1. Chemical Equilibrium
Chapter 15

2. Homework problems
pages
1, 4, 5, 6, 8, 12, 15, 16, 19, 20,
28, 30, 32, 34, 36 – 40, 42 – 44,
46, 48
EH Unit 17 Due Feb ?
Study Spring 2001 Test #3
1999 Test #2 (1-17) Test 4(26-46)

3. Physical equilibrium Chemical equilibrium
Equilibrium is a state in which 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. N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4
14.1

5. constant
14.1

6. 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
Equilibrium Expression
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
14.1

7. 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

8. Homogeneous Equilibrium
(Aqueous solution of acetic acid.)
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

9. Ex 15.1
Write expressions for Kcand Kp for
2 NO(g) + O2(g) NO2(g)
CH3COOH(aq) + CH3OH(aq)
CH3COOCH3(aq) + H2O(l)

10. Our experiment: mix acetic acid and propyl alcohol
CH3COOH + CH3CH2CH2OH
CH3COOCH2CH2CH3 + H2O(l)

11. 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
14.2

12. 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

13. 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

14. 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

15. Consider the following equilibrium at 295 K:
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = P
NH3
H2S P
= x =
14.2

16. ‘ ‘ ‘ ‘ ‘ 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

17. Writing the Reaction Quotient for an Overall Reaction
Problem: Oxygen gas combines with nitrogen gas in the internal
combustion engine to produce nitric oxide, which when out in the
atmosphere combines with additional oxygen to form nitrogen dioxide.
(1) N2 (g) + O2 (g) NO(g) Kc1 = 4.3 x 10-25
(2) 2 NO(g) + O2 (g) NO2 (g) Kc2 = 6.4 x 109
(a) Show that the overall Kc for this reaction sequence is the same as the
product of the Kc’s for the individual reactions.
(b) Calculate Kc for the overall reaction.

18. ‘ 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

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. 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

21. [N2O4] = 0.12 M and [NO2] = 0.55 M. Is the
Sample Problem
Problem:
For the reaction
N2O4(g) NO2 (g)
Kc = 0.21 at 100°C. At a point during the reaction
[N2O4] = 0.12 M and [NO2] = 0.55 M. Is the
reaction at equilibrium? If not, in which direction
is it progressing?

22. Predicting Reaction Direction and Calculating
Equilibrium Concentrations
Problem: Two components of natural gas can react according to the
following chemical equation:
CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g)
In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and
2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature,
Kc = (a) In which direction will the reaction go?
(b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of the
other substances?
[H2S] = 8.00 M, [CS2] = 4.00 M
and [H2 ] = 8.00 M
1.00 mol
0.250 L
[CH4] = = M

23. [I2] = 0.0884 M (at equilibrium)
Calculating Kc from Concentration Data
Problem: Hydrogen iodide decomposes at moderate temperatures by the
reaction below:
When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibrium
mixture was found to contain mol I2. What is the value of Kc ?
Set up a reaction table (also called a change table).
2 HI(g) H2 (g) + I2 (g)
[HI] = M (initial)
[I2] = M (at equilibrium)

24. Determining Equilibrium Concentrations from Kc
Problem: One laboratory method of making methane is from carbon
disulfide reacting with hydrogen gas, and Kc this reaction at 900°C is 27.8.
CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g)
At equilibrium the reaction mixture in a 4.70 L flask contains mol
CS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane was
formed?
[CS2] = M
[H2] = M
[H2S] = M

25. 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

26. Determining Equilibrium Concentrations from Initial
Concentrations and Kc
Problem: Given that the reaction to form HF from molecular
hydrogen and fluorine has a Kc of 115 at a certain temperature.
If mol of each component is added to a L flask,
calculate the equilibrium concentrations of each species.
H2 (g) + F2 (g) HF(g)
3.000 mol
[H2] = = M
1.500 L
Kc = = 115
[HF]2
[H2] [F2]
3.000 mol
[F2] = = M
1.500 L
3.000 mol
[HF] = = M
1.500 L

