Chemical Equilibrium
N2(g) + 3 H2(g) --> 2 NH3(g)
Many reactions do not go to completion - under the given
conditions it is possible that not all of the reactants are
consumed.
Instead the extent of the reaction is determined by the
equilibrium point.
A + B --> C + D
As the concentrations of C and D increase, C and D could
react to form A and B - the REVERSE reaction.
A+B
N2(g) + 3H2(g)
C+D
2NH3(g)
The forward and reverse directions “oppose” one another.
At some point in time, the rate of the forward reaction will
equal the rate of the reverse reaction - this point corresponds
to EQUILIBRIUM.
Hence, when equilibrium has been reached, the
concentration of A, B, C and D stay constant, as long as the
conditions are held the same.
At the equilibrium point: A and B combine to form C and D; C
and D combine to form A and B; but both occur at the same
rate.
There is no NET change in the concentrations of A, B, C & D.
Equilibrium
When opposing forces acting on a system are equal in
magnitude, the system is said to be in a state of equilibrium.
A dynamic equilibrium is one at which changes to the system
do occur at the microscopic level, but at the macroscopic
level these changes are not observed.
In general: Processes not at equilibrium will act or react to
reach equilibrium.
A
B
Characteristics of equilibrium
1) The attainment of equilibrium is spontaneous; i.e. it is a
natural tendency
2) At equilibrium there is no macroscopic evidence of any
changes in the system
3) A dynamic balance is established between opposing forces
4) Equilibrium is reached from either direction
The Equilibrium Expression
For a general reaction:
aA+bB
cC+dD
the ratio:
(concentration C)c (concentration D)d
(concentration
A)a
(concentration
B)b
= K (constant)
K is called the EQUILIBRIUM CONSTANT
Note: K has a fixed value for a particular reaction and
varies with temperature
If all reactants and products are gases, the relationship
between the partial pressures of all gases at equilibrium is:
PCc PDd
=K
PAa PBb
If all reactants and products are in solution, the relationship
between the concentrations of all species at equilibrium is:
[C]c [D]d
=K
[A]a [B]b
Where [X] is the concentration (example molarity) of
species X at equilibrium
Homogenous reactions; reactants and products in the same
phase
Heterogeneous reaction: reactants and products are not
in the same phase
a A(aq) + b B(aq)
[C]c PDd
[A]a [B]b
c C(aq) + d D(g)
=K
K is a dimensionless quantity.
2A(g)
B(g)
K = PB/P2A
is actually
K=
PB/Pref
(PA/Pref)2
Pref is set to 1 atm
K is dimensionless
For solutions, if concentration is M; [A]ref = 1M
HCl(aq)
H+(aq) + Cl-(aq)
[H+] [Cl-]
=K
[HCl]
CH3COOH(aq)
K ~ 107 at 25oC
H+(aq) + CH3COO-(aq)
[H+] [CH3COO-]
=K
[CH3COOH]
K ~ 10-5 at 25oC
NH3(g)
H2(g) + N2(g)
The reaction of SO2(g) and O2(g) forming SO3(g)
2 SO2(g)+ O2(g)
P2SO3
2
2SO3(g)
=K
PSO2 PO2
Equilibrium can be reached for different partial pressures of
SO2, O2, and SO3, depending on the starting conditions, but
at 25oC, the value of K is the same.
The Magnitude of the Equilibrium Constant
The magnitude of the equilibrium constant reveals the extent
to which the reaction will proceed in the desired direction.
aA+bB
cC+dD
Reactions that have K values > 1 are favored in the direction
written; i.e. forward direction.
Reactions that have K values < 1 are favored in the reverse
direction
Reactions for which K is near I have substantial amounts of
both reactants and products when equilibrium is established.
Applying the Equilibrium Expression to Gas Phase Reactions
PCl5(g)
PCl3(g) + Cl2 (g)
K = 2.15
What are the equilibrium partial pressures of all three gases
in a closed container containing only PCl5 at 0.100 atm and
held at 250oC?
