1. Chemical Equilibrium
Chapter 16

2. Equilibrium: the extent of a reaction
In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower.
Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce.
Equilibrium looks at the extent of a chemical reaction.

3. Physical equilibrium Chemical equilibrium
Equilibrium is a state in which two opposing processes are occurring at the same rate so there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
the rates of the forward and reverse reactions are equal and
the concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)
14.1

4. Rate of sale of cookies = Rate of replacing cookies

5. The Concept of Equilibrium
Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2:
N2O4(g)  2NO2(g).
At some time, the color stops changing and we have a mixture of N2O4 and NO2.
Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant.
Using the collision model:
as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4.
At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g)  N2O4(g)) does not occur.

6. The Concept of Equilibrium
As the substance warms it begins to decompose:
N2O4(g)  2NO2(g)
When enough NO2 is formed, it can react to form N2O4:
2NO2(g)  N2O4(g).
At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4
The double arrow implies the process is dynamic.

7. The Concept of Equilibrium
As the reaction progresses
[A] decreases to a constant,
[B] increases from zero to a constant.
When [A] and [B] are constant, equilibrium is achieved.

8. N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4
14.1

9. constant
14.1

10. The Equilibrium Constant
No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.
For a general reaction
the equilibrium constant expression is
where Kc is the equilibrium constant.

11. Equilibrium Will N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3
aA + bB cC + dD
K =
[C]c[D]d
[A]a[B]b
Law of Mass Action
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
14.1

12. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g)
Kp =
NO2 P2
N2O4 P
Kc =
[NO2]2
[N2O4]
In most cases
Kc  Kp
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
14.2

13. Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
[CH3COO-][H3O+]
[CH3COOH][H2O]
Kc = ‘
[H2O] = constant
[CH3COO-][H3O+]
[CH3COOH] =
Kc [H2O] ‘
Kc =
General practice not to include units for the equilibrium constant.
14.2

14. The Equilibrium Expression
Write the equilibrium expression for the following reaction:

15. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
[COCl2]
[CO][Cl2] =
0.14
0.012 x 0.054
Kc =
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R =
T = = 347 K
Kp = 220 x ( x 347)-1 = 7.7
14.2

16. The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm?
2NO2 (g) NO (g) + O2 (g) 2
Kp = 2
PNO PO
PNO PO 2
= Kp
PNO PO 2
= 158 x (0.400)2/(0.270)2
= 347 atm
14.2

17. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
Kc = ‘
[CaO][CO2]
[CaCO3]
[CaCO3] = constant
[CaO] = constant
Kc x ‘
[CaCO3]
[CaO]
Kc = [CO2] =
Kp = PCO 2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
14.2

18. does not depend on the amount of CaCO3 or CaO
CaCO3 (s) CaO (s) + CO2 (g)
PCO 2
= Kp
PCO 2
does not depend on the amount of CaCO3 or CaO
14.2

19. 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

20. 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.
14.4

21. ICE At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.
Br2 (g) Br (g)
Let x be the change in concentration of Br2
Br2 (g) Br (g)
Initial (M)
0.063
0.012
ICE
Change (M) -x
+2x
Equilibrium (M) x x
[Br]2
[Br2]
Kc =
Kc =
( x)2 x
= 1.1 x 10-3
Solve for x
14.4

22. Kc =
( x)2 x
= 1.1 x 10-3
4x x = – x
4x x = 0
-b ± b2 – 4ac  2a
x =
ax2 + bx + c =0
x =
x =
Br2 (g) Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063
0.012 -x
+2x x x
At equilibrium, [Br] = x = M
or M
At equilibrium, [Br2] = – x = M
14.4

23. Example Problem: Calculate Concentration
Note the moles into a L vessel stuff ... calculate molarity.
Starting concentration of HI: 2.5 mol/10.32 L = M
2 HI H I2
Initial:
Change:
Equil:
0.242 M 0 0
-2x +x +x
x x x
What we are asked for here is the equilibrium concentration of H2 ...
... otherwise known as x. So, we need to solve this beast for x.

24. Example Problem: Calculate Concentration
And yes, it’s a quadratic equation. Doing a bit of rearranging:
x = or –
Since we are using this to
model a real, physical system,
we reject the negative root.
The [H2] at equil. is M.

25. Example Problem: Calculate Keq
This type of problem is typically tackled using the “three line” approach:
2 NO + O2 2 NO2
Initial:
Change:
Equilibrium:

26. Approximating
If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants.
0.20 – x is just about 0.20 is x is really dinky.
If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

27. Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I
Example
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I I
More than 3
orders of mag.
between these
numbers. The
simplification will
work here.
Initial
change
equil:
-x x
0.20-x x
With an equilibrium constant that small, whatever x is, it’s near
dink, and 0.20 minus dink is 0.20 (like a million dollars minus a
nickel is still a million dollars).
0.20 – x is the same as 0.20
x = 3.83 x 10-6 M

28. Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I
Example
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I I
These are too close to
each other ...
0.20-x will not be
trivially close to 0.20
here.
Initial
change
equil:
-x x
0.20-x x
Looks like this one has to proceed through the quadratic ...

29. ‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] Kc A + B C + D Kc
C + D E + F
[E][F]
[A][B]
Kc =
A + B E + F Kc
Kc = Kc ‘ x
If a reaction can be expressed as the 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

30. ‘ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4]
= 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
14.2

31. Qc > Kc system proceeds from right to left to reach equilibrium
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
14.4

32. Le Châtelier’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.
Changes in Concentration
N2 (g) + 3H2 (g) NH3 (g)
Add
NH3
Equilibrium shifts left to offset stress
14.5

33. Le Châtelier’s Principle
Changes in Concentration continued
Remove
Add
Add
Remove
aA + bB cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
left
Decrease concentration of product(s)
right
Increase concentration of reactant(s)
right
Decrease concentration of reactant(s)
left
14.5

34. Le Châtelier’s Principle
Changes in Volume and Pressure
A (g) + B (g) C (g)
Change
Shifts the Equilibrium
Increase pressure
Side with fewest moles of gas
Decrease pressure
Side with most moles of gas
Increase volume
Side with most moles of gas
Decrease volume
Side with fewest moles of gas
14.5

35. Le Châtelier’s Principle
Changes in Temperature
Change
Exothermic Rx
Endothermic Rx
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
colder
hotter
14.5

36. Le Châtelier’s Principle
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
uncatalyzed
catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
14.5

37. Example

38. Life at High Altitudes and Hemoglobin Production
Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Hb (aq) + O2 (aq) HbO2 (aq)
Kc =
[HbO2]
[Hb][O2]

39. Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) NH3 (g) DH0 = kJ/mol

40. Le Châtelier’s Principle
Change Equilibrium
Constant
Change
Shift Equilibrium
Concentration
yes no
Pressure
yes no
Volume
yes no
Temperature
yes
yes
Catalyst no no
14.5