled to 25 °C?

17. THERMOCHEMISTRY
EXAMPLE:
• The temperature of a 74 g sample
of material increases from 15 °C
to 45 °C when it absorbs 2.0 kJ of
heat. What is the specific heat of
this material?

18. THERMOCHEMISTRY
EXAMPLE:
• How much heat is needed to raise
the temperature of 5.0 g of gold
by 25 °C?

19. THERMOCHEMISTRY
EXAMPLE:
• What mass of liquid water at
room temperature (25 °C) can be
raised to its boiling point with the
addition of 24 kJ of heat energy?

20. THERMOCHEMISTRY
EXAMPLE:
• Heat in the amount of 420 J is
added to a 35 g sample of water
at a temperature of 10 °C. What
will be the final temperature of
the water?

21. THERMOCHEMISTRY
• The heat of reaction is the
quantity of heat released or
absorbed during a chemical
reaction.
• You can think of heat of reaction
as the difference between the
stored energy of the reactants
and the products.

22. THERMOCHEMISTRY
EXAMPLE:
2H2(g) + O2(g) --> 2H2O(g)
• If a mixture of hydrogen and
oxygen is ignited, water will form
and heat energy will be released.
• The heat released comes from the
formation of the product (water).

23. THERMOCHEMISTRY
• Chemical equations do not tell
you that heat is evolved during
the reaction.
• Experiments show that 483.6 kJ
of heat are evolved when 2 mol of
gaseous water are formed at
298.15 K from its elements.

24. THERMOCHEMISTRY
• Thermochemical equations
include the quantity of heat
released or absorbed during the
reaction.
2H2(g) + O2(g) --> 2H2O(g) + 483.6 kJ
• The quantity of heat for any
reaction depends on the amounts
of reactants and products.

25. THERMOCHEMISTRY
• Producing twice as much water
vapor would require twice as
many moles of reactants, and
would release twice as much
heat.
4H2(g) + 2O2(g) --> 4H2O(g) + 967.2 kJ

26. THERMOCHEMISTRY
• Producing one-half as much water
would require one-half as many
moles of reactants and would
release only one-half as much
heat.
H2(g) + 1/2O2(g) --> H2O(g) + 241.8 kJ

27. THERMOCHEMISTRY
• Fractional coefficients are
sometimes used in
thermochemical equations.
• In chemical reactions, heat can
either be released or absorbed.
–Exothermic = release of heat
–Endothermic = absorption of heat

28. THERMOCHEMISTRY
• In writing thermochemical
equations for endothermic
reactions, the situation is
reversed because the products
have a higher heat content than
the reactants.
• Decomposition of water:
2H2O(g) + 483.6 kJ --> 2H2(g) + O2(g)

29. THERMOCHEMISTRY
• The physical states of the
reactants and products must
always be included in
thermochemical equations
because they influence the overall
amount of heat exchanged
(Ex: Ice).

30. THERMOCHEMISTRY
• The heat absorbed or released
during a chemical reaction at
constant pressure is represented
by the symbol H.
• The H is the symbol for a quantity
called enthalpy.
• Enthalpy is the heat content of a
system at constant pressure.

31. THERMOCHEMISTRY
• Most chemical reactions are run in
open vessels, so the pressure is
constant and equal to
atmospheric pressure.
• Only changes in enthalpy can be
measured ( H = Enthalpy
change).

32. THERMOCHEMISTRY
• Enthalpy change is the amount of
heat absorbed or lost by a system
during a process at constant
pressure.
H = Hproducts - Hreactants

33. THERMOCHEMISTRY
• Thermochemical equations are
usually written by designating the
value for H rather than writing
the heat as a reactant or product.
• Exothermic reactions have a H
that is always negative, because
the system loses energy.
2H2(g) + O2(g) --> 2H2O(g) H = -483.6 kJ

34. THERMOCHEMISTRY
• The H for endothermic reactions
is always positive, because the
system gains energy.
2H2O(g) --> 2H2(g) + O2(g) H = +483.6 kJ

35. THERMOCHEMISTRY
• Keep in mind the following
concepts when using
thermochemical equations:
1. The coefficients in a balanced
thermochemical equation
represent the number of moles,
not molecules, which allows us to
write these as fractions.

