Transcript 8-3-2011 - Professor Monzir Abdel

8-3-2011
It is clear from the first two experiments that when
the concentration of O3 was doubled, the rate was
doubled as well. Therefore, the reaction is first
order with respect to O3
From experiments 2 and 3 keeping the
concentration of O3 constant at 2.0x10-5 M,
decreasing the concentration of NO2 by one-half
results in a decrease of the initial rate by the same
value. The reaction is therefore first order with
respect to NO2
The rate law can be written as:
Rate = k [NO2][O3]
The rate constant can be calculated by
substitution for the concentrations of
reactants and corresponding rate
value, taking the first experiment, for
instance, will give:
0.022 mol L-1s-1 = k * 5.0x10-5 (mol L-1)*
1.0x10-5 (mol L-1)
k = 4.4x107 L mol-1 s-1
Example
In the reaction:
2 NOCl g 2 NO + Cl2
The following data was collected at 27
oC:
What is the rate law of the reaction?
What is the value of the rate constant?
Determination of Rate Law
F2 (g) + 2ClO2 (g)
Experiment
1
2
3
4
2FClO2 (g)
[F2] [ClO2] Rate (M/s)
0.04 0.03
1.0x10-2
0.04 0.04
1.3x10-2
0.02
0.04
0.04
0.06
6.7x10-3
2.0x10-2
Determination of Rate Law
Experiments 1 & 4
As [F2] doubles, so does the rate
Experiments 2 & 3
As [ClO2] doubles, so does the rate
2:2 ratio…..1:1 ratio
x = 1 and y = 1
Rate = k [F2] [ClO2]
Determine the rate law and calculate the rate
constant for the following reaction from the
following data:
S2O82- (aq) + 3I- (aq)
2SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
3
0.08
0.16
0.017
0.017
1.1 x 10-4
2.2 x 10-4
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]
k = rate
=
2[S2O8 ][I ]
2.2 x 10-4 M/s = 0.08/M•s
(0.08 M)(0.034 M)
Relation between Reactant
Concentration and Time
First Order Reaction- a reaction whose rate
depends on the reactant concentration raised to
the first power.
Reaction Type: A→B
Rate of: -Δ [A]/Δt or
k[A]
Combining and simplifying these equations
brings us to the following rate equation:
ln[A]t = -kt + ln[A0]
Relation between Reactant
Concentration and Time
Reaction Time
The reaction 2A
B is first order in A with a
rate constant of 2.8 x 10-2 s-1 at 800C. How long
will it take for A to decrease from 0.88M to 0.14M?
ln[A] = ln[A]0 - kt
kt = ln[A]0 – ln[A]
[A]0 = 0.88 M
[A] = 0.14 M
0.88 M
[A]0
ln
ln
ln[A]0 – ln[A]
[A] = 0.14 M
= 66 s
=
t=
-2
-1
2.8 x 10 s
k
k
Example
At 400 oC, the first order conversion of
cyclopropane into propylene has a rate
constant of 1.16x10-6 s-1. What will the
concentration of cyclopropane be after
24 hours of reaction if the initial
concentration was 0.0100 M.
Cyclopropane g propylene
ln [cyclopropane]0/[cyclopropane]t = kt
ln 0.0100/[cyclopropane]t = 1.16x10-6 s-1 *
(24 hr *3600 s/hr)
ln (0.0100 M)/[cyclopropane]t = 0.100
0.0100/[cyclopropane]t = e0.100
[cyclopropane]t = 9.01x10-3 M
The isomerization reaction
CH3NC g CH3CN
obeys the first order rate law
rate = k [CH3NC]
Measurements at 500 K reveal that in 520
seconds the concentration of CH3NC has
decreased to 71% of its original value.
Calculate the rate constant k of the
reaction at 500 K.
ln[At] = -kt + ln[A0]
It is important to remember that for first order
reactions, one does not need to know the
absolute values of [At] and [A0] to determine
the rate constant (if time is known) or the
time (if the rate constant is given).
