Transcript Equilibrium Conditions

Equilibrium
L. Scheffler
Lincoln High School
2010.
1
Equilibrium Systems



Many chemical reactions are reversible.
– Such reactions do not go to completion.
A state of balance exists between the
products and the reactants
When the concentration of neither the
reactants nor the products is changing, the
system is in equilibrium.
– Note: this does NOT mean the reactions
have stopped. Just the rates are equal.
2
Chemical Equilibrium
Chemical equilibrium occurs in chemical
reactions that are reversible.
In a reaction such as:
CH4(g) + H2O(g)  CO(g) + 3H2 (g)
The reaction can proceed in both directions
CO(g) + 3H2 (g)  CH4(g) + H2O(g)
3
Equilibrium Conditions

At equilibrium, the rate of reaction in the
forward direction and the rate in the
reverse direction are equal.
4
An Equilibrium System
CH4(g) + H2O(g)


CO(g) + 3H2 (g)
 After some of the products are created,
the products begin to react to form the
reactants.
 At equilibrium, there is NO net change in
the concentrations of the reactants and
products.
 The concentrations do not change but
they are not necessarily equal.
5
Dynamic Equilibrium



At equilibrium, two opposing
processes are taking place at equal
rates.
In other words, the rate in the
forward direction = the rate in the
reverse direction
Examples
H2O (l)  H2O (g)
NaCl (s)
H2O


NaCl (aq)
CO (g) + 2 H2 (g)


CH3OH (g)
6
Equilibrium Conditions
H2O + CO


H2 + CO2
7
Law of Mass Action

Given the reaction
aA + bB  cC + dD
The rate in the forward direction is
rate forward = kf [A]a [B]b
The rate in the reverse direction is
rate reverse = kr [C]c [D]d
k is the rate constant. It is unique for each reaction @ a
certain temperature.
8
Law of Mass Action

At equilibrium these rates are equal
rate forward = rate reverse
kf [A]a [B]b = kr [C]c [D]d

The ratio of the rate constants is
Keq
=
[C]c [D]d
[A]a [B]b
Keq = kf/kr
9
Writing Equilibrium Expressions

1. N2 (g) + 3 H2 (g)
2. 2 SO2 (g) +
3. H2 (g) +
4. 2N2O (g)
O2 (g)
Br2 (g)







2NH3 (g)
2SO3 (g)
2 HBr(g)
2 N2 (g) + O2 (g)
10
Answers
Check your work against the following:

1. N2 (g) + 3 H2 (g)
2. 2 SO2 (g) +
3. H2 (g) +
4. 2N2O (g)
O2 (g)
Br2 (g)







2NH3 (g)
2SO3 (g)
2 HBr(g)
2 N2 (g) + O2 (g)
Keq =
[NH3]2
[N2] [H2]3
Keq =
[SO3]2
[SO2]2 [O2]
Keq
Keq
=
[HBr]2
[H2] [Br2]
=
[N2 ]2 [O2]
[N2O]2
11
Calculating
Equilibrium Constants
Nitrogen dioxide decomposes at high temperatures
according to this equation:
2 NO2 (g)  2 NO (g) + O2 (g)
If the equilibrium concentrations are as follows:
[NO2]= 1.20 M, [NO] = 0.160 M, and [O2] = 0.080 M;
calculate the equilibrium constant.
Keq =
[NO]2 [O2]
[NO2] 2
(0.160)2 (0.080)
So Keq =
(1.20)2
= 0.00142
12
Calculating
Equilibrium Constants
The equilibrium equation for the oxidation of sulfur
dioxide is as follows:
2SO2 (g) + O2 (g)  2 SO3 (g)
If the equilibrium concentrations are as follows:
[SO2 ]= 0.44 M, [O2] = 0.22 M, and [SO3] = 0.78 M,
Calculate the equilibrium constant
Keq=
[SO3]2
[SO2]2 [O2]
(0.78)2
So
Keq =
(0.44)2
(0.22)
= 14.3
13
Practice Problem 1
The equilibrium equation for the carbon monoxide with
steam to produce hydrogen gas is as follows:
CO2 (g) + H 2 (g)  CO (g) + H2O (g)
If the equilibrium concentrations are as follows: [CO] = 1.00
M, [H2O] = 0.025 M, [CO2] = 0.075 M and [H2] = 0.060 M,
calculate the equilibrium constant.
Keq =
So Keq =
[CO] [H2O]
[CO2] [H2]
(1.00) (0.025)
(0.075) (0.060)
= 5.6
14
The Meaning of the
Equilibrium Constant

Kc>>1: The reaction is product-favored;
equilibrium concentrations of products are
greater than equilibrium concentrations of
reactants.

Kc<<1: The reaction is reactant-favored;
equilibrium concentrations of reactants are
greater than equilibrium concentrations of
products.
15
The Meaning of the
Equilibrium Constant
So, what do those Keq values mean?
2 NO2 (g)  2 NO (g) + O2 (g)
Kc = 0.00142
Kc<<1: The reaction is reactant-favored
2SO2 (g) + O2 (g)  2 SO3 (g)
Kc = 14.3
Kc>>1: The reaction is product-favored
CO2 (g) + H 2 (g)  CO (g) + H2O (g) Kc = 5.6
Kc~1: The reaction is slightly product-favored
16
Reaction Quotient

The equilibrium constant (Keq or Kc) is a constant
ratio only when the system is in equilibrium.
– Similar Equation: Q =[prod]/[reactants]

If the system it not at equilibrium, the ratio is
known as a Reaction Quotient. (while reactions
are “in progress”).