36. THERMOCHEMISTRY
2. The physical state of the
product/reactant involved in a
reaction is an important factor
and must be included in the
thermochemical equation.

37. THERMOCHEMISTRY
3. The change of energy
represented by a thermochemical
equation is directly proportional to
the number of moles of
substances undergoing a change.

38. THERMOCHEMISTRY
4. The value of the energy change,
H, is usually not significantly
influenced by changing
temperature.

39. THERMOCHEMISTRY
• The formation of water from
hydrogen and oxygen is a
composition reaction - the
formation of a compound from its
elements.
• The amount of heat released or
absorbed when one mole of a
compound is formed from its
elements is called the molar heat
of formation.

40. THERMOCHEMISTRY
• To make comparisons, heats of
formations are given for the
standard states of reactants and
products.
• These states are found at
atmospheric pressure and room
temperature (298.15 K).
–Example: Water is a liquid at
room temperature, not a solid.

41. THERMOCHEMISTRY
• To signify that a value represents
measurements on substances in
their standard states, a 0 sign is
added to the enthalpy symbol
( H0).
• Adding the subscript f, further
indicates a standard heat of
formation ( H0f).

42. THERMOCHEMISTRY
• Heats of Formation, Appendix
Table A-14.
• Each entry in the table is the heat
of formation for the synthesis of
one mole of the compound listed
from its elements in their
standard states.

43. THERMOCHEMISTRY
• Elements in their standard states
are defined as having a H0f = 0.
• Compounds with a high negative
heat of formation are very stable,
which means that they will not
decompose back into their
respective elements.

44. THERMOCHEMISTRY
• Compounds with relatively
positive values, or slightly
negative values, are relatively
unstable and will spontaneously
decompose into their elements if
the conditions are appropriate.
–Ex: Hydrogen Iodide (HI), ethyne
(C2H2), and Mercury Fulminate
(HgC2N2O2).

45. THERMOCHEMISTRY
• Combustion reactions produce a
considerable amount of energy in
the form of light and heat when a
substance is combined with
oxygen.
• The heat released by the
complete combustion of one mole
of a substance is called the heat
of combustion of the substance.

46. THERMOCHEMISTRY
• Heat of combustion is defined in
terms of one mole of reactant
(opposite of the heat of formation
= one mole of product).
• Heat of combustion is given the
notation Hc.
• Heats of Combustion, Appendix
Table A-5.

47. THERMOCHEMISTRY
• Remember that CO2 and H2O are
the products of the complete
combustion of organic
compounds.
Example:
• Combustion of propane.
C3H8(g) + 5O2(g) --> 3CO2 + 4 H2O(l)
H0c = -2219.2 kJ/mol

48. THERMOCHEMISTRY
• Thermochemical equations can be
rearranged, terms can be
canceled, and the equations can
be added to give enthalpy
changes for reactions not included
in the data tables.
• The basis for calculating heats of
reaction is known as Hess’s Law.

49. THERMOCHEMISTRY
• Hess’s Law
–The overall enthalpy change in a
reaction is equal to the sum of the
enthalpy changes for the
individual steps in the process.
• Measured heats of reaction can be
combined to calculate heats of
reaction that are difficult or
impossible to actually measure.

50. THERMOCHEMISTRY
Calculating Heats of Reaction:
• Suppose you want to know the
heat of reaction for the
decomposition of ice to give
oxygen and hydrogen gases.
Given the information below:
H0f (liquid water) = -285.83 kJ
Heat of Fusion (melting of ice) = 6.0 kJ/mol

51. THERMOCHEMISTRY
Given:
H2(g) + 1/2O2(g) --> H2O(l) H0f = -285.83
H2O(s) --> H2O(l) H = 6.0 kJ
Want:
H2O(s) --> H2(g) + 1/2O2(g) H = ?????

