Transcript Chapter 17_18 Notes

Chapter 17
Reaction Rates
Chapter 18
Chemical
Equilibrium
Reaction Rate
• Change in concentration of reactant
or product per a unit of time.
Concentration of A at t2  Concentration of A t1
Rate 
t2  t1
Rate 
[A]t2  [A]t1
t2  t1
[A]
Rate 
t
[A] means concentration in mol/L
Collision theory
•
•
•
•
Molecules must collide to react.
Must collide hard enough.
Must collide with the right orientation.
Only a small number of collisions
produce reactions.
P
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t
e
n
t
i
a
l
Reactants
E
n
e
r
g
y
Products
Reaction Coordinate
P
o
t
e
n
t
i
a
l
Activation
Energy Ea
Reactants
E
n
e
r
g
y
Products
Reaction Coordinate
P
o
t
e
n
t
i
a
l
Activated
complex
Reactants
E
n
e
r
g
y
Products
Reaction Coordinate
P
o
t
e
n
t
i
a
l
Reactants
E
n
e
r
g
y
}
Products
Reaction Coordinate
E
Collision Theory
Breaking of
Covalent Bond
O
O
N
N
Br
Br
Forming of
Covalent Bond
O
O
N
N
N
Br
Br
Br---NO
P
o
t
e
n
t
i
a
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Br---NO
Transition
State
2BrNO
E
n
e
r
g
y
2NO + Br2
Reaction Coordinate
Terms
• Activation energy - the minimum energy
needed to make a reaction happen.
• Activated Complex or Transition State The arrangement of atoms at the top of
the energy barrier.
Conditions that Affect Reaction
Rates
1. Temperature
– Increase in temperature increases
reaction rate.
2. Catalyst
– a substance that speeds up a reaction
without being used up.
• Enzymes – catalysts found in our bodies.
3. Inhibitor
– slows down a reaction.
4. Concentration
– increases with increase in concentration.
5. Surface area
– increases with increase of surface area.
Activation Energy with Catalyst
Heterogeneous Reactions
• Homogeneous reaction – reactions
involving only one phase.
2BrNO(g)  2NO(g) + Br2(g)
• Heterogeneous reactions – reactions
that involve reactant in two phases.
Zn(s) + 2HCl(aq)  H2(g) + ZnCl2(aq)
Reaction Rate
The change in concentration of a reactant or
product per unit of time
[ A] at time t 2  [ A] at time t1
Rate 
t 2  t1
[ A]
Rate 
t
2NO2(g)  2NO(g) + O2(g)
Reaction Rates:
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically
2NO2(g)  2NO(g) + O2(g)
[NO2]
t
Reaction Rates:
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
if the rate is
dependent upon
concentration
[ NO2 ]
 constant
t
Calculate the average reaction rate of the reaction
[H2] (mol/L)
t(s)
below.
H2 ( g )  F2 ( g )  2HF ( g )
Average reaction rates:
0.090
100
0.080
200
 [H2 ]  [F2 ] [HF ]
,
,
,
t
t
t
Start with
(0.080 mol / L  0.090 mol / L )
 [H2 ]
 
t
200s  100s
4 mol
 1.0x 10
Ls
Now use stoich to calculate the rest:
 [F2 ]
1molF2
 4 mol
 1.0x 10
H2x
t
Ls
1molH2
 1.0x 10 4 M s 1
[HF ]
t
 1.0x 10
 4 mol
 2.0x 10 4 M s  1
2molHF
H2x
Ls
1molH2
Day 2
RATE LAWS
Rate Laws
Differential rate laws express (reveal) the
relationship between the concentration of
reactants and the rate of the reaction.
The differential rate law is usually just
called “the rate law.”
Writing a (differential) Rate Law
aA  bB  cC  dD
Rate  k [ A] [ B]
m
•
•
•
•
n
k is the rate constant
m,n represents the reaction order
Overall reaction order = m+n
All are determined experimentally
Reaction Orders
• What does the reaction order tell us?
– Reaction order tells us how the reaction
rate is affected by the concentration of
the reactants.
– If the reaction order is zero, then the
reaction rate is independent of the
reactant concentration.
