Integrated Rate Laws

Download Report

Transcript Integrated Rate Laws 

Integrated Rate Laws
Finally a use for calculus!
Text 692019 and Questions to 37607
1
What is a rate?
It’s a β€œdelta/delta”!
Rate of reaction =
Ξ”π‘π‘œπ‘›π‘π‘’π‘›π‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›
Ξ”π‘‘π‘–π‘šπ‘’
In other words, it is a differential.
As you MAY recall from calculus, if you take a small
enough delta (difference) you end up with a
derivative!
Text 692019 and Questions to 37607
2
A rate as a derivative
Rate of reaction =
Ξ”π‘π‘œπ‘›π‘π‘’π‘›π‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›
Ξ”π‘‘π‘–π‘šπ‘’
If Ξ”time is small enough, we have:
βˆ’π‘‘[π‘Ÿπ‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘‘]
π‘…π‘Žπ‘‘π‘’ π‘œπ‘“ π‘Ÿπ‘’π‘Žπ‘π‘‘π‘–π‘œπ‘› =
𝑑𝑑
Why β€œ-”? Because you are losing reactants and the rate
should always be positive.
Text 692019 and Questions to 37607
3
Let’s look at the rate law
Rate = k[A]
βˆ’π‘‘[𝐴]
π‘…π‘Žπ‘‘π‘’ π‘œπ‘“ π‘Ÿπ‘’π‘Žπ‘π‘‘π‘–π‘œπ‘› =
= π‘˜[𝐴]
𝑑𝑑
This is actually an integrable equation.
[Don’t worry, this isn’t a math class…it’s just
masquerading as one!]
Text 692019 and Questions to 37607
4
Solving the equation
I’ll show you how to solve it, but it is only the
solution that you need to know.
βˆ’π‘‘[𝐴]
= π‘˜[𝐴]
𝑑𝑑
We collect the [A] on one side and get:
𝑑[𝐴]
= βˆ’π‘˜π‘‘π‘‘
[𝐴]
Text 692019 and Questions to 37607
5
Solving the equation
𝑑[𝐴]
= βˆ’π‘˜π‘‘π‘‘
[𝐴]
Now you can integrate both sides:
𝐴 π‘“π‘–π‘›π‘Žπ‘™
𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
𝑑[𝐴]
=βˆ’
[𝐴]
π‘“π‘–π‘›π‘Žπ‘™ π‘‘π‘–π‘šπ‘’
π‘˜π‘‘π‘‘
π‘‘π‘–π‘šπ‘’=0
Text 692019 and Questions to 37607
6
Solving the equation
𝐴 π‘“π‘–π‘›π‘Žπ‘™
𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
ln 𝐴
𝑑[𝐴]
=βˆ’
[𝐴]
π‘“π‘–π‘›π‘Žπ‘™
βˆ’ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™ π‘‘π‘–π‘šπ‘’
π‘˜π‘‘π‘‘
π‘‘π‘–π‘šπ‘’=0
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
= βˆ’π‘˜π‘‘
This is the only equation we really need. This is
called the β€œintegrated rate law”…well, because
we integrated the rate law. 
Text 692019 and Questions to 37607
7
What it means…
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
βˆ’ ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
= βˆ’π‘˜π‘‘
What it means is that the concentration at any time
decays logarithmically from the initial concentration. If I
rearrange the equation a little:
ln 𝐴 π‘“π‘–π‘›π‘Žπ‘™ = βˆ’π‘˜π‘‘ + ln 𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
What does this look like to you?
Yes, it is the equation of a straight line (y=mx+b)!
Text 692019 and Questions to 37607
8
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
If you know k and the initial concentration, you could
calculate the concentration at any time.
For example, if I know k=0.015 s-1 and I start with 0.250
M A, how much A is left after 1 minute?
Beware the units. 1 minutes = 60 seconds. Since k is in s1, I need my time to be in seconds.
Plug and chug, baby!
