Chapter 6
Energy
Changes,
Reaction
Rates and
Equilibrium
Prepared by
Andrea D. Leonard
University of Louisiana at Lafayette
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1
Introduction to Chemical Reactions
A chemical change—a chemical reaction—converts
one substance into another.
Chemical reactions involve:
•breaking bonds in the reactants (starting materials)
•forming new bonds in the products
CH4 and O2
CO2 and H2O
2
Introduction to Chemical Reactions
A chemical equation is an expression that uses
chemical formulas and other symbols to illustrate
what reactants constitute the starting materials in a
reaction and what products are formed.
•The reactants are written on the left.
•The products are written on the right.
•Coefficients show the number of molecules of
a given element or compound that react or are
formed.
3
Introduction to Chemical Reactions
4
Balancing Chemical Equations
HOW TO Balance a Chemical Equation
Example Write a balanced chemical equation for
the reaction of propane (C3H8) with
oxygen (O2) to form carbon dioxide (CO2)
and water (H2O).
Step [1] Write the equation with the correct formulas.
C3H8 + O2
CO2 + H2O
•The subscripts in a formula can never be changed
to balance an equation, because changing a
subscript changes the identity of a compound.
5
Balancing Chemical Equations
HOW TO Balance a Chemical Equation
Step [2] Balance the equation with coefficients one
element at a time.
•Finally, balance the O’s:
6
Mass Calculations in Chemical Equations
HOW TO Convert Grams of Reactant to Grams of Product
Example
Ethanol (C2H6O, molar mass 46.1 g/mol) is
synthesized by reacting ethylene (C2H4,
molar mass 28.1 g/mol) with water.
How many grams of ethanol are formed
from 14 g of ethylene?
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Mass Calculations in Chemical Equations
HOW TO Convert Grams of Reactant to Grams of Product
Grams of
reactant
molar mass
conversion
factor
mole–mole
conversion
factor
molar mass
conversion
factor
1 mol C2H4
1 mol C2H6O
46.1 g C2H6O
14 g C2H4 x 28.1 g C H x 1 mol C H x 1 mol C H O
2 4
2 4
2 6
Grams C2H4
cancel.
=
Moles C2H4
cancel.
23 g C2H6O
Moles C2H6O
cancel.
Grams of
product
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Percent Yield
•The theoretical yield is the amount of product
expected from a given amount of reactant based
on the coefficients in the balanced chemical
equation.
•Usually, however, the amount of product formed
is less than the maximum amount of product
predicted.
•The actual yield is the amount of product isolated
from a reaction.
9
Percent Yield
Calculating Percent Yield
Sample Problem 5.14
If the reaction of ethylene with water to form
ethanol has a calculated theoretical yield of 23 g
of ethanol, what is the percent yield if only 15 g
of ethanol are actually formed?
Percent yield
actual yield (g)
= theoretical yield (g) x 100%
=
15 g
23 g x
100% =
65%
10
Energy
•Energy is the capacity to do work.
•Potential energy is stored energy; kinetic energy
is the energy of motion.
•The law of conservation of energy states that the
total energy in a system does not change. Energy
cannot be created or destroyed.
•Chemical bonds store potential energy.
•A compound with lower potential energy is more
stable than a compound with higher potential energy.
•Reactions that form products having lower potential
energy than the reactants are favored.
11
Energy
The Units of Energy
•A calorie (cal) is the amount of energy needed to
raise the temperature of 1 g of water by 1 oC.
•A joule (J) is another unit of energy.
1 cal = 4.184 J
•Both joules and calories can be reported in the
larger units kilojoules (kJ) and kilocalories (kcal).
1,000 J = 1 kJ
1,000 cal = 1 kcal
1 kcal = 4.184 kJ
12
Energy
Energy Changes in Reactions
•When molecules come together and react, bonds
are broken in the reactants and new bonds are
formed in the products.
•Bond breaking always requires an input of energy.
•Bond formation always releases energy.
To cleave this bond,
58 kcal/mol must be
added.
Cl
Cl
To form this bond,
58 kcal/mol is
released.
13
Energy Changes in Reactions
H is the energy absorbed or released in a
reaction; it is called the heat of reaction or
the enthalpy change.
•When energy is absorbed, the reaction is said to
be endothermic and H is positive (+).
•When energy is released, the reaction is said to
be exothermic and H is negative (−).
To cleave this bond,
H = +58 kcal/mol.
Cl
Cl
To form this bond,
H = −58 kcal/mol.
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Energy Changes in Reactions
Bond Dissociation Energy
The bond dissociation energy is the H for breaking
a covalent bond by equally dividing the e− between
the two atoms.
•Bond dissociation energies are positive values,
because bond breaking is endothermic (H > 0).
H
H
H
+
H
H = +104 kcal/mol
•Bond formation always has negative values,
because bond formation is exothermic (H < 0).
H
+
H
H
H
H = −104 kcal/mol
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Energy Changes in Reactions
Bond Dissociation Energy
The stronger the bond, the higher its bond
dissociation energy.
In comparing bonds formed from elements in the
same group, bond dissociation energies generally
decrease going down the column.
16
17
Energy Changes in Reactions
Calculations Involving H Values
H indicates the relative strength of the bonds
broken and formed in a reaction.
When H is negative:
•More energy is released in forming bonds than is
needed to break the bonds.
•The bonds formed in the products are stronger
than the bonds broken in the reactants.