According to the ideal gas laws, the partial pressures of
gases is proportional to the number of moles of each gas, as
long as the volume and temperature are kept fixed.
The stoichiometry of this reaction is 1 : 1 : 1
If the partial pressure of PCl5 decreases by x at equilibrium,
the partial pressures of PCl3 and Cl2 increases by x at
equilibrium.
PCl5(g)
PCl3(g) + Cl2 (g)
Initial P (atm)
0.100
0
0
Change in P (atm)
-x
x
x
Equilibrium P (atm)
0.100-x
x
x
At equilibrium:
PPCl3 PCl2
PPCl5
=K
(x) (x)
= 2.15
(0.100-x)
x2 = 2.15 (0.100-x)
x2 + 2.15x - 0.215 = 0
This is a quadratic equation of the form
ax2+ bx + c = 0
and the solution of this equation is of the form
x = (-b ± √b2 - 4ac)
2a
Using this expression and solving for x, the roots of the
equation are x = 0.0957 and -2.25 atm.
At equilibrium, the partial pressures of Cl2 and PCl3 are
0.0957 atm, and that of PCl5 is (0.100 - 0.0957) = 0.004atm
4 NO2(g)
N2O(g) + 3 O2 (g)
The three gases are introduced into a container at partial
pressures of 3.6 atm NO2, 5.1 atm N2O and 8.0atm O2 and
react to reach equilibrium at a fixed temperature. The
equilibrium partial pressure of NO2 is measured to be 2.4 atm.
Calculate the equilibrium constant of the reaction at this
temperature, assuming that no competing reactions occur.
4 NO2(g)
2 N2O(g) + 3 O2 (g)
Initial P (atm)
3.6
5.1
8.0
Change in P (atm)
-x
2x/4
3x/4
Change in P (atm)
- 4x
2x
3x
Equilibrium P (atm)
2.4
5.1+2x
8.0+3x
At equilibrium, the partial pressure of NO2 is 2.4 atm
3.6 - 4x = 2.4 => x = 0.3 atm
Hence PN2O at equilibrium = 5.7 atm; PO2 = 8.9 atm
K = (5.7)2 (8.9)3
2
=
6.9
x
10
(2.4)4
In applying the equilibrium expression the following must be
considered.
1) The equilibrium constant for a reverse reaction is the
reciprocal of the equilibrium constant for the corresponding
forward reaction.
2 H2(g) + O2 (g)
2H2O(g)
2
PH2O
= K1
2
PH2 PO2
2H2O(g)
2 H2(g) + O2 (g)
PH2 2 PO2
2
PH2O
-1
K1 = K2
= K2
2) When the coefficients in a balanced chemical equation are
multiplied by a constant factor, the corresponding
equilibrium constant is raised to the power equal to that
factor.
H2(g) +
1
2
O2 (g)
PH2O
1/2
PH2 PO2
2 H2(g) + O2 (g)
H2O(g)
= K2
2H2O(g)
2
PH2O
2
PH2 PO2
K2 = √K1
= K1
3) When chemical equations are added or subtracted to
obtain a net equation, the corresponding equilibrium
constants are multiplied or divided to obtain the equilibrium
constant of the net equation.
2 BrCl(g)
Cl2 (g) + Br2(g)
PCl2 PBr2
2
= K1 = 0.45 at 25oC
PBrCl
Br2(g) + I2 (g)
2 IBr(g)
2
PIBr
PBr2 PI2
= K2 = 0.051 at 25oC
Adding the two chemical equations gives:
2 BrCl(g) + Br2 (g) + I2 (g)
2 BrCl(g) + I2 (g)
2
PCl2 PIBr
2
2 IBr(g) + Cl2 (g) + Br2(g)
2 IBr(g) + Cl2 (g)
= K3
PBrCl PI2
Looking at the expressions for K1, K2 and K3
K1 K2 = K3
Hence, K3 = 0.023 at 25oC
The Ideal Gas Equation and Chemical Equilibrium
For gaseous reactants or products, the concentration may
be in moles/liter.
