Chemical Equilibrium
The state where the concentrations of
all reactants and products remain
constant with time
How is “equilibrium” different than
chemical reactions we have studied so
far?
• For stoichiometry calculations, we
assume reaction goes to completion.
• In calling a compound “soluble,” we
assume that ions remain dissociated in
water: NaCl + H2O  Na+(aq) + Cl-(aq)
• In a closed vessel, a chemical reaction
achieves a state of equilibrium, where
concentrations of both reactants and
products remain constant over time.
Equilibrium = Dynamic
• Reactants convert continually to
products AND
• Products continually revert to
reactants
• Due to molecular collisions
• Forward and reverse RATES are
equal
The Equilibrium Expression
•
•
•
•
Given the following reaction at equilibrium:
2A(g) + 3B(g)  2C(g) + 4D(g)
This is the equilibrium expression:
Keq = [C]2[D]4
[A]2[B]3
• Keq is the equilibrium constant.
• Include concentrations for products or reactants
ONLY for gases and aqueous solutions.
•
Liquids and solids are not included because their
concentrations do not change.
Most Familiar Equilibrium
Example: Water
• Water molecules split other water molecules
into hydrogen ions and hydroxide ions to a
small extent.
• This is called auto-ionization, or selfionization of water.
• This reaction is reversible and does not go
to completion in the forward direction. It
reverses itself after only a few ions are
formed.
• The equilibrium expression for this equation:
Kw = [H3O+][OH-]
Most Familiar Equilibrium
Example: Water
H2O   H3O+(aq) + OH- (aq)
• Kw = 1.0 x 10-14
[OH-][H3O+] = 1.0 x 10-14
*
• Since the K value is so small, this indicates that not very
many ions form before the equation reverses itself and
begins to form the molecules of water again.
• In pure water and in aqueous solutions, there are both H+
and OH• ACIDS: [H3O+] > [OH- ]
• BASES: [OH-] > [H3O+]
.
*Includes concentrations for products or reactants
ONLY for gases and aqueous solutions
Equilibrium Positions
• At a given temp, there is only one value of K
but an infinite number of possible
equilibrium positions.
• Eq. position defined by the concentrations
that satisfy the equilibrium expression.
• Equilibrium position
– Depends on initial concentrations (aqueous or
gaseous mixtures)
– Never depends on amount of pure solid or liquid
in the system
Problem: Calculating K
At 127º C,
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
Find K for the Haber process.
N2(g) + 3H2(g)   2NH3 (g)
K = [NH3]2 =
(3.1 x 10-2 M)2
[N2][H2]3 (8.5 x 10-1 M)(3.1 x 10-3 M)3
K = 3.8 x 104
(no units)
Problem: Calculating K for
Reverse Reaction
At 127º C,
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
Find K for 2NH3 (g)   N2(g) + 3H2(g)
K’ = [N2][H2]3 =(8.5 x 10-1 M)(3.1 x 10-3 M)3
[NH3]2
(3.1 x 10-2 M)2
K’ = 2.6 x 10-5 (no units)
Visualizing Equilibrium
• Try this activity to see effects of
changing concentrations on the
equilibrium of a system:
http://cheminfo.chem.ou.edu/~mra/CC
LI2004/ERGBN.htm
Important Points about K
•
1.
2.
3.
•
For a balanced equation, Kc:
is constant at a given temperature
changes if the temperature changes
does not depend on initial
concentrations
Kc’ is the equilibrium constant for the
equation written in reverse, and is
the reciprocal of Kc
Reaction Quotient, Q
• Is a measure of
reaction progess
Q < Kc
• Uses same form as Kc
• Also called mass
action expression
Q > Kc
• Concentrations not
necessarily at
equilibrium
Q = Kc
Forward reaction
dominates until
equilibrium is achieved
Reverse reaction
dominates until
equilibrium is achieved
System at
equilibrium
Application: Using Kc to
find concentrations
PCl3(g) + Cl2 ( g)
PCl5 (g)
At a given temp, this reaction occurs in a 1.0-L
container with 0.25 mol PCl5 and 0.16 mol PCl3. At
equilibrium, what is the concentration of Cl2?