Writing a (differential) Rate
Law
Problem - Write the rate law, determine the value of
the rate constant, k, and the overall order for the
following reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
1
2
3
[NO]
(mol/L)
0.250
0.500
0.250
[Cl2]
(mol/L)
0.250
0.250
0.500
Rate
Mol/L·s
1.43 x 10-6
5.72 x 10-6
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]m[Cl2]n
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Writing a Rate Law
Part 1 – Determine the values for the exponents
in the rate law: R = k[NO]2[Cl2]n
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Writing a Rate Law
Part 2 – Determine the value for k, the rate constant,
by using any set of experimental data:
R = k[NO]2[Cl2]
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
Writing a Rate Law
Part 3 – Determine the overall order for
the reaction.
R = k[NO]2[Cl2]
2+ 1= 3
 The reaction is 3rd order
Overall order is the sum of the
exponents, or orders, of the
reactants
Method of Initial Rates
NO2 ( g )  CO( g )  NO( g )  CO2 ( g )
1. Determine the rate law using method of initial
rates
2. Calculate the rate constant WITH UNITS
Exp
Initial rate
mol/L‧s
[NO2]0
mol/L
[CO]
mol/L
1
2
0.0050
0.080
0.10
0.40
0.10
0.10
3
0.0050
0.10
0.20
Method of initial rates
Step 1: write the generic rate law
Step 2:
compare the ratio of two
experiments where one of the
concentrations are constant and
solve for m, then repeat to solve
for n
Step 3: calculate k with units
Method of Initial Rates
NO2 ( g )  CO( g )  NO( g )  CO2 ( g )
Exp
Initial rate
mol/L‧s
[NO2]0
mol/L
[CO]
mol/L
1
2
0.0050
0.080
0.10
0.40
0.10
0.10
3
0.0050
0.10
0.20
Method of Initial Rates
NO2 ( g )  CO( g )  NO( g )  CO2 ( g )
Exp
Initial rate
mol/L‧s
1
2
3
0.0050
0.080
0.0050
[NO2]0
mol/L
0.10
0.40
0.10
[CO]
mol/L
0.10
0.10
0.20
Method of Initial Rates
NO2 ( g )  CO( g )  NO( g )  CO2 ( g )
Exp
Initial rate
mol/L‧s
1
2
3
0.0050
0.080
0.0050
[NO2]0
mol/L
0.10
0.40
0.10
[CO]
mol/L
0.10
0.10
0.20
Method of Initial Rates
NO2 ( g )  CO( g )  NO( g )  CO2 ( g )
Exp
Initial rate
mol/L‧s
1
2
3
0.0050
0.080
0.0050
[NO2]0
mol/L
0.10
0.40
0.10
[CO]
mol/L
0.10
0.10
0.20
Day 3
REACTION MECHANISMS
Reaction Mechanisms
The overall progress of a chemical reaction can be
represented at the molecular level by a series of simple
elementary steps or elementary reactions.
The sequence of elementary steps that leads to product
formation is the reaction mechanism.
2NO (g) + O2 (g)
2NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Intermediates are species that appear in a reaction mechanism
but not in the overall balanced equation.
An intermediate is always formed in an early elementary step
and consumed in a later elementary step.
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
The molecularity of a reaction is the number of molecules
reacting in an elementary step.
•
Unimolecular reaction – elementary step with 1 molecule
•
Bimolecular reaction – elementary step with 2 molecules
•
Termolecular reaction – elementary step with 3 molecules
Rate Laws and Elementary Steps
Unimolecular reaction
A
products
Bimolecular reaction
A+B
products
Bimolecular reaction
A+A
products
Writing plausible reaction mechanisms:
•
The sum of the elementary steps must give the overall
balanced equation for the reaction.
•
The rate-determining step should predict the same rate
law that is determined experimentally.
The rate-determining step is the slowest step in the
sequence of steps leading to product formation.
The experimental rate law for the reaction between NO2
and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1:
NO2 + NO2
NO + NO3
Step 2:
NO3 + CO
NO2 + CO2
What is the equation for the overall reaction?
What is the intermediate?
What can you say about the relative rates of steps 1 and 2?
Determining Molecularity and Rate Laws for Elementary Steps
PROBLEM:
(1)
The following two reactions are proposed as elementary steps in
the mechanism of an overall reaction:
NO2Cl(g)
NO2(g) + Cl (g)
(2)
NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Determine the molecularity of each step.
