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AP CHEMISTRY
MRS. MOLCHANY
EQUILIBRIUM
Chemical equilibrium is the condition in which the concentrations of all reactants
and products in a closed system cease to change with time.
Chemical equilibrium occurs when opposing reactions are proceeding at equal
rates.
Consider the reaction
Forward reaction
Reverse reaction
A→B
A→B
B→A
Rate = kf [A]
Rate = kr [B]
kf is rate constant forward rxn
kr is rate constant reverse rxn
If the reaction is in equilibrium:
kf [A] = kr [B]
rearrange:
[B] = kf = a constant
[A] kr
Once equilibrium is established, the concentrations of A and B do not change.
The equilibrium is dynamic, the forward and reverse reactions continue occurring
at the same rate.
To indicate that the reaction proceeds in both the forward and reverse directions,
use a double arrow.
EQUILIBRIUM
Uses of equilibrium constants:
1. Calculate the concentrations of the various substances at equilibrium.
2. Predict the initial direction of a reaction.
The Haber Process is used to convert N2(g) and H2(g) to NH3(g)
** At some point the reaction seems to stop and all three components of the
reaction mixture are present at the same time.
EQUILIBRIUM
For any equilibrium reaction, the ratio of concentrations of the substances on the
right to the concentrations of those on the left equals a constant appropriate for
that specific reaction.
The Law of Mass Action:
aA + bB
(See Example #1)
↔ pP + qQ
[P]p [Q]q = Kc
[A]a [B]b
Kc is the equilibrium constant (using molar conc.)
a,b,p,q are numerical coefficients
[P] denotes the molar concentration of P
K depends on the particular reaction, the temperature, and the units used to
describe concentration (molarity, pressure or others).
Use the following expression when the reactants and products are gases and
partial pressures (in atm.) are used.
[PP]p [PQ]q = Kp
[PA]a [PB]b
Kp is the equilibrium constant (using partial press.)
Use PV = nRT to convert between concentration (in M) and press. (in atm)
rearrange:
P = n RT
V
substitute in: Kp =
Kp =
remove Kc:
use sci. not.:
Kp =
n = Molarity
V
P = MRT
PA = [A] (RT)
([P] RT)p ([Q] RT)q
([A] RT)a ([B] RT)b
[P] p (RT)p [Q] q (RT)q
[A] a (RT)a [B] b (RT)b
Kc (RT)p (RT)q
(RT)a (RT)b
Kp =
general express: Kp =
Kc (RT)(p+q) – (a+b)
Kc (RT)∆n
∆n = prod. moles of gas – reac. moles of gas
EQUILIBRIUM
The magnitude of the equilibrium constant can show whether products or
reactants are favored in a reaction.
K >> 1:
Equilibrium lies to the right; products favored.
K << 1:
Equilibrium lies to the left; reactants favored.
(See Example #2)
The direction in which you write the chemical equation for an equilibrium is
arbitrary, because equilibrium can be approached from either direction.
The equilibrium constant expression for a reaction written in one direction is the
reciprocal of the one for the reaction in the reverse direction.
The equilibrium constant allows you:
1. to predict the direction in which a reaction mixture will proceed to achieve
equilibrium.
2. to calculate the concentrations of reactants and products once equilibrium
has been established.
Homogeneous equilibria involve substances all in the same phase.
Heterogeneous equilibria involve substances all in different phases.
Express the concentration of a solid by using its density.
Density
M
= g/cm3 = mol
g/mol
cm3
M = molarity
The density of a pure liquid or solid is a constant at any given temperature.
CO2(g) + H2(g) ↔ CO(g) + H2O(l)
Kc =
[CO]___
[CO2] [H2]
Kp = PCO____
PCO2 PH2
H2O(l) is considered a pure liquid with a specific density
EQUILIBRIUM
Calculating Equilibrium Constants
1. Tabulate the known initial and equilibrium concentrations of all species
involved in the equilibrium.
2. For those species for which both the initial and equilibrium concentrations
are known, calculate the change in concentration that occurs as the system
reaches equilibrium.
3. Use the stoichiometry of the reaction (this is, use the coefficients in the
balanced chemical equation) to calculate the changes in concentration for
all the other species in the equilibrium.
4. From the initial concentrations and the changes in concentration, calculate
the equilibrium concentrations. These are used to evaluate the equilibrium
constant.
Or, think of it this way:
1.
2.
3.
4.
initial and equil. conc.
change in conc.
stoichiometry to calculate changes in conc.
calculate equil. conc.
(See Example #3)
When the reactant and product concentrations are substituted into the equilibrium
constant expression, the result is the reaction quotient (Q).