The concentrations of the gases in moles/liter must be
converted to partial pressures.
Concentration of a species A in moles/lit = [A]
[A] = n = PA
V
RT
This equation can be used to convert concentration of a
gas in moles/lit to partial pressure of the gas.
Hence, for a general gas phase reaction
a A(g) + b B(g)
[C]c [D]d
[A]a [B]b
c C(g) + d D(g)
(a+b-c-d)
R
T
=K (
)
Pref
where Pref is the reference pressure = 1 atm and ensures
that K is unitless.
Reaction Quotient
For the general reaction:
a A(g) + b B(g)
c C(g) + d D(g)
Define the reaction quotient, Q:
PCc PDd
=Q
PAa PBb
where P is the partial pressure of a species at any point in
time.
a A(g) + b B(g)
c C(g) + d D(g)
If Q = K, the reaction is at equilibrium
If Q ≠ K, the reaction is not at equilibrium.
If Q > K , reaction proceeds from right to left
If Q < K, reaction proceeds from left to right
The equilibrium constant for the reaction
CH4(g) + H2O(g)
CO(g) + 3 H2(g)
Equals 0.172 at 900K. The concentrations of H2(g), CO(g),
and H2O(g) in an equilibrium mixture of gases all equal
0.00642 mol/L. Calculate the concentration of CH4(g) in the
mixture, assuming that this is the only reaction taking place.
[CO] [H2 ]3
-2
R
T
=K (
)
[CH4] [H2O]
Pref
[0.00642 mol/L] [0.00642 mol/L ]3
[CH4] [0.00642 mol/L]
= 0.172 [(0.08206 L atm mol-1) 900 K]-2
[CH4] = 0.00839 mol/L
What happens if a system at equilibrium undergoes a
change in conditions?
The tendency of a system to achieve equilibrium is
spontaneous.
Once a system is at equilibrium it will remain at equilibrium.
However, if conditions change the system will respond to
this change in a way to achieve equilibrium again.
Note: the concentrations of species when equilibrium is reestablished need not be the same as the ones established at
the previous equilibrium.
N2(g) + 3H2(g)
2NH3(g)
LeChatelier’s Principle
If a stress is applied to a system at equilibrium, the system
tends to react so that the stress is minimized
a A(g) + b B(g)
c C(g) + d D(g)
PCc PDd
=Q
PAa PBb
A) Changing the concentration of a reactant or product
If reactant added: Q < K, reaction proceeds from left to right
If product added: Q > K, reaction proceeds from right to left
If a product is removed from the equilibrium mixture, Q also
decreases, and the reaction once again proceeds to the right
to increase the concentration of the product.
N2(g) + 3H2(g)
2NH3(g)
B) Changing the Volume
Decreasing the volume, increases the total pressure of the
reaction mixture.
The reaction will then proceed in the direction which reduces
the total pressure.
2 NO2(g)
N2O4(g)
If the volume is decreased, the above reaction will move to
the right, to decrease the total number of molecules.
C) Changing Temperature
The effect of changing the temperature of a system at
equilibrium depends on whether the reaction proceeds by
absorbing energy (endothermic) or by releasing energy
(exothermic).
An endothermic reaction lowers the temperature of the system
and an exothermic reaction raises the temperature of the
system.
a A(g) + b B(g)
c C(g) + d D(g) + heat
Forward reaction is exothermic; reverse reaction is
endothermic.
If a reaction is exothermic, raising the temperature causes
the equilibrium to shift to the left.
Heterogeneous Reactions
CaCO3(s) + SO3 (g)
[CaSO3] PCO2
[CaCO3] PSO3
CaSO3 (s) + CO2(g)
=K
In all equilibrium expressions, the concentrations of all
pure solids and liquids are set to 1.
