Chapter 14 Chemical Equilibrium
I.
Equilibrium Conditions
A.
Example Reaction: NO2 + NO2
N2O4
(brown)
(colorless)
1. This reaction proceeds to the right at the start
2. But, it never goes all the way to completion. It reaches a point where the
forward and reverse reactions are going at the same rate.
a) D[NO2] = 0
b) D[N2O4] = 0
c) Chemical Equilibrium = The state where the concentrations of all
reactants and products remain constant with time.
B.
3.
Some reactions almost reach completion before equilibrium is achieved.
We say the equilibrium lies to the right.
2H2 + O2
2H2O
4.
Some reactions barely get started before reaching equilibrium. We say
the equilibrium lies to the left.
2CaO
2Ca + O2
Equilibria are Dynamic
1. Even though D[NO2] = 0, the forward and reverse reactions are both
occurring.
2. At equilibrium, there is no net change in concentration, even though
individual molecules are constantly reacting.
C.
Sample Reaction: N2 + 3H2
2NH3
1. When we mix any concentrations of these gases at room temperature,
there are no changes in concentration.
a) We might be at equilibrium
b) The reaction rates might be very slow at these conditions.
i. The N≡N bond is very strong (432 kJ/mol) and hard to break
ii. Entropy favors no forward reaction
2.
When a catalyst is added and the reaction is heated, the concentrations do
change until a real equilibrium is reached.
a)
H2 disappears three
times as fast as N2
b)
NH3 appears twice as
fast as N2 disappears.
II.
Equilibrium Constants
A.
The Law of Mass Action
1. This is an empirical law discovered in 1864
2. Every reaction has a constant associated with it telling us where the
equilibrium position is.
c
d
[C]
[D]
K
aA + bB
cC + dD
K
[A]a [B] b
3.
K = Equilibrium Constant = tells us where the equilibrium position is
a) K > 1 tells us the equilibrium lies to the right If K = 1, halfway
b) K < 1 tells us the equilibrium lies to the left
4.
5.
If we know the concentrations, we can find K from its equation
K is written without units, even in cases where there are units left not
cancelled. This is correct for nonideal behavior of molecules.
Example: Write K for: 4NH3 + 7O2
4NO2 + 6H2O
6.
III. The Reaction Quotient
1.
2.
We can write K only for a reaction at equilibrium
How can we describe a reaction at other times? N2 + 3H2
2NH3
2
Q
3.
4.
[NH3 ]0
3
[N 2 ]0 [H 2 ]0
Describes the reaction when not at equilibrium
Q = reaction quotient has the same form as K, but is at non-equilibrium times.
Q can tell us how a reaction will change to get to equilibrium
1) If Q = K, we are at equilibrium
2) If Q > K, then products > reactants and the reaction will shift left
3) If Q < K, then reactants > products and the reaction will shift right
Q
K
Predicted Direction of Reaction
0.55
1.45
To the right (towards products)
2.55
1.45
To the left (towards reactants)
1.45
1.45
No change (at equilibrium)
Le Chatelier’s Principle
A.
We can change the position of an equilibrium by changing conditions
1. N2 + 3H2
2NH3
2. Le Chatelier’s Principle: When a chemical system at equilibrium is
disturbed, the system shifts in a direction that minimizes the disturbance.
B.
Effect of change in concentrations N2 + 3H2
2NH3 K = 0.0596
1) At equilibrium, [N2] = 0.399 M, [H2] = 1.197 M, [NH3] = 0.202 M
2) Let’s add 1 M N2. How does the equilibrium shift?
a) Too much N2 for equilibrium
b) Reaction must adjust to get back to equilibrium
c) Reaction must shift to the right
Eq. shifts to the right
2
2
[NH3 ]
(0.202)
Q

 0.017  K
3
2
[N 2 ][H 2 ]
(1.399)(1.197)
3)
We could do the x calculation and find the new concentrations if we
wanted, but sometimes you just want to know which way the reaction will
shift. Le Chatelier’s principle lets us do that simply.
3.
Another Example:
What would happen if we add N2O4?
C.
Effect of Change in Pressure/Volume
1. If we add/remove reactant/product, we change concentration (see above)
2. Changing the size of the flask: shifts to appropriately fill the flask
a. More volume (less pressure) results in more gas molecules to fill it
b. Less volume (more pressure) results in less molecules to reduce pressure
V↓ = P↑
V↑ = P↓
D.
Effect of change in Temperature
1. The above changes (P, V, Concentration) effect the equilibrium position, but not
the equilibrium constant K
2. Changing Temperature changes the equilibrium constant K
3. We can predict the changes by treating heat as a product or reactant of every
reaction
4.
Exothermic reactions produce heat as a product
a) Adding heat (increasing temp.) shifts away from heat, to left, K decreases
b) Removing heat (decreasing temp.) shifts towards heat, to right, K increases
c) N2 + 3H2
2NH3 + 93 kJ/mol (DH = - 93 kJ/mol)
5.
Endothermic reactions require heat as a reactant
a) Add heat = shift to right, K increases
b) Remove heat = shift to left, K decreases
c) 556 kJ/mol + CaCO3(s)
CaO(s) + CO2(g) (DH = + 556 kJ/mol)
6.
Examples:
N2(g) + O2(g)
2SO2(g) + O2(g)
2NO(g) DH = 181 kJ/mol
2SO3 DH = - 198 kJ/mol)
E.
Today’s Reactions
1) 2CrO4-(a) + 2H3O+(aq)
Cr2O72-(aq) + 3H2O(l) “Chromate/Dichromate”
a) Chromate is yellow; Dichromate is orange
b) We will add and remove acid and observe the result
2)
Co(H2O)62+(aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
a) Cobalt(II) coordination number effects its color
b) Six-coordinate cobalt(II) is pink; Four-coordinate cobalt(II) is blue
c) We will add Cl- and H2O and see the effect (takes lots of Cl-)
3)
NH4Cl(s)
NH4+(aq) + Cl-(aq)
a) Temperature effects the solubility of Ammonium Chloride
b) We will change concentrations
c) We will change temperature and decide if dissolving is endo- or exothermic