(c) Write the rate law for each step.
PLAN:
(a) The overall equation is the sum of the steps.
(b) The molecularity is the sum of the reactant particles in the step.
SOLUTION:
NO2Cl(g)
NO2(g) + Cl (g)
(2) NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
(a) (1)
2NO2Cl(g)
2NO2(g) + Cl2(g)
(b) Step(1) is unimolecular.
Step(2) is bimolecular.
(c) rate1 = k1 [NO2Cl]
rate2 = k2 [NO2Cl][Cl]
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Reaction energy diagram for the two-step NO2-F2 reaction
Day 4
EQUILIBRIUM
The Equilibrium Condition
• Equilibrium – the exact balancing of two
processes, one of which is the opposite
of the other.
20lb
20lb
• Chemical Equilibrium – a dynamic state where
the forward and the reverse reactions
balance each other because they take place at
equal rates.
H H
Br
H
H
Br
Br
Br
The Equilibrium Constant
• Law of Chemical Equilibrium – describes the
equilibrium condition.
aA + bB
cC + dD
• Equilibrium expression – represents the law of
equilibrium.
• Give the equilibrium expression for the
following reaction:
H2(g) + F2(g)
2HF(g)
Heterogeneous Equilibria
• Heterogeneous Equilibrium – does not depend
on the amounts of pure solids or liquids.
• The concentration of pure solids and liquids is
constant.
CaCO3(s)
CaO(s) + CO2(g)
• Give the equilibrium expression for the
following reaction:
2H2O(l)
2H2(g) + O2(g)
Calculating K
•Equilibrium concentrations are not always
the same.
•Equilibrium Position – a set of equilibrium
concentrations.
•The equilibrium constant remains the same.
N2(g) + 3H2(g)
Initial
InitialConditions
Conditions
Trial
1
2
3
[N2]0
[H2]0
1.0M 1.0M
0
0
2.0M 1.0M
[NH3]0
2NH3(g)
Equilibrium Conditions
[N2]
[H2]
[NH3]
K
0
.92M .76M
.16M
.060
1.0M
.39M 1.2M
.20M
.060
3.0M
2.6M 2.8M
1.8M
.060
Applications Involving the
Equilibrium Constant
• K > 1, Equilibrium lies to the right;
the forward reaction is predominate.
• K< 1, Equilibrium lies to the left; the
reverse reaction is predominate.
• This information allows us to know
how successful a reaction will be.
2SO2(g) + O2(g)
[SO3] = 3.50M
[SO2] = 1.50M
[O2] = 1.25M
2SO3(g)
Day 5
REACTION QUOTIENT AND
EQUILIBRIUM CONCENTRATIONS
The Reaction Quotient
For some time, t, when the system is not at
equilibrium, the reaction quotient, Q takes
the place of K, the equilibrium constant, in
the law of mass action.
jA + kB  lC + mD
l
m
[C ] [ D]
Q
j
k
[ A] [ B]
Q is changing as the
reaction proceeds to
equilibrium
Significance of the Reaction
Quotient
 If
K = Q, the system is at equilibrium
 If
K < Q then….
[Products]

is too large
[reactants]
the system shifts to the left, consuming
products and forming reactants until
equilibrium is achieved
Significance of the Reaction
Quotient

If K > Q then….
[Products]

is too small
[reactants]
the system shifts to the right,
products must be formed until
equilibrium is achieved
Significance of the Reaction
Quotient
N2O4 ( g )  2 NO2 ( g ) K  0.21 at 100C
 At a particular point in this reaction [N2O4] =
0.12 M and [NO2] = 0.55 M. Is the reaction at
equilibrium?
Calculate Q
Q 
NO2  2
N2O4 

0.55 2
0.12
 2.52
Q ≠ K so the reaction is not at equilibrium
Significance of the Reaction
Quotient
N2O4 ( g )  2 NO2 ( g ) K  0.21 at 100C
 Which
direction must the reaction proceed
to reach equilibrium?