52. THERMOCHEMISTRY
General Principles for combining
thermochemical equations:
1. Reverse the direction of the
known equations so that when
added together they give the
desired thermochemical equation.
• When reversing equations, the
sign of H is also reversed.

53. THERMOCHEMISTRY
2. Multiply the coefficients of the
known equations so that when
added together they give the
desired thermochemical equation.
• When multiplying the
coefficients, you must also
multiply the value for H.

54. THERMOCHEMISTRY
• To solve this example, we must
reverse the direction of the
equation for the formation of H2O
as a liquid, then add the
equations and heats together.
H2O(l) --> H2(g) + 1/2O2(g) H0f = 285.83
+ H2O(s) --> H20(l) H = 6.0
H2O(s) --> H2(g) + 1/2O2(g) H = 291.83

55. THERMOCHEMISTRY
Example:
• What is the heat of reaction used
to prepare ultra-pure silicon for
the electronics industry?
Given:
Si(s) + 2Cl2(g) --> SiCl4(l) H = -687.01
Mg(s) + Cl2(g) --> MgCl2(s) H = -641.32

56. THERMOCHEMISTRY
Want:
2Mg(s) + SiCl4(l) --> Si(s) + 2MgCl2(s) H=?
• In this example we must reverse
the first equation and multiply the
second equation by 2, then add
them together.

57. THERMOCHEMISTRY
SiCl4(l) --> Si(s) + 2Cl2(g) H = 687.01
+ 2Mg(s) + 2Cl2(g) --> 2MgCl2(s) H = -1282.64
2Mg(s) + SiCl4(l) --> Si(s) + 2MgCl2(s) H = -595.63

58. THERMOCHEMISTRY
Example:
• Calculate the heat of reaction for
the combustion of of methane
gas, CH4, to form CO2(g) and
H2O(l). Write all equations
involved.

59. THERMOCHEMISTRY
• To find the heat of formation of a
compound, you would use the
same method as finding the heat
of reaction.

60. THERMOCHEMISTRY
Example:
• When carbon is burned in a
limited supply of oxygen, carbon
monoxide and is produced. In this
reaction, carbon is probably first
oxidized to carbon dioxide. Then
part of the carbon dioxide is
reduced with carbon to give some
carbon monoxide.Find the heat of
formation of carbon monoxide.

61. THERMOCHEMISTRY
• We know the heat of formation of
carbon dioxide and the heat of
combustion of carbon monoxide
from the tables in the book.
C(s) + O2(g) --> CO2(g) H0f = -393.5 kJ/mol
CO(g) + 1/2O2(g) --> CO2(g) H0c = -283.0 kJ/mol

62. THERMOCHEMISTRY
• We want to find the heat of
formation of carbon monoxide.
C(s) + 1/2O2(g) --> CO(g) H0f = ?
• From the information given we
should now be able to solve this
problem by rearranging and
combining equations.

63. THERMOCHEMISTRY
Example:
• Calculate the heat of formation of
pentane, C5H12, using the
information on heats of formation
in Appendix Table A-14 and the
information of heats of
combustion in Appendix Table A-
5. Solve by combining known
thermochemical equaitons.

64. THERMOCHEMISTRY
• Given:
C(s) + O2(g) --> CO2(g) H0f = -393.5 kJ/mol
H2(g) + O2(g) --> H2O(l) H0f = -285.8
kJ/mol
C5H12(g) + 8O2(g) --> 5CO2(g) + 6H2O(l)
H0c = -3535.6 kJ/mol
• Want:
5C(s) + 6H2(g) --> C5H12(g) H0f = ?

