Le Chatelier`s Principle

Report
Mark S. Cracolice
Edward I. Peters
www.cengage.com/chemistry/cracolice
Chapter 18
Chemical Equilibrium
Mark S. Cracolice • The University of Montana
Collision Theory of Reactions
Collision Theory of Gas-phase Reactions
A chemical reaction can occur only when two molecules collide
with a kinetic energy at least equal to certain energy Ea ,
called activation energy of the reaction.
The success of a collision also depends the relative orientation of
molecules. This direction-dependence is called the steric
requirement of the reaction.
Collision Theory of Reactions
Chemical reaction is the overall effect of collisions
between reacting molecules
a) Sufficient energy
proper orientation
b) Proper orientation
not sufficient energy
c) Sufficient energy
poor orientation
Collision Theory of Reactions
A conversion of kinetic energy to potential energy
occurs during formation of an intermediate complex that
can either go on to form products or fall apart into the
unchanged reactants.
This can be shown by a graph that traces the energy of
the system before, during, and after the collision.
Energy Changes During a Reaction
Energy Changes During a Reaction
Transition State Complex
In the transition state complex, the original bonds have
weakened, whereas the new bonds are only partially
formed .
Activation Energy Ea:
The difference between the
energy of the transition state
complex and the reactant
energy.
Rate of a Chemical Reaction
Three important factors influence
the speed of chemical reactions:
Temperature
The higher the temperature, the faster the rate of reaction.
Catalysis
A catalyst increases the rate of reaction.
Concentration of Reactants
The greater the concentration, the greater the rate of reaction.
Effect of Temperature on the Distribution of Energy
Effect of Temperature on Reaction Rates
Kinetic energy distribution curves at two temperatures explain
the effect of temperature on reaction rates.
Ea, the activation energy,
is the same at both temperatures.
Only the fraction of the particles
in the sample represented by the
area beneath the curve to the
right of Ea is able to react.
The fraction of molecules that is
able to react increases rapidly as the temperature is raised.
Effect of Catalyst on the Rate of a Chemical
Reaction: Change in Activation Energy
A catalyst speeds up a reaction by providing a new pathway that
has a lower activation energy.
Effect of Change in Activation Energy on the
Rate of a Chemical Reaction
Effect of Change in Activation Energy on the
Rate of a Chemical Reaction
The fraction of molecules that collide with kinetic
energy that is at least equal to the activation energy,
Ea , is bigger in a catalyzed reaction because the
activation energy barrier is lowered.
Since Ea‘ < Ea
The catalyzed
reaction rate
is faster.
Effect of Concentration on Reaction Rate
Reaction rate depends on the frequency of effective collisions:
The more particles there are in a given volume,
the more frequently collisions will occur and
the more rapidly the reaction will take place.
Development of Equilibrium
For a reversible reaction in a closed system,
the equilibrium is established when
the forward reaction rate is
equal to the reverse reaction rate.
Development of Equilibrium
If the system is not in equilibrium, the concentration
of the species in the faster reaction will decrease,
and thus the reaction will become slower; the
concentration of the species in the slower reaction
will increase, and thus the reaction will become
faster.
Opposite rates
will eventually
become equal,
and an equilibrium
will be established.
The Equilibrium Constant
Consider the reaction H2(g) + I2(g)
2 HI(g)
At equilibrium the following ratio is a constant
