Chapter 6_Chemical-Equilibrium_Le Chateliers Principle-1.pdf

1
Chemical Equilibrium
Rates of Reactions
General, Organic, and Biological Chemistry

2
Collision Theory of Reactions
A chemical reaction occurs when
collisions between molecules have sufficient energy to break
the bonds in the reactants
molecules collide with the proper orientation
bonds between atoms of the reactants (N2 and O2) are
broken and new bonds (NO) form
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.

3
Collision Theory of Reactions (continued)
A chemical reaction does not take place if the
collisions between molecules do not have sufficient
energy to break the bonds in the reactants
molecules are not properly aligned

4
Activation Energy
The activation energy
is the minimum energy needed for a reaction to take
place upon proper collision of reactants

5
Reaction Rate and Temperature
Reaction rate
is the speed at which reactant is used up
is the speed at which product forms
increases when temperature rises because
reacting molecules move faster, thereby
providing more colliding molecules with energy of
activation

6
Reaction Rate and Concentration
Increasing the
concentration of reactants
increases the number of
collisions
increases the reaction rate

7
Reaction Rate and Catalysts
A catalyst
• speeds up the rate of a reaction
• lowers the energy of activation
• is not used up during the reaction

8
Factors that Increase Reaction Rate

9
Learning Check 1:
State the effect of each on the rate of reaction as:
I) increases D) decreases N) no change
A. increasing the temperature
B. removing some of the reactants
C. adding a catalyst
D. placing the reaction flask in ice
E. increasing the concentration of a reactant

10
Learning Check 2:
Indicate the effect of each factor listed on the rate of the
following reaction as I) increases, D) decreases, or
N) none:
2CO(g) + O2(g) 2CO2 (g)
A. raising the temperature
B. removing O2
C. adding a catalyst
D. lowering the temperature

11
Reversible Reactions
In a reversible reaction, there are both forward and reverse
reactions.
• Suppose SO2 and O2 are present initially. As they collide, the
forward reaction begins.
2SO2(g) + O2(g) 2SO3(g)
• As SO3 molecules form, they also collide in the reverse reaction
that forms reactants. This reversible reaction is written with a
double arrow.
forward
2SO2(g) + O2(g) 2SO3(g)
reverse

12
At equilibrium,
the rate of the forward reaction
becomes equal to the rate of
the reverse reaction
the forward and reverse
reactions continue at equal
rates in both directions

13
Chemical Equilibrium (continued)
When equilibrium is reached,
there is no further change in the amounts of
reactant and product

14
Equilibrium
At equilibrium,
the forward reaction of N2 and O2 forms NO
the reverse reaction of 2NO forms N2 and O2
the amounts of N2, O2, and NO remain constant
N2(g) + O2(g) 2NO(g)

15
Learning Check 3:
Write the forward and reverse reactions for the following:
CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

16
Learning Check 4:
Complete with
1) equal 2) not equal 3) forward
4) reverse 5) changes 6) does not change
A. Reactants form products in the ________ reaction.
B. At equilibrium, the reactant concentration _______.
C. When products form reactants, it is the _______ reaction.
D. At equilibrium, the rate of the forward reaction is ______ to the rate of
the reverse reaction.
E. If the forward reaction is faster than the reverse, the amount of
products ________.

17
Equilibrium Constants

18
Equilibrium Constants
For the reaction
aA bB
• The equilibrium constant expression, Kc, gives the
concentrations of the reactants and products at equilibrium:
Kc = [B]b = [Products]
[A]a [Reactants]
• The square brackets indicate the moles/liter of each
substance.
• The coefficients b and a are written as superscripts that raise
the moles/liter to a specific power.

19
Guide to Writing the Kc Expression

20
Writing a Kc Expression
Write the Kc expression for the following:
STEP 1 Write the balanced equilibrium equation:
2CO(g) + O2(g) 2CO2(g)
STEP 2 Write the product concentrations in the numerator and the reactant
concentration in the denominator:
Kc = [CO2] [products]
[CO][O2] [reactants]
STEP 3 Write the coefficients as superscripts:
Kc = [CO2]2
[CO]2 [O2]