27. 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

28. 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

29. Test on Chapters 14 and 15
Monday June 2

30. 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

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

32. None of these changes affect Kc.
Predicting the Effect of a Change in Concentration on the Position of the Equilibrium
Problem: Carbon will react with water to yield carbon monoxide and
and hydrogen, in a reaction called the water gas reaction that was used
to convert coal into a fuel that can be used by industry.
C(s) + H2O (g) CO(g) + H2 (g)
What happens to:
(a) [CO] if C is added? (c) [H2O] if H2 is added?
(b) [CO] if H2O is added? (d) [H2O] if CO is removed?
None of these changes affect Kc.

33. FeSCN2+(aq) Fe3+(aq) + SCN-(aq)
red pale yellow colorless
What happens when some NaSCN is added?
What happens when some Na2C2O4 is added?
(C2O42- reacts with Fe3+ to form a complex ion.)
1. What was stress?
2. Which direction does rxn shift?
3. How does this relieve stress?
4. How does this explain observations?

34. The Effect of a Change in Pressure (Volume)
Pressure changes are mainly involving gases as liquids and solids
are nearly incompressible. For gases, pressure changes can occur in
three ways:
Changing the concentration of a gaseous component
Adding an inert gas (one that does not take part in the reaction)
Changing the volume of the reaction vessel
An increase in pressure by a decrease in volume favors the net
reaction that decreases the total number of moles of gases.
WHY??
Kc does not change.

35. Consider the following equilibrium systems.
2 PbS(s) + 3 O2(g) PbO(s) + 2 SO2(g)
PCl5(g) PCl3(g) + Cl2(g)
H2(g) + CO2(g) H2O(g) + CO(g)
Predict the direction of the net reaction as a
result of increasing the pressure by decreasing
the volume.

36. The Effect of a Change in Temperature
Only temperature changes will alter the equilibrium constant, and that is
why we always specify the temperature when giving the value of Kc.
When you raise the temperature you are adding heat. In order to use
up some of this heat the reaction will shift in the endothermic direction.
If you have an exothermic reaction, it will shift in the reverse direction
when the temperature is increased. (What about Kc?)
O2 (g) + 2 H2 (g) H2O(g) + Energy = Exothermic
A temperature increase favors the endothermic direction and a
temperature decrease favors the exothermic direction. WHY??
A temperature rise will increase Kc for a system with a + Δ H0rxn
A temperature rise will decrease Kc for a system with a - ΔH0rxn

37. Sample problem
How does an increase in temperature affect the
equilibrium concentration of the underlined
substance and Kc?
ΔH(kJ)
CaO(s) + H2O Ca(OH)2(aq)
CaCO3(s) CaO(s) + CO2(g)
SO2(g) S(s) + O2(g)

38. Predicting the Effect of Temperature and Pressure
Problem: How would you change the volume (pressure) and temperature
in the following reactions to increase the chemical yield of the products?
(a) 2 SO2 (g) + O2 (g) SO3 (g); Ho = 197 kJ
(b) CO(g) + 2 H2 (g) CH3OH(g); Ho = kJ
(c) C(s) + CO2 (g) CO(g); Ho = kJ

39. The Haber process is used to produce
110 million tons of ammonia each year.
N2(g) + 3 H2(g) NH3(g)
ΔH = kJ
What conditions of temperature and pressure
favor the maximum yield of ammonia?

40. Effect of Temperature for Ammonia Synthesis
T (K) Kc
x 1015
x 108
x 104
x 102
x 100
x 10 -1
x 10 -2

41. Percent Yield of Ammonia vs. Temperature (°C)
at five different operating pressures.

42. 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

43. Le Châtelier’s Principle
Changes in Temperature
Change
Exothermic Rx
Endothermic Rx
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
14.5

44. Catalyst lowers Ea for both forward and reverse reactions.
uncatalyzed
catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
14.5

45. 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