[HI]2
K
[H2 ] [I2 ]
K is called equilibrium constant
The Equilibrium Constant
For the general equilibrium
aA + bB
cC + dD
K=
[C] c [D]d
[A]a [B]b
When writing an equilibrium constant expression,
use only the concentrations of gases, (g),
or dissolved substances, (aq).
Do not include solids, (s), or liquids, (l).
The Equilibrium Constant
Equilibrium Constant, K
For any equilibrium at a given temperature, the ratio of
the product of the concentrations of the species on
the right side of the equilibrium equation, each
raised to a power equal to its coefficient in the
equation, to the corresponding product of the
concentrations on the left side of the equation, each
raised to a power equal to its coefficient in the
equation, is a constant.
The equilibrium constant is both
equation-dependent and temperature-dependent.
Significance of the Value of K
Example:
Is the forward reaction favored, the reverse reaction favored, or
are appreciable quantities of all species present at equilibrium
in the following reaction?
HC2H3O2(aq)
H+(aq) + C2H3O2–(aq)
K = 1.8 × 10–5.
Solution:
Since K is very small, the reverse reaction is favored.
Significance of the Value of K
Consider the general reaction: Reactants
Products
[P roduct s]
K=
[React ant ]s
If the equilibrium constant is very large (K > 100),
[Products] > [Reactants], so the forward reaction is favored.
If the equilibrium constant is very small (K < 0.01),
[Products] < [Reactants], so the reverse reaction is favored.
If the equilibrium constant is neither larger nor small,
[Products] ≈ [Reactants], so appreciable quantities of all species
are present at equilibrium.
Le Chatelier’s Principle
Le Chatelier’s Principle
If a system is in equilibrium, any change imposed on
the system tends to shift the equilibrium in a direction
that tends to counteract the initial change.
Le Chatelier’s principle only suggests an outcome; it
does not provide an explanation.
Le Chatelier’s Principle
The Pressure (Volume) Effect
A gas-phase equilibrium responds to compression-a
reduction in volume of the reaction vessel.
If a gaseous equilibrium is compressed, the equilibrium
will be shifted in the direction of formation of fewer
molecules, thus minimizes the increase in pressure.
If the system is expanded, the shift will be in the
direction of formation of more molecules.
3 H2 (g)
+
N2 (g)
More molecules
↔
2 NH3 (g)
less molecules
Le Chatelier’s Principle
The Pressure (Volume) Effect
3 H2 (g)
+
N2 (g)
↔
2 NH3 (g)
If the pressure increases, the equilibrium will be
shifted to the right (less molecules).
If the pressure decreases, the shift will be to the left.
To increase the yield of ammonia, industrial process
uses pressures of 250 atm or higher.
Le Chatelier’s Principle
The Temperature Effect
If a reaction is exothermic, the reverse reaction is
endothermic.
If the temperature increases, the equilibrium will be
shifted to the direction of consuming heat
(endothermic, to the left for NH3 reaction lelow).
If the temperature decreases, the shift will be in
direction of producing heat (exothermic, to the right).
3 H2 (g)
+
N2 (g)
↔
2 NH3 (g) +
92kJ
Le Chatelier’s Principle
The Temperature Effect
+ heat
colorless
↔
2 NO2 (g)
brown
The left tube at 25 0C contains very little brown gas
compared to the tube on the right at 80 0C
Le Chatelier’s Principle
The Concentration Effect
Let us consider the reaction equilibrium:
3H2 (g)
+
N2 (g)
↔
2 NH3 (g)
If H2 is added to the reaction chamber, the shift will
be in the forward direction to counteract the increase
in the number of hydrogen molecules thus producing
more NH3.
If H2 is removed, the equilibrium will shift to the
reverse direction to increase the H2 concentration.
Le Chatelier’s Principle