21
Learning Check 5:
Which of the following is the correctly written Kc expression
for the reaction shown below?
N2(g) + 3Cl2(g) 2NCl3(g)
1) [NCl3] 2) [N2][Cl2]3
[N2][Cl2] [NCl3]2
3) [NCl3]2 4) [NCl3]2
[N2]3 [Cl2] [N2][Cl2]3

22
Heterogeneous Equilibrium
In heterogeneous equilibrium,
• Gases and solid and/or liquid states are part of the
reaction.
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)
• The concentration of solids and liquids is constant.
• The Kc expression is written with only the
compounds that are gases.
Kc = [CO2][H2O]

23
Learning Check 6:
The Kc expression for the following reaction is:
2KClO3(s) 2KCl(s) + 3O2(g)
1) [KCl][O2] 2) [KCl]2[O2]3
[KClO3] [ KClO3]2
3) [KCl]2[O2]3 4) [O2]3

Guide to Calculating the Kc Value
24

25
Example of Calculating Equilibrium Constants
What is the Kc for the following reaction?
H2(g) + I2(g) 2HI(g)
Equilibrium concentrations: [H2] = 1.2 moles/L
[I2] = 1.2 moles/L [HI] = 0.35 mole/L
STEP 1 Write the Kc expression:
Kc = [HI]2
[H2][I2]
STEP 2 Substitute the equilibrium concentrations in Kc:
Kc = [0.35]2 = 8.5 x 10-2
[1.2][1.2]
General, Organic, and Biological Chemistry Copyright © 2010
Pearson Education, Inc.

26
Learning Check 7:
Calculate the Kc for the following reaction:
CH4(g) + H2O(g) CO(g) + 3H2(g)
with the following equilibrium concentrations:
[CH4] = 0.13 M [H2O] = 0.53 M
[CO] = 0.81 M [H2] = 0.28 M
1) 0.26
2) 3.3
3) 3.9

27
Reaching Chemical Equilibrium
A container initially filled with SO2(g) and O2(g) or only
SO3(g)
contains mostly SO3(g) and small amounts of O2(g) and SO3(g)
at equilibrium
reaches equilibrium in both situations

28
Equilibrium Can Favor Product
If equilibrium is reached after most of the forward
reaction has occurred,
the system favors the products

29
Equilibrium with a Large Kc
At equilibrium,
• a reaction with a large Kc produces a large amount of
product; very little of the reactants remain
Kc = [NCl3]2 = 3.2 x 1011
[N2][Cl2]3
• a large Kc favors the products
N2(g) + 3Cl2(g) 2NCl3(g)
When this reaction reaches equilibrium, it will essentially
consist of the product NCl3.

30
Equilibrium Can Favor Reactant
If equilibrium is reached when very little of the forward reaction
has occurred, the reaction favors the reactants

31
Equilibrium with a Small Kc
At equilibrium,
• a reaction that produces only a small amount of product has a
small Kc
Kc = [NO]2 = 2.3 x 10-9
[N2][O2]
• a small Kc favors the reactants
N2(g) + O2(g) 2NO(g)
When this reaction reaches equilibrium, it will essentially
consist of the reactants N2 and O2.

32
Summary of Kc Values
A reaction
that favors products has a large Kc
with about equal concentrations of products and reactants
has a Kc close to 1
that favors reactants has a small Kc

Summary of Kc Values (continued)
33

34
Large and Small Kc Values

35
Learning Check 8:
For each Kc, indicate whether the reaction at equilibrium
contains mostly of (R) reactants or (P) products.
__A. H2(g) + F2(g) 2HF(g)
Kc = 1 x 1095
__B. 3O2(g) 2O3(g)
Kc = 1.8 x 10-7

Guide to Using the Kc Value
36

37
Using Kc to Solve for an Equilibrium Concentration
At equilibrium, the reaction
PCl5(g) PCl3(g) + Cl2(g)
has a Kc of 4.2 x 10–2 and contains [PCl3] = [Cl2] = 0.10 M.
What is the equilibrium concentration of PCl5?

