Announcements
For the projects: answer “what chemistry needs to be
done” to get my product to market.
You are then make a plan that your team will
implement in the lab. You are to describe what each
member in your group will do in the lab.
Identify chemistry that needs to be done but eliminate
that which can not be done (explain why).
Provide a detailed costing analysis of your product.
Yep, somebody has to find out pricing and availability.
Effort has to go way way way up or expect grades to be
low across the board.
A chemical reaction rate is the change in the
concentration (molarity) of a reactant or a product
with time. By convention, the reaction rate it is
always a positive number.
A
B
rate = - Δ[A]
Δt
Δ[A] = [A]t - [A]t=0 = change
in concentration of [A] over
a period of time Δt = t - t0
Because [A] is a reactant it decreases with time therefore
Δ[A] is a negative value so we place a minus sign in the expression!
Δ[B]
rate =
Δt
Δ[B] = change in concentration
of B over time period Δt
Because [B] is a product it increases with time: Δ[B] is a positive
value and so does the rate!
The half-life, t1/2 is the
time for the half of the
initial concentration [A]t=0
falls to 1/2 [A]t=0.
for a first-order process
k
=
0.693
Experiments show this is a 1st
t=0
order reaction
t1/2
0.5
k
[N2O5]
t1/2 =
ln 2
0.6
2N2O5 ==> 2N2 + 5O2
2t1/2
0.4
3t1/2
0.3
0.2
0.1
0.0
0
24
48
Time (min)
72
t1/2 can be obtained for all the integrated rate laws by
substituting 1/2[A]o for [A]t in each integrated
equation.
[A]t - [A]0 = -kt
0th Order 1/2[A]0 - [A]0 = -kt
1/2[A]0 = kt
t1/2 = [A]0/2k
1st
Order
2nd Order
ln[A]t - ln[A]0 = -kt
ln(1/2[A]0) - ln[A]0 = -kt
ln(1/2) = -kt
t1/2 = 0.693/k
t1/2 ([A]0)
independent
of [A]
1/[A]t - 1/[A]0 = kt
1/1/2[A]0 - 1/[A]0 = kt
t1/2 =1/k[A]0
t1/2 ([A]0)
Summary of Kinetic Reaction Equations
A→C
rate law
integrated rate law in
straight-line form
0th Order
rate = k
[A]t = -k t + [A]0
1st Order
rate = k[A]
1/[A]t = k t + 1/[A]0
ln[A]t vs t
1/[A]t vs t
[A]t vs t
slope, y-intercept
-k, [A]0
-k, ln[A]0
t1/2 half-life
[A]0/2k
0.693/k
mol/L.s
rate = k[A]2
ln[A]t = -k t + ln[A]0
straight line plot
units for k
2nd Order
1/s
k, 1/[A]0
1/k[A]0
1/M sec
Imagine that the rate law for the reaction of A → B
is zero order in reactant A and that the rate
constant, k, is known to be 0.02 M/s. If the
reaction begins with 1.50 M A, how much [A] is
present 15 seconds after the reaction starts?
Imagine that the rate law for the reaction of A → B
is zero order in reactant A and that the rate
constant, k, is known to be 0.02 M/s. If the
reaction begins with 1.50 M A, how much [A] is
present 15 seconds after the reaction starts?
Because the problem states it is 0th Order we must use the
0th Order Integral Rate Law to determine [A] after 15 seconds
elapses.
[A]t = [A]0 - k t
= 1.50M - (0.02M/sec) (15 sec)
[A]t = 1.3M
The rate law for the reaction of A→2B is zero
order in A and has a rate constant of 0.12 M/s.
If the reaction starts with 1.50 M A, after what
time will the concentration of A be 0.90M?
The rate law for the reaction of A→2B is zero
order in A and has a rate constant of 0.12 M/s.
If the reaction starts with 1.50 M A, after what
time will the concentration of A be 0.90M?
In this case: [A]0 = 1.50M
[A]t = 0.90M
k = 0.12 M/sec
[A]t = [A]0 - kt
t = ([A]0 - [A]t)/k
t = (1.50M - 0.90M)/0.12M/sec
t = 5 sec
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 0.88 M to 0.14
M?
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 0.88 M to 0.14
M?
1st order and time!
We are given: [A] = 0.88 M
0
k = 0.028 sec-1
[A]t = 0.14 M
ln[A] - ln[A]0 = - k t
1st order integrated law
k t = ln[A]0 – ln[A]
ln[A]0 – ln[A]
=
t=
k
ln
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x
10-2
s-1
= 66 s
The first order rate constant is 1.87 x 10-3 min-1 at 37˚C
for a reaction of cisplatin (a cancer drug) with water.
Suppose that the concentration of cisplatin in the blood
stream in a cancer patient is 4.73 x 10-4 mol/L. What
will be the concentration of cisplatin 24 hours later?
The first order rate constant is 1.87 x 10-3 min-1 at 37˚C
for a reaction of cisplatin (a cancer drug) with water.
Suppose that the concentration of cisplatin in the blood
stream in a cancer patient is 4.73 x 10-4 mol/L. What
will be the concentration of cisplatin 24 hours later?
ln[A]t = −k t + ln[A]0
[A]0 = 4.73 X 10-4 M
k
= 1.87 x 10-3 min-1
ln[A]t − ln[A]0 = −k t
ln[A]t
= −k t
exp(-1.87 x 10-3 min-1 x 24 hr x 60 min/hr)
ln[A]0
!
