2. Collision Theory
The speed of a chemical reactions can vary from
instantaneous (strike a match) to extremely slow (coal)
Speed is measured as a change in distance in a given
interval of time. Rate = distance/time
Rate is a measure of the speed of any change that
occurs within an interval of time.
In chemistry, the rate of chemical change (the reaction
rate) is usually expressed as the amount of reactant
changing per unit time.
3. Collision Theory
According to the collision theory, atoms, ions, and
molecules can react to form products when they collide
with one another, provided that the colliding particles
have enough kinetic energy.
Particles lacking the necessary kinetic energy to react,
bounce apart unchanged when they collide.
To illustrate the collision theory, If soft balls of clay are
thrown together with great force, they will stick tightly
together. (analogous to colliding particles of high energy
that react)
Balls of clay thrown together gently, don’t stick to one
another. (analogous to colliding particles of low energy
that fail to react)
4. Collision Theory
If you roll clay into a rope and begin to shake one end
more and more vigorously, eventually it will break.
If enough energy is applied to a molecule, the bonds
holding it together can break.
The minimum energy that colliding particles must have in
order to react is called the activation energy.
When two reactant particles with the necessary activation
energy collide, a new entity called the activated complex
may form.
An activated complex is an unstable arrangement of
atoms that forms momentarily at the peak of the
activation energy barrier.
6. Collision Theory
The lifetime of an activated complex is typically about
10-13 s. The reactants either re-form or the products
form.
Both cases are equally likely, thus the activated complex
is sometimes called the transition state.
High activation energies explain the slow reaction of
some natural substances at room temperature.
The collisions are not great enough to break the bonds,
thus the reaction rate is essentially zero or very slow.
7. Factors Affecting Reaction Rates
Every chemical reaction proceeds at its own rate. Some
fast, some slow under the same conditions.
By varying the conditions of a reaction, you can modify
the rate of almost any reaction.
The rate of a chemical reaction depends upon:
• temperature
• concentration
• particle size
• the use of a catalyst.
8. Temperature
Usually, raising the temperature speeds up reactions,
while lowering the temperature slows down reactions.
At higher temperatures, the motions of the reactant
particles are faster and more chaotic than they are at
lower temperatures.
Increasing the temperature increases both the frequency
of collisions and the number of particles that have
enough KE to slip over the activation energy barrier to
become products.
An increase in temperature causes products to form
faster.
9. Concentration
The number of particles in a given volume affects the rate
at which reactions occur.
Cramming more particles into a fixed volume increases
the frequency of collisions.
Increased collision frequency leads to a higher reaction
rate.
10. Particle Size
Surface area plays an important role in determining the
rate of reaction.
The smaller the particle size, the larger the surface area
for a given mass of particles.
An increase in surface area increases the amount of the
reactant exposed for reaction, which increases the
collision frequency and the reaction rate.
One way to increase the surface area of solid reactants is
to dissolve them. In solution, particles are separated
and more accessible to other reactants.
You can also increase the surface area by grinding it into
a fine powder.
11. Catalysts
Increasing the temperature is not always the best way to
speed up a reaction. A catalyst is often better.
A catalyst is a substance that increases the rate of a
reaction without being used up during the reaction.
Catalysts permit reactions to proceed along a lower
energy path.
The activation energy barrier for a catalyzed reaction is
lower than that of a uncatalyzed reaction.
With a lower activation energy barrier, more reactants
have the energy to form products within a given time.
Because a catalyst is not consumed during a reaction, it
does not appear as a reactant or product in the
chemical equation.
12. Catalysts
Enzymes are biological catalysts that increase the rates
of biological reactions.
For example, without catalysts, digesting protein would
take years.
An inhibitor is a substance that interferes with the action
of a catalyst.
The inhibitor reduces the amount of functional catalyst
available.
Reactions slow or even stop when a catalyst is poisoned
by an inhibitor.
14. Reversible Reactions
A reversible reaction is one in which the conversion of
reactants to products and the conversion of products to
reactants occur simultaneously.
2SO2 (g) + O2 (g) 2SO3 (g)
The double arrow tells you that this reaction is reversible.
When the rates of the forward and reverse reactions are
equal, the reaction has reached a state of balance
called chemical equilibrium.
At chemical equilibrium, no net change occurs in the
actual amounts of the components of the system.
The amount of SO3 in the equilibrium mixture is the
maximum amount that can be produced by this reaction
under the conditions of the reaction.
15. Reversible Reactions
Chemical equilibrium is a dynamic state.
Both the forward and reverse reactions continue, but
because their rates are equal, no net change occurs in
their concentrations.
Even though the rates are equal at equilibrium, the
concentrations of the components on both side of the
equation are not necessarily the same.
The relative concentrations of the reactants and products
at equilibrium constitute the equilibrium position of a
reaction.
The equilibrium position indicates whether the reactants
or products are favored.
16. Reversible Reactions
A B Product is favored. Equilibrium mixture
contains more product than reactant.
A B Reactant is favored. Equilibrium mixture
contains more reactant than product.