Kc = [PCl5]
= 0.25
=1.9
[Cl2][PCl3]
[Cl2]0.16
[Cl2]=0.25
(0.16)(1.9)
= 0.82 M
Q: Problem provides
concentrations of species and K
1. Write the equilibrium expression,
using the letter Q instead of K
2. Substitute concentrations (or
pressures ) into the equation to find Q
3. Compare Q to K:
Put these two values on a number line to
easily state the direction of equilibrium
Q is used for specific instances of
known concentrations.
Predicting shift based on Q:
If Q < K, there are more reactants
than there will be at equilibrium, so
reaction shifts right.
Number line trick:
Q value …….K value
→ Point to K, (right shift)
More predicting equilibrium shift based
on Q:
If Q > K, there are more products than
there will be at equilibrium, so
reaction shifts left.
Number line trick:
K value
…….Q value
Point to K ← (left shift)
RICE Table
• A more common application gives initial
concentrations and Kc and you must find
the equilibrium concentrations.
• Make a table showing the reaction, initial
concentrations, change and equilibrium
concentrations to solve this problem.
• You will often need to use the quadratic
formula to solve these, so refresh your
memory on your calculator.
a
If Kc = 49.0 at a given temp, and 0.400
mol of each reactant are placed in a
2.0 L container, what concentrations
of all species are present at
equilibrium for A + B   C + D?
[A]o = [B]o = 0.400 mol/2L = 0.200 M
[C]0 = [D]0 = 0 M
R
A+
B
I
0.200 M
0.200 M 0
0
C
-X
-X
+X
+X
E
(0.200 –x)M
(0.200 –x)M
+X M
+X M
KC = 49.0 =
C+
(x)(x)
(0.200 –x) (0.200 –x)
D
A rare occasion without
quadratic formula
Kc = 49.0 =
(x)(x)
(0.200 –x) (0.200 –x)
7.00 = x/(0.200-x)
x = 1.40 – 7.00 x
8.00x = 1.40
X = 0.175
[A] = [B] = 0.200 – 0.175 = 0.025 M
[C] = [D] = 0.175 M
LeChâtelier’s Principle
When a system at equilibrium is
disturbed by application of stress, it
attains a new equilibrium position
that minimizes the stress.
 "If stress is applied to a system at
equilibrium,
the system will tend to readjust so
that the stress is reduced."
Pressure and volume
effects on equilibrium
• Pressure and volume do not affect
concentrations of solids and liquids
because they are not particularly
compressible.
• Recall the universal gas law for ideal
gases:
PV = nRT
so P = nRT
V
Pressure and volume at constant temperature
• If volume decreases, partial pressure of a
gas increases and therefore concentration
(n/V) increases.
• If number of moles of gas are the same on
both sides of equation, then equilibrium is
unaffected.
• If volume decreases, equilibrium shifts
toward side of equation with FEWER moles
gas because the concentration of that side
has increased:
• A(g)<->2B(g) Kc = [B]2 so Q would be >Kc
[A]
Proof
• If decrease V, shift to side of FEWER
moles gas:
• A(g)<->2B(g)
If equilibrium V = 2 L and V2 = 1L,
[A]eq = .5 M and [B] eq = 1 M
Kc = [B]2 = 1 = 2
[A] .5
At V2, [A]eq = 1 mol/1L and [B] eq = 2mol/1L
Q = [B]2 = 22= 4
[A] 1
Q >Kc so equilibrium shifts back to side of
reactants, WHICH IS SIDE WITH FEWER
MOLES!
Pressure and volume
effects on equilibriumSUMMARY
Change
Q vs K
Volume ↓
Pressure ↑
Q > Kc
Volume ↑
Pressure ↓
Q < Kc
Shift
Toward
smaller # gas
moles
Toward larger
# gas moles