Chapter 14
Chemical
Equilibrium
Hill, Petrucci, McCreary & Perry 4th Ed.
Chemical Equilibrium
•Many Reactions seem to STOP before all
the reactants are used up.
•The Concentrations of Reactants and
Products reach constant values.
•A State of Chemical Equilibrium is said to
have been established, but nothing stopped!
•Chemical Equilibrium is a Dynamic State
RateForward Reaction = RateReverse Reaction
The Hydrogen Iodide Decomposition
2 HI(g)
H2(g) + I2(g)
The equilibrium
constant is actually
the ratio of the rate
constants for the
forward and reverse
reactions!
Kc =
[H2][I2]
[HI]2
The exponents here
are the same as the
coefficients in the
chemical equation.
Only this form of the equilibrium expression gives a constant ratio.
Kc = 1.84 x 10-2 at 698 K
This value of the equilibrium constant is valid only at 698 K!
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The Kinetic View of Equilibrium
H2(g) + I2(g)
2 HI(g)
Rate of the forward reaction:
Rate = kf [HI]2
Rate of the reverse reaction:
Rate = kr [H2][I2]
At Equilibrium the rates are equal:
kf [HI]2 = kr [H2][I2]
kf
kr
=
[H2][I2]
Kc =
[HI]2
The Value of an Equilibrium Constant
2 HI(g)
H2(g) + I2(g)
Kc =
[H2][I2]
[HI]2
= 1.84 x 10-2 at 698 K
A value for any one concentration may be calculated
if the others are known:
[HI]2 =
[HI] =
[H2][I2]
1.84 x 10-2
[H2][I2]
1
2
1.84 x 10-2
General Equation for Equilibria
For the General Equation:
a A + b B + ...
g G + h H + ...
The Equilibrium Expression will have the form:
Kc =
[G]g[H]h ...
[A]a[B]b ...
The Equilibrium Expression will always have the form:
Kc =
[products]n
[reactants]m
2
Calculating an Equilibrium Constant
Need:
•Balanced Chemical Equation
•Initial Concentrations of all species
•One Equilibrium Concentration
Alternatively the initial moles of all species
and one equilibrium mole plus the volume
of the container Ù Concentrations
Decomposition of Hydrogen Iodide at 731 K
4.00 moles of HI (g) in a 5.00 L container was decomposed
at 731 K. The amount of I 2(g) at equilibrium was found
experimentally to be 0.442 mole. Calculate value of K c.
balanced chemical equation:
2 HI(g)
H2(g) + I2(g)
from the stoichiometry:
-2x
Initial concentrations from data:
[HI]o =
+x
+x
4.00 mol
= 0.800 M [H2]o = [I2]o = 0
5.00 L
[I2]eq =
0.442 mol
= 0.0884 M
5.00 L
Can You Solve This Problem?
balanced chemical equation:
[initial]
from the stoichiometry:
[equilibrium]
2 HI(g)
H2(g)
0.800 M
0M
0M
-2x
+x
+x
+
I2(g)
0.0884 M
3
Variations of Equilibrium Expressions
balanced chemical equation:
H2(g)
2 HI(g)
Kc =
[H2][I2]
[HI]2
+
I2(g)
= 0.0201
If the equation is reversed;
H2(g)
K'c =
+
I2(g)
[HI]2
=
[H2][I2]
2 HI(g)
1
= 49.7
0.0201
K'c is the reciprocal of Kc
Equilibrium Expressions other than Kc
balanced chemical equation:
Kc =
[H2][I2]
[HI]2
2 HI(g)
H2(g)
+
I2(g)
= 0.0201
Quantities proportional to moles/liter can be substituted for Molarity:
From PV = nRT the ideal gas equation
n
= P
M =
V
RT
PH2 PI2
PH2 P I
2
[H2][I2]
RT RT
= Kp
=
=
Kc =
2
2
P
[HI]
HI
P HI 2
RT
Kp = Kc only when # moles reactants = # moles products.
