Transcript Document

Chemistry 216
Instructor: Dr. Dennis Pederson
Please pick up a copy of the following:
Course Syllabus
Course Study Guide
List of Key Terms and Equations
First two Homework Assignments (on one sheet)
Course Questionnaire
Make the following corrections to the dates given in the Syllabus:
Page 3: Add 7 days to each of the examination dates listed.
Example - First Examination change Jan 23 to Jan 30
Page 5: Last two dates should be M March 23 and W Mar 25
Chemistry 216: Principles of Chemical Reactions
Key questions for the course.
1. Is a given reaction or process possible?
Goal: Find a criterion that predicts that a reaction or
process can occur at a given set of conditions.
2. If a given reaction or process is possible, how long will it take?
Goal: Identify the conditions that affect the rate of the reaction.
3. If a given reaction or process is possible, what pathway
will if follow?
Goal: Identify the steps that occur during a reaction.
4. If a given reaction or process is possible, what will be the
distribution of products and reactants at the end?
Goal: Identify the conditions that affect product-reactant
distribution.
Important Terms
Energy: capacity to do work or transfer heat.
Work: energy transferred when a force
moves an object.
Heat: energy transferred from a hotter object
to a colder one.
Classes of energy:
Potential Energy: energy of position
Kinetic Energy: energy of motion
Units of energy:
Joule = 1 kg-m2/s2
Calorie = 4.184 J (energy to change temperature
of 1 g water by 1oC)
System: portion of the universe being studied.
Surroundings: everything else.
Types of Systems
Open System:
Can exchange both
matter and energy
with surroundings
Closed System:
Can exchange energy
but not matter
with surroundings
Isolated System:
Can exchange neither
energy nor matter
with surroundings
Exothermic
Endothermic
Exothermic and Endothermic Phase Changes
5
First Law of Thermodynamics:
Energy can be neither created nor destroyed.
Internal energy (E) of a system = sum of all kinetic
and potential energies of the components of the system.
Change in internal energy (E) = Efinal – E initial
Meaning of E:
E > O, system gained energy
E > O, system lost energy
Relation to heat (q) and work (w)
E is the sum of the energy transferred as heat and
that transferred as work: E = q + w
Signs of q and w:
q > 0: system gained heat, q < 0: system lost heat
w > 0: work done on system, w < 0: work done by system
First Law Problems
A system absorbs 430 J of heat energy and has 530 J of work
done on it by the surroundings. What is E for the system?
Use E = q + w
E = (+430 J) +
(+530 J) = +960 J
q = +430 J (system absorbed heat)
w = +530 J (surroundings did
work on the system)
A system that does 380 J of work on the surroundings has
an internal energy change of 260 J. What is q for the system?
Rearrange E = q + w to solve for q: q = E – w
E = +260 J, w = –380 J (system did work on the surroundings)
q = E – w = (+260 J) – (–380 J) = + 640 J
Specific Heat Sample Problems
Calculate the energy required to raise the temperature of
140.0 g Cu from 24.0 oC to 95.0 oC.
(specific heat of Cu = 0.386 J/g oC)
0.386 J/g oC = (energy) / (140.0 g)(95.0oC – 24.0oC)
energy = (0.386 J/goC) (140.0 g)(71.0oC) = 3800 J
Determine the specific heat of aluminum.
Add 50.0 g Al at 99.0oC with 70.0 g water at 24.5oC.
Final temperature = 34.4oC
qwater = (70.0g)(9.9oC)(4.18 J/goC ) = 2900 J
qwater + qAl = 0 so qAl = –3510 J
specific heat Al = (-2900J)/(50.0 g)(–64.6oC)
= 0.90 J/goC
Bomb
Calorimeter
Bomb Calorimetry Sample Problem
A 1.608 g sample of cymene (C10H14) was burned in a bomb
calorimeter whose heat capacity is 3.640 kJ/oC. The temperature
Of the calorimeter increased from 25.50 to 44.85oC.
