Chemical Equilibrium: Ch. 15
15-1 Dynamic Equilibrium
15-2 The Equilibrium Constant Expression
15-3 Relationships Involving Equilibrium
Constants
15-4 The Magnitude of an Equilibrium Constant
15-5 The Reaction Quotient, Q: Predicting the
Direction of a Net Change
15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
15-7 Equilibrium Calculations: Some Illustrative
Examples
15-1 Dynamic Equilibrium
• Equilibrium – two
opposing processes
taking place at equal
rates.
H2O(l)
NaCl(s)
Prepared
Equilibrium
I2(H2O)
I2(H2O)
I2(CCl4)
I2(CCl4)
H2O(g)
H2O
NaCl(aq)
CO(g) + 2 H2(g)
I2(H2O)
Dynamic Equilibrium
I2(CCl4)
CH3OH(g)
Reversible reactions: Both forward and
reverse reactions take place
15-2 The Equilibrium Constant Expression
• Methanol synthesis is a reversible
reaction.
CO(g) + 2 H2(g)
CH3OH(g)
k-1
CO(g) + 2 H2(g)
k1
CH3OH(g)
Three Approaches to Equilibrium
H2(g) + 2 ICl(g)
I2(g) + 2 HCl(g)
ICl
CO(g) + 2 H2(g)
k1
k-1
CH3OH(g)
As is Ammonia production (Haber-Bosch process)
N2(g) + 3H2(g)
2NH3(g)
H2
I2
The Equilibrium Constant: A kinetic perspective
H2(g) + 2 ICl(g)
d[P]
I2(g) + 2 HCl(g)
dt
Mechanism:
= k[H2][ICl]
Briggs Raucher reaction:Oscillating
reactions
At Equilibrium
Step 1
H2(g) + ICl(g)
HI(g) + HCl(g)
k1[H2][ICl] = k-1[HI][HCl]
Step 2
HI(g) + ICl(g)
I2(g) + HCl(g)
step1
[ HI ] =
k2[HI][ICl] = k-2[I2][HCl]
k1 [ H 2 ][ ICl ] step 2 k − 2 [ I 2 ][ HCl ]
=
k −1 [ HCl ]
k 2 [ ICl ]
kk
[ I 2 ][ HCl ]2
= 1 2 = Kc
2
k −1k − 2
[ H 2 ][ ICl ]
So,
The Equilibrium Constant and Activities; a
General Expressions
• Activity
a A + b B …. → g G + h H ….
– Thermodynamic concept introduced by Lewis.
– Dimensionless ratio referred to a chosen
reference state.
aB = γ B
[ B]
cB0
coefficient of activity
and cB0 is a standard reference state
= 1 mol L-1
[G]g[H]h ….
Equilibrium constant = Kc= [A]a[B]b ….
Thermodynamic
(a )g(a )h ….
Equilibrium constant = Keq= G a H b ….
(aA) (aB)
=
[G]g[H]h …. ( γG)g(γH)h ….
(a+b…)-(g+h…)
×
×
[A]a[B]b …. (γA)a(γB)b …. (co)
•γB accounts for non-ideal behaviour in solutions and gases.
•γB[B] can be considered an effective or active concentration.
15-3 Relationships Involving
the Equilibrium Constant
• When 2 chemical equations (with K1 and K2) are
added, the overall equilibrium constant is K1×K2
• Reversing an equation causes inversion of K.
• Multiplying all coefficients by a common factor
raises the equilibrium constant to the
corresponding power.
• Dividing the coefficients by a common factor
causes the equilibrium constant to be taken to that
root.
≈1
under ideal conditions
= 1 mol L-1
Combining Equilibrium Constant
Expressions
N2O(g) + ½O2
N2(g) + ½O2
N2(g) + O2
2 NO(g)
N2O(g)
2 NO(g)
Kc= ?
[N2O]
Kc(2)= 2.7×10-18 =[N ][O ]½
2
2
2
[NO]
Kc(3)= 4.7×10-31 =[N ][O ]
2
2
[NO]2
1
[NO]2 [N2][O2]½
Kc= [N O][O ]½ =
= Kc(3) K
[N
][O
]
[N
O]
2
2
c(2)
2
2
2
= 1.7×10-13
Gases: The Equilibrium Constant, KP
• Mixtures of gases are solutions just
as liquids are.
