Rates of Reaction and
Chemical Equilibrium
Chemistry
Ms. Piela
Rates of Reaction
 A rate is any change that occurs within an interval of
time
 Collision theory states that particles will react to
form products when they collide

This requires a certain amount of kinetic energy
As an example, imagine throwing clay balls at one another
 Without enough force, they will ricochet off from one another.
When they have enough kinetic energy, they will smash and fuse
into something entirely new.
 This is the same in chemical reactions

Activation Energy
 Activation energy (Ea) is the minimum amount
of energy required for a chemical reaction to
occur
Activation Energy Diagram
 Diagram can be used to
determine in reaction is
exothermic (- ∆H) or
endothermic (+∆H)
 Activated complex (or
transition state) is the
arrangement of atoms at
the peak of the Ea barrier

Unstable and only appears
briefly
Activation Energy Diagram
Factors Affecting Rates of Reactions
Four Major Factors
Temperature
Concentration
Particle Size
Catalysts
Factors Affecting Rates of Reaction
 Temperature


Generally, an increase in temperature will increase the rate of
a reaction, and vice versa
Explanation:
Increasing temperature speeds up particles, increasing kinetic
energy
 This allows a reaction to overcome its activation energy

 Example
 Popcorn
Factors Affecting Rates of Reaction
 Concentration
Increasing the number of particles (thereby increasing
concentration) will increase the reaction rate
 Explanation:

 More
particles will increase the collision frequency
 More collisions will mean more chances for a reaction to
occur
 Example
 Wooden splint in pure oxygen
Factors Affecting Rates of Reaction
 Particle Size (aka Surface Area)



The smaller the particle, the larger the surface area for a given
mass of particles
An increase in surface area will increase the rate of reaction
Explanation:

With more exposed area to react, this increases the rate of reaction
 Example
 Fine Dust Explosions (1902 Chicago Fire)
 Lycopodium dust explosion
Factors Affecting Rates of Reaction
 Catalysts



Any substance that
increases the rate of a
reaction without being
consumed
Catalysts allow a reaction
to occur at lower
activation energy
Enzymes are biological
catalysts
Activation Energy Diagram
Factors Affecting Rates of Reaction
Review of Factors Affecting Rates
Temperature
• Increase causes higher kinetic energy
Concentration
• Higher concentration causes more collisions
Particle Size
• More surface area leads to more exposed spaces to react
Catalysts
• Substances that increase the rates of reaction without being spent
Chemical Equilibrium
 Reversible reactions are reactions that
simultaneously occur in both directions
 Chemical equilibrium is the state in which
the forward and reverse reactions take
place at the same rate
 Sample
Reaction:
2 SO2 (g) + O2 (g)  2 SO3 (g)
Chemical Equilibrium
 Differences between rates and concentration
One doesn’t imply the other!
 Although rates of forward and reverse reactions are equal
at chemical equilibrium, the concentrations are not
necessarily the same

 Example

AB
1% 99%
If A reacts to give B and the equilibrium mixture contains 99%
B and only 1% A, the formation of B is said to be favored
Effect of Catalyst on Equilibrium
 Although catalysts speed up a reaction rate, they
will not affect the concentration of reactants and
products at equilibrium

Catalysts can only decrease the time required to reach
equilibrium
Affects speed,
not
concentration!
Meep meep!
Le Châtelier's Principle
 When a stress is applied to a system
at equilibrium, the equilibrium will
shift in order to relieve the stress
Factors Affecting Equilibrium
Three Major Factors
Concentration
Temperature
Pressure
Le Châtelier's Principle
 Concentration
 Changing the amount, or concentration, of any reactant or
product in a system at equilibrium disturbs that equilibrium
Example equilibrium:
2 H2CO3 (g)  2 H2O (l) + 2 CO2 (g)

Adding CO2
Shifts equilibrium to the left
 Concentration of H2O will decrease
 Concentration of H2CO3 will increase

Le Châtelier's Principle
Example equilibrium:
2 H2CO3 (g)  2 H2O (l) + 2 CO2 (g)
 Removing
CO2
Shifts equilibrium to the right
Concentration of H2O will increase
Concentration of H2CO3 will decrease
Le Châtelier's Principle
 Temperature
 Increasing the temperature causes an equilibrium position of a
reaction to shift in the direction that will absorb the heat.