"
ln[A]t
exp
= exp(−k t)
ln[A]0
[A]t = A0 exp(−k t)
Cis24 = 3.20 x 10-5 M
At 10000C, cyclobutane (C4H8) decomposes in a
first-order reaction, with the very high rate constant
of 87s-1, to two molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M,
what is the concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed
in this time?
PLAN:
We need to find the [C4H8] at time, t. We have to use
the integrated rate law for a 1st order reaction. Once
that value is found, divide the amount decomposed by
the initial concentration.
Summary of Kinetic Reaction Equations
A→C
zero
order
rate law
rate = k
integrated rate law in
straight-line form
[A]t = -k t + [A]0
first
order
rate = k[A]
second
order
rate = k[A]2
ln[A]t = -k t + ln[A]0
1/[A]t = k t + 1/[A]0
ln[A]t vs t
1/[A]t vs t
straight line plot
[A]t vs t
slope, y-intercept
-k, [A]0
-k, ln[A]0
k, 1/[A]0
t1/2 half-life
[A]0/2k
0.693/k
1/k[A]0
units for k
mol/L.s
1/s
L/mol.s
At 10000C, cyclobutane (C4H8) decomposes in a
2nd-order reaction, with the very high rate constant
of 87s-1, to two molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M, what
is the concentration after 0.010 s?
2nd order rate equation: 1/[A]t = k t + 1/[A]0
Given: [A]0 = 2.00 M, k = 87, t = 0.010 sec
Rearranging:
Rearranging:
[A]t = 1/(kt + 1/[A]0)
[A]t = 1/[(87 s-1 (0.01 s)) + 1/2M] = 0.7299M
(b) What fraction of C4H8 has decomposed in this
time?
Fraction decomposed = ([A]0 - [A]t) /[A]0
Fraction decomposed = (2M - 0.7299) /2M =
0.635 ~ 0.64
At 10000C, cyclobutane (C4H8) decomposes in a
first-order reaction, with the very high rate constant
of 87s-1, to two molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M,
what is the concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed
in this time?
(a)
ln
(b)
[C4H8]0
[C4H8]t
= kt
ln
[C4H8] =
0.83mol/L
[C4H8]0 - [C4H8]t
[C4H8]0
=
2.00
= (87s-1)(0.010s)
[C4H8]t
2.00M - 0.87M
2.00M
= 0.58
At 10000C, cyclobutane (C4H8) decomposes in a
first-order reaction, with the very high rate constant
of 87s-1, to two molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M, what is
the concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed in
this time?
[C4H8]0
ln
= kt
[C4H8]t
(a)
2.00
ln
= (87s-1)(0.010s)
[C4H8]t
[C4H8] = 0.83 mol/L
(b)
[C4H8]0 - [C4H8]t
[C4H8]0
=
2.00M - 0.87M
2.00M
= 0.58
The rate constant for the second order reaction
2A→B is 5.3×10-5 M-1s-1. What is the original
amount present if, after 2 hours, there is
0.35M available?
A0=0.40 M
The rate constant for a second order reaction is
4.5×10-4 M-1s-1. What is the half-life if we start
with a reactant concentration of 5.0 M?
t1/2 =440 s
=7.4 min
The Chernobyl nuclear reactor accident occurred in
1986. At the time that the reactor exploded some 2.4
MCi of radioactive 137Cs was released into the
atmosphere. The half-life of 137Cs is 30.1 years. In
what year will the amount of 137Cs released from
Chernobyl finally decrease to 100 Ci? A Ci is a unit of
radioactivity called the Curie.
Cyclopropane is the smallest cyclic hydrocarbon.
Because its 60o bond angles allow poor orbital
overlap, its bonds are weak. As a result, it is
thermally unstable and rearranges to propene at
1000 oC via the following first-order reaction:
CH2
H2 C
CH2
( g)
Δ
H3 C
The rate constant is 9.2 s-1. (a) What is the half-life of the
reaction? (b) How long does it take for [cyclopropane] to
reach one-quarter of its initial value?
0.693
, to find the half-life.
k
One-quarter of the initial value means two half-lives have passed.
PLAN: Use the half-life equation, t1/2 =
SOLUTION:
(a)
t1/2 = 0.693 / 9.2 s-1 = 0.075 s
(b)
2 t1/2 = 2 (0.075 s) = 0.15 s
That plot that gives a straight
line tells us what the order is!
Plot [A] vs t.
Time (s) [A] (mM)
Plot ln[A] vs t.
Plot 1/[A] vs t.
0.0
5.0
1.00 M
0.62 M
10.0
0.44 M
15.0
0.38 M
20.0
0.24 M
25.0
0.21 M
30.0
0.20 M
35.0
0.18 M
40.0
0.17 M
We can use the logarithmic form or the
exponential form of the integrated rate law.
ln[A]t = −k t + ln[A]0
1st order Integrated Rate
Equation (log form)
ln[A]t − ln[A]0 = −k t
Moving ln[A]0 to the left side
ln[A]t
= −k t
ln[A]0
ln A - ln B = ln (A/B)
exp
!
ln[A]t
ln[A]0
"
= exp(−k t)
[A]t = A0 exp(−k t)
Take the exp of each side
Exponential Form of
Integrated Rate Equation