Kp = Kc(RT)∆n
Decomposition of PCl5
PCl3(g) +
PCl5(g)
Cl2(g)
Kc = 3.26 x 10 -2 at 191 o; Calculate K p
Kp = Kc (RT)∆n
∆n = # moles product - # moles reactant
∆n = 2 - 1 = 1
TK = 191 + 273 = 464 K
Kp = Kc (RT)∆n
Kp = (3.26 x 10-2 M) [ (0.0821 L atm )( 464 K)]1
K mol
Kp = 1.24 atm
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Value of Kc & The Extent of Reaction
Kc >> 1 Reaction goes to completion!
Kc << 1 Very little Product Forms!
Kc =
[products]n
[reactants]m
Value of Kc for Reactions Added Together
Kctotal = Kc1 x Kc2 x …
Homogeneous versus Heterogeneous
Equilibria
•Homogeneous Equilibria – All species
have the same phase, all gas, liquid etc.
•Heterogeneous Equilibria – Some species
have different phases - Only the
Principal Phase, usually the most mobile
is in the Kc expression.
3 Fe (s) + 4 H 2O (g)
Kc =
Fe 3O 4(s) + 4 H 2(g)
[H 2]4
[H 2O] 4
Can You Write Equilibrium Expressions
for the Following Reactions?
CaCO3(s)
H2O(l)
C(s) + H2O(g)
CaO(s) + CO2(g)
H2O(g)
CO(g) + H2(g)
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The Reaction Quotient, Q
Predicting the Direction of Net Change
2 HI(g)
H2(g) + I2(g)
For any real system at equilibrium:
Kc =
[H2][I2]
[HI]2
= 1.84 x 10-2 at 698 K
For any real system NOT at equilibrium:
Qc =
[H2][I2]
[HI]2
= Any value
If Qc > Kc the reaction will go toward more reactants
If Qc < Kc the reaction will go toward more products
Le Chatelier’s Principle
•When a change (concentration, temperature,
pressure or volume) is imposed on a system at
equilibrium, the system responds by attaining a new
equilibrium condition that minimizes the impact of
the imposed change.
•Adding reactants will cause more product to form, and
conversely.
•Removing Products will cause more product to form,
and conversely.
•Increasing temperature will cause a shift in the
endothermic direction, and conversely.
Le Chatelier’s Principle
2 NH3(g)
N2(g) + 3 H2(g)
At equilibrium:
Kc =
[NH 3 ] 2
[N 2 ][H 2 ] 3
New equilibrium:
[NH3]2
= Kc
[N2] [H2]3
Change:
Add hydrogen!
[NH3]2
= Qc < K
c
[N2] [H2]3
The reaction will react to reduce the
excess H 2 by combining it with N 2
to produce more NH 3. From the new
concentrations N 2 & H 2 will decrease
and NH 3 will increase until Q c = K c.
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Le Chatelier’s Principle
2 NH3(g)
N2(g) + 3 H2(g)
At equilibrium:
Kc =
[NH 3]2
[N 2][H 2]3
The reaction will react to produce
more N 2 by decomposing NH 3 ,
the concentrations of N 2 & H 2
will increase until Q c = K c.
Change:
Remove Nitrogen!
[NH 3]2
=
[N2] [H 2]3
Qc
> Kc
[NH3]2
= Kc
[N2] [H2]3
New equilibrium:
Le Chatelier’s Principle
2 NH3(g)
N2(g) + 3 H2(g)
At equilibrium:
Kc =
[NH 3]2
[N 2][H 2]3
Change:
Remove ammonia!
The reaction will react to to produce
more NH 3 by combining H 2 with N 2.
The N 2 & H2 concentrations will decrease
and NH 3 will increase until Q c = K c.
