Equilibrium ‫שווי משקל‬
Reactants
Products
In an equilibrium, the forward and reverse processes
continue to occur – but at equal rates!
The reactant and
product
concentrations
remain constant
We are usually concerned
with the situation after
equilibrium is reached.
1
Dynamic Equilibrium Illustrated
NaCl containing radioactive Na+ is
added to a saturated NaCl solution.
After a time, the
solution contains
radioactive Na+ …
NaCl dissolves
and
recrystallizes
continuously.
… and the salt
now contains
some
radioactive Na+.
2
1
Concentration vs. Time
If we begin with only 1
M HI, the [HI]
decreases and both [H2]
and [I2] increase.
2 HI(g)
Regardless of the
starting concentrations;
once equilibrium is
reached …
2
Beginning with 1 M H2
and 1 M I2, the [HI]
increases and both [H2]
and [I2] decrease.
H2(g) + I2(g)
Beginning with 1 M each
of H2, I2, and HI, the [HI]
increases and both 3[H2]
and [I2] decrease.
… the expression with products
in numerator, reactants in
denominator, where each
concentration is raised to the
power of its coefficient, appears
to give a constant.
4
The Equilibrium Constant .‫מ‬.‫קבוע ש‬
For the general reaction:
aA + bB
yY + zZ
The equilibrium expression is:
Keq =
Each concentration is
raised to the power of its
stoichiometric coefficient
[Y]y[Z]z
[A]a[B]b
Keq:
• is independent on the initial concentrations
• depends on temperature
• depends on the way the equation is written!
• has units of Mx
Products in
numerator.
Reactants in
denominator.
5
If the equilibrium concentrations of COCl2 and Cl2 are the same at
395 °C, find the equilibrium concentration of CO in the reaction:
COCl2(g)
CO(g) + Cl2(g)
Keq = 1.2 x 103 M-1 at 395 °C
6
3
Keq of the Reverse Reaction
Consider the reaction:
2 NO(g) + O2(g)
2 NO2(g)
[NO2]2
Keq = –––––––––
= 4.67 x 1013 M-1 (at 298 K)
[NO]2 [O2]
Now consider the reaction: 2 NO2(g)
2 NO(g) + O2(g)
What will be the equilibrium constant K'eq for the new
reaction?
[NO]2 [O2]
1
1
1
K'eq = –––––––––
=
––
=
–––––––––
= 2.14 x 10–14 M
=
–––––––––––
2
13
2
[NO2]
Keq
4.67 x 10
[NO2]
–––––––––
2
[NO] [O2]
7
Keq of a Multiplied Equation
Consider the reaction:
Keq
2 NO(g) + O2(g)
2 NO2(g)
[NO2]2
= –––––––––
= 4.67 x 1013 M-1 (at 298 K)
[NO]2 [O2]
Now consider the reaction:
NO(g) + ½ O2(g)
NO2(g)
What will be the equilibrium constant K"eq for the new reaction?
[NO2]
K"eq = ––––––––– = Keq½
[NO][O2]½
=
4.67 x 1013 = 6.83 x 106 M-½
8
4
Keq and the Modifying Equation: Summary
• For the reverse reaction, K is the reciprocal of K for the
forward reaction.
‫הופכי‬
• When an equation is multiplied by a common factor n to
produce a new equation, Keq is raised to the power of n to
obtain the new equilibrium constant.
• It should be clear that Keq always refers to the way the
balanced chemical equation is written.
• The equilibrium concentrations do not depend on the way
the equation is written!
9
The equilibrium constant for the reaction:
½ H2(g) + ½ I2(g)
at 718 K is 7.07.
HI(g)
(a) What is the value of Keq at 718 K for the reaction
HI(g)
½ H2(g) + ½ I2(g)
(b) What is the value of Keq at 718 K for the reaction
H2(g) + I2(g)
2 HI(g)
10
5
The Equilibrium Constant
for an Overall Reaction
Consider the following reactions:
N2O(g) + ½ O2(g)
2 NO(g) + O2(g)
N2O(g) + 3/2 O2(g)
2 NO(g)
K1 = 1.7 x 10–13 M½
2 NO2(g)
K2 = 4.67 x 1013 M-1
2 NO2(g)
K3 = ??
What would be the equilibrium constants for the overall reaction?
⇒
K3 = K1 ⋅ K2
= 7.9 M-½
• Adding the given equations gives the desired equation.
• Multiplying the given values of K gives the equilibrium
constant for the overall reaction.
11
Equilibria Involving Pure
Solids and Liquids
• The concentration of pure solid and liquid phases remains
constant during the reaction (although their amount changes).
– Why?