EQUILIBRIUM
CONCENTRATIONS
Solving for Equilibrium
Concentration
Consider this reaction at some temperature:
H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0
Assume you start with 8 M H2O and 6 M CO. What
are concentrations of H2O, CO, H2, and CO2 are
present at equilibrium?
Here, we learn about “ICE” tables – the most
important problem solving technique in the
second semester.
Solving for Equilibrium
Concentration
H2O(g) + CO(g)  H2(g) + CO2(g)
K = 2.0
Step #1: We write the equilibrium expression for the
reaction:
[ H 2 ][CO2 ]
2.0 
[ H 2O][CO]
Solving for Equilibrium
Concentration
Step #2: We “ICE” the problem, beginning with the
Initial concentrations
M
Initial:
Change:
Equilibrium:
H2O(g) + CO(g)  H2(g) + CO2(g)
Solving for Equilibrium
Concentration
Step #3: We plug equilibrium concentrations into our
equilibrium expression, and solve for x (quadratic or
perfect square)
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium:
8-x
6-x
x
x
Solving for Equilibrium
Concentration
Step #4: Substitute x into our equilibrium concentrations
to find the actual concentrations
Equilibrium:
8-x
6-x
x
x
Equilibrium:
8-4=4
6-4=2
4
4
Day 6
LE CHATELIER’S PRINCIPLE
Le Chatelier’s Principle
• Le Chatelier’s Principle – If a stress
is added to a system at equilibrium,
the system counteracts that change
and reaches a new state of
equilibrium. (Don’t rock the boat.)
Stresses (change in):
Concentration
2SO3(g) ⇌ 2SO2(g) + O2(g)
Add SO3, which way will the
equilibrium shift?
Concentration
2SO3(g) ⇌ 2SO2(g) + O2(g)
Remove O2, which way will the equilibrium
shift?
Temperature
N2(g) + O2(g)
2NO(g)
This is an endothermic reaction, which way will
an increase in temperature shift the equilibrium?
N2(g) + O2(g) + energy ⇌ 2NO(g)
Temperature
2H2 + O2
2H2O
This is an exothermic reaction. Will the
concentration of the hydrogen gas increase or
⇌
decrease when the temperature
is increased?
2H2 + O2 ⇌ 2H2O + energy
The concentration of hydrogen will
increase.
Pressure
• Only GASES are affected.
1N2(g) + 3H2(g)
2NH3(g)
• Increase in Pressure – causes a shift
towards the lesser amount of gas
moles to relieve pressure.
• Decrease in Pressure – opposite
• When pressure is increased which way
will the equilibrium shift for the
reaction above?
Pressure
2SO3(g)
2SO2(g) + O2(g)
Will the concentration of SO3 increase or
decrease when the pressure is decreased?
SO3 will decrease.
Catalysts
• Increased rate of both the forward and
reverse reaction.
• Favors neither.
• Does not create a stress
Shift Right
• Products
favored.
• Forward rxn
speeds up.
Shift Left
• Reactants
favored.
• Reverse rxn
speeds up
Solubility Equilibria
Solubility Product Constant (Ksp) equilibrium constant for the dissolving
of sparingly soluble ionic compounds.
• Write the Ksp expression for PbCl2(s)
1. Write the equation for the dissolution
of the solid.
•
PbCl2(s) ⇌ Pb+2(aq) + 2Cl-1(aq)
2. Write the Ksp expression just as you did the
equilibrium expression.
CuBr has a solubility of 2.0 x 10-4 mol/L at
25OC. Calculate the solids Ksp.
CuBr(s) ⇌ Cu+1(aq) + Br-1(aq)
Ksp=
[Cu+1][Br-1]
Solubility
• The solubility product constant can be
used to calculate the solubility of an
ionic compound.
• Solubility - the amount of substance
that will dissolve in a given amount of
water.
• Find the solubility of AgI. Ksp = 8.5x1017.
Predicting Precipitation
Ion Product, Q =[M+]a[Nm-]b
If Q>Ksp a precipitate forms.
If Q<Ksp No precipitate, unsaturated.
If Q = Ksp saturated, no change will
occur.
• A solution is prepared by mixing equal
volumes of 1.00 x 10-2M Mg(NO3)2 and
1.00 x 10-1M NaF. Will a MgF2
precipitate form? (Ksp = 6.4x10-9).
•
•
•
•