Text 692019 and Questions to 37607
9
Using the integrated rate law
ln 𝐴 π‘“π‘–π‘›π‘Žπ‘™ = βˆ’π‘˜π‘‘ + ln 𝐴
ln[A]final = -0.015s-1*60 s + ln(0.250 M)
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
ln[A]final = -2.286
[A]final = e-2.286 = 0.102 M
You can see the power of the integrated rate law. I
can determine the remaining concentration of
reactants at any second in time! (And, using
stoichiometry, I could determine the concentration
of products at any second in time!)
Text 692019 and Questions to 37607
10
Compare the integrated rate law to the
rate law
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Rate = k[A]
For the same problem, the rate law only allows
me to calculate the initial rate of the reaction:
Rate = (0.015 s-1)[0.250 M) = 0.00375 M/s
I could also calculate the RATE for any specific
concentration. But I can’t know how long it
takes me to get to that new concentration.
Text 692019 and Questions to 37607
11
Other uses of the integrated rate law
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
It’s a straight line. Scientists LOVE LOVE LOVE straight lines!
If you have a reaction that you KNOW is 1st order, you could
measure the [A] at a number of different times and plot the
data and you’ll get a straight line where the slope=-k. So you
could use the equation to find the rate constant.
Text 692019 and Questions to 37607
12
For example, suppose I monitor [A]
Time (seconds)
[A] (M)
0.25
0.20
0.17
0.075
0
10
20
60
Since this is a first order reaction, the data should obey my
integrated rate law.
So I plot the ln[A] vs time and I should get a straight line.
Text 692019 and Questions to 37607
13
For example, suppose I monitor [A]
Time (seconds)
0
10
20
60
[A] (M)
ln[A]
0.25
0.20
0.17
0.075
-1.386
-1.609
-1.772
-2.590
Now, I plot the last column against the first column and put
the best fit straight line on it.
Text 692019 and Questions to 37607
14
LOOK! IT’S A POINT! IT’S A PLANE!
NO!!! IT’S A STRAIGHT LINE!
-1
0
10
20
30
40
50
60
70
-1.2
-1.4
y = -0.02x - 1.3863
R² = 1
-1.6
-1.8
-2
-2.2
-2.4
-2.6
-2.8
Text 692019 and Questions to 37607
15
So, what’s the rate constant?
y = -0.02x - 1.3863
ln[A]final = - kt + ln[A]t=0
m= slope=-0.02
m=-k
k=-(-0.02)=0.02 s-1
So, if I KNOW it’s a 1st order reaction, I can make a graph
to find the rate constant. I can also make a graph to find
out IF it is 1st order.
Text 692019 and Questions to 37607
16
Different reaction
2 H2 + O2 β†’ 2 H2O
Time (seconds)
0
10
20
60
[H2] (M)
ln[H2]
0.500
0.300
0.200
0.100
-0.69315
-1.20397
-1.60944
-2.30259
Now, I plot the last column against the first column and put
the best fit straight line on it to see IF IF IF it is actually a
straight line.
Text 692019 and Questions to 37607
17
NOT a straight line – NOT a 1st order
reaction!
0
0
10
20
30
40
50
60
70
-0.5
y = -0.0249x - 0.8919
R² = 0.9288
-1
-1.5
-2
-2.5
-3
Text 692019 and Questions to 37607
18
Is it or isn’t it a straight line?
A.
B.
C.
D.
E.
It is a straight line
It is NOT a straight line
I can’t tell without error bars
I really don’t care it’s Monday
Your mother!
Text 692019 and Questions to 37607
19
This works for other orders of reaction
also.
For a second order reaction:
Rate = k[A]2
You get an integrated rate law
1
𝐴
= π‘˜π‘‘ +
π‘“π‘–π‘›π‘Žπ‘™
1
𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Same idea, it’s a straight line (y = mx+b) where:
Slope = k
Intercept =
1
𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Text 692019 and Questions to 37607
20
Hey! It’s 2nd order!