The Concentration Effect
Let us consider the reaction equilibrium:
3H2 (g)
+
N2 (g)
↔
2 NH3 (g)
[NH3 ]2
K
[N2 ] [H2 ]3
If H2 is added (increase of [H2]), the shift will be in the
forward direction to increase in the concentration
[NH3] and decrease the concentrations [H2] and [N2].
Adding an inert gas has no effect on the equilibrium,
although the total pressure increases.
Solubility Equilibria
The equation for dissolving AgCl , a low-solubility
compound, is
AgCl (s)
↔
Ag+ ( aq) +
Cl- (aq)
This equilibrium is characterized by the solubility
product constant Ksp
Ksp = [Ag+ ] [Cl-]
Solubility Equilibria
Calculation of solubility product from the solubility:
The chloride ion concentration of a saturated solution
of silver chloride is 1.3 x 10-5 M . Calculate the
solubility product for silver chloride .
Ksp = [Ag+ ] [Cl-]
In saturated solution of pure silver chloride the
concentration of [Ag+ ] and [Cl-] are equal. Therefore
Ksp = [Ag+ ] [Cl-] =(1.3 x 10-5) x(1.3 x 10-5)
= 1.7 x 10-10
Solubility Equilibria
Solubility and Solubility Product.
For compounds of similar structure, the smaller the
solubility product, the smaller the solubility. For
example the solubility of silver bromide ( Ksp = 5.2 x
10-13 ) is lower than the solubility of silver chloride.
Common Ion Effect
Suppose that a soluble chloride, such as NaCl were
to be added to the saturated solution of silver
chloride. According to the Le Chatelier’s principle the
equilibrium would shift the equilibrium in the reverse
direction, reducing the solubility of silver chloride.
.
Ionization Equilibria: Weak acid
Weak acids ionize only slightly when dissolved in water.
For a general weak acid HA,
HA(aq)
H+(aq) + A–(aq)
+
Ka
–
[H ] [A ]
=
[HA]
Major species: HA(aq)
Minor Species: H+(aq) + A–(aq)
Ionization Equilibria: Weak acid
The ionization of a weak acid is usually so small that it is
negligible compared with the initial concentration of the acid.
We assume that all ionization concentrations are negligible
when subtracted from the initial concentration.
In other words, the initial concentration of the weak acid is also
the final concentration after the acid ionizes.
Ionization Equilibria: Weak acid
Example:
Find the pH of 0.1 M nitrous acid. Ka = 4.5 × 10–4.
Solution:
HNO2(aq)
+
Ka
H+(aq) + NO2–(aq)
[NO 2 ]
[H ]
=
[HNO 2 ]
= 4.5 ´ 10 -4
Let x = [H+] = [NO2–]; [HNO2] = 0.1 M
+
Ka =
-
[H ] [NO2 ]
(x) (x)
= 4.5 10-4 
[HNO2 ]
0.1
Ionization Equilibria: Weak acid
Find the pH of 0.1 M nitrous acid. Ka = 4.5 × 10–4.
Solution:
+
[H ] [NO 2 ]
(x) (x)
Ka =
= 4.5 ´ 10 -4 =
[HNO 2 ]
0.1
x2 = (0.1) (4.5 × 10–4)
x =
(0.1) (4.5 ´ 10 -5 )
x = [H+] = 7 × 10–3
pH = – log [H+] = – log (7 × 10–3) = 2.2
Ionization Equilibria: Buffer Solution
Buffer Solution
A solution that resists changes in pH because
it contains relatively high concentrations of both
a weak acid and a weak base.
The acid reacts with any added OH–;
The base reacts with any added H+.
Ionization Equilibria: Buffer Solution
Determine the pH of a solution that is 0.25 M in HAc and 0.35 M
in NaAc. Ka = 1.8 × 10–5.
NaAc(aq) 
Na+(aq) + Ac–(aq)
HAc(aq)
H+(aq) + Ac–(aq)
[H+ ] [Ac— ]
Ka =
[HAc]
[HAc]
0.25M
—5
—5
[H ] = K a 
=
1.8

10

=
1.3

10
[Ac- ]
0.35M
+
pH = – log [H+] = – log (1.3 × 10–5) = 4.89
Ionization Equilibria: Buffer Solution
Example:
Determine the acid-to-base concentration ratio that will yield a
buffer solution with a pH of 4.50 if the acid has
Ka = 1.0 × 10–5.
Solution:
HA(aq)
H+(aq) + A–(aq)
[H+] = antilog (–pH) = antilog (–4.50) = 3.2 × 10–5 M
[H+ ] [A— ]
Ka =
[HA]
Ka
[A– ]
=
+
[H ] [HA]
[HA] [H+ ] 3.2 10—5
=
=
= 3.2
—
—5
[A ] K a
1.0 10
Homework
• Homework: 25, 27, 29, 31, 33, 35, 37, 41, 43, 47, 61, 63, 71,
75

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