65. THERMOCHEMISTRY
• Practice Problems, pg. 524, 1-3.

66. THERMOCHEMISTRY
• The change in heat content of a
reaction system is one of two
factors that allow chemists to
predict whether a reaction will
occur spontaneously and to
explain how it occurs.
• The randomness of the particles
in a system is the second factor
affecting spontaneity.

67. THERMOCHEMISTRY
• A great majority of chemical
reactions in nature are exothermic.
• As these reactions proceed, energy
is liberated and the products have
less energy than the original
reactants.
• The products are also more
resistant to change, more
stable, than the original reactants.

68. THERMOCHEMISTRY
• The tendency throughout nature
is for a reaction to proceed in a
direction that leads to a lower
energy state.
• In endothermic reactions, energy
is absorbed and the products are
at a higher potential energy and
are less stable than the original
reactants.

69. THERMOCHEMISTRY
• So one might think that
endothermic reactions do not take
place spontaneously, but they do.

70. THERMOCHEMISTRY
• Consider the melting of ice.
–An ice cube melts spontaneously
as heat is transferred from the
warm air to the ice.
–Well-ordered crystal arrangement
to a less orderly liquid phase.
–A system that can go from one
state to another without an
enthalpy change does so by
becoming more disordered.

71. THERMOCHEMISTRY
• A disordered system is one that
lacks a regular arrangement of its
parts.
• This factor is called Entropy.
• Entropy (S):
–A measure of the degree of
randomness of the particles, such
as molecules.

72. THERMOCHEMISTRY
• The more disordered or the more
random a system, the higher its
entropy.
• Entropy increases when a gas
expands, a solid or liquid
dissolves, or the number of
particles in a system increases.

73. THERMOCHEMISTRY
• Processes in nature are driven in
two directions:
–Towards lowest enthalpy.
–Towards highest entropy.
• When these two oppose each
other, the dominant factor
determines the direction of
change.

74. THERMOCHEMISTRY
• To predict which will dominate for
a given system, a function has
been defined to relate the
enthalpy and entropy factors at a
given temperature.
• Free Energy (G):
–The combined enthalpy-entropy
function of a system.

75. THERMOCHEMISTRY
• Natural processes proceed in the
direction that lowers the free
energy of a system.
• Only the change in free energy
can be measured.
• The change in free energy can be
defined in terms of the changes in
enthalpy and entropy.

76. THERMOCHEMISTRY
• Free-Energy Change ( G):
–The difference between the
change in enthalpy, H, and the
product of the Kelvin temperature
and the entropy change, which is
defined as T S.
–Equation for Free Energy Change:
G 0 = H 0 - T S0

77. THERMOCHEMISTRY
• Each of the variables in the
equation can have positive or
negative values, this leads to four
possible combinations of terms.
• The more negative H is, the
more negative G is likely to be.
• The more positive S is, the more
negative G is likely to be.

78. THERMOCHEMISTRY
• Reactions systems that change
from a high enthalpy state to a
low one tend to proceed
spontaneously.
• Also, systems that change from a
well-ordered state to a highly
disordered state also tend to
proceed spontaneously.

79. THERMOCHEMISTRY
• G can either be positive or
negative, depending on the
temperature (T).
–If the temperature of the system
is the dominant factor that
determines the relative
importance of the tendencies
toward lower energy and higher
entropy.

80. THERMOCHEMISTRY
EXAMPLE:
• Consider the reaction between
water and carbon to produce
carbon monoxide and hydrogen.
H2O(g) + C(s) --> CO(g) + H2(g)
H = +131.3 kJ
S = +0.134 kJ/(mol•K)
T = 298 K

81. THERMOCHEMISTRY
• Calculate free energy change and
tell whether or not this reaction is
spontaneous or not.
G = 131.3 - (298)(0.134)
G = 131.4 - 39.9
G = +91.4 kJ/mol

82. THERMOCHEMISTRY
• Since G has a positive value, the
reaction that produces produces
hydrogen gas is not spontaneous
at a relatively low temperature of
298 K.