Reaction: 2C10H14(s) + 27O2(g)
20CO2(g) + 14H2O(l)
Calculate the heat of combustion of cymene in kJ/g and kJ/mol.
qcalorimeter = (3.640 kJ/oC)(44.85oC – 25.50oC) = 70.43 kJ
Since: qcalorimeter + qrxn = 0 (isolated system)
qrxn = –70.43 kJ
Heat of combustion = (–70.43 kJ/1.608 g) = –43.80 kJ/g
One mole cymene = (10)(12.0) + (14)(1.0) = 134 g
Heat of combustion = (–43.80 kJ/g)(134.0 g/mol)= –5870 kJ/mol
Hess’s Law
For a reaction or process that can be obtaining by adding two
or more other reactions or processes, the enthalpy change of
the reaction is the sum of the enthalpy changes of the combined
reactions or processes.
Example:
A + B
A + X
2A + B
X
Y
Y
HI
HII
HIII = HI + HII
Hess’s Law Problem
Given the following data:
C(s) + O2(g)
CO2(g)
CO(g) + 1/2 O2(g)
CO2(g)
H = -393.5 kJ
H = -283.0 kJ
Calculate H for the formation of carbon monoxide
Cs) + 1/2 O2(g)
CO(g)
Hess’s Law Problem
Given the following data:
C2H2(g) + 5/2 O2(g)
2 CO2(g) + H2O(l) H = -1305 kJ
C2H6(g) + 7/2 O2(g)
H2(g) + 1/2 O2(g)
2 CO2(g) + 3 H2O(l) H = -1560 kJ
H2O(l)
H = -286 kJ
Calculate H for the hydrogenation of acetylene
C2H2(g) + 2 H2(g)
C2H6(g)
Examples of Standard Enthalpies of Formation
Chemistry: The Science in Context, 4th edition. © 2014 W. W. Norton Co., Inc.
14
Using Standard Enthalpies of Formation
o
Find H for C2H5OH(l) + 3O2(g)
C2H5OH(l)
2C(s) + 3H2(g) + 1/2O2(g)
2C(s) + 2O2(g)
2CO2(g)
3H2(g) + 3/2O2(g)
C2H5OH(l) + 3O2(g)
o
2CO2(g) + 3H2O(l)
3H2O(l)
–(–277.7 kJ)
2(–393.5 kJ)
3(–285.8 kJ)
2CO2(g) + 3H2O(l)
H = (277.7 kJ) + (2)(–393.5 kJ) + (3)(–285.8 kJ)
= –1922.1 kJ
Determination of Enthalpy of Formation
o
Find Hf
for C2H5OC2H5 (diethyl ether)
Use data from the combustion reaction.
C2H5OC2H5 + 6O2
o
4CO2 + 5H2O
o
Hrxn = –2723.7 kJ
o
o
Hrxn = [ 4 Hf (CO2) + 5 Hf (H2O)] – Hfo(C2H5OC2H5)
o
Rearrange to solve for Hf (C2H5OC2H5)
o
o
o
–
[
]
+
Hf (C2H5OC2H5) = 4Hf (CO2) 5Hf (H2O)
Hrxn
o
o
Hf (C2H5OC2H5) = 4(–393.5 kJ) + 5(–285.8 kJ)] – (–2723.7 kJ)
= –279.3 kJ
Indicate whether the heat energy change (q) accompanying each of the following processes is positive, negative, or zero:
a.
C6H 6(l)
C6H6(g)
b.
Mixing 50 mL of water at 25.0 oC with 50 mL of ethyl alcohol also at 25.0 oC.
The temperature of the mixture is 28.5 oC.
Indicate whether the work (w) accompanying each of the following processes is positive, negative, or zero:
a.
The reaction N2(g) + 3 H2(g)
2 NH 3(g) is carried out in a closed cylinder fitted with a piston on the top
that can freely move up or down.
b. The same reaction is carried out in a sealed container that cannot expand or contract.
Chemistry: The Science in Context, 4th edition. © 2014 W. W. Norton Co., Inc.
18
Table of Standard Molar Entropies
Calculate So for the following reaction:
CH4(g) + 2O2(g)
CO2(g) +
2H2O(l)
So = [(1 mol)(213.8 J/mol K) + (2 mol)(69.9 J/mol K)]
–[(1 mol)(186.2 J/mol K) + (2 mol)(205 J/mol K)] = –242.6 J/K
Chemistry: The Science in Context, 4th edition. © 2014 W. W. Norton Co., Inc.
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Entropy and Phase Changes for Water
Chemistry: The Science in Context, 4th edition. © 2014 W. W. Norton Co., Inc.
