Energy Changes, Reaction Rates, and
Equilibrium

The capacity to do work
◦ The ability to move or change something
 Change in position
 Change in speed
 Change in state (form of matter)


Stored energy = potential energy
Moving energy = kinetic energy
potential
kinetic

The total energy in a system does not
change. Energy cannot be created or
destroyed. It can only be changed from one
form into another or transferred from one
object to another.

Stored energy
◦ Exists in natural attractions and repulsions

Chemical energy
◦
◦
◦
◦
◦

PE possessed by chemicals
Stored in chemical bonds
Breaking bonds requires energy
Forming bonds releases energy
PE may be released and converted to heat
 Heat is a form of kinetic energy due to motion of
particles
A compound with lower potential energy is
more stable than a compound with higher
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 a SI 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

A gummy bear is 9.000 Calories (nutritional
calories). How much energy is stored in a
gummy bear in units of Joules?
9.000 Cal = 9.000 kcal x 1000 cal = 9000. cal
1 kcal
9000. cal x 4.184 J = 37660. J = 37.66 kJ
1 cal


Breaking bonds requires energy
Forming bonds releases energy
To cleave this bond, 58
kcal/mol must be added.
H = +58 kcal/mol
Endothermic
Cl
Cl
To form this bond, 58
kcal/mol is released.
H = −58 kcal/mol
Exothermic
H is the energy absorbed or released in a
reaction; it is called the heat of reaction or
the enthalpy change.


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

The stronger the bond, the higher its bond
dissociation E.

H indicates the relative strength of the
bonds broken and formed in a reaction:
• H negative: Exothermic reaction: more
energy is released in forming bond than is
need to break bonds: products have stronger
bonds.
• H positive: Endothermic reaction: more
energy is needed to break bonds than is
released in forming bonds: products have
weaker bonds.

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.

Ea, the energy of activation, is the difference in
energy between the reactants and the
transition state. It can be thought of as the
energy barrier that must be overcome for the
reaction to occur.


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.


ΔH is negative,
the reaction is exothermic:

If H is positive, the reaction is endothermic

Increasing the concentration of the reactants:
o Increases the number of collisions
o Increases the reaction rate

Increasing the temperature of the reaction:
o Increases the kinetic energy of the molecules
o Increases the reaction rate

A catalyst is a substance that speeds up the
rate of a reaction and can be recovered
unchanged.
o Catalysts lower Activation Energy, Ea




A catalyst is a substance that speeds up the
rate of reaction
A catalyst is recovered unchanged in a
reaction, and does not appear in the product
A catalyst lowers Ea
With a catalyst H is the same


The uncatalyzed reaction (higher Ea) is slower
The catalyzed reaction (lower Ea) is faster




Enzymes (usually protein molecules) are
biological catalysts held together in a very
specific three- dimensional shape
The active site binds a reactant, which then
under- goes 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.



A reversible reaction can occur in either
direction
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
The forward reaction
proceeds to the right.
CO(g) + H2O(g)
CO2(g) + H2(g)
The reverse reaction
proceeds to the left.

The relationship between the concentration of
the products and the concentration of the
reactants is the equilibrium constant, K.
aA + bB
equilibrium
constant = K =
cC + dD
[products]
[reactants] =
[C]c [D]d
[A]a [B]b
*Brackets, [ ], are used to symbolize concentration
in moles per liter (mol/L).
N2(g) + O2(g)
equilibrium
constant
2 NO(g)
= K
=
[NO]2
[N2] [O2]
*The coefficient becomes the exponent.


Magnitude of the equilibrium constant
When K is much greater than 1, equilibrium
lies to the right and favors the products
[products]
The numerator is larger.
[reactants]

When K is much less than 1, equilibrium lies
to the left and favors the reactants
[products]
[reactants]
The denominator is larger.

When K is around 1 (0.01 < K < 100), both
reactants and products are present in
similar amounts
[products]
[reactants]
Both are similar
in magnitude.
HOW TO Calculate the Equilibrium
Constant for a Reaction
Step [1]
A2
+ B2
Step [2]
K =
Write the expression for the equilibrium
constant from the balanced equation.
2 AB
K =
[AB]2
[A2][B2]
Substitute the given concentrations in
the equilibrium expression and calculate K.
[AB]2
[A2][B2]
=
[0.50]2
[0.25][0.25]
=
0.25
0.0625
=
4.0
If a chemical system at equilibrium is
disturbed or stressed, the system will
react in a direction that counteracts
the disturbance or relieves the stress
 Some of the possible disturbances:
◦ Concentration changes
◦ Temperature changes
◦ Pressure changes

2 CO(g) + O2(g)

2 CO2(g)
What happens if [CO(g)] is increased?
◦ The concentration of O2(g) will decrease.
◦ The concentration of CO2(g) will increase.
2 CO(g) + O2(g)

2 CO2(g)
What happens if [CO2(g)] is increased?
◦ The concentration of CO(g) will increase.
◦ The concentration of O2(g) will increase.
•What happens if a product is removed?
•The concentration of ethanol will decrease.
•The concentration of the other product (C2H4)
will increase.
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•When the temperature is increased, the reaction
that absorbs heat is favored.
•An endothermic reaction absorbs heat, so increasing
the temperature favors the forward reaction.
32
•An exothermic reaction releases heat, so increasing
the temperature favors the reverse reaction.
•Conversely, when the temperature is decreased,
the reaction that adds heat is favored.
33
•When pressure increases, equilibrium shifts in
the direction that decreases the number of moles
in order to decrease pressure.
34
•When pressure decreases, equilibrium shifts in the
direction that increases the number of moles in
order to increase pressure.
35