38
Using Kc to Solve for an Equilibrium Concentration
(continued)
STEP 1 Write the Kc expression:
Kc = [PCl3][Cl2]
[PCl5]
STEP 2 Solve for the unknown concentration:
[PCl5] = [PCl3][Cl2]
Kc
STEP 3 Substitute the known values and solve:
[PCl5] = [0.10][0.10] = 0.24M
4.2 x 10–2
STEP 4 Check answer by placing concentrations in Kc:
Kc = [0.10][0.10] = 4.2 x 10–2
[0.24]

39
Learning Check 9:
The Kc is 2.0 for the reaction:
2NOBr(g) 2NO(g) + Br2(g)
If the equilibrium concentrations are [NOBr] = 0.50 M and [NO]
= 0.80 M, what is the equilibrium concentration of Br2?
1) 0.39 M
2) 0.78 M
3) 1.3 M

40
Le Chatelier’s Principles:
Changing Equilibrium Conditions:
Le Châtelier’s Principle

41
Le Châtelier’s Principle
Le Châtelier’s principle states that
• any change in equilibrium conditions upsets the equilibrium
of the system
• a system at equilibrium under stress will shift to relieve the
stress
• there will be a change in the rate of the forward or reverse
reaction to return the system to equilibrium

42
Adding Reactant
For the reaction A + B C at equilibrium,
 adding more A upsets the equilibrium
 the rate of forward reaction increases to re-establish Kc
A + B C

43
Effect of Adding Reactant
Consider the following reaction at equilibrium.
H2(g) + F2(g) 2HF(g)
• If more reactant (H2 or F2) is added, there is an increase in
the number of collisions.
• The rate of the forward reaction increases and forms more
HF product until new equilibrium concentrations equal Kc
again.
• The effect of adding a reactant shifts the equilibrium toward
the products.

44
Effect of Adding Product
Consider the following reaction at equilibrium.
H2(g) + F2(g) 2HF(g)
• When more HF is added, there is an increase in
collisions of HF molecules.
• The rate of the reverse reaction increases and
forms more H2 and F2 reactants.
• The effect of adding a product shifts the equilibrium
toward the reactants.

45
Adding Reactant or Product
The equilibrium shifts toward
• products when H2(g) or F2(g) is added
• reactants when HF(g) is added
H2(g) + F2(g) 2HF(g)
Add H2 or F2
Add HF

46
Effect of Removing Reactant
Removing reactant, H2(g) or F2(g), from the following
reaction at equilibrium:
H2(g) + F2(g) 2HF(g)
• decreases the collisions between reactants
• decreases the rate of the forward reaction
• shifts the equilibrium toward the reactants
H2(g) + F2(g) 2HF(g)
Remove H2 or F2

47
Effect of Removing Product
Removing HF(g) from the following reaction at
equilibrium:
H2(g) + F2(g) 2HF(g)
• decreases the number of collisions between products
• decreases the rate of the reverse reaction
• shifts equilibrium toward the products
H2(g) + F2(g) 2HF(g)
Remove HF

48
Concentration Changes and Equilibrium

Effect of a Catalyst
Adding a catalyst
• lowers the activation energy of the forward reaction
• increases the rate of the forward reaction
• lowers the activation energy of the reverse reaction
• increases the rate of the reverse reaction
• decreases the time to reach equilibrium
• has no effect on the equilibrium
49

50
Learning Check 10:
Predict any shift in equilibrium for each of the following
changes on the reaction
NH4HS(s) H2S(g) + NH3(g)
1) to products 2) to reactants 3) no change
A. H2S(g) is added.
B. NH4HS(s) is added.
C. NH3(g) is removed.
D. A catalyst is added.

51
Effect of Decreasing the Volume
When a reaction at equilibrium contains different numbers
of moles of reactants than products, a decrease in
volume
• increases the concentrations (mole/L), upsetting the
equilibrium
• shifts the equilibrium toward the fewer number of moles
N2(g) + 3H2(g) 2NH3(g)
Decrease volume
More moles Fewer moles

52
Volume Decrease and Equilibrium
A volume decrease
shifts the equilibrium toward the side (A) with the smaller
number of moles

53
Effect of Increasing the Volume
When a reaction at equilibrium contains different
numbers of moles of reactants than products, an
increase in volume
• decreases the concentrations (mole/L), upsetting
the equilibrium
• shifts the equilibrium toward the greater number of
moles
N2(g) + 3H2(g) 2NH3(g)
Increase volume
More moles Fewer moles

54
Volume Increase and Equilibrium
A volume increase
shifts the equilibrium toward the side (B and C) with
the greater number of moles