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(l)
H = −213 kcal/mol
Heat is released.
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Energy Changes in Reactions
Calculations Involving H Values
When H is positive:
•More energy is needed to break bonds than is
released in the formation of new bonds.
•The bonds broken in the reactants are stronger
than the bonds formed in the products.
6 CO2(g) + 6 H2O(l)
C6H12O6(aq) + 6 O2(g)
ΔH = +678 kcal/mol
Heat is absorbed.
19
Energy Changes in Reactions
Calculations Involving H Values
20
Sample problem 6.5
21
Energy Diagrams
For a reaction to occur, two molecules must collide
with enough kinetic energy to break bonds.
The orientation of the two molecules must be correct
as well.
22
Energy Diagrams
•Ea, the energy of activation, is the difference in
energy between the reactants and the transition
state.
23
Energy Diagrams
•The Ea is the minimum amount of energy that the
reactants must possess for a reaction to occur.
•Ea is called the energy barrier and the height of
the barrier determines the reaction rate.
•When the Ea is high, few molecules have enough
energy to cross the energy barrier, and the reaction
is slow.
•When the Ea is low, many molecules have enough
energy to cross the energy barrier, and the reaction
is fast.
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Energy Diagrams
The difference in energy between the reactants
and the products is the H.
•If H is negative, the reaction is exothermic:
25
Energy Diagrams
•If H is positive, the reaction is endothermic:
26
Reaction Rates
How Concentration and Temperature
Affect Reaction Rate
Increasing the concentration of the reactants:
•increases the number of collisions
•increases the reaction rate
Increasing the temperature:
•increases the kinetic energy of the molecules
•increases the reaction rate
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Reaction Rates
Catalysts
•A catalyst is a substance that speeds up the rate
of a reaction.
•A catalyst is recovered unchanged in a reaction,
and does not appear in the product.
•In the following reaction, Pd acts as a catalyst:
•Catalysts accelerate a reaction by lowering Ea
without affecting H.
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Reaction Rates
Catalysts
•The uncatalyzed reaction (higher Ea) is slower.
•The catalyzed reaction (lower Ea) is faster.
H is the same for both reactions.
29
Reaction Rates
Focus on the Human Body: Biological Catalysts
•Enzymes (usually protein molecules) are
biological catalysts held together in a very
specific three-dimensional shape.
•The active site binds a reactant, which then undergoes a very specific reaction with an enhanced rate.
•The enzyme lactase converts the carbohydrate
lactose into the two sugars glucose and galactose.
•People who lack adequate amounts of lactase suffer
from abdominal cramping and diarrhea because
they cannot digest lactose when it is ingested.
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Figure 6.3
31
Equilibrium
A reversible reaction can occur in either direction,
from reactants to products or from products to
reactants.
The forward reaction
proceeds to the right.
CO(g) + H2O(g)
CO2(g) + H2(g)
The reverse reaction
proceeds to the left.
The system is at equilibrium when the rate of the
forward reaction equals the rate of the reverse reaction.
The net concentrations of reactants and products
do not change at equilibrium.
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Figure 6.5
33
Equilibrium
The Equilibrium Constant
•The relationship between the concentration of the
products and the concentration of the reactants is
the equilibrium constant, K.
•Brackets, [ ], are used to symbolize concentration
in moles per liter (mol/L).
•For the reaction:
aA + bB
cC + dD
equilibrium
[products]
[C]c [D]d
constant = K = [reactants] = [A]a [B]b
34
Figure 6.4
35
Equilibrium
The Equilibrium Constant
•For the following balanced chemical equation:
N2(g) + O2(g)
equilibrium
constant
2 NO(g)
= K
=
[NO]2
[N2] [O2]
•The coefficient becomes the exponent.
36
Equilibrium
The Magnitude of the Equilibrium Constant
•When K is much greater than 1 (K > 1):
[products]
[reactants]
The numerator is
larger.
Equilibrium favors the products and lies to the left.
•When K is much less than 1 (K < 1):
[products]
[reactants]
The denominator is
larger.
Equilibrium favors the reactants and lies to the right.
37
Equilibrium
The Magnitude of the Equilibrium Constant
•When K is around 1 (0.01 < K < 100):
[products]
Both are similar
[reactants]
in magnitude.
Both reactants and products are present.
•For the reaction:
2 H2(g) + O2(g)
2 H2O(g)
K = 2.9 x 1082
The product is favored because K > 1.
The equilibrium lies to the right.
38
Equilibrium
39
Calculating the Equilibrium Constant
HOW TO Calculate the Equilibrium Constant for
a Reaction
Example Calculate K for the reaction between the
general reactants A2 and B2. The
equilibrium concentrations are as follows:
[A2] = 0.25 M
A2
Step [1]
[B2] = 0.25 M
+ B2
[AB] = 0.50 M
2 AB
Write the expression for the equilibrium
constant from the balanced equation.
[AB]2
K =
[A2][B2]
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Calculating the Equilibrium Constant
HOW TO Calculate the Equilibrium Constant for
a Reaction
Step [2] Substitute the given concentrations in
the equilibrium expression and calculate K.
K =
[AB]2
[A2][B2]
=
[0.50]2
[0.25][0.25]
=
0.25
=
0.0625
4.0
•Since the concentration is always reported in
mol/L, these units are omitted during the
calculation.
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Sample Problem 6.9
Question: Calculate K for the following equation:
The equilibrium concentrations are as follows:
Solution
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