83. THERMOCHEMISTRY
• Increases in temperature tend to
favor increases in entropy.

84. THERMOCHEMISTRY
EXAMPLE:
• For the reaction NH4Cl(s) -->
NH3(g) + HCl(g), at 25°C, H =
176 kJ/mol and S = 0.285
kJ/(mol•K). Calculate G and
decide if this reaction will be
spontaneous in the forward
direction.

85. THERMOCHEMISTRY
G = 176 - (298)(0.285)
G = 176 - 84.9
G = 91 kJ/mol
• The positive value of G shows
that this reaction is not
spontaneous at 25°C.

86. THERMOCHEMISTRY
EXAMPLE:
• For the vaporization reaction
Br2(l) --> Br2(g), H = 31.0
kJ/mol and S = 93.0 kJ/(mol•K).
At what temperature will this
process be spontaneous?

87. THERMOCHEMISTRY
• No matter how simple a balanced
equation, the reaction pathway
may be complicated and difficult
to determine.
• Chemists believe that simple, one
step reaction mechanisms are
unlikely, and that nearly all
reactions are complicated.

88. THERMOCHEMISTRY
EXAMPLE:
• A reaction between colorless H2
gas and violet colored I2 gas at
elevated temperatures produces
hydrogen iodide, a colorless gas.
HI molecules, in turn, tend to
decompose and re-form hydrogen
and iodine.

89. THERMOCHEMISTRY
H2(g) + I2(g) --> 2HI(g)
2HI(g) --> H2(g) + I2(g)
• These types of reactions do not
show the reaction pathway by
which either reaction proceeds.

90. THERMOCHEMISTRY
• Reaction Mechanism:
–The step-by-step sequence of
reactions by which the overall
chemical change occurs.
• Most reactions take place in a
sequence of simple steps.

91. THERMOCHEMISTRY
• This reaction is also a
Homogenous reaction.
• Homogenous Reaction:
–A reaction whose reactants and
products exist in a single phase.
–In such a reaction, all reactants
and products in intermediate
steps are in the same phase.

92. THERMOCHEMISTRY
• The real reaction has 2 possible
mechanisms:
First Reaction Mechanism
I2 <--> 2I
2I + H2 <--> 2HI
I2 + H2 <--> 2HI

93. THERMOCHEMISTRY
Second Reaction Mechanism
I2 <--> 2I
I + H2 <--> H2I
H2I + I <--> 2HI
I2 + H2 <--> 2HI
• Some species that appear in
some steps, but not in the net
equation are known as
intermediates.

94. THERMOCHEMISTRY
• Collision Theory:
–A set of assumptions regarding
collisions and reactions.
–In order for reactions to occur
between substances, their
particles (molecules, atoms, or
ions) must collide.

95. THERMOCHEMISTRY
EXAMPLE:
HI + HI --> H2 + 2I
• According to the collision theory,
2 HI molecules must collide in
order to react.

96. THERMOCHEMISTRY
• The HI molecules must collide
while favorably orientated and
with enough energy to disrupt the
bonds of the molecules.

97. THERMOCHEMISTRY
• If the collision
is too
gentle, the old
bonds are not
disrupted and
new ones are
not formed.
They simply
bounce off each
other.

98. THERMOCHEMISTRY
• If the reactant
molecules are
poorly
orientated, the
collision has
little effect.

99. THERMOCHEMISTRY
• If the reactants
collide with
enough energy
and proper
orientation, a
reshuffling of
bonds leads to
the formation of
1 H2 molecule
and 2 I atoms.

100. THERMOCHEMISTRY
• The collision theory provides two
reasons why a collision between
reactants molecules may fail to
produce a new chemical species:
–Collision is not energetic enough.
–Colliding molecules are not
orientated properly.

101. THERMOCHEMISTRY
• Activation Energy (Ea):
–Energy required to transform the
reactants into the activated
complex.
• Activated Complex:
–A transitional structure that
results from an effective collision
and that persists while old bonds
are breaking and new bonds are
forming.