20
Molecular Complexity and Molar Entropy
Chemistry: The Science in Context, 4th edition. © 2014 W. W. Norton Co., Inc.
21
Interplay of H, S, T, and Spontaneity
Chemistry: The Science in Context, 4th edition. © 2014 W. W. Norton Co., Inc.
22
Photochemical Smog
N2(g) + O2(g)
2NO(g)
2NO(g) + O2(g)
NO2(g)
h
O2(g) + O(g)
H = +186.6 kJ
S = +0.025 J/K
2NO2(g) H = –114.2 kJ S = –0.147 J/K
NO(g) + O(g)
O3(g)
Factors the Affect Rate
1. Physical state and nature of the reactants.
2. Concentration of the reactants.
3. Temperature.
4. Presence of catalysts.
Concentration-Time Data
Effect of Concentration on Reaction Rate
Rate Law Determination
Method of Initial Rates
F2(g) + 2 ClO2(g)
Experiment
[F2(g)] (mol/L)
2 FClO2(g)
[ClO2(g)] (mol/L)
Initial Rate
(mol/Ls)
1.
0.10
0.10
0.0012
2.
0.20
0.10
0.0024
3.
0.10
0.40
0.0048
Rate Law Determination
Method of Initial Rates
2 H2(g) + 2 NO(g)
Experiment
[H2(g)] (mol/L)
2 H2O(g) + N2(g)
[NO(g)] (mol/L)
Initial Rate
(mol/Ls)
1.
0.212
0.136
0.0248
2.
0.424
0.136
0.0495
3.
0.424
0.408
0.446
4.
0.636
0.270
0.298
(3)1
(2)2
Compare Exp.4
With Exp. 1
(3x4)x0.0248
Rate Law Determination
Method of Initial Rates
2 HgCl2(aq) + C2O42–(aq)
2 Cl–(g) + CO2(g) + Hg2Cl2(s)
Experiment
[HgCl2(aq)]
(mol/L)
[C2O42–(aq)]
(mol/L)
Initial Rate
(mol/Ls)
1.
0.164
0.15
0.000032
2.
0.164
0.45
0.00029
3.
0.082
0.45
0.00014
4.
0.246
0.15
0.000048
Summary of Results - First Examination
Grade Scale
A’s 85 - 100
B’s 70 - 84
C’s 50 - 69
D’s 35 - 49
F’s 0- 34
Examination Average: 69.7
Decomposition of N2O5
2 N2O5
4 NO2(g) + O2(g)
rate = k[N2O5]
Integrated rate law:
ln[N2O5]t = –kt + ln[N2O5]o
time(m)
[N2O5] ln[N2O5] %Decrease
(mol/L)
0
0.0150
–4.20
200
0.0096
–4.65
400
0.0062
–5.08
600
0.0040
–5.52
800
0.0025
–5.99
1000
0.0016
–6.44
36%
35%
35%
37%
36%
Decomposition of NO2
2NO2
2NO + O2
Rate = k[NO2]2
Integrated rate law:
1/[NO2]t = kt + 1/[NO2]o
Determination of Reaction Order
Time(s)
[A] mol/L
[A] mol/L
[A] mol/L
0
20
0.200
0.175
0.200
0.142
0.200
0.100
40
0.150
0.100
0.067
60
0.125
0.071
0.050
80
0.100
0.050
0.040
100
0.075
0.036
0.033
Orientation Factor and Rate
O3(g) + NO(g)
O2(g) + NO2(g)
Energy Distribution of Molecules
Activation Energy Representations
Ea(rev)
Exothermic
Exothermic
Endothermic
Ea(for) = 25kJ/mol
Ea(rev) = 40kJ/mol
H = Ea(for) – Ea(rev) = (25 kJ/mol – 40 kJ/mol) = –15 kJ/mol
Determination of the Activation Energy
H3C N C
H3C C N
=
Ea = –(slope)(R)
(-6.6) - (-10.4)
(1.95 - 2.15) x 10 -3
= –1.9 x 10 4 K
= –(–1.9 x 10 4 K)(8.31 J/mol K)(1 kJ/1000 J) = 160 kJ/mol
Chemical Kinetics: Elementary Reaction
Terminology and Rate Laws
Example Reaction
2NO2(g)
Proposed
Mechanism
Rate = k[NO2]2
2NO(g) + O2(g)
NO2(g) + NO2(g)
NO(g) + NO3(g)
slow
rate = k1[NO2]2
NO3(g)
NO(g) + O2(g)
rate = k2[NO3]
fast
Representation of Rate-Determining Step
NO2(g) + NO2(g)
NO3(g)
k2
k1
NO(g) + NO3(g)
NO(g) + O2(g)
slow
fast
The rate is controlled by the rate-determining step.
