13 Dec. 2010
Take Out Homework: Lab Notebook and Kinetics
AP Questions.
Objective: SWBAT
Do now:
1
Agenda
I. Do now
II. Kinetics Problem Set Solutions
III. Introduction to Equilibrium
IV. Demo
Homework: p. #
2
Chemical Equilibrium
Chapter 14
3
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
•
However, there is a lot of activity at the molecular level!
•
Reactants continue to form products, while products
continue to yield reactants!
4
Physical vs. Chemical Equilibrium
Physical equilibrium
H2O (l)
NO2
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)
5
N2O4 (g)
2NO2 (g)
equilibrium
equilibrium
Start with N2O4
Start with NO2 & N2O4
equilibrium
Start with NO2
6
constant
7
N2O4 (g)
Equilibrium
Constant
K=
[NO2]2
[N2O4]
aA + bB
K=
[C]c[D]d
2NO2 (g)
= 4.63 x 10-3
cC + dD
Law of Mass Action
[A]a[B]b
8
Law of Mass Action
• For a reversible reaction at equilibrium and a
constant temperature, a certain ratio of
reactant and product concentration has a
constant value, K (the equilibrium constant)
– concentrations may vary
– but as long as temperature stays the same and the
reaction is at equilibrium, K will not change.
9
K=
[C]c[D]d
aA + bB
cC + dD
[A]a[B]b
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
10
Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g)
Kc =
2NO2 (g)
[NO2]2
Kp =
[N2O4]
2
PNO
2
PN2O4
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)
11
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l)
[CH3COO-][H3O+]
Kc′ =
[CH3COOH][H2O]
CH3COO- (aq) + H3O+ (aq)
[H2O] = constant
[CH3COO-][H3O+]
= Kc′ [H2O]
Kc =
[CH3COOH]
General practice not to include units for the equilibrium
constant.
12
Example 1
Write expressions for Kc, and KP if applicable, for the
following reversible reactions at equilibrium:
a) HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
b) 2NO(g) + O2(g)  2NO2(g)
c) CH3COOH(aq) + C2H5OH(aq)  CH3COOC2H5(aq) + H2O(l)
Hints: KP applies only to gaseous reactions; the
concentration of the solvent (usually water) does not
appear in the equilibrium constant expression.
13
Problem 1
Write Kc and KP for the decomposition of
dinitrogen pentoxide:
2N2O5(g)  4NO2(g) + O2(g)
14
Example 2
The following equilibrium process has been
studied at 230oC:
2NO(g) + O2(g)  2NO2(g)
In one experiment, the concentrations of the
reacting species at equilibrium are found to be
[NO]=0.0542 M, [O2]=0.127 M and [NO2]=15.5
M. Calculate the equilibrium constant (Kc) of the
reaction at this temperature.
15
Problem 2
Carbonyl chloride (COCl2), also called phosgene,
was used in WWI as a poisonous gas. The
equilibrium concentrations for the reaction
between carbon monoxide and molecular
chlorine to form carbonyl chloride
CO(g) + Cl2(g)  COCl2(g)
at 74oC are [CO]=1.2x10-2 M, [Cl2]=0.054 M and
[COCl2]=0.14 M. Calculate the equilibrium
constant (Kc).
16
Example 3
• At the equilibrium constant KP for the
decomposition of phosphorus pentachloride
to phosphorus trichloride and molecular
chlorine
PCl5(g)  PCl3(g) + Cl2(g)
is found to be 1.05 at 250oC. If the equilibrium
partial pressures of PCl5 and PCl3 are 0.875
atm and 0.463 atm, respectively, what is the
equilibrium partial pressure of Cl2 at 250oC?
17
Problem 3
The equilibrium constant KP for the reaction
2NO2(g)  2NO(g) + O2(g)
is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400
atm and PNO = 0.270 atm.
18
Example 4
Methanol (CH3OH) is manufactured industrially
by the reaction
CO(g) + 2H2(g)  CH3OH(g)
The equilibrium constant Kc for the reaction is
10.5 at 220oC. What is the value of KP at this
temperature?
19
Problem 4
For the reaction
N2(g) + 3H2(g)  2NH3(g)
KP is 4.3x10-4 at 375oC. Calculate Kc for the
reaction.
20
14 Dec. 2010
• Objective: SWBAT write equilibrium constant
expressions for reactions with heterogeneous
equilibria, and calculate reaction quotients.