In principle, almost all reactions are reversible to some
extent under the right conditions.
In practice, one set of components is often so favored
that the other set cannot be detected.
If one set of components (reactants) is completely
converted to new substance (products), you can say
that the reaction has gone to completion, or is
irreversible.
17. Reversible Reactions
You can mix chemicals expecting to get a reaction but no
products can be detected, you can say that there is no
reaction.
Reversible reactions occupy a middle ground between
the theoretical extremes of irreversibility and no
reaction.
A catalyst speeds up both the forward and the reverse
reactions equally.
The catalyst lowers the activation energy of the reaction
by the same amount in both the forward and reverse
directions.
Catalysts do not affect the amounts of reactants and
products present, just the time it takes to get to
equilibrium.
18. Factors Affecting Equilibrium
Changes of almost any kind can disrupt the balance of
equilibrium
When the equilibrium of a system is disturbed, the
system makes adjustment to restore equilibrium.
The equilibrium position of the restored equilibrium is
different from the original equilibrium position.
The amount of products and reactants may have
increased or decreased. This is called a shift in the
equilibrium system.
19. Factors Affecting Equilibrium
LeChatelier’s principle states that if a stress is applied
to a system in equilibrium, the system changes in a way
that reflects the stress.
Stresses that upset the equilibrium include:
• Changes to the concentration of reactants or products
• Changes to temperature
• Changes in pressure
21. Change in Concentration
Adding a product to a reaction at equilibrium pushes a
reversible reaction in the direction of reactants.
Removing a product always pushes a reversible reaction
in the direction of products.
Farmers use this to increase yield of eggs
If product are continually removed, the reaction will shift
equilibrium to produce more product until the reactants
are all used up. (will never reach equilibrium)
22. Changes in Temperature
Increasing the temperature causes the equilibrium
position of a reaction to shift in the direction that
absorbs heat.
The heat absorption reduces the applied temperature
stress.
add heat
direction of shift
2SO2 (g) + O2 (g) 2SO3 (g) + heat
remove heat
direction of shift
Heat can be considered a product, just like SO3.
Cooling, pulls equilibrium to right, and product yield
increases. Heating pushed equilibrium to left and
product yield decreases.
23. Changes in Pressure
A change in pressure affects only gaseous equilibria that
have an unequal number of moles of reactants and
products.
add Pressure
direction of shift
N2 (g) + 3H2 (g) 2NH3 (g)
reduce pressure
direction of shift
When pressure is increased for gases at equilibrium, the
pressure momentarily increases because the same
number of molecules is contained in a smaller volume.
System immediately relieves some of the pressure by
reducing the number of gas molecules.
24. Change in Concentration
N2 (g) + 3H2 (g) → 2NH3 (g)
Removing a product always pushes a reversible reaction
in the direction of products.
Farmers use this to increase yield of eggs
If product are continually removed, the reaction will shift
equilibrium to produce more product until the reactants
are all used up. (will never reach equilibrium)
25. Le Châtelier’s Principle
If an external stress is applied to a system at
equilibrium, the system adjusts in such a way that the
stress is partially offset as the system reaches a new
equilibrium position.
• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
Equilibriu
Add
m shifts
NH3
left to
offset
stress
14.5
26. Le Châtelier’s Principle
• Changes in Concentration continued
Remove
Add Remove
Add
aA + bB cC + dD
Change Shifts the Equilibrium
ncrease concentration of product(s) left
Decrease concentration of product(s) right
ncrease concentration of reactant(s) right
Decrease concentration of reactant(s) left
14.5
27. Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g) C (g)
Change Shifts the Equilibrium
Increase pressureSide with fewest moles of gas
Decrease pressureSide with most moles of gas
Increase volume Side with most moles of gas
Decrease volume Side with fewest moles of gas
14.5
29. The Concept of Equilibrium
Equilibrium is a state in which there are no
observable changes as time goes by.
Chemical equilibrium is achieved when:
• the rates of the forward and reverse reactions
are equal
• the concentrations of the reactants and products
remain constant
30. The Concept of Equilibrium
As the reaction progresses
• [A] decreases to a constant,
• [B] increases from zero to a constant.
• When [A] and [B] are constant, equilibrium is
achieved.
A B
31. The Equilibrium Constant
• No matter the starting composition of reactants and
products, the same ratio of concentrations is
achieved at equilibrium.
• For a general reaction
aA + bB(g) pP + qQ
the equilibrium constant expression is
K eq =
[ P ] [ Q]
p q
[ A ] [ B]
a b
where Keq is the equilibrium constant. The square
brackets indicate the concentrations of the species.
32. The Equilibrium Constant Expression
For the general reaction:
aA + bB → gG + hH
Each concentration
The equilibrium expression is: is simply raised to
the power of its
coefficient
[G]g[H]h
Kc = Products in
numerator.
[A]a[B]b
Reactants in
denominator.