12
y = 0.1329x + 2.0924
R² = 0.9977
10
1/[H2}
8
6
4
2
0
0
10
20
30
40
50
60
70
Time (s)
Text 692019 and Questions to 37607
21
Also, there’s the rare zeroth order
βˆ’π‘‘[𝐴]
π‘…π‘Žπ‘‘π‘’ =
= π‘˜
𝑑[𝑑]
If you integrate
[A]t = -kt + [A]
Text 692019 and Questions to 37607
22
Those are the easy ones
For more complicated mixed orders like:
Rate = k[A][B]
The math gets much more complicated, so we’ll
ignore them until you become a chemistry
major. But you can do a similar thing.
Text 692019 and Questions to 37607
23
But a lot of reactions fall into those
three categories.
0th order
[A]t = -kt + [A]
1st order
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2nd order
1
𝐴
π‘“π‘–π‘›π‘Žπ‘™
= π‘˜π‘‘ +
1
𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Text 692019 and Questions to 37607
24
How do we use this?
N2 (g)+ 3 Cl2(g)β†’ 3 NCl3(g)
Given the following data, determine the rate law.
Time
[N2(g)] (M)
0 min
0.40
5 min
0.25
10 min
0.17
30 min
0.04
60 min
0.005
Text 692019 and Questions to 37607
25
GRAPH IT!
Text 692019 and Questions to 37607
26
Graph It!
N2 (g)+ 3 Cl2(g)β†’ 3 NCl3(g)
Given the following data, determine the rate law.
Time
[N2(g)] (M)
Ln([N2])
1/[N2]
0 min
0.40
-0.916
2.5
5 min
0.25
-1.386
4.0
10 min
0.17
-1.772
5.88
30 min
0.04
-3.219
25
60 min
0.005
-5.298
200
Text 692019 and Questions to 37607
27
Try all 3 and see which one fits!
Text 692019 and Questions to 37607
28
Not 0th order – unless it was a sloppy
experiment
0.45
y = -0.0107x + 0.3348
R² = 0.8641
0.4
0.35
[H2] (M)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
5
10
15
20
25
30
35
time (s)
Text 692019 and Questions to 37607
29
Maybe 2nd order – it’s not a horrible
fit.
30
y = 0.787x + 0.4915
R² = 0.9664
25
1/[H2}
20
15
10
5
0
0
5
10
15
20
25
30
35
Time (s)
Text 692019 and Questions to 37607
30
Hey, Goldilocks! It Fits 1st order!
0
0
10
20
30
40
50
60
70
-1
y = -0.0722x - 1.0024
R² = 0.9989
ln([H2]
-2
-3
-4
-5
-6
time (s)
Text 692019 and Questions to 37607
31
What if I don’t want to or can’t make a
graph?
A. Find someone who can make a graph.
B. Copy the answer from the person next to me.
C. Calculate the rate of the reaction and see if
the rate is constant or if the ln(rate) is
constant or 1/rate is constant.
D. Calculate the slope between data points and
see if they are constant.
Text 692019 and Questions to 37607
32
What if I don’t want to make a graph?
N2 (g)+ 3 Cl2(g)β†’ 3 NCl3(g)
Given the following data, determine the rate law.