• Do now: The equilibrium constant KP for the
reaction
2SO3(g) ⇄ 2SO2(g) + O2(g)
is 1.8x10-5 at 350oC. What is KC for this
reaction?
21
Agenda
I. Do now
II. Homework solutions
III. Heterogeneous equilibrium constants
IV. Reaction quotients
p. 649 #23, 25, 27, 29, 31, 35, 37
22
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc′ =
[CaCO3]
Kc = [CO2] = Kc′ x
[CaCO3]
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
23
CaCO3 (s)
CaO (s) + CO2 (g)
PCO 2 = Kp
PCO 2 does not depend on the amount of CaCO3 or CaO
24
Example 5
Write the equilibrium constant expression Kc,
and KP if applicable, for each of the following
heterogeneous sytems:
a) (NH4)2Se(s)  2NH3(g) + H2Se(g)
b) AgCl(s)  Ag+(aq) + Cl-(aq)
c) P4(s) + 6Cl2(g)  4PCl3(l)
25
Problem 5
Write equilibrium constant expressions for Kc
and KP for the formation of nickel
tetracarbonyl, which is used to separate nickel
from other impurities:
Ni(s) + 4CO(g)  Ni(CO)4(g)
26
Example 6
Consider the following heterogeneous
equilibrium:
CaCO3(s)  CaO(s) + CO2(g)
At 800oC, the pressure of CO2 is 0.236 atm.
Calculate
a) Kp
b) Kc
for the reaction at this temperature.
27
Problem 6
Consider the following equilibrium at 395 K:
NH4HS(s)  NH3(g) + H2S(g)
The partial pressure of each gas is 0.265 atm.
Calculate KP and Kc for the reaction.
28
For which of the following reactions is Kc equal
to KP?
a) 4NH3(g) + O2(g) ⇄ 4NO(g) + 6H2O(g)
b) 2H2O2(aq) ⇄ 2H2O(l) + O2(g)
c) PCl3(g) + 3NH3(g) ⇄ 3HCl(g) + P(NH2)3(g)
29
• What if the product molecules of one
reversible reaction are involved in a second
reaction as the products?
A+B
C+D
C+D
E+F
30
A+B
C+D
Kc′
C+D
E+F
Kc′′
A+B
E+F
[C][D]
Kc′ =
[A][B]
Kc
[E][F]
Kc′′=
[C][D]
[E][F]
Kc =
[A][B]
Kc = Kc′ x Kc′′
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.
31
N2O4 (g)
K=
[NO2]2
[N2O4]
2NO2 (g)
= 4.63 x
10-3
2NO2 (g)
N2O4 (g)
[N2O4]
1
= 216
K′ =
=
2
K
[NO2]
When the equation for a reversible reaction is written in
the opposite direction, the equilibrium constant becomes
the reciprocal of the original equilibrium constant.
32
Writing Equilibrium Constant Expressions
1. 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.
2. The concentrations of pure solids, pure liquids and solvents
do not appear in the equilibrium constant expressions.
3. The equilibrium constant is a dimensionless quantity.
4. In quoting a value for the equilibrium constant, you must
specify the balanced equation and the temperature.
5. 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.
33
15 Dec. 2010
• Objective: SWBAT calculate reaction quotient
and calculate an equilibrium concentration.
• Do now: If the equilibrium constant for the
reaction
N2O4 (g) ⇄ 2NO2 (g) is 4.63 x 10-3
calculate the equilibrium constant for the
reaction
2NO2(g) ⇄ N2O4(g)
34
Agenda
I. Do now
II. Homework solutions
III. Reaction quotient
IV. Equilibrium concentrations
Homework: p. 650 #23, 30, 36, 39
35
The Relationship between
Chemical Kinetics and Chemical Equilibrium
A + 2B
kf
kr
ratef = kf [A][B]2
AB2
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2]
kf
[AB2]
= Kc =
kr
[A][B]2
What does the Equilibrium
Constant tell us?
• We can use the equilibrium constant to
calculate unknown equilibrium concentrations
of products or reactants (at a constant
temperature)
• We can predict the direction in which a
reaction will proceed to achieve equilibrium
37
What if the reaction is not yet at
equilibrium?
• Calculate reaction quotient (Qc) by
substituting the initial concentrations into the
equilibrium constant expression.
