Reaction Rates and
Equilibrium
Chapter 17
Collision Theory or Model
Molecules react by colliding with each
other with enough energy and proper
orientation to break bonds, rearrange
and form new bonds
 Explains why increasing concentration of
reactants (solutions) increases rate of
reaction (more particles collide)
 Explains why increasing temperature
increases reaction rate (increases the
energy (speed)) of the particles

Collision Theory
Collision Theory or Model
Why does increasing concentration or
temperature increase the rate of
reaction?
 More particles have collisions with
enough energy and proper orientation to
reach activation energy

Collision Theory or Model
Increasing pressure of gases increases
rate of reaction. Why?
 It increases the number of collisions that
occur by reducing the volume.

Collision Theory or Model
Increasing surface area increases rate of
heterogeneous reaction. Why?
 More places for reactions to occur
(locations for collisions)

Catalyst and Reaction Rate
Catalyst is not used up in the reaction
 Lowers the activation energy of the
reaction
 Provides a “shortcut” for reaction
 Enzymes are examples of catalysts

Activation Energy
Homogeneous Reactions
Reactions that involve only one phase.
 Ex. N2(g) + 3H2(g)  2NH3 (g)

Heterogeneous Reactions
Reactions involving more than one
phase
 Ex. Mg(s) + 2HCl(aq)  MgCl2 (aq) + H2 (g)

Equilibrium
When forward and backward reactions
are occurring at the same rate, the
system is said to be at equilibrium.
 Formation of reactants and products
occur at the same rate, but the quantities
are not necessarily the same.

Assignment
Read Chapter 17.1
 Answer Section17.1 Review Questions
1-7

Answers – page 604
1. Molecules must collide for reactions to
occur.
 2. A catalyst provides a new pathway for
the reaction at a lower activation energy.
 3. At lower temperatures, molecules
have less kinetic energy. Therefore,
fewer molecules will have the necessary
activation energy for the reaction making
it slower.

Answers con’t.
4. Grinding increases surface area which
increases locations for collisions to
occur. More collisions increases the rate.
 5. 2NaCl(s) + H2SO4(aq)  Na2SO4 (aq) + 2HCl (aq)


2H2(g) + O2(g)  2H2O(g)
6. The rates of the forward and reverse
reactions are equal at equlibrium.
 7. Changes are occurring at the
molecular level.

LeChatelier’s Principle
When a stress is applied to a system, it
will react in such a fashion as to counter
or offset the stress.
 Most reactions are reversible so stress
may favor formation of either reactant(s)
or product(s). It depends on the type of
“stress”.

LeChatelier’s Principle
What kind of stressors may be applied to
a system?
 Addition or removal of a reactant
 Addition or removal of a product
 Increase or decrease temperature
 Increase or decrease pressure
 Pure solids and liquids do not affect
equilibrium. Why?

Pressure
Increasing pressure tends to favor the
side of the reaction with the fewest
number of gas molecules.
 Decreasing pressure tends to favor the
side of the reaction with the largest
number of gas molecules.

LeChatelier’s Principle
If more of a reactant is added, which way
do you think the equilibrium will shifttowards the reactants or towards the
product?
 Why?
 If the reaction is exothermic and the
temperature is increased, which way will
the equilibrium shift? Why?

Equilibrium Constant





Equilibrium is expressed in terms of concentrations of
the products and reactants
From the balanced chemical equation, the coefficients
become exponents (superscripts)
If the actual concentrations (moles per liter) of the
reactants and products is known, K can be calculated
Remember, pure solids and liquids ( such as water)
do not affect equilibrium
K = [Prod]x[ucts]y
[Reac]x[tants]y
Ex. 2O3(g)  3O2(g) K = [O2]3
[O3]2
Assignment
Read 17.2
 Complete Practice Problem 17.3 (yellow
box) a-d, bottom of page 610
 Complete Section 17.2 Section Review
(page611) 1-5

Answers p.610
A. K = [O2]3
 B. K = [NO2] [H2O]2
 C. K = __1__
[ CO2]
 D. K = __1__
[SO3]

Answers p. 611
1. K= __[NO2]2__
[NO]2 [O2]
 2. K =
[0.141]2
[0.000104] [0.000201]3
K = 2.35 X 1013
 3. Equilibrium position is a set of
equilibrium concentrations. The
equilibrium constant is a specific ratio of
these concentrations. Ex. .5 = 1/2, 2/4

Answers con’t. p. 611
4. Their concentrations do not change.
 5. a) K = ___1___
[NH3] [HCl]
b) K = [H2O]2
[H2]2 [O2]
c) K = __1__
[H2]2 [O2]
