Entry Task: Nov 5th Wednesday
Sign off on Expression and Law ws
Grab Integrated Law notes by door
MAYHAN
Agenda:
• Go through Expression and Law ws
• Walkthrough NOTES Ch. 14 sec 3 – The change
in concentration with time (integrated –
graph)
• Integrated Rate Law ws
MAYHAN
MAYHAN
1. Given the reaction: 2A (g) --->B (g) + C (g)
a. Express the rate of reaction in terms of the change in
concentration of each of the reactants and products.
- 1 Δ[A] = 1 Δ[B] = 1 Δ[C]
1 Δt
1 Δt
2 Δt
b. When [C] is increasing at 2.0 mol.L–1.s–1 , how fast is [A]
decreasing? SHOW WORK!
- 1 Δ[A] = 1 Δ[2.0]
1 Δt
2 Δt
4 mol.L-1s-1 =
A decreases twice as fast
Δ[A] = 2 Δ[2.0]
1 Δt
Δt
MAYHAN
2. Given the reaction: 2D (g) + 3E (g) + F (g) ----> 2G (g) + H (g)
a. Express the rate of reaction in terms of the change in
concentration of each of the reactants and products.
- 1 Δ[D] = -1 Δ[E] = -1 Δ[F] = 1 Δ[G] = 1 Δ[H]
2 Δt
1 Δt
1 Δt
3 Δt
2 Δt
b. When [E] is decreasing at 0.10 mol.L–1.s–1, how fast is [G]
increasing? SHOW WORK!
1 Δ[G] = - 1 Δ[0.10]
3 Δt
2 Δt
0.067 mol.L-1s-1 =
2/3 as fast
Δ[G] = 2 Δ[0.10]
3 Δt
Δt
MAYHAN
3. For the reaction 3 ClO – (aq) → ClO3 – (aq) + 2 Cl – (aq) doubling
the concentration of ClO – quadruples the initial rate of
formation of ClO3-. What is the rate law for the reaction?
Rate = k [ClO –]2
4. The reaction C6H5N2Cl (aq) + H2O (l) → C6H5OH (aq) + N2 (g) +
HCl (aq) is first order in C6H5N2Cl and zero order in H2O. What is
the rate law for this reaction?
Rate = k [C6H5N2Cl]
5. For the reaction H3PO4 (aq) + 3I – (aq) + 2 H+ (aq) → H3PO3 (aq) + I3- (aq) + H2O (l)
the rate law under certain conditions is given by Rate =
k[H3PO4][I -][H+]2. What method(s) could be used if you want
to double the reaction rate?
Double the concentration of H3PO4 or IMAYHAN
6. What is the overall order of reaction for each of the
following?
2nd
Zero
a) Rate = k[NO2]2 _______
b) Rate = k _______
1st
3rd
c) Rate = k[Br2] _______
d) Rate = k[NO]2[O2] ______
7. For the reaction 2 NO (g) + Cl2 (g) → 2 NOCl (g) If the
concentration of NO is tripled, the rate of the reaction
increases by a factor of nine. If the concentration of Cl2 is cut
in half, the rate of the reaction is decreased to half the
original rate. Find the order of reaction for each reactant and
write the rate law for the reaction.
NO is a second order Cl2 is 1st order
Rate = k [NO]2[Cl2]
MAYHAN
8. In the decomposition of ammonia on a platinum surface at
856°C , 2 NH3 (g) → N2 (g) + 3 H2 (g) , changing the concentration
of NH3 has no effect on the rate. Write the rate law for the
reaction.
Rate = k
9. A reaction has the experimental rate law of Rate = k[A]2.
a) What happens to the rate if the concentration of A is
tripled?
Rate = k [3]2
Rate would increase by a factor of 9
b) What happens to the rate if the concentration if A is
reduced to one third the initial concentration?
Rate = k [1/3 ]2
Rate would decrease by a factor of 1/9
MAYHAN
10. For the reaction 2 A + B → C + D, if the concentration of A
is doubled, the reaction rate doubles. If the concentration of
B is halved, there is no change in the reaction rate. Determine
the order of reaction with respect to each reactant and the
overall order of reaction. Write the rate law for the reaction.
A would be 1st order and B is zero order
Rate = k [A]
MAYHAN
MAYHAN
I can…
• Graph the relationship of time with amount
of reactant concentrations and integrate
this with rates of reactions.
• Determine the graphical relationship
between time and the rate order.
MAYHAN
Chemical
Kinetics
Equation Sheet
Under thermochemistry and kinetics
1st order
2nd order
MAYHAN
Chemical
Kinetics
Two Types of Rate Laws
1. Differential- Data table contains
RATE AND CONCENTRATION DATA.