New equilibrium:
[NH3]2
=
[N2][H2]3
[NH3]2
Qc
< Kc
= Kc
[N2][H2]3
Le Chatelier’s Principle
2 NH3(g) ∆H < 0 (exothermic)
N2(g) + 3 H2(g)
At equilibrium:
[NH3]2
Kc =
[N2][H2]3
The forward reaction will speed
up less than the endothermic
reverse reaction. This change
actually changes Kc. The new
K'c will be smaller than the old.
Kc =
kf
kr
K'c =
kf
kr
Change:
Increase Temperature!
[NH3]2
= Qc
[N2][H2]3
> K'c
New equilibrium:
[NH3]2
[N2][H2]3
= K'c < Kc
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Changing the Pressure by Compression of Volume
N2O4(g)
2 NO2(g)
∆n = 1
Increasing pressure by
[2x]2
[x]2
[NO2]2
compression causes the
= [y] < [2y]
Kc =
numerator to become
[N2O4]
larger than denominator.
The position of equilibrium will shift to the side having fewer molecules.
H2(g) + I2(g)
2 HI(g)
Kc =
[H2][I2]
[HI]2
=
[x][x]
[y]2
=
∆n = 0
[2x][2x]
[2y]2
Increasing pressure by compressing the volume has no effect on the
ratio, numerator and denominator increase to the same extent. There
will be no shift in the position of equilibrium when ∆n = 0.
Optimum Conditions
2 NH3(g) ∆H < 0 (exothermic)
N2(g) + 3 H2(g)
4 moles reactants
]2
[NH3
Kc =
[N2][H2]3
2 moles products
∆n = -2
Increasing pressure shifts to right, products.
Decreasing pressure shifts to left, reactants.
Increasing temperature shifts to left, reactants.
Decreasing temperature shifts to right, products.
Optimum Conditions:
highest feasible pressure
lowest feasible temperature
The converse of the equilibrium shifts also apply.
See the last two problems for the pressure effect.
Effect of a Catalyst on Equilibria
•A catalyst changes the mechanism and
lowers activation energies. Both forward
and reverse reactions are affected precisely
the same amount.
•A catalyst will not change an equilibrium
composition
•A catalyst will change the rate that any
equilibrium is achieved.
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Using Kc to Calculate Equilibrium
Concentrations
Need:
•Balanced Chemical Equation -- Or –
the Kc Expression.
•Value of Kc at a Specified Temperature.
•All Initial Concentrations -- Or –
All but one Equilibrium Concentrations
Calculating Equilibrium Concentrations
Calculate the equilibrium concentrations for the
following reaction given the initial concentrations.
H2(g)
Kc' =
+
I2(g)
2 HI(g)
[HI]2
= 49.7 at 731 C
[H2][I2]
[H2]
[initial]
[I2]
0.500 M
[HI]
0.500 M
0.0 M
Calculating Equilibrium Concentrations
Calculate the equilibrium concentrations for
the reaction below given all but one [equil].
PCl5(g)
Kc =
PCl3(g) +
[PCl3] [Cl2]
[PCl5]
Cl2(g)
= 0.415 at 250 o
Calculate [PCl5]eq
Given that [PCl 3] = [Cl 2] = 0.020 M
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Which way will this Reaction Go?
A 50.0 L reaction vessel contains 1.00 mol N2; 3.00 mol H2
and 0.500 mol NH3. If Kc = 0.500 at 400EC, determine the
direction of reaction. Note: Ptotal = ~5 atm.
N2(g) + 3 H2(g)
2 NH3(g)
•First we need to calculate the concentrations of each
reactant and product.
•Then we need to write the equilibrium expression.
•Then we will set the equilibrium expression equal to Qc
since the system is not at equilibrium and calculate Qc
from the values.
•From the value of Qc we shall determine the direction
the reaction will go.
Which way will this Reaction Go?
A 5.0 L reaction vessel contains 1.00 mol N2; 3.00 mol H2
and 0.500 mol NH3. If Kc = 0.500 at 400EC, determine the
direction of reaction. Note: Ptotal = ~50 atm.
N2(g) + 3 H2(g)
2 NH3(g)
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