rate = k1[NO2]2
This mechanism agrees with the overall stoichiometry and with
The experimentally-determined rate law and is therefore consistent.
Energy Diagram for Reaction
2NO2(g)
2NO(g) + O2(g)
Representation of Rate-Determining Step
Second reaction is the rate-detemining step.
Kinetics of the Reverse Reaction
2NO(g) + O2(g)
Rate = k[NO]2 [O2]
2NO2(g)
Follows same pathway - Principal of Microscopic Reversibility.
NO(g) + O2(g)
NO(g) + NO3(g)
k1
k–1
k2
fast
NO3(g)
NO2(g) + NO2(g)
slow
rate = k2[NO][NO3]
However, NO3 is an intermediate and its concentration has to be replaced.
Pre-equilibrium approximation: Prior fast reactions are said to be fast and
reversible so rate of formation of products = rate of formation of reactants.
k1[NO][O2] = k–1[NO3]
[NO3] = k1[NO][O2]
rate = k2[NO][NO3] = k2[NO] k1[NO][O2] = k[NO]2[O2]
k–1
k–1
Another Example
2NO(g) + 2H2(g)
Rate = k[NO]2 [H2]
N2(g) + 2H2O(g)
Possible mechanism 1
k1
2NO(g) + H2(g)
k2
NO2(g) + H2(g)
NO2(g) + H2O(g)
slow
N2(g) + H2O(g)
fast
rate = k1[NO]2 [H2]
Problem: Involves a termolecular elementary reaction.
Possible mechanism 2
NO(g) + NO(g)
k1
k–1
N2O2(g) + H2(g) k2
N2O(g) + H2(g)
rate = k[NO]2 [H2]
k3
N2O2(g)
fast
N2O(g) + H2O(g)
slow
N2(g) + H2O(g)
fast
Another Possible Mechanism and Summary
Possible mechanism 3
NO(g) + H2(g)
k1
NOH2(g)
k–1
NOH2(g) + NO(g)
N2O(g) + H2(g)
k2
k3
N2O(g) + H2O(g)
N2(g) + H2O(g)
fast
slow
fast
rate = k[NO]2 [H2]
Each of these mechanisms is consistent with the overall
stoichiometry and the experimental rate law.
Catalysis
Example of Catalysis
I– -catalyzed decomposition of H2O2
2 H2O2
2 H2 O + O2
Rate = k[H2O2][I–]
Proposed mechanism:
k1
–
H2O + IO–
H2O2 + I
slow
k2
–
fast
IO + H2O2
H2O + O 2
From the mechanism: rate = k[H2O2][I–]
Now look at the reverse reaction; follows the same mechanism.
k–1
fast
H2O + O2
IO– + H2O2
k1
k–2
–
slow
H2O + IO
H 2 O 2 + I–
From the mechanism: rate = k[H2O][O2][I–] / [H2O2]
Important: Catalyst acts on reaction in both directions.
Guidelines for Writing a Mechanism
1.
All elementary steps are connected - an intermediate in one step is
a reactant in the next step.
2.
There is only one step that is the rate-determining (slow) step. All
of the other steps are faster and labeled fast.
3.
All species that do not cancel out up to the slow step will be in the rate law.
a. Any reactant species will be in the numerator of the rate law.
b. Any product species will be in the denominator of the rate law.
4.
A catalyst will appear as a reactant in the mechanism before or during
the slow step, and again as a product in the slow step or a later step.
5.
An inhibitor will appear as a product in the mechanism before
the slow step, and again as a reactant of a step after the slow step.
6.
The total steps must add up to give the overall reaction and the rate
law derived from the mechanism must agree with the experimentally–
determined rate law.
Practice Mechanisms
1.