Uses “table logic” or algebra to find the
order of reaction and rate law
2. Integrated- Data table contains TIME
AND CONCENTRATION DATA. Uses
graphical methods to determine the
order of the given reactant. K=slope of
best fit line found through linear
Chemical
Kinetics
regressions
MAYHAN
Integrated Rate Law
• Can be used when we want to know
how long a reaction has to proceed to
reach a predetermined concentration of
some reagent
MAYHAN
Chemical
Kinetics
Graphing Integrated Rate Law
• Time is always on x axis
• Plot concentration on y axis of 1st graph
• Plot ln [A] on the y axis of the second
graph
• Plot 1/[A] on the y axis of third graph
• You are in search of a linear graph
MAYHAN
Chemical
Kinetics
Zero order ReactionsUse A  B as an example. What happens when
we double [A], what happens to the rate of
reaction that is zero order?
The rate of reaction does not change
So does the concentration affect rate? Y/N____
No
What would the rate law be for a zero order?
Rate = k
MAYHAN
Chemical
Kinetics
Sketch a graph with rate on Y and
concentration on X axis- Label axis!!
As concentration
increases, the
rate of reaction
remains the
same.
MAYHAN
Chemical
Kinetics
Sketch a graph with concentration on Y
and time on X axis- Label axis!!
Integrated Rate laws
We look for straight
lines. This provides
a “clean” visual about
the relationship of
concentration and
time.
MAYHAN
Chemical
Kinetics
Sketch a graph with concentration on Y
and time on X axis- Label axis!!
So when we plot our
data table and get a
negative straight line
it is
Zero
____________order!
Slope is negative (-k)
MAYHAN
Chemical
Kinetics
First Order Reactions
AB in a reaction.
① Write the rate expression for reactant A.
(sec. 1 stuff)
Rate = -
∆[A]
∆t
MAYHAN
Chemical
Kinetics
14.3- The Change of Concentration with Time
② Write the rate law for reactant A. (sec 2
stuff)
Rate = k[A]1
MAYHAN
Chemical
Kinetics
14.3- The Change of Concentration with Time
Describe a First Order reaction.
Double the concentration the reaction doubles.
Low amount of reactant = low rate of reaction
MAYHAN
Chemical
Kinetics
Sketch a graph with rate on Y and
concentration on X axis- Label axis!!
As we double our
concentration , the
rate doubles. It’s a
direct relationship.
MAYHAN
Chemical
Kinetics
Sketch a graph with concentration on Y
and time on X axis- Label axis!!
Integrated Rate laws
We look for straight
lines. This provides
a “clean” visual about
the relationship of
concentration and
time. This does not
provide a straight line
MAYHAN
Chemical
Kinetics
Sketch a graph with concentration on Y
and time on X axis- Label axis!!
Integrated Rate laws
We can manipulate
the data to provide a
straight line plot.
Change how we plot
concentration.
Natural log x Concentration
In [A]
Chemical
Slope is negative (-k)
Kinetics
MAYHAN
14.3- The Change of Concentration with Time
Take equations ① and ② and smash
them together.
Rate = -
∆[A]
∆t
MAYHAN
= k[A]
Chemical
Kinetics
14.3- The Change of Concentration with Time
How do you use this equation to solve for
concentration?
Using calculus to integrate the rate law
for a first-order process gives us
[A]t
ln
= −kt
[A]0
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t, Chemical
Kinetics
MAYHAN
during the course of the reaction.
Integrated Rate Laws
Manipulating this equation produces…
[A]t
ln
[A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y
= mx + b
MAYHAN
Chemical
Kinetics
First-Order Processes
Relate this equation to the slope.
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a
plot of ln [A] vs. t will yield a straight
line, and the slope of the line will be -k.
MAYHAN
Chemical
Kinetics
Sample Exercise 14.5 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of
1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of
5.0  107 g/cm3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of
the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to
decrease to 3.0  10–7 g/cm3? PLUG & CHUG
k = 1.45 yr-1
ln [A]0 = [5.0 x 10-7g/cm3]
SET IT UP
ln [insecticide]t-1yr = [X]
t = 1 year
ln [insectacide]t-1yr = -(1.45 yr-1) (1 year) + ln [5.0 x 10-7g/cm3]
ln[insecticide]t -1 yr = 1.45 + (14.51)
ln[insecticide]t - 1 yr = 15.96
Get rid of ln by ex on both sides
7 g/cm3
[insecticide]t = 1 yr = e15.96 = 1.2  10MAYHAN
Chemical
Kinetics
Sample Exercise 14.5 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of
1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of
5.0  107 g/cm3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of
the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to
decrease to 3.0  10–7 g/cm3? PLUG & CHUG
k = 1.45 yr-1
ln [A]0 = [5.0 x 10-7g/cm3]
ln [3.0 10-7]t =
t = X
SET IT UP
ln [3.0 10-7]t = -(1.45 yr-1) X + ln [5.0 x 10-7g/cm3]
Get X by itself- move to left side
ln [3.0 10-7]t - ln [5.0 x 10-7g/cm3] =X
-1.45
yr-1
MAYHAN
-15.02 - -14.51
=0.35 years
-1.45 yr
Chemical
Kinetics
Sample Exercise 14.5 Using the Integrated First-Order Rate Law
Continued
Practice Exercise
The decomposition of dimethyl ether, (CH3)2O, at 510 ºC is a first-order process with a rate constant of 6.8  10–4 s–1:
(CH3)2O(g)  CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is its pressure after 1420 s?