Time
[N2(g)] (M)
0 min
0.40
5 min
0.25
10 min
0.17
30 min
0.04
60 min
0.005
Text 692019 and Questions to 37607
33
3 possibilities
Rate = k
Rate = k[N2]
Rate = k[N2]2
βˆ’βˆ†[𝑁2 ]
π‘…π‘Žπ‘‘π‘’ =
βˆ†π‘‘
βˆ’{ 𝑁2 π‘™π‘Žπ‘‘π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’ βˆ’ 𝑁2 π‘’π‘Žπ‘Ÿπ‘™π‘–π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’}
=
π‘™π‘Žπ‘‘π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’ βˆ’ π‘’π‘Žπ‘Ÿπ‘™π‘–π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’
Text 692019 and Questions to 37607
34
k is the rate CONSTANT and it’s the
slope of the line
0th order
[A]t = -kt + [A]
1st order
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™ = βˆ’ π‘˜π‘‘
Or 𝑙𝑛
[𝐴]π‘“π‘–π‘›π‘Žπ‘™
[𝐴]π‘–π‘›π‘–π‘‘π‘Žπ‘™
+ ln[𝐴] π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
= βˆ’π‘˜π‘‘
2nd order
1
𝐴
π‘“π‘–π‘›π‘Žπ‘™
= π‘˜π‘‘ +
1
𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Text 692019 and Questions to 37607
35
Slope is all over the place except 1st
order
N2 (g)+ 3 Cl2(g)β†’ 3 NCl3(g)
Given the following date, determine the rate law.
Time
(min)
[N2(g)]
(M)
slope– 0th order
slope - 1st order
0
0.40
=
βˆ’(0.25 βˆ’ 0.40)
5 min βˆ’ 0 π‘šπ‘–π‘›
= 0.03 𝑀/π‘šπ‘–π‘›
1
1
βˆ’[ln 0.25 βˆ’ ln(0.40)]
βˆ’
0.25 0.40 = 0.30
5 min βˆ’ 0 π‘šπ‘–π‘›
5 π‘šπ‘–π‘›
= 0.094
5
0.25
0.016
0.077
0.376
10
0.17
0.0065
0.0723
0.96
30
0.04
0.00117
0.069
5.83
60
0.005
Text 692019 and Questions to 37607
K – 2nd order
36
Problem recognition
What’s the tell?
How do I know how to handle the problem?
Text 692019 and Questions to 37607
37
Method of initial rates – Rates
measured for different initial mixes
The reaction:
2 I-(aq) + S2O82-(aq) β†’ 6 I2 (aq) + 2 SO42-(aq)
was studied at 25° C. The following results were2obtained for the rate of disappearance of S2O8
[I-]0 (M)
0.080
0.040
0.080
0.032
0.060
[S2O82-]0 (M)
0.040
0.040
0.020
0.040
0.030
Initial rate (M/s)
12.5x10-6
6.25x10-6
6.25x10-6
5.00x10-6
7.00x10-6
Text 692019 and Questions to 37607
38
Integrated rate law – concentration at
different times
N2 (g)+ 3 Cl2(g)β†’ 3 NCl3(g)
Given the following date, determine the rate law.
Time
[N2(g)] (M)
0 min
0.40
5 min
0.25
10 min
0.17
30 min
0.04
60 min
0.005
Text 692019 and Questions to 37607
39
Does that make sense?
A. Yes
B. No
C. Maybe
Text 692019 and Questions to 37607
40
Once I know the order, how’s it
work…?
Once I know the order of the reaction, I can use
the integrated rate law to determine the
concentration at any time.
Text 692019 and Questions to 37607
41
The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) β†’ 2 HCl(g)
2 M H2 and 2 M Cl2 was placed in a 5 L flask at
298 K. The initial rate was 3.82x10-3 M/s. What
was the rate after 10 minutes? How much HCl
had been made after 10 minutes?
Text 692019 and Questions to 37607
42
Do I know the rate constant?
A.
B.
C.
D.
E.
Yes
No
Not directly but implicitly
I have no clue
You look beautiful today
Text 692019 and Questions to 37607
43
As soon as I’m talking about TIME, it’s an
integrated rate law problem.
The order of the reaction was given. This
actually tells me two things:
The Rate Law
The Integrated Rate Law
Text 692019 and Questions to 37607
44
The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) β†’ 2 HCl(g)
Rate=k[Cl2]
Once I know that, the I.R.L. is automatic:
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=𝑑
𝑙𝑛
= βˆ’π‘˜π‘‘
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=0
Ln[Cl2]time = - kt + ln[Cl2]time=0
Text 692019 and Questions to 37607
45
Rate=k[Cl2]
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=𝑑
𝑙𝑛
= βˆ’π‘˜π‘‘
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=0
Does this help me? What do I need to know?
