Chemical Equilibrium
Chapter 14
14.1-14.5
Equilibrium
Equilibrium is a state in which there are
no observable changes as time goes by.
Chemical equilibrium is achieved when:
1.) the rates of the forward and reverse
reactions are equal and
2.) the concentrations of the reactants
and products remain constant
Equilibrium
• There are two types of equilibrium:
Physical and Chemical.
– Physical Equilibrium
• H20 (l) ↔ H20 (g)
– Chemical Equilibrium
• N2O4 (g) ↔ 2NO2
Physical Equilibrium
Chemical Equilibrium
Chemical Equilibrium
N2O4 (g) ↔ 2NO2 (g)
Law of Mass Action
• Law of Mass Action- For a reversible reaction
at equilibrium and constant temperature, a
certain ratio of reactant and product
concentrations has a constant value (K).
• The Equilibrium Constant (K)- A number equal
to the ratio of the equilibrium concentrations of
products to the equilibrium concentrations of
reactants each raised to the power of its
stoichiometric coefficient.
Law of Mass Action
• For the general reaction:
aA (g) + bB (g)
K=
[C]c[D]d
[A]a[B]b
cC (g) + dD (g)
Equilibrium Constant
N2O4 (g) ↔ 2NO2 (g)
Chemical Equilibrium
• Chemical equilibrium is defined by K.
• The magnitude of K will tell us if the
equilibrium reaction favors the reactants or
the products.
• If K » 1……..favors products
• If K « 1……..favors reactants
Equilibrium Constant Expressions
• Equilibrium constants can be expressed
using Kc or Kp.
• Kc uses the concentration of reactants and
products to calculate the eq. constant.
• Kp uses the pressure of the gaseous
reactants and products to calculate the eq.
constant.
Equilibrium Constant Expressions
• Equilibrium Constant Equations
aA (g) + bB (g)
Kc =
[NO2
]2
[N2O4]
cC (g) + dD (g)
Kp =
2
PNO
2
PN2O4
Homogeneous Equilibrium
• Homogeneous Equilibrium- applies to
reactions in which all reacting species are
in the same phase.
N2O4 (g) ↔ 2NO2 (g)
Kc =
[NO2]2
Kp =
[N2O4]
In most cases
Kc  Kp
2
PNO
2
PN2O4
Equilibrium Constant Expressions
• Relationship between Kc and Kp
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
Equilibrium Constant Calculations
The equilibrium concentrations for the reaction between carbon monoxide and
molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] =
0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and
Kp .
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
= 220
=
Kc =
[CO][Cl2]
0.012 x 0.054
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R = 0.0821
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
Equilibrium Constant Calculations
• The equilibrium constant Kp for the reaction is 158 at
1000K. What is the equilibrium pressure of O2 if the PNO
= 0.400 atm and PNO = 0.270 atm?
Kp =
2
PNO
PO2
2
PNO
2
PO2 = Kp
2
PNO
2
2
PNO
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
Heterogeneous Equilibrium
• Heterogeneous Equilibrium- results from
a reversible reaction involving reactants
and products that are in different phases.
• Can include liquids, gases and solids as
either reactants or products.
• Equilibrium expression is the same as that
for a homogeneous equilibrium.
• Omit pure liquids and solids from the
equilibrium constant expressions.
Heterogeneous Equilibrium
Constant
CaCO3 (s)
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
Heterogeneous Equilibrium
Constant
Equilibrium Constant Calculations
Consider the following equilibrium at 295 K:
NH4HS (s)
NH3 (g) + H2S (g)
The partial pressure of each gas is 0.265 atm. Calculate Kp
and Kc for the reaction.
Kp = PNH PH S = 0.265 x 0.265 = 0.0702
3
2
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2
T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
Multiple Equilibria
• Multiple Equilibria- Product molecules of one equilibrium
constant are involved in a second equilibrium process.
A+B
C+D
A+B
C+D
E+F
E+F
Kc = Kc‘ x Kc‘‘
Kc‘
Kc‘‘
Kc
[C][D]
Kc‘ =
[A][B]
[E][F]
Kc‘‘ =
[C][D]
[E][F]
Kc =
[A][B]
Writing Equilibrium Constant Expressions
•
The concentrations of the reacting species in the condensed phase are
expressed in M. In the gaseous phase, the concentrations can be expressed in
M or in atm.
•
The concentrations of pure solids, pure liquids and solvents do not appear in
the equilibrium constant expressions.
•
The equilibrium constant is a dimensionless quantity.
•
In quoting a value for the equilibrium constant, you must specify the balanced
equation and the temperature.
•
If a reaction can be expressed as a sum of two or more reactions, the
equilibrium constant for the overall reaction is given by the product of the
equilibrium constants of the individual reactions.