Propose a mechanism consistent with the following information. Use
only unimoleular or bimolecular elementary steps.
Reaction: 3 A + 2 B
2.
M + 2 N; rate = k[A]2[B][C]
Shown below is the proposed mechanism for a reaction
between two substances X and Y.
Y+Y
Y2
fast
X + Y2
R
fast
R+B
S+Z
slow
Z+X
P+W
fast
W+X
S+B
fast
Answer the following concerning this reaction and mechanism.
a. Write the balanced equation for the reaction. 2Y + 3X
2S + P
b. Write the rate law for this mechanism.
rate = k[X][Y]2[B]
c. Give the symbol for two intermediates in the mechanism. Y2, R, Z, W
d. Identify the species acting as a catalyst in the reaction. B
Practice Mechanisms
1.
Propose a mechanism consistent with the following information. Use
only unimoleular or bimolecular elementary steps.
Reaction: 3 A + 2 B
2.
M + 2 N; rate = k[A]2[B][C]
Shown below is the proposed mechanism for a reaction
between two substances X and Y.
Y+Y
Y2
fast
X + Y2
R
fast
R+B
S+Z
slow
Z+X
P+W
fast
W+X
S+B
fast
Answer the following concerning this reaction and mechanism.
a. Write the balanced equation for the reaction.
b. Write the rate law for this mechanism.
c. Give the symbol for two intermediates in the mechanism.
d. Identify the species acting as a catalyst in the reaction.
Chemical Equilibrium
N2O4(g)
2 NO2(g)
-x
+2x
Chemical Equilibrium: Empirical Study
Reaction being studied: CO(g) + 2 H2(g)
Experiment
[CO] mol/L
[H2] mol/L
CH3OH(g)
[CH3OH] mol/L
1
initial
change
equilibrium









2
initial
change
equilibrium

5
5

5
5

5
4
3
initial
change
equilibrium









Chemical Equilibrium: Empirical Study
Trial relationships of the equilibrium data:
Experiment
[CH3OH ]
[CO][H2]
[CH3OH ]
[CO]2[H2]
[CH3OH ]
[CO][H2]2
1
(0.00892)
= 1.19
(0.0911)(0.0822)
(0.00892)
= 0.596
(0.0911)2(0.0822)
(0.00892)
= 14.5
2
(0.0911)(0.0822)
2
(0.0247)
= 2.17
(0.0753)(0.151)
(0.0247)
= 1.09
(0.0753)2(0.151)
(0.0247)
= 14.4
(0.0753)(0.151)2
3
(0.0620)
= 2.55
(0.138)(0.176)
(0.0620)
= 1.28
(0.138)2(0.176)
(0.0620)
= 14.5
2
(0.138)(0.176)
Chemical Equilibrium: Kinetic Analysis
Chemical reaction is at equilibrium when forward rate = reverse rate.
Dynamic not static.
Kinetic analysis of sample reaction:
Iodide ion catalyzed decomposition of hydrogen peroxide.
2 H2O2(aq)
Proposed mechanism:

2 H2O + O2(g)
H2O2 + I –
H2O + IO–
H2O2 + IO–
slow
H2O + I– + O2 fast
Rate law: rate forward = kforward [H2O2][I–]
Reverse reaction: 2 H2O + O2(g)
Mechanism will be same
except all steps reversed.
2 H2O2(aq
H 2 O + I– + O 2
H2O + IO–
H2O2 + IO– fast
H2O2 + I– slow
Rate law: ratereverse = kreverse [H2O]2[I–][O2]/[H2O2]
At equilibrium: rateforward = ratereverse kforward [H2O2][I–] = kreverse [H2O]2[I–][O2]/[H2O2]
Rearranging: kforward [H2O]2[I–][O2] [H2O]2[O2]
kreverse = [H2O2]2 ][I–] = [H2O2]2
= Kc (equilibrium constant)
Equilibrium Constants in Terms of Pressure, Kp
CO(g) + 2 H2(g)
Kp = (P CH3 OH)
(PCO )(PH )2
CH3OH(g)
2
Does Kp = Kc ?
From the ideal gas law: PCO = nCORT/VCO = [CO] RT
Replacing the pressures gives
Kp = (P CH3 OH)
(PCO )(PH )2
=
2
[CH3OH] (RT)
[CO] (RT) [H2]2 (RT)2
=
Kc (RT)–2
When does Kp = Kc ?