Consider heat as if it were a part of the reaction
2 SO2 (g) + O2 (g)  2 SO3 (g) + heat
 Adding heat
 Shifts equilibrium to the left.
 Concentration of SO2 and O2 will increase.
 Concentration of SO3 will decrease.
Le Châtelier’s Principle
2 SO2 (g) + O2 (g)  2 SO3 (g) + heat
 Removing heat
 Shifts
equilibrium to the right.
 Concentration of SO2 and O2 will decrease.
 Concentration of SO3 will increase.
Le Châtelier's Principle
 Pressure
 Changing
the pressure only affects an
equilibrium with an uneven number of reactants
and products
 Increasing pressure will shift an equilibrium in
the direction of the least number of moles
 Be aware that a change in volume also affects
pressure
Reducing volume will increase the pressure,
and vice versa
Le Châtelier's Principle
Example Equilibrium:
CO + 3 H2 (g)  CH4 (g) + H2O (g)
 Increasing pressure (reducing volume) will
shift the equilibrium to the…
right
 Decreasing pressure (increasing volume)
will shift the equilibrium to the…
left
Haber Bosch Process & Le Châtelier’s Principle
Equilibrium Constant Expressions
 A mathematical relationship exists between
concentration of the reactants and products
once equilibrium has been reached that is
independent of the initial concentration of
the participants
 Given the symbol Keq
 Can be used to determine whether products
or reactants are favored in an equilibrium
Equilibrium Constant Expressions
Steps for Determining Keq
 Make sure the chemical reaction is balanced and at
equilibrium
 Place the product concentrations in the numerator,
and the reactant concentrations in the denominator
[products]
Keq = [reactants]
 Concentrations of any solid or liquid is left out because
the concentrations never change

Water is always omitted in aqueous solution. Water is almost
constant during the reaction
Steps for Determining Keq
 To complete the expression, raise each substance’s
concentration to the power equal to the substance’s
coefficient in the balanced chemical equation
 For a general reaction: aA+ bB  cC + dD
 The equilibrium constant expression can be written
as
Keq =
[C]c[D]d
[A]a[B]b
 The equilibrium constant is unitless
Example #1
1.
An aqueous solution of carbonic acid reacts to
reach equilibrium described below:
H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq)
Write the equilibrium expression, Keq, for this reaction


[ HCO3 ][ H 3O ]
K eq 
[ H 2CO3 ]
Example #2
2. The brown gas, NO2, is in equilibrium with the
colorless gas N2O5. Write the equilibrium
expression for the following reaction:
N2O4 (colorless)  2 NO2 (brown)
Keq 
[ NO2 ]2
[ N 2O4 ]
Practice Problems
 Practice #1
2
[ NH 3 ]
K eq 
3
[H 2 ] [N2 ]
 Practice #2
2
[ SO3 ]
K eq 
2
[ SO2 ] [O2 ]
Calculating using Keq
 Example #1
H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq)
The solution contains the following solute
concentrations: H2CO3; 3.3 x 10-2 M; HCO3- ion, 1.19 x
10-4 M; and H3O+, 1.19 x 10-4 M. Determine the Keq
expression, and the value of Keq
Example #1


[ HCO3 ][ H 3O ]
K eq 
[ H 2CO3 ]
[1.19 x10 4 M ][1.19 x10 4 M ]
6
K eq 
 1.1x10
2
[3.3x10 M ]
Example #2
 Keq for the equilibrium below is 1.8 x 10-5 at a
temperature of 25 °C. Calculate [NH4+] when
[NH3] = 6.82 x 10-3 M and [OH-] = 3.45 x 10-3
M.
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
Example #2


[ NH 4 ][OH ]
K eq 
[ NH 3 ]

3
[ NH 4 ][3.45 x10 M ]
1.8 x10 
3
[6.82 x10 M ]
5
5
 3.56 x10 M
Practice Problems
 Problem #1
2
[ NO]
1.65 x10 
[1.8 x103 M ][ 4.2 x104 M ]
3
1.25 x10
9
 [ NO]
5
3.54 x10 M  [ NO]
 Problem #2
2
2
2
[1.98 x10 M ]
K eq 
 32752
3
2
3
[1.5x10 M ] [5.32 x10 M ]
Magnitude of Keq
 The meaning of Keq

If Keq is very large, products will be favored
 This
implies the numerator is large in the Keq
expression compared to denominator

If Keq is very small, reactants will be favored
 This
implies the denominator is large in the Keq
expression compared to numerator

If Keq is close to 1, then roughly equal amounts of
reactants and products are present at equilibrium