102. THERMOCHEMISTRY
Reading Energy Diagrams:
• Symbols:
– H = Change in Enthalpy (Heat)
H = HPRODUCTS - HREACTANTS
–Ea = Activation Energy of the
forward reaction.
–Ea’ = Activation Energy of the
reverse reaction.

103. THERMOCHEMISTRY

104. THERMOCHEMISTRY
• Reaction Rate:
–The change in concentration of
reactants per unit time as a
reaction proceeds.
• Chemical Kinetics:
–The area of chemistry that is
concerned with reaction rates and
reaction mechanisms.

105. THERMOCHEMISTRY
• There are 5 important factors that
influence the rate of a chemical
reaction:
1. Nature of Reactants:
• Substances vary greatly in their
tendencies to react.
• The rate of reaction depends on the
particular reactants and bonds
involved.

106. THERMOCHEMISTRY
2. Surface Area:
• The more surface area an object
has the faster the reaction will
occur.

107. THERMOCHEMISTRY
3. Concentration:
• Increasing the concentration of a
substance could increase the rate
of reaction or it could have no
effect on the reaction. The effect
depends on the particular
reaction.

108. THERMOCHEMISTRY
• Reactions take place in a series of
steps. The step that proceeds the
slowest determines the overall
rate of reaction.
• The slowest step is called the
Rate-Determining Step.

109. THERMOCHEMISTRY
4. Temperature:
• The rates of many reactions
roughly double or triple with a
10° C rise in temperature.
• The increase in temperature
increases the energy of the
particles in the reaction, thus
increasing the number of
collisions between those particles.

110. THERMOCHEMISTRY
5. Presence of a Catalyst:
• A substance that changes the rate
of a chemical reaction without
itself being permanently
consumed.
• The action of a catalyst is called
catalysis.
• Catalysts do not appear among
the final products of reactions the
accelerate.

111. THERMOCHEMISTRY
• A catalyst may be effective in
forming an alternative activated
complex that requires a lower
activation energy.

112. THERMOCHEMISTRY
• A catalyst that is in the same
phase as all the reactants and
products in a reaction system is
called a homogeneous catalyst.
• When its phase is different from
that of the reactants, it is called a
heterogeneous catalyst.

113. THERMOCHEMISTRY
• Sometimes chemists need to slow
down the rate of reaction, to do
so they add an inhibitor.
• Inhibitor:
–A substance that slows down a
process.
–A negative catalyst.

114. THERMOCHEMISTRY
• The relationship between rate of
reaction and the concentration of
one reactant is determined
experimentally by first keeping
the concentrations of other
reactants and the temperature of
the system constant.

115. THERMOCHEMISTRY
• Then the reaction rate is
measured for various
concentrations of the reactant in
question.
• A series of such experiments
reveals how the concentration for
each reactant affects the reaction
rate.

116. THERMOCHEMISTRY
EXAMPLE:
• Hydrogen gas reacts with nitrogen
monoxide gas at constant volume
and at an elevated constant
temperature, according to the
following equation:
2H2(g) + 2NO(g) --> N2(g) + 2 H2O(g)

117. THERMOCHEMISTRY
• Four moles of reactant gases
produce three moles of product
gases; thus, the pressure of the
system diminishes as the reaction
proceeds.
• The rate of reaction can,
therefore, be determined by
measuring the change of pressure
in the vessel with time.

118. THERMOCHEMISTRY
• Suppose a series of experiments
is conducted using the same
initial concentration of NO but
different initial concentrations of
H2.

119. THERMOCHEMISTRY
• The initial reaction rate is found
to vary directly with H2
concentration:
–Doubling the concentration of H2
doubles the rate.
–Tripling the concentration of H2
triples the rate.