Text 692019 and Questions to 37607
46
The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) β†’ 2 HCl(g)
2 M H2 and 2 M Cl2 was placed in a 5 L flask at
298 K. The initial rate was 3.82x10-3 M/s. What
was the rate after 10 minutes? How much HCl
had been made after 10 minutes?
Text 692019 and Questions to 37607
47
Rate=k[Cl2]
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=𝑑
𝑙𝑛
= βˆ’π‘˜π‘‘
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=0
Time=10 minutes
[H2]initial = 2M
[Cl2]initial = 2M
Rateinitial = 3.82x10-3 M/s
Text 692019 and Questions to 37607
48
3.82x10-3 M/s = k[2M]
k=1.91x10-3 s-1
This allows me to use the I.R.L.
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=𝑑
𝑙𝑛
= βˆ’π‘˜π‘‘
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=0
Text 692019 and Questions to 37607
49
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=𝑑
𝑙𝑛
= βˆ’π‘˜π‘‘
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=0
[𝐢𝑙2 ]π‘‘π‘–π‘šπ‘’=𝑑
𝑙𝑛
= βˆ’(1.92 × 10βˆ’3 𝑠 βˆ’1 ) (600 𝑠)
2𝑀
([𝐢𝑙2 ]10 π‘šπ‘–π‘› )
ln
= βˆ’1.152
2𝑀
[𝐢𝑙2 ]10 π‘šπ‘–π‘›
= 𝑒 βˆ’1.152 = 0.316
2𝑀
[Cl2]10 min = 0.632 M
Text 692019 and Questions to 37607
50
A.
B.
C.
D.
E.
Yes
No
Maybe
You look like crap
You look beautiful
Text 692019 and Questions to 37607
51
The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) β†’ 2 HCl(g)
2 M H2 and 2 M Cl2 was placed in a 5 L flask at
298 K. The initial rate was 3.82x10-3 M/s. What
was the rate after 10 minutes? How much HCl
had been made after 10 minutes?
Text 692019 and Questions to 37607
52
In a β€œword”…
A.
B.
C.
D.
E.
Exploitation
Death
Life
Stoichiometry
Integration
Text 692019 and Questions to 37607
53
Rate = k[Cl2]
Rate = 1.92x10-3 s-1 (0.632 M) = 1.2135x10-3 M/s
Text 692019 and Questions to 37607
54
How much HCl?
Just stoichiometry folks…
I started with 10 moles Cl2 :
2 π‘šπ‘œπ‘™
𝐿
× 5 𝐿 = 10 π‘šπ‘œπ‘™π‘’π‘  π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
I end up with:
0.632 π‘šπ‘œπ‘™
× 5𝐿 = 3.16 π‘šπ‘œπ‘™ 𝑙𝑒𝑓𝑑
𝐿
So…
10 moles initial – 3.16 mol left = 6.84 mol reacted!
Text 692019 and Questions to 37607
55
Cl2 (g) + H2 (g) = 2 HCl (g)
2 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
6.84 π‘šπ‘œπ‘™ 𝐢𝑙2 π‘Ÿπ‘’π‘Žπ‘π‘‘π‘’π‘‘
1 π‘šπ‘œπ‘™ 𝐢𝑙2
= 13.68 π‘šπ‘œπ‘™ 𝐻𝐢𝑙
Text 692019 and Questions to 37607
56
Another fun little rate thing…
Half-life!
For a reaction, you start with a lot of reactants
and you end up with less reactants and more
product. The amount of reactants should always
be decreasing.
Let’s look at our earlier example…
Text 692019 and Questions to 37607
57
Not 0th order – unless it was a sloppy
experiment
0.45
y = -0.0107x + 0.3348
R² = 0.8641
0.4
0.35
[H2] (M)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
5
10
15
20
25
30
35
time (s)
Text 692019 and Questions to 37607
58
Hey, Goldilocks! It Fits 1st order!