Hence,
PCO2
PSO3
=K
In general, to write the equilibrium expression for a reaction
1) Concentration of gases are expressed as partial pressures
2) Concentration of dissolved species in solution are
expressed as moles/liter
3) Concentrations of pure solids and pure liquids are set to 1
(for a solvent taking part in a reaction, its concentration is
also set to 1 providing the solution is dilute)
Problem: The reaction between Ni(s) and CO(g) to form
Ni(CO)4(g) is as follows:
Ni(s) + 4CO(g)
Ni(CO)4(g)
A quantity of Ni(s) is added to a vessel containing CO at a
partial pressure of 1.282 atm and 354 K. At the equilibrium
point of this reaction, the partial pressure of CO is 0.709
atm. Calculate the equilibrium constant of this reaction at
354 K.
PNiCO4
K=
4
PCO
Ni(s) + 4CO(g)
Ni(CO)4(g)
Initial P (atm)
1.282
0
Change in P (atm)
0.573
0.143
Equilibrium P (atm)
0.709
0.143
K=
PNiCO4
4
PCO
0.143
K=
(0.709)4
= 0.567
Applications of Chemical Equilibria
Product yields can be increased by adjusting conditions
N2(g) + 3H2(g)
2NH3(g) + heat
Gas phase reaction: 4 moles of gas --> 2 moles of gas
Exothermic reaction
Reaction conditions that favor NH3(g) production
high pressure (~ 250 atm)
low temperature (use a catalyst)
Hemoglobin/O2 Equilibrium
Hemoglobin (Hb) carries oxygen from the lungs to the body
tissue, transporting oxygen from a region of high
concentration to low concentration.
The oxygen-hemoglobin complex, oxyhemoglobin (HBO2)
transports O2
Hb(aq) + O2(g)
HbO2 (aq)
[HbO2]
KO2=
[Hb]PO2
Level of O2 in blood is increased by ~ 70 times because of
hemoglobin
Because of the formation of the HbO2 complex, the amount of
O2 in blood is increased by a factor of 70.
Hb(aq) + O2(g)
HbO2 (aq)
LeChatelier’s principle predicts that in regions of high O2
partial pressure, the Hb-HbO2 equilibrium is shifted to the
right, which is the case in the lungs
In regions of low O2 partial pressure, the equilibrium shifts to
the left, resulting in a breakup of the HbO2 complex, releasing
O2 to the body’s tissues.
Why is CO lethal?
Hb(aq) + CO(g)
KCO=
HbCO (aq)
[HbCO]
[Hb]PCO
[HbO2]
KO2=
[Hb]PO2
When Hb is exposed to both O2 and CO, there is competition
for the Hb, and the following reaction takes place:
HbO2 (aq) + CO(g)
HbCO(aq) + O2 (g)
The Kcompetition for this reaction is:
Kcompetition =
[HbCO]
=
[HbO2]
[HbCO] PO2
[HbO2] PCO
KCO
=
KO2
PCO
KCO
K
competition =
PO2
KO2
HbO2 (aq) + CO(g)
HbCO(aq) + O2 (g)
Since KCO > KO2 Kcompetition is >1
At 38oC, the value of Kcompetition is 210 strongly favoring the
formation of the HbCO complex
CO displaces O2 from the HbO2 complex, resulting in
asphyxiation.
The process is reversible - from LeChatelier’s principle, a
large partial pressure of O2 will shift the reaction above from
right to left.
Extraction and Separation
A solute dissolved in a solvent A can be extracted using
another solvent B.
Condition: The solute must dissolve in both solvents A & B
and solvent B must be immiscible with the solvent A.
CCl4 and H2O are immiscible.
I2(s) dissolves in both solvents.
If to an aqueous solution of I2(aq) CCl4 is added, and the
flask shaken, some of the I2 in the water layer will be
extracted into the CCl4 layer
The following equilibrium is established:
I2(aq)
K=
I2(CCl4)
[I2]CCl4
[I2]H2O
H2O layer
K: partition coefficient
CCl4 layer
Chromatography
kA =
time in mobile phase
time in stationary phase
time in stationary phase
signal
kB =
time in mobile phase
time