120. THERMOCHEMISTRY
• If R represents the reaction rate
and [H2] is the concentration of
hydrogen in moles per liter, the
mathematical relationship
between rate and concentration
can be written as:
R [H2]
• = “is proportional to”

121. THERMOCHEMISTRY
• Now suppose the same initial
concentration of H2 is used but
the initial concentration of NO is
varied.
• The initial reaction rate is found
to increase fourfold when the NO
concentration is doubled and
ninefold when the concentration
of NO is tripled.

122. THERMOCHEMISTRY
• Thus, the reaction rate varies
directly with the square of the NO
concentration.
R [NO]2

123. THERMOCHEMISTRY
• Because R is proportional to [H2]
and to [NO]2, it is proportional to
their product:
R [H2][NO]2

124. THERMOCHEMISTRY
• By introduction of an appropriate
proportionality constant, k, the
expression becomes an equality:
R = k [H2][NO]2

125. THERMOCHEMISTRY
• An equation that relates reaction
rate and concentration of
reactants is called the rate law.
• It is applicable for a specific
reaction at a given temperature.
• A rise in temperature increases
the reaction rates of most
reactions.

126. THERMOCHEMISTRY
• The value of k usually increases
as the temperature increases, but
the relationship between reaction
rate and concentration almost
always remains unchanged.

127. THERMOCHEMISTRY
• The form of rate law depends on
the reaction mechanism.
• For a reaction that occurs in a
single step, the reaction rate of
that step is proportional to the
product of the reactant
concentrations, each of which is
raised to its stoichiometric
coefficient.

128. THERMOCHEMISTRY
EXAMPLE:
• Suppose one molecule of a gas A
collides with one molecule of gas
B to form two molecules of
substance C.
A + B --> 2C

129. THERMOCHEMISTRY
• One particle of each reactant is
involved in each collision.
Thus, doubling the concentration
of either reactant will double the
collision frequency.
• It also will double the reaction
rate for this step.

130. THERMOCHEMISTRY
• Therefore, the rate for this step is
directly proportional to the
concentration of A and B.
• The rate law for this one-step
reaction follows:
R = k[A][B]

131. THERMOCHEMISTRY
• Now suppose the reaction is
reversible.
• In the reverse step, 2 molecules
of C must decompose to form one
molecule of A and one of B.
2C --> A + B

132. THERMOCHEMISTRY
• Thus, the reaction rate for this
reverse step is directly
proportional to [C] × [C].
• The rate law for this step is:
R = k[C]2

133. THERMOCHEMISTRY
• The power to which the molar
concentration of each reactant is
raised in the rate laws above
corresponds to the coefficient for
the reactant in the balanced
chemical equation.

134. THERMOCHEMISTRY
• Such a relationship holds only if
the reaction follows a simple one-
step path.
• If a chemical reaction proceeds in
a series of steps, the rate law is
determined from the slowest step
because it has the lowest rate.
–Rate Determining Step.

135. THERMOCHEMISTRY
EXAMPLE:
• Consider the following reaction:
NO2(g) + CO(g) --> NO(g) + CO2(g)

136. THERMOCHEMISTRY
• The reaction is believed to be a
two step process represented by
the following mechanism:
Step 1: NO2 + NO2 --> NO3 + NO slow
Step 2: NO3 + CO --> NO2 + CO2 fast

137. THERMOCHEMISTRY
• The first step is the slower of the
two and is therefore the rate-
determining step.
• We can write the rate law from
this rate-determining step:
R = k[NO2]2

138. THERMOCHEMISTRY
• CO does not appear in the rate
law because it reacts after the
rate-determining step, so the
reaction rate will not depend on
[CO].

139. THERMOCHEMISTRY
• The general form for the rate law
is given by the following
equation:
R = k[A]n[B]m….

140. THERMOCHEMISTRY
• The reaction rate is represented
by R, k is the rate constant, and
[A] and [B]… represent the molar
concentrations of reactants.
• The n and m are the respective
powers to which the
concentrations are raised. They
must be determined from
experimental data.