One can work with relative values since the
above equation can be recast in the following
way: ln([At]/[A0]) = -kt
Since the relationship between [At] and
[A0] is known then the problem can be
solved.
The problem states 71% of A is left after
520 seconds:
ln([At]/[A0]) = -kt
ln(0.71[A0]/[A0) = -k (520 seconds)
k = 0.00066 s-1
Reaction Half-life
As a reaction proceeds, the concentrations
of the reactants decreases.
Another way to measure [reactant] over time
is to use the half-life.
Half-life, t1/2 – the time required for the
concentration of a reactant to decrease to
half of its initial concentration.
Reaction Half-life
Considering the condition where one-half of A
is consumed ([A]t = ½ [A]0), the time can be
denoted as t½;
ln [A]0 / ½ [A]0 = kt½
ln 2 = kt½
t½ = (ln 2)/k
t½ = 0.693/k
First-order reaction
A
product
# of
half-lives
1
[A] = [A]0/n
2
4
3
8
4
16
2
13.3
Reaction Half-life
The decomposition of RX has a rate law:
rate = k[RX]. If, at 550 oC, k = 0.032 s-1,
find:
1. The half life of the reaction
2. The initial concentration of RX
provided that the concentration of RX
after one min of reaction was 0.010 M.
a. Calculation of the half life of the reaction
t½ = (ln 2)/k
t½ = 0.693/0.032 s-1
t½ = 22 s
b. ln [A]0/[A]t = kt
ln [RX]0/(0.010 M) = 0.032 s-1 * (1 min * 60
s/min)
ln [RX]0/(0.010 M) = 1.92
[RX]0 = 0.068 M
Reaction Half-life
What is the half-life of N2O5 if it
decomposes with a rate constant of
5.7 x 10-4 s-1?
t½= ln2
k
t½ = 0.693 -4 -1
5.7 x 10 s
t½
= 1200 s
t½ = 20 minutes
Second-Order Reactions
•
•
Second-order reaction- a reaction whose rate
depends on the concentration of one reactant
raised to the second power OR on the
concentrations of two different reactants, each
raised to the first power.
Simple Type: A→B
rate = k[A]2
• Complex Type: A + B→C
rate = k[A][B]
Second-order Reactions
•
•
•
•
For A→B, the following expression is used:
rate = - D[A]/Dt = k[A]2
k = - D[A]/([A]2 * Dt)
M/M2s = 1/s = M-1s-1
Integration gives:
1 = 1 + kt
[A]t [A]0
• we can obtain t1/2, by setting [A]t = [A]o/2, therefore,
• 1/([A]o/2) = 1/[A]o + kt1/2
• t1/2 = 1/([A]o* k).
Half-life of a Second-order
Reaction
Equation for half-life
t½ =
1
k[A]0
What is the difference between this equation
and the equation for half-life of first-order
reactions?
At 25oC with CCl4 as solvent, the reaction:
I + I g I2
is second order with respect to the concentration of
iodine atoms. The rate constant has been measured
as 8.2 x 109 L mol-1s-1. Suppose the initial
concentration of I is 1.00 x 10-4 mol L-1. Calculate [I]
after 2.0 x 10-6 s.
In this particular problem, k is given as well as t and
[A0] while [At] is unknown. This is usually the case,
the integrated form of the second order rate law has
4 variables so three should be given in order to solve
for one .
The solution to this problem is straightforward,
it merely involves substituting the known
values for t, k and [A0] and then solving for
[At].
(1/[At]) = kt + (1/[A0])
(1/[At]) = (8.2 x 109 L mol-1 s-1) (2.0 x 10-6 s) +
(1/1.00 x 10-4 mol L-1)
(1/[At]) = (2.6 x 104 L mol-1)
[At] = 3.8 x 10-5 mol L-1
Zero-order Reactions
• Very rare reactions (dissociation of
ammonia)
• Usually occur on metallic surfaces
• Half-life Equation:
[A]0
t½ = 2k
• Reaction rate is described by:
Rate = k