14.2
What does the Equilibrium
Constant tell us?
• We can:
– Predict the direction in which a reaction
mixture will proceed to reach equilibrium
– Calculate the concentration of reactants and
products once equilibrium has been reached
Predicting the Direction of a
Reaction
• The Kc for hydrogen iodide in the following equation is
53.4 at 430ºC. Suppose we add 0.243 mol H2, 0.146 mol
I2 and 1.98 mol HI to a 1.00L container at 430ºC. Will
there be a net reaction to form more H2 and I2 or HI?
H2 (g) + I2 (g) → 2HI (g)
[HI]02
Kc =
[H2]0 [I2]0
[1.98]2
Kc =
[0.243] [0.146]
Kc = 111
Reaction Quotient
The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into
the equilibrium constant (Kc) expression.
IF
• Qc > Kc system proceeds from right to left to
reach equilibrium
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds from left to right to
reach equilibrium
Reaction Quotient
Calculating Equilibrium
Concentrations
• If we know the equilibrium constant for a
reaction and the initial concentrations, we can
calculate the reactant concentrations at
equilibrium.
• ICE method
Reactants
Products
Initial (M):
Change (M):
Equilibrium (M):
Calculating Equilibrium
Concentrations
• At 1280ºC the equilibrium constant (Kc) for the reaction
is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063
M and [Br] = 0.012 M, calculate the concentrations of
these species at equilibrium.
Br2 (g)
2Br (g)
Let x be the change in concentration of Br2
Br2 (g)
2Br (g)
0.012
0.063
Initial (M)
-x
+2x
Change (M)
Equilibrium (M)
0.063 - x
0.012 + 2x
Calculating Equilibrium
Concentrations
[Br]2
Kc =
[Br2]
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
ax2
+ bx + c = 0
-b ± b2 – 4ac
x=
2a
x = -0.0105
x = -0.00178
Calculating Equilibrium
Concentrations
Initial (M)
Change (M)
Equilibrium (M)
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.063 – x = 0.0648 M
Calculating Equilibrium
Concentrations
• Express the equilibrium concentrations of all
species in terms of the initial concentrations and
a single unknown x, which represents the
change in concentration.
• Write the equilibrium constant expression in
terms of the equilibrium concentrations.
Knowing the value of the equilibrium constant,
solve for x.
• Having solved for x, calculate the equilibrium
concentrations of all species.
Factors that Affect Chemical
Equilibrium
• Chemical Equilibrium represents a
balance between forward and reverse
reactions.
• Changes in the following will alter the
direction of a reaction:
– Concentration
– Pressure
– Volume
– Temperature
Le Châtlier’s Principle
• Le Châtlier’s Principle- if an external
stress is applied to a system at
equilibrium, the system adjusts in such a
way that the stress is partially offset as the
system reaches a new equilibrium
position.
• Stress???
Changes in Concentration
• Increase in concentration of reactants
causes the equilibrium to shift to the
________.
• Increase in concentration of products
causes the equilibrium to shift to the
________.
Changes in Concentration
Change
Increase in [Products]
Decrease in [Products]
Increase in [Reactants]
Decrease in [Reactants]
Shift in Equilibrium
left
right
right
left
Changes in Concentration
FeSCN2+(aq) ↔ Fe3+(aq) +
SCN-(aq)
a.) Solution at equilibrium
b.) Increase in SCN-(aq)
c.) Increase in Fe3+(aq)
d.) Increase in FeSCN2+(aq)
Changes in Volume and Pressure
• Changes in pressure primarily only
concern gases.
• Concentration of gases are greatly
affected by pressure changes and volume
changes according to the ideal gas law.
PV = nRT
P = (n/V)RT
Changes in Pressure and Volume
Change
Increase in Pressure
Decrease in Pressure
Increase in Volume
Decrease in Volume
Shift in Equilibrium
Side with fewest moles
Side with most moles
Side with most moles
Side with fewest moles
Changes in Pressure and Volume
Changes in Temperature
• Equilibrium position vs. Equilibrium
constant
• A temperature increase favors an
endothermic reaction and a temperature
decrease favors and exothermic reaction.
Change
Increase T
Decrease T
Endo. Rx
K decreases
K increases
Exo. Rx
K increases
K decreases
Changes in Temperature
Consider: N2O4(g) ↔ 2NO2(g)
The forward reaction absorbs heat; endothermic
heat + N2O4(g) ↔ 2NO2(g)
So the reverse reaction releases heat; exothermic
2NO2(g) ↔ N2O4(g) + heat
Changes in temperature??
Effect of a catalyst
• How would the presence of a catalyst
affect the equilibrium position of a
reaction?