If the reaction quotient is equal to the equilibrium
constant then the system is at equilibrium.
17
Le Chatelier’s Principle
18
Le Chatelier’s Principle
– Le Chatelier's Principle states: When a
system in chemical equilibrium is
disturbed by a change of temperature,
pressure, or a concentration, the
system shifts in equilibrium
composition in a way that tends to
counteract this change of variable.
– A change imposed on an equilibrium
system is called a stress
– The equilibrium always responds in
such a way so as to counteract the
stress
19
Le Chatelier’s Principle
• A stress is any sudden change in conditions that
drives the system out of equilibrium.
• When a stress is placed on an equilibrium system, the
system will shift in such a way so as to lessen or
mitigate the stress and restore equilibrium.
• A stress usually involves a change in the
temperature, pressure, or in the concentration of one
or more of the substances that are in equilibrium.
• Le Chatelier's principle predicts the direction of the
change in the equilibrium.
20
Applications of
Le Chatelier’s Principle
N2 (g) +3 H2 (g)  2NH3 (g)
DH = - 92 kJ
Haber’s process for the production of ammonia is an
example of an industrial equilibrium system. We will
use this equilibrium as a model to explain the how Le
Chatelier’s principle operates with the following
stresses:
• Change in the concentration of one of the components
• Changes in pressure
• Changes in temperature
• Use of a catalyst
21
Applications of
Le Chatelier’s Principle
N2 (g) +3 H2 (g)  2NH3 (g) DH = - 92 kJ
The equilibrium constant for this reaction is
[NH3]2
Keq =
[N2] [H2]3
Any change in the concentration (or partial pressure)
of any component will cause the equilibrium to shift in
such a way so as to return the equilibrium constant to
its original value.
22
Le Chatelier’s Principle –
The Concentration Effect
N2 (g) +3 H2 (g)  2NH3 (g)
Keq =
DH = - 92 kJ
[NH3]2
[N2] [H2]3
An increase in the concentration of N2 results in
a decrease H2 and an increase in NH3 in such a way to
keep the equilibrium constant the same
23
Le Chatelier’s Principle –
The Concentration Effect
N2 (g) +3 H2 (g)  2NH3 (g)
Keq =
DH = - 92 kJ
[NH3]2
[N2] [H2]3
Likewise an increase in the concentration of H2 results
in a decrease in N2 and an increase in NH3 in such a
way to keep the equilibrium constant the same
24
Le Chatelier’s Principle –
The Concentration Effect
N2 (g) +3 H2 (g)  2NH3 (g)
Keq =
DH = - 92 kJ
[NH3]2
[N2] [H2]3
An increase in the concentration of NH3 results in a
increase in N2 and an increase in H2 in such a way
to keep the equilibrium constant the same.
25
Le Chatelier’s Principle –
The Temperature Effect
N2 (g) +3 H2 (g)  2NH3 (g)
DH = - 92 kJ
 The reaction is exothermic in the forward direction.
 An increase in the temperature would trigger a
response in the heat consuming (endothermic)
direction.
 An increase in temperature therefore causes the
reaction to shift in the reverse direction. Some NH3
decomposes to N2 and H2.
26
Le Chatelier’s Principle –
The Pressure Effect
N2 (g) +3 H2 (g)  2NH3 (g)
DH = - 92 kJ
 All molecules in the equilibrium are gases
 When the reaction proceeds in the forward direction the
number of moles are reduced from 4 to 2.
 Pressure is proportional to the number of moles of gas
 An increase in pressure therefore causes the reaction to
move in the forward direction. Some N2 and H2 combine
to form more NH3
27
Le Chatelier’s Principle –
The Effect of Catalysts
N2 (g) +3 H2 (g)  2NH3 (g) DH = - 92 kJ
 Catalysts lower the activation energy
28
Le Chatelier’s Principle –
The Effect of Catalysts
N2 (g) +3 H2 (g)  2NH3 (g) DH = - 92 kJ
 Catalysts lower the activation energy (prior knowledge)
 A catalyst affects the forward and the reverse direction
equally
 There is no change in the equilibrium position from a
catalyst
 A catalyst decreases the time required for the system
to achieve equilibrium
29
Le Chatelier’s Principle –
Summary
Change
Effect on Equilibrium
Change in Kc?
Increase concentration
Shifts to opposite side
No
Decrease concentration
Shifts to same side
No
Increase pressure
Shifts to side with least
moles of gas
No
Decrease pressure
Shifts to side with most
moles of gas
No
Increase temperature
Shifts in endothermic
direction
Yes
Decrease temperature
Shifts in exothermic
direction
Yes
Add a catalyst
No change
No
30
Different types of
equilibrium constants







Keq General designation for an equilibrium constant
Kc Equilibrium constant based on concentration
Kp Equilibrium constant based on pressure (gases)
Ksp Solubility product
Ka Acid equilibrium constant
Kb Base equilibrium constant
Kw Ion product of water
31