33. N2O4 (g) 2NO2 (g)
[NO2]2
K= = 4.63 x 10-3
[N2O4]
aA + bB cC + dD
[C]c[D]d Law of Mass Action
K= a b
[A] [B]
Equilibrium Will
K >> 1 Lie to the right Favor products
K << 1 Lie to the left Favor reactants
14.1
34. Le Châtelier’s Principle
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
uncatalyzed catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift
equilibrium.
35. Le Châtelier’s Principle
Change Equilibrium
Change Shift Equilibrium Constant
Concentration yes no
Pressure yes no
Volume yes no
Temperature yes yes
Catalyst no no
14.5
36. Write the equilibrium expression for Keq for the
following reactions:
Write the equilibrium-constant expression, K c for
37. Calculation of the Equilibrium
Constant
At 454 K, the following reaction takes place:
3 Al2Cl6(g) = 2 Al3Cl9(g)
At this temperature, the equilibrium concentration of
Al2Cl6(g) is 1.00 M and the equilibrium
concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute
the equilibrium constant at 454 K.
39. Ksp Solubility Product Constant
• Ksp is the equilibrium constant between an ionic
solute and its ions in a saturated solution.
• A very small Ksp indicates that only a small
amount of solid will dissolve in water.
• Ksp is equal to the product of the concentration of
the ions in the equilibrium, each raised to the
power of its coefficient in the equation.
• The smaller Ksp the lower the solubility of the
compound.
40. Solubility Products
• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass
of solute dissolved in 1 L (g/L) or 100 mL
(g/mL) of solution, or in mol/L (M).
43. What is the concentration of lead ions and chromate ions
in a saturated lead chromate solution at 25ºC
(Ksp = 1.8 x 10-14)
PbCrO4 (s) ⇔ Pb2+ (aq) + CrO42- (aq)
Ksp = [Pb2+] x [CrO42-] = 1.8 x 10-14
At equilibrium [Pb2+] = [CrO42-]
Ksp = [Pb2+] x [Pb2+] = 1.8 x 10-14
[Pb2+]2 = 1.8 x 10-14
[Pb2+] = [CrO42-] = 1.8 x 10-14
[Pb2+] = [CrO42-] = 1.3 x 10-7 M
44. Ksp and Solubility
1. NaCl has a solubility of 35.7 g/100 mL. What
is the molar solubility and Ksp? (Ans: 6.10 M,
37.2)
2. CaCl2 has a solubility of 74.5 g/100 mL. What
is the molar solubility and Ksp? (Ans: 6.71 M,
1.21 X 103)
45. Ksp and Solubility
1. A saturated soln of AgCl is found to have an
eq. concentration of Ag+ 1.35 X 10-5 M.
Calculate Ksp. (Ans: 1.82 X 10-10.)
2. A saturated soln of MgF2 is prepared. At
eq. the concentration of Mg2+ is measured
to be 0.0012 M. Calculate Ksp.
(Ans: 7.0 X 10-9)
46. Explaining the Common Ion Effect
The presence of a common ion in a solution will
lower the solubility of a salt.
• LeChatelier’s Principle:
The addition of the common ion will shift the
solubility equilibrium backwards. This means
that there is more solid salt in the solution and
therefore the solubility is lower
CaF2(s) ⇔ Ca2+(aq) + 2F-(aq)
a) Add Ca2+ (shifts to reactants)
b) Add F- (shifts to reactants)
47. The Ksp of silver bromide is 5.0 x 10-13. What is the
bromide ion concentration of a 1.00L saturated solution
of AgBr to which 0.020 mol of AgNO2 is added?
AgBr (s) ⇔ Ag+ (aq) + Br- (aq)
Ksp = [Ag+] x [Br-] = 5.0 x 10-13
[Ag+] x [X] = 5.0 x 10-13
X = 5.0 x 10-13 / [Ag+]
X = 5.0 x 10-13 / 0.020
X = 2.5 x 10-11
48. What is the concentration of sulfide ion in a 1.0 L solution
of iron (II) sulfide to which 0.04 mol of iron (II) nitrate is
added. The Ksp of FeS is 8 x 10-19
FeS (s) ⇔ Fe2+ (aq) + S2- (aq)
Ksp = [Fe2+] x [Sr2-] = 8 x 10-19
[Fe2+] x [X] = 8 x 10-19
X = 8 x 10-19 / [Fe2+]
X = 8 x 10-19 / 0.04
X = 2.0 x 10-17
49. Will a ppt form?
Q = Reaction Quotient = product of a
concentration of the ions
Q < Ksp Shifts to prod (no ppt)
Q = Ksp Eq. (ppt)
Q > Ksp Shifts to reac.(ppt)
49
50. Will a ppt form?
Will a ppt form if a solution is made from 0.50L of
0.002M Ba(NO3)2 and mix it with 0.50L of
0.008M Na2SO4? Ksp of BaSO4 is 1.1 x 10-10
Q = [Ba(NO3)2] x [Na2SO4] = (0.002/2)(0.008/2)
Q = 4 x 10-6
A precipitate will form because 4 x 10-6 is larger
than 1.1 x 10-10
50