0
0
10
20
30
40
50
60
70
-1
y = -0.0722x - 1.0024
R² = 0.9989
ln([H2]
-2
-3
-4
-5
-6
time (s)
Text 692019 and Questions to 37607
59
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
So, it is 1st order. It must obey the first order
rate equation.
The concentration of the reactants should be
asymptotically approaching zero. So if I start
with the maximum A, soon I have 90% left, then
80% left, then 70% left…eventually 50% left.
The time it takes for ½ (50%) of the A to react is
called the β€œhalf-life”.
Text 692019 and Questions to 37607
60
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Let’s do a little algebra.
I start with [A]initial.
I end up with ½ [A]initial.
1
ln 𝐴
2
1
ln 𝐴
2
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
βˆ’ ln 𝐴
= βˆ’π‘˜π‘‘1/2 + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
1
𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
1
2
= ln
= ln
= βˆ’π‘˜π‘‘1/2
𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2
Or:
βˆ’0.693 = βˆ’π‘˜π‘‘1/2
0.693
𝑑1/2 =
π‘˜
The half-life of a reaction (in this case 1st order) is just another way of
specifying k.
Text 692019 and Questions to 37607
61
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
I start with [A]initial. I end up with ½ [A]initial.
NOTICE, I DIDN’T USE ANY PARTICULAR AMOUNT
1
ln 𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ = βˆ’π‘˜π‘‘1/2 + ln 𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2
1
ln 𝐴
2
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
βˆ’ ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
1
𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
1
2
= ln
= ln
= βˆ’π‘˜π‘‘1/2
𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2
Or:
βˆ’0.693 = βˆ’π‘˜π‘‘1/2
0.693
𝑑1/2 =
π‘˜
The half-life is always the same (for a given k) no matter how
much you start with.
Text 692019 and Questions to 37607
62
t1/2 = 2 hours
So, let’s say I start with 1 mol of Cl2.
In 2 hours, how much Cl2 is left?
A. 1 mol
B. 0.75 mol
C. 0.50 mol
D. 0.25 mol
E. I don’t know enough to calculate it.
Text 692019 and Questions to 37607
63
t1/2 = 2 hours
Suppose I come back the next morning and find
that there is only 0.016 mol Cl2 left.
In 2 hours, how much Cl2 is left?
A. 0.016 mol
B. 0.008 mol
C. 0.004 mol
D. It depends on how much the rate has slowed
down as the Cl2 decreased.
E. You look FAB-ulous!
Text 692019 and Questions to 37607
64
1st order is special…
Radioactive decays show 1st order kinetics.
That’s why you hear β€œhalf-life” when people are
talking about reactivity. But β€œhalf-life” actually
applies to any reaction: it’s the time it takes for
½ the reactants to react!
Text 692019 and Questions to 37607
65
It also doesn’t have to be ½ life.
Suppose I’m arrogant, obstinate, and just a
general pain in the patootie…
I insist on using t9/10 – the time it takes for 90%
of the reactants to react.
Again, if it is first order….
Text 692019 and Questions to 37607
66
ln 𝐴
π‘“π‘–π‘›π‘Žπ‘™
= βˆ’π‘˜π‘‘ + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
Let’s do a little algebra.
I start with [A]initial.
I end up with 1/10 [A]initial.
1
ln
𝐴
10
1
ln
𝐴
10
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
βˆ’ ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
= βˆ’π‘˜π‘‘1/2 + ln 𝐴
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
1
𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
1
10
= ln
= ln
= βˆ’π‘˜π‘‘9/10
𝐴 π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
10
Or:
βˆ’2.303 = βˆ’π‘˜π‘‘9/10
3.3033
𝑑9/10 =
π‘˜
The 9/10th life of a reaction (in this case 1st order) is just another way of
specifying k.
Text 692019 and Questions to 37607
67