In an equilibrium involving gases when the total moles of gases on the
reactant side equals the total moles of gases on the product side.
2 NO(g) + 2 H2(g)
2 H2O(g) + N2(g)
2 NH3(g) + 2 O2(g)
3 H2O(g) + N2O(g)
Kp = Kc ?
No
4 mol ≠ 3 mol
Yes
4 mol = 4 mol
Equilibrium Constant Relationships
Significance of magnitude of K
If K >> 1 Equilibrium lies to the right, products predominate
If K << 1 Equilibrium lies to the left, reactants predominate
Change in direction or stoichiometry
CO(g) + 2 H2(g)
CH3OH(g)
Kc = [CH3OH]
[CO][H2]2
CH3OH(g)
Kc = [CO][H2]2 = 1/14.5 = 0.069
[CH3OH]
= 14.5
Reverse the reaction
CO(g) + 2 H2(g)
Double the reaction coefficients
2 CO(g) + 4 H2(g)
2CH3OH(g) Kc = [CH3OH]2
[CO]2[H2]4
= (14.5)2 = 210
Equilibrium Constant Relationships
Adding equilibrium reactions: to find equilibrium constant of the net reaction
2 NOBr(g)
2 NO(g) + Br2(g)
Br2 (g) + Cl2(g)
2 BrCl(g)
Kc = [NO2] 2[Br2] = 0.014
[NOBr] 2
Kc =
[BrCl]2
[Br2][Cl2]
= 7.2
Find Kc for the sum of these two reactions:
2 NOBr(g) + Cl2(g)
Kc = [NO2] 2[BrCl]2
[NOBr]2[Cl2]
2 NO(g) + 2 BrCl(g)
=
[BrCl]2
[NO2] 2[Br2]
X [Br ][Cl ]
[NOBr]2
2
2
= (0.014)(7.2) = 0.10
Equilibrium Constant Relationships
2 CH4(g)
2 C2H6(g) + 2 H2(g)
CH4(g) + H2O(g)
Find Kc for
CH3OH(g) + H2(g)
2CH3OH(g) + H2(g)
Kc = 9.5 x 10 –13
Kc = 2.8 x 10 –21
C2H6(g) + 2 H2O (g)
Le Châtelier’s Principle
If a system at equilibrium is stressed, the position of
equilibrium shifts in the direction that relieves that stress.
2 NO2(g)
2NO(g) + O2(g)
or 114 kJ + 2 NO2(g)
H = 114 kJ
2NO(g) + O2(g)
Predict effect on the equilibrium: shift direction, effect on K
1. Addition of some NO (temperature, volume constant)
Q > K, shift left to use some of added NO, K unchanged
2. Removal of some O2 (temperature, volume constant)
Q < K, shift right to replace some of removed O2, K unchanged
Le Châtelier’s Principle
2 NO2(g)
2 NO(g) + O2(g)
or 114 kJ + 2 NO2(g)
H = 114 kJ
2NO(g) + O2(g)
Predict effect on the equilibrium: shift direction, effect on K
3. Increase the volume (temperature constant)
Q < K, shift right (greater moles of gas) to fill volume, K unchanged
4. Increase the temperature (volume constant)
Q < K, shift right use added heat, Q unchanged, K increased
5. Add a catalyst (temperature, volume constant)
Q = K no effect on Q or K
6. Add an unreactive gas (temperature, volume constant)
Q = K no effect on Q or K
Determination of K
From determined equilibrium concentrations:
N2(g) + 3 H2(g)
2NH3(g)
Equilibrium Concentrations at 500 K
[N2] = 0.115 M
Kc =
[NH3]2
[N2][H2]3
[H2] = 0.105 M
[NH3] = 0.439 M

(4)
=
(5)(5)
= 45
Determination of K
Given all initial amounts and one equilibrium amount:
2 SO2(g) + O2(g)
[SO2]mol/L
2 SO3(g)
[O2]mol/L
[SO3]mol/L
Initial:
Change:
1.000
1.000
0.000
–0.926
–0.463
+0.926
Equilibrium:
0.074
0.547
0.926
Kc =
[SO3]2
[SO2]2[O2
]
()
=
(4)(54)
= 
Finding Equilibrium Concentrations
N2(g) + 3 H2(g)
2NH3(g)
Kp = 41 at 400 K
If PN2 = 0.10 atm and PH2 = 0.20 atm at equilibrium, find PNH3
Kp =
(PNH3)2
(PN2) (PH2)3
(PNH3)2
= Kp (PN ) (PH )3
2
2
(PNH3)2
=
(PNH3)
= (0.0328)1/2 = 0.18 atm
(4)()(0.20)3 = 0.0328
Finding Equilibrium Concentrations
Br2(g) + Cl2(g)
2 BrCl(g) Kc = 7.00 at 373 K
If initial [Br2] = 0.200 M and [Cl2] = 0.200 M , find
[BrCl] at equilibrium.