38
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
39
Example 7
At the start of the reaction, there are 0.249 mol
N2, 3.21x10-2 mol H2 and 6.42x10-4 mol NH3 in
a 3.50 L reaction vessel at 375oC. If the
equilibrium constant Kc for the reaction
N2(g) + 3H2(g) ⇄ 2NH3(g)
is 1.2 at this temperature, decide whether the
system is at equilibrium. If it is not, predict
which way the net reaction will proceed.
40
Problem 7
The equilibrium constant (Kc) for the formation
of nitrosyl chloride, an orange-yellow
compound, from nitric oxide and molecular
chlorine
2NO(g) + Cl2(g) ⇄ 2NOCl(g)
is 6.5x104 at 35oC. In a certain experiment,
2.0x10-2 mole of NO, 8.3x10-3 mole of Cl2, and
6.8 moles of NOCl are mixed in a 2.0 L flask. In
which direction will the system proceed to
reach equilibrium?
41
What do you do with the Kc and KP?
• If we know the equilibrium constant for a
particular reaction, we can calculate the
concentrations in the equilibrium mixture
from the initial concentrations!
42
Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
43
Example 8
At 1280oC the equilibrium constant (Kc) for the reaction
Br2 (g)
2Br (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] =
0.063 M and [Br] = 0.012 M, calculate the
concentrations of these species at equilibrium.
At 1280oC the equilibrium constant (Kc) for the reaction
Br2 (g)
2Br (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] =
0.063 M and [Br] = 0.012 M, calculate the
concentrations of these species at equilibrium.
Let x be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
[Br]2
Kc =
[Br2]
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
Solve for x
45
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
-b ± b2 – 4ac
2
x=
ax + bx + c =0
2a
x = -0.0105 x = -0.00178
Initial (M)
Change (M)
Equilibrium (M)
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
46
Example 9
A mixture of 0.500 mol H2 and 0.500 mol I2 was
placed in a 1.00 L stainless steel flask at 430oC.
The equilibrium constant KC for the reaction
H2(g) + I2(g) ⇄ 2HI(g)
is 54.3 at this temperature. Calculate the
concentrations of H2, I2 and HI at equilibrium.
47
Example 10
For the reaction below, the initial concentrations
of each species are H2 = 0.00623 M, I2 =
0.00414 M and HI = 0.224 M at 430oC.
The equilibrium constant KC for the reaction is
still 54.3. Calculate the concentrations of
these species at equilibrium.
H2(g) + I2(g) ⇄ 2HI(g)
48
Homework
• p. 650 #23, 30, 36, 39
49
Problem 8
Consider the reaction below:
H2(g) + I2(g) ⇄ 2HI(g)
with an equilibrium constant of 54.3.
Starting with a concentration of 0.040 M for HI,
calculate the concentrations of HI, H2 and I2 at
equilibrium.
50
Problem 9
At 1280oC the equilibrium constant (Kc) for the
reaction
Br2(g) ⇄ 2Br(g)
is 1.1x10-3. If the initial concentrations are
[Br2] = 6.3x10-2 M and [Br] = 1.2x10-2 M,
calculate the concentrations of these species
in equilibrium.
51
3 January 2011
• Take Out AP Problem Set: Equilibrium
• Objective: SWBAT review equilibrium
concentration calculations and describe
factors that affect chemical equilibrium.
• Do now: What does a very large Kc indicate?
What does it mean if Qc is much smaller than Kc?
Agenda
I. Do now
II. AP Problem Set Questions?
III. Review ICE boxes and equilibrium
concentrations
IV. Factors that affect chemical equilibrium:
Demo
V. Notes
Homework: p. 651 #45, 49, 51, 52, 53-61 odds
Demo
• Solid copper metal reacts with nitric acid to
produce dinitrogen tetraoxide gas, water
vapor and a solution of copper (II) nitrate.
• Dinitrogen tetraoxide gas forms nitrogen
dioxide gas.
54
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)
2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
55
Le Châtelier’s Principle
• Changes in Concentration continued
Remove
Add
Remove
Add
aA + bB
cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
left
right
right
left
56
Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g)
C (g)
Change
Shifts the Equilibrium
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Side with fewest moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewest moles of gas
57
Example 1
At 720oC, the equilibrium constant Kc for the reaction
N2(g) + 3H2(g) ⇄ 2NH3(g)
is 2.37x10-3. In a certain experiment, the equilibrium
concentrations are [N2]=0.683 M, [H2]=8.80M and
[NH3]=1.05 M. Suppose some NH3 is added to the
mixture so that its concentration is increased to
3.65 M.
a) Use Le Châtelier’s principle to predict the shift in
direction of the net reaction to reach a new
equilibrium.
b) Confirm your prediction by calculation the reaction
quotient Qc and comparing its value to Kc.