⇒ Concentrations of pure solid and liquid phases get incorporated
into the value of Keq and do not appear in the expression
Example:
CaCO3(s)
[CaO] [CO2]
Keq = ––––––––––
[CaCO3]
CaO(s) + CO2(g)
Keq = [CO2]
12
6
Equilibrium Constants: When Do We Need
Them and When Do We Not?
• A very large numerical value of Keq signifies that a
reaction goes (essentially) to completion.
• A very small numerical value of Keq signifies that the
forward reaction, as written, occurs only to a slight extent.
• An equilibrium constant expression applies only to a
reversible reaction at equilibrium.
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The Reaction Quotient, Q ‫מנת הריכוזים‬
For non-equilibrium conditions, the expression
having the same form as Keq is called the reaction
quotient, Q.
The reaction quotient is useful for predicting the
direction in which a net change must occur to
establish equilibrium
Q < Keq ⇒ reaction moves in the forward direction
Q > Keq ⇒ reaction moves in the reverse direction
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7
The Reaction Quotient, Q
15
Le Châtelier’s Principle
When any change in concentration,
temperature, pressure, or volume is
imposed on a system at equilibrium, the
system responds by attaining a new
equilibrium condition that minimizes the
impact of the imposed change.
Henri Le Châtelier
1850-1936
16
8
Changing the Amounts of Reactants
•
•
•
•
At equilibrium, Q = Keq.
If [reactant]↑ ⇒ Q↓
Q is now less than Keq.
This condition is only temporary, however, because the
concentrations of all species must change in such a way so
as to make Q = Keq again.
• In order to do this, the concentrations of the products
increase; the equilibrium is shifted to the right.
17
C8H17OH + CH3COOH
… the acetic acid
concentration
first increases …
When acetic acid (a
reactant) is added
to the equilibrium
mixture …
9
CH3COO-C8H17 + H2O
… then the
concentrations of
both reactants
decrease …
… and the
concentrations of
both products
increase, until a
new equilibrium
is established.
18
Equilibrium Calculations
When we have:
1. initial amounts of the reactants
2. equilibrium amount of the product(s)
We can calculate:
a) amounts of the species in equilibrium
b) Keq
How?
Use the "ICE" table!
(Initial/Change/Equilibrium)
19
Equilibrium Quantities from Keq Values
When we have:
1. initial amounts of the reactants
2. Keq
We can calculate:
Amounts of the species
in equilibrium
Consider the reaction:
H2(g) + I2(g)
2 HI(g)
Keq = 54.3 at 698 K
If we start with 0.500 mol I2(g) and 0.500 mol H2(g) in a 5.25-L vessel at 698
K, how many moles of each gas will be present at equilibrium?
20
10
Carbon monoxide and chlorine react to form phosgene, COCl2, which is
used in the manufacture of pesticides, herbicides, and plastics:
CO(g) + Cl2(g)
COCl2(g) Keq = 1.2 x 103 M-1 at 668 K
How much of each substance, in moles, will there be at equilibrium in a
reaction mixture that initially has 0.0100 mol CO, 0.0100 mol Cl2, and
0.100 mol COCl2 in a 10.0-L flask?
21
Summary of Concepts
• At equilibrium, the concentrations of all reactants
and products remain constant with time
• Equilibrium constant is expressed as the ratio of
concentration of products to reactants
• For the general reaction:
aA + bB
yY + zZ
The expression for Keq is:
Keq =
11
[Y]y[Z]z
[A]a[B]b
22
Summary of Concepts
• If the chemical equation for a reversible reaction is
modified, the equilibrium constant expression
must also be modified
– If the equation is reversed, the Keq expression is
inverted
– If the coefficients of the equation are multiplied by a
common factor, the Keq expression is raised to the
corresponding power
• The expression for Keq does not include pure
liquids and solids
23
Summary of Concepts
• In general, if Keq is very large, the forward reaction goes to
completion; if Keq is very small, the forward reaction
occurs to a very limited extent
– Usually, calculations based on the equilibrium constant expression
are necessary only when Keq values lie between these extremes
• The reaction quotient is a ratio of concentrations (Q)
having the same form as the equilibrium constant
expression but using non-equilibrium concentrations
– If Q < Keq ⇒ reaction will proceed in the forward direction
– If Q > Keq ⇒ reaction will proceed in the reverse direction
24
12
Summary of Concepts
• Qualitative predictions about the effect of various changes
(amounts of reactants or products, volume, external
pressure, or temperature) can be based on Le Châtelier’s
principle
• When a change is imposed on a system at equilibrium, the
system responds by attaining a new equilibrium in which
the impact of the change is minimized
• The most common types of equilibrium calculations are
– determining the value of an equilibrium constant from
initial and equilibrium conditions
– using initial conditions and the equilibrium constant to
determine equilibrium conditions
25
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