Set up an equilibrium table: Let x = change in [Br2]
[Cl2]
[Br2]
Initial
Change
Equilibrium
Kc =
2
0.200
–x
0.200–x
0.200
–x
0.200–x
[2x]2
[BrCl]
0
+2x
2x
4x2
[BrCl]
=
=
[Br2][Cl2]
[0.200–x][0.200–x] 0.0400 – 0.400x + x2
7.00 = (4x2)/(0.0400 – 0.400x + x2)
(7.00)(0.0400 – 0.400x + x2) = 4x2 = 0.280 –2.80 x + 7x2
3x2 – 2.80x + 0.280 = 0 (quadratic equation)
Finding Equilibrium Concentrations
x = –(–2.80)±[(–2.80)2 –(4)(3)(0.280)]1/2
(2)(3)
x = 2.80 ± (4.48)1/2 = 2.80 ± 2.12 = 0.820 or 0.113
6
6
[Cl2]
[BrCl]
[Br2]
Initial
0.200
0.200
0
+2x
Change
–
–
0.200–x
2x
Equilibrium 0.200–x
x
x
Only the x = 0.113 will work, x = 0.820 will give negative concentration values.
Initial
Change
Equilibrium
[Br2]
[Cl2]
[BrCl]
0.200
–0.113
0.087
0.200
–0.113
0.087
0
+0.226
0.226
[BrCl] = 0.226 M
An Alternative to Using the Quadratic Equation
(Works only when both the numerator and denominator are exact squares)
Kc =
[BrCl]2
[Br2][Cl2]
=
[2x]2
[0.200–x]2
= 7.00
2x
= 2.65
0.200–x
2x = (2.65)(0.200 – x) = 0.530 – 2.65x
2x + 2.65x = 4.65x = 0.530; x = 0.530/4.65 = 0.114
Taking the square root of both sides gives:
Solving for x gives:
Initial
Change
Equilibrium
Initial
Change
Equilibrium
[Br2]
[Cl2]
0.200
–x
0.200–x
0.200
–x
0.200–x
[BrCl]
0
+2x
2x
[Br2]
[Cl2]
[BrCl]
0.200
–0.114
0.200
–0.114
0.086
0.086
0
+0.228
0.228
[BrCl] = 0.228 M
Acid-Base Titration: Strong Acid with Strong Base
Titration of 50.0 mL0.100 M HCl with 0.100 M NaOH
excess strong acid
excess strong base
Acid-Base Titration: Strong Base with Strong Acid
Titration of 50.0 mL 0.100 M NaOH with 0.100 M HCl
excess strong base
excess strong acid
Acid-Base Indicators
Acid-Base Indicators: Use in Acid-Base Titration
Strong Acid-Strong Base
The color change of an indicator can be used to detect the
equivalence point for an acid-base titration.
Acid-Base Titration: Weak Acid with Strong Base
Titration of 0.100 M HC2H3O2 (CH3COOH)with 0.100 M NaOH
Buffer Region
weak acid
mixture:
weak base
weak acid
conjugate base
excess strong base
Acid-Base Titration: Weak Acid with Strong Base
Use of Indicators
Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
At equivalence point, product is a weak base: pH > 7
Acid-Base Titration: Weak Base with Strong Acid
Use of Indicators
Titration of 50.0 mL of 0.100 M NH3 with 0.100 M HCl
At equivalence point, product is a weak acid: pH < 7
Acid-Base Titration: Polyprotic Acid (H3PO4) with Strong Base