The reaction quotient will equal the equilibrium constant, K, only if the system
is at equilibrium: Q = K only at equilibrium.
When Q > K, substances on the right side of the equation will react to form
more reactants.
When Q < K, the reaction will achieve equilibrium by forming more products.
Q>K
forms
reactants
Q=K
at equilibrium
Q<K
forms
products
EQUILIBRIUM
Le Chatelier’s Principle – If a system at equilibrium is disturbed by a change in
temperature, pressure or the concentration of one of the components, the
system will shift its equilibrium position so as to counteract the effect of the
disturbance.
3 ways that a chemical equilibrium can be disturbed:
1. adding or removing a reactant or product
2. changing the pressure
3. changing the temperature
EQUILIBRIUM
EXAMPLES
#1
What is the equilibrium expression for the following equation?
CH4(g) + O2(g)
↔ CO2(g) + 2H2O(g)
#2
At 1000C, the halogen reaction
Br2(g) + Cl2(g) ↔ 2BrCl(g)
has an equilibrium constant of 0.15 when the concentrations are expressed in
either moles or atmospheres. If a reaction vessel is filled with 100 g of Br2 and
50 g each of Cl2 and BrCl, calculate to predict the direction of the initial reaction.
ANS:
moles BrCl
moles Br2
moles Cl2
= 50/115.35
= 100/159.80
= 50/70.90
[BrCl]2 = __(0.433)2___
[Br2] [Cl2]
(0.626) (0.705)
= 0.433
= 0.626
= 0.705
= 0.425 > 0.15
The reaction 2BrCl(g) ↔ Br2(g) + Cl2(g) is the initial reaction.
#3
A mixture of 5.00 x 10-3 mol of H2 and 1.00 x 10-2 mol of I2 is placed in a 5.00 L
container at 448 0C and allowed to come to equilibrium. Analysis of the equilibrium
mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 0C
for the reaction.
H2(g) + I2(g) ↔ 2HI (g)
1. initial and equil. conc.:calculate the initial conc. of H2 and I2
[H2]initial = 5.00 x 10-3 mol = 1.00 x 10-3 M
5.00 L
[ I2]initial = 1.00 x 10-2 mol = 2.00 x 10-3 M
5.00 L
H2(g)
+
I2(g)
Initial
Change
Equilibrium
1.00 x 10-3 M
2. change in conc.:
calculate the change in conc. of HI
↔
2.00 x 10-3 M
2HI (g)
0M
1.87 x 10-3 M
∆HI = 1.87 x 10-3 M - 0 M
= 1.87 x 10-3 M
Initial
Change
Equilibrium
H2(g)
1.00 x 10-3 M
+
I2(g)
↔
-3
2.00 x 10 M
3. stoichiometry to calculate changes in conc.
(1.87 x 10-3 mol HI ) ( 1 mol H2 ) = 0.935 x 10-3 mol H2
L
( 2 mol HI )
L
Since one mole of I2 was also consumed, the change of I2 is also
0.935 x 10-3 mol H2
L
2HI (g)
0M
+ 1.87 x 10-3 M
1.87 x 10-3 M
Example #3
Initial
Change
Equilibrium
H2(g)
1.00 x 10-3 M
- 0.935 x 10-3 M
+
I2(g)
↔
-3
2.00 x 10 M
- 0.935 x 10-3 M
2HI (g)
0M
+ 1.87 x 10-3 M
1.87 x 10-3 M
4. calculate equil. conc.
[H2] = 1.00 x 10-3 M - 0.935 x 10-3 M = 0.065 x 10-3 M
[ I2] = 2.00 x 10-3 M - 0.935 x 10-3 M = 1.065 x 10-3 M
Initial
Change
Equilibrium
H2(g)
1.00 x 10-3 M
- 0.935 x 10-3 M
0.065 x 10-3 M
+
I2(g)
↔
-3
2.00 x 10 M
- 0.935 x 10-3 M
1.065 x 10-3 M
2HI (g)
0M
+ 1.87 x 10-3 M
1.87 x 10-3 M
Finally solve the equilibrium constant equation.
Kc = [ HI ]2 = ______(1.87 x 10-3 M)2________ = 51
[H2] [ I2]
(0.065 x 10-3 M) (1.065 x 10-3 M)
Example #4
Sulfur trioxide decomposes at high temperature in a sealed container:
2SO3(g)
↔ 2SO2(g) + O2(g). Initially the vessel is charged at 1000K with
SO3(g) at a concentration of 6.09 x 10-3 M. At equilibrium the SO3 concentration is
2.44 x 10-3M. Calculate the value of Kc at 1000K.
ANS: 4.08 x 10-3