Problem 1
At 430oC, the equilibrium constant (KP) for the
reaction
2NO(g) + O2(g) ⇄ 2NO2(g)
is 1.5x105. In one experiment, the initial
pressures of NO, O2 and NO2 are 2.1x10-3 atm,
1.1x10-2 atm and 0.14 atm respectively.
Calculate QP and predict the direction that the
net reaction will shift to reach equilibrium.
59
Le Châtelier’s Principle
• Changes in Temperature (gases only!)
Change
Increase temperature
Decrease temperature
N2O4 (g)
Exothermic Rx
Endothermic Rx
K decreases
K increases
K increases
K decreases
2NO2 (g)
colder
hotter
60
Example 2
Consider the following equilibrium systems:
a) 2PbS(s) + 3O2(g) ⇄ 2PbO(s) + 2SO2(g)
b) PCl5(g) ⇄ PCl3(g) + Cl2(g)
c) H2(g) + CO2(g) ⇄ H2O(g) + CO(g)
Predict the direction of the net reaction in each
case as a result of increasing the pressure
(decreasing the volume) on the system at
constant temperature.
61
Problem 2
Consider the equilibrium reaction involving
nitrosyl chloride, nitric oxide and molecular
chlorine:
2NOCl(g) ⇄ 2NO(g) + Cl2(g)
Predict the direction of the net reaction as a
result of decreasing the pressure (increasing
the volume) on the system at a constant
temperature.
62
5 January 2011
• Objective: SWBAT describe how stress on a
system shifts equilibrium.
• Do now: Complete this table:
Stress to system (exothermic)
Increase in temperature
Increase in pressure
increase in concentration of
reactant
Effect
increase in volume
63
Agenda
I.
II.
III.
IV.
Do now
Homework answers
Finish Le Chatlier’s examples
Equilibrium problem set (work time
tomorrow, too)
Test Monday, 2nd half of class
64
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
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift
equilibrium.
65
Example 3
Consider the following equilibrium process between
dinitrogen tetrafluoride and nitrogen difluoride:
N2F4(g) ⇄ 2NF2(g)
∆Ho=38.5 kJ/mol
Predict the changes in the equilibrium if
a) the reacting mixture is heated at a constant
volume.
b) some N2F4 gas is removed from the reacting
mixture at constant temperature and volume
c) the pressure on the reacting mixture is decreased
at constant temperature
d) a catalyst is added to the reacting mixture.
Problem 3
Consider the equilibrium between molecular
oxygen and ozone:
3O2(g) ⇄ 2O3(g)
ΔHo=284 kJ/mol
What would be the effect of
a) increasing the pressure on the system by
decreasing the volume?
b) adding O2 to the system at constant volume?
c) decreasing the temperature?
d) adding a catalyst?
Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Hb (aq) + O2 (aq)
Kc =
HbO2 (aq)
[HbO2]
[Hb][O2]
68
Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g)
2NH3 (g) DH0 = -92.6 kJ/mol
69
Le Châtelier’s Principle - Summary
Change Equilibrium
Constant
no
Change
Shift Equilibrium
Concentration
yes
Pressure
yes*
no
Volume
yes*
no
Temperature
yes
yes
Catalyst
no
no
*Dependent on relative moles of gaseous reactants and products
70
Homework
p. 651 #45, 49, 51, 52, 53-61 odds
71
20 Dec. 2010
Objective: SWBAT determine the
stoichiometric relationship between
reactants and products in a chemical
reaction.
Do now: Grab your lab notebook and
complete the pre-lab.
72
Changes to the lab
• Use HALF the volume of each solution called
for in the lab.
• Complete reactions in a 50 mL graduated
cylinder.
• Measure reagents in a 25 mL graduated
cylinder.
73
Notes
• Pour or transfer pipette OUT of reagent
bottles.
• Never pour or transfer pipette INTO reagent
bottles.
• Pour excess down the sink while running the
water.
• Pour products into the waste beaker.
74
Homework
• Post lab calculations and conclusions due
tomorrow.
75
Equilibrium Review Problem Set
• p. 648 #8d, 9a, 16, 18, 21, 28, 31, 40, 42, 48,
54, 58, 65, 71, 84, 88, 98
• In-class work time: today and Thursday.
• Due Monday
• Test on Equilibrium: Monday
• Next unit: Acid-Base Reactions