k = 6.8 x 10-4-s-1
ln [A]0 = [135 torr]
SET IT UP
ln [X torr]t = [X]
t = 1420 s
ln [X torr]t = -(6.8 x 10-4) (1420 s) + ln [135 torr]
ln[torr]t = 0.9656 + (4.91)
ln[torr]t = 3.94
Get rid of ln by ex on both sides
[torr]t = e3.94= 51.4 torr
MAYHAN
Chemical
Kinetics
14.3- The Change of Concentration with Time
Describe a second-order reaction.
When you double the reactant the rate
increases by a power of 2, to quadruple the
rate
MAYHAN
Chemical
Kinetics
Sketch a graph with rate on Y and
concentration on X axis- Label axis!!
The relationship is more
pronounced.
Double your
concentration and the
rate goes up by the
power of 2.
Hence- second order.
MAYHAN
Chemical
Kinetics
Sketch a graph with concentration on Y
and time on X axis- Label axis!!
Integrated Rate laws
We look for straight
lines. This provides
a “clean” visual about
the relationship of
concentration and
time. This does not
provide a straight line
MAYHAN
Chemical
Kinetics
Sketch a graph with concentration on Y
and time on X axis- Label axis!!
Integrated Rate laws
We can manipulate
the data to provide a
straight line plot.
Change how we plot
concentration.
1 divided by Concentration
1/[A]
Chemical
And the slope is positive
Kinetics
MAYHAN
(k)
Second-Order Processes
Provide the second order equation.
Similarly, integrating the rate law for a
process that is second-order in reactant
A, we get
1
1
= kt +
[A]t
[A]0
also in the form
y = mx + b
MAYHAN
Chemical
Kinetics
Second-Order Processes
1
1
= kt +
[A]t
[A]0
So if a process is second-order in A, a
plot of 1/[A] vs. t will yield a straight line,
and the slope of that line is k.
MAYHAN
Chemical
Kinetics
14.3- The Change of Concentration with Time
What does second order reactions depend
on?
A second order reaction is one whose rate
depends on the initial reactant concentration
MAYHAN
Chemical
Kinetics
Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
MAYHAN
Chemical
Kinetics
Second-Order Processes
• Graphing ln [NO2] vs. t
yields:
• The plot is not a straight
line, so the process is not
first-order in [A].
Time (s)
0.0
50.0
[NO2], M
0.01000
0.00787
ln [NO2]
−4.610
−4.845
100.0
200.0
300.0
0.00649
0.00481
0.00380
−5.038
−5.337
−5.573MAYHAN
Chemical
Kinetics
Second-Order Processes
• Graphing ln
1/[NO2] vs. t,
however, gives this
plot.
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
1/[NO2]
100
127
154
208
MAYHAN
263
• Because this is a
straight line, the
process is secondorder in [A].
Chemical
Kinetics
[A]  [A]0  kt
1
1
ln[A] 
 kt 
[A]0
 kt  ln[A] 0 [A]
M
s
[A] vs. t
1
s
ln[A] vs. t
MAYHAN
1
M s
1
vs. t
[A]
Chemical
Kinetics
Practice with graphs- After creating regression
graphs of various reactions, provide the rate
order for each graph.
What order is this reaction and what formula would I use to calculate
various times/concentrations?
First order and
ln[A]   kt  ln[A] 0
MAYHAN
Chemical
Kinetics
Practice with graphs- After creating regression
graphs of various reactions, provide the rate
order for each graph.
What order is this reaction and what formula would I use to calculate
various times/concentrations?
Zero order and
[A]  [A]0  kt
MAYHAN
Chemical
Kinetics
Practice with graphs- After creating regression
graphs of various reactions, provide the rate
order for each graph.
What order is this reaction and what formula would I use to calculate
various times/concentrations?
Second order and
1
1
 kt 
[A]
[A]0
MAYHAN
Chemical
Kinetics
MAYHAN
Chemical
Kinetics