• Define KP, based on partial pressure
(or activities) of gasses, rather than
concentration
2 SO2(g) + O2(g)
aSO3 =
PSO3
P°
KP =
2 SO3(g)
aSO2 =
KP =
PSO2
(PSO3 )2
2 SO2(g) + O2(g)
=
(aSO2 )2(aO 2)
PO 2
P°
= P°
RT
P°
2 SO2(g) + O2(g)
2 SO3(g)
PSO3 2
2
2
PSO3
]
[SO
RT
3
=
Kc =
RT
=
2
[SO2]2[O2] PSO 2 PO
P
P
SO
O
2
2
2
2
RT RT
Kc = KP(RT)
;PX= n_xRT = [X] RT
V
([SO3] RT)2
P°
([SO2] RT)2([O2] RT)
[SO3]2
[SO2]2[O2]
=
KC
RT
Where P° = 1 bar
(PSO2 )2(PO2)
Gases: The Equilibrium Constant, KC
2 SO3(g)
(PSO3)2
(a )2
KP = SO3
= P°
2
(PSO2)2(PO 2)
(aSO2) (aO 2)
(aSO3 )2
aSO3 =
P°
Gases: The Equilibrium Constant, KC
Heterogeneous Reactions:
Pure Liquids and Solids
• Equilibrium constant expressions do not
contain concentration terms for solid or
liquid phases of a single component
(that is, pure solids or liquids).
C(s) + H2O(g)
CO(g) + H2(g)
KP = Kc(RT)-1
In general:
KP = Kc(RT)Δn
15-4 The Significance of the
Magnitude of the Equilibrium Constant.
Kc =
P P
[CO][H2]
= CO H2 (RT)1
[H2O]
PH2O
15-4 The Significance of the
Magnitude of the Equilibrium Constant.
• Does a large equilibrium constant ensure that a
reaction will take place if only reactants are
present?
– No!! The rate can be so low that no product is formed
Significant quantities of both reactants and
products are likely to be present at equilibrium
only if ~10-10 < K < ~1010
• Thermodynamic control: concentration of
products and reactants depends on relative stability
of the species.
• Kinetic control: concentration of products and
reactants depends rates of reaction (i.e. k and Ea)
15-5 The Reaction Quotient, Q:
Predicting the Direction of Net Change.
CO(g) + 2 H2(g)
k1
k-1
Three Approaches to Equilibrium
CO(g) + 2 H2(g)
[G]tg[H]th…
[A]ta[B]tb…
k-1
CH3OH(g)
CH3OH(g)
H2
• Equilibrium can be approached various ways.
• What changes in concentration do we expect
as equilibrium is approached from the
conditions prepared at some time, t?
• The reaction quotient helps us predict:
Qc =
k1
CO
CH3OH
At equilibrium Qc = Kc
Reaction Quotient
15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
• When an equilibrium system is subjected
to a change in temperature, pressure, or
the concentration of a reacting species,
the system responds by attaining a new
equilibrium that partially offsets the impact
of the change.
What happens if we add some ‘extra’ SO3
to the reaction system at equilibrium?
Le Châtelier’s Principle
2 SO2(g) + O2(g)
k1
k-1
2 SO3(g)
Kc =
2.8×102
at 1000K
Effect of Changes in ‘P’ or ‘V’
• Adding a gaseous reactant or product changes
Pgas. Equivalent to a change in concentration
• Adding an inert gas changes the total pressure,
but relative partial pressures are
unchanged.
[SO3]2
Q=
[SO2]2[O2] = Kc
Q > Kc
• Changing the volume of the system causes a
change in the equilibrium position.
nSO32
2
V
nSO3
Kc = [SO3] =
=
V
[SO2]2[O2] nSO22 nO2 n 2n
SO2 O2
V V
Effect of Change in Volume
g
h
g
h
Kc = [G] [H] = nG nH V(a+b)-(g+h)
[A]a[B]b
nAa nBa
2 SO2(g) + O2(g)
=
nAa
naB
V-Δn
• When the volume of an equilibrium mixture
of gases is reduced, a net change occurs in
the direction that produces fewer moles of
gas. When the volume is increased, a net
change occurs in the direction that produces
more moles of gas.
Effect of a Catalyst on Equilibrium
• A catalyst changes the reaction
mechanism to one with a lower activation
energy.
• A catalyst has no effect on the condition of
equilibrium: the relative thermodynamic
stability of reactants and products in
unchanged.
A catalyst does affect the rate at
which equilibrium is attained.
Synthesis of
Ammonia
Haber-Bosch process:
N2(g) + 3H2(g)
k1
k-1
2 SO3(g)
ΔH° = -197.8 kJ/mol
• Raising the temperature of an equilibrium
mixture shifts the equilibrium condition in
the direction of the endothermic reaction.
g
nG nHh
Effect of Temperature on Equilibrium
2NH3(g)
The optimum conditions are
only for the equilibrium
position and do not take into
account the rate at which
equilibrium is attained.
•Lowering the temperature causes a shift in
the direction of the exothermic reaction.
What does this behaviour say about the
temperature dependence of k1 compared to k-1?
15-7 Equilibrium Calculations:
Some Illustrative Examples.
• Five numerical examples are given in the
text that illustrate ideas that have been
presented in this chapter.
• Refer to the “comments” which describe
the methodology. These will help in
subsequent chapters.
• Exercise your understanding by working
through the examples with a pencil and
paper.
Problem Set: Ch15
7,9,13,21,23,27,36,47,51,55,56,69,79,83