Effect of Pressure
• Affects gases only.
• For unequal number of moles of
reactants and products, if pressure is
increased, the equilibrium will shift to
reduce the number of particles.
• For equal number of moles of
reactants and products, no shift
occurs.
2NO2 (g) N2O4 (g)

Ex: Effect of Pressure
2NO2 (g) N2O4 (g)
2 moles 1 mole
Stress: increasing the pressure
Shift: to the right (side of less moles)
Stress: decreasing the pressure
Shift : to the left (side of more moles )

57
Heat and Endothermic Reactions
For an endothermic reaction at equilibrium,
• a decrease in temperature (T) removes heat, and the
equilibrium shifts toward the reactants
• an increase in temperature adds heat, and the equilibrium
shifts toward the products.
CaCO3(s) + 133 kcal CaO(s) + CO2(g)
Decrease T
Increase T

58
Equilibrium Shift in Endothermic Reactions with
Temperature

59
Heat and Exothermic Reactions
For an exothermic reaction at equilibrium,
• a decrease in temperature removes heat, and the equilibrium
shifts toward the products
• an increase in temperature adds heat, and the equilibrium
shifts toward the reactants.
N2(g) + 3H2(g) 2NH3(g) + 22 kcal
Decrease T
Increase T

60
Equilibrium Shifts with Temperature in
Exothermic Reactions

61
Learning Check 11:
Indicate if each change on a reaction at equilibrium shifts
2NO2(g) + heat 2NO(g) + O2(g)
1) towards products 2) towards reactants
3) no change
A. adding NO(g) F. decreasing pressure
B. adding O2(g) G. decreasing the tempt. (Endo)
C. raising the temperature(Exo)
D. removing O2(g)
E. increasing the volume

62
Equilibrium in Saturated Solutions

63
Saturated Solution
A saturated solution
 contains the maximum amount of dissolved solute
 contains solid solute
 is an equilibrium system:
rate of dissolving = rate of recrystallization
solid ions in solution

64
Solubility Product Constant
The solubility product constant for a saturated solution
 gives the ion concentrations at constant temperature
 is expressed as Ksp
 does not include the solid, which is constant
Fe(OH)2(s) Fe2+(aq) + 2OH−(aq)
Ksp = [Fe2+] [OH−]2

65
Seatwork:
Write the Ksp expression for each of the
following:
A. Ag2CrO4(s) --> 2 Ag+(aq) + CrO4
2-(aq)
B. BaSO4(s) --> Ba2+(aq) + SO4
2-(aq)
C. PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)

66
Guide to Calculating Ksp

67
Example of Calculating Solubility Product Constant
Calculate the Ksp of PbSO4 (solubility 1.4 x 10–4 M).
STEP 1 Write the equilibrium equation for dissociation:
PbSO4(s) Pb2+(aq) + SO4
2−(aq)
STEP 2 Write the Ksp expression:
Ksp = [Pb2+][SO4
2−]
STEP 3 Substitue molarity values and calculate:
Ksp = (1.4 x 10–4) x (1.4 x 10–4)
= 2.0 x 10–8

68
Examples of Solubility Product Constants

69
Learning Check 13:
What is the Ksp value of PbF2 if the solubility at
25 C is 2.6 x 10–3 M?
1) 1.4 x 10–5
2) 6.8 x 10–6
3) 7.0 x 10–8

70
Molar Solubility (S)
The molar solubility (S) is
 the number of moles of solute that dissolve in 1 L of
solution
 determined from the formula of the salt
 calculated from the Ksp

71
Calculating Molar Solubility (S)

72
Solubility Calculation
Determine the solubility (S)2 of SrCO3 (Ksp = 5.4 x 10–10).
STEP 1 Write the equilibrium equation for dissociation:
SrCO3(s) Sr2+(aq) + CO3
2−(aq)
STEP 2 Write the Ksp expression:
Ksp = [Sr2+][CO3
2−]
STEP 3 Substitute S for the molarity of each ion into Ksp:
Ksp = [Sr2+][CO3
2−] = [S][S]
= S2 = 5.4 x 10−10
STEP 4 Calculate the solubility, S:
S = = 2.3 x 10−5 M
-10
5.4 10


73
Seatwork:
1. Calculate the solubility (S) of CaF2, if the
Ksp =3.9 x 10 -11.
2. Calculate the solubility (S) of PbSO4 , if the
Ksp = 1.6 x 10–8.

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