Integrated Rate Laws
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Transcript Integrated Rate Laws
Integrated Rate Laws
Finally a use for calculus!
Text 692019 and Questions to 37607
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What is a rate?
Itβs a βdelta/deltaβ!
Rate of reaction =
Ξπππππππ‘πππ‘πππ
Ξπ‘πππ
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!
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A rate as a derivative
Rate of reaction =
Ξπππππππ‘πππ‘πππ
Ξπ‘πππ
If Ξtime is small enough, we have:
βπ[πππππ‘πππ‘]
π
ππ‘π ππ πππππ‘πππ =
ππ‘
Why β-β? Because you are losing reactants and the rate
should always be positive.
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Letβs look at the rate law
Rate = k[A]
βπ[π΄]
π
ππ‘π ππ πππππ‘πππ =
= π[π΄]
ππ‘
This is actually an integrable equation.
[Donβt worry, this isnβt a math classβ¦itβs just
masquerading as one!]
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Solving the equation
Iβll show you how to solve it, but it is only the
solution that you need to know.
βπ[π΄]
= π[π΄]
ππ‘
We collect the [A] on one side and get:
π[π΄]
= βπππ‘
[π΄]
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Solving the equation
π[π΄]
= βπππ‘
[π΄]
Now you can integrate both sides:
π΄ πππππ
π΄
ππππ‘πππ
π[π΄]
=β
[π΄]
πππππ π‘πππ
πππ‘
π‘πππ=0
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Solving the equation
π΄ πππππ
π΄
ππππ‘πππ
ln π΄
π[π΄]
=β
[π΄]
πππππ
βln π΄
πππππ π‘πππ
πππ‘
π‘πππ=0
ππππ‘πππ
= βππ‘
This is the only equation we really need. This is
called the βintegrated rate lawββ¦well, because
we integrated the rate law. ο
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What it meansβ¦
ln π΄
πππππ
β ln π΄
ππππ‘πππ
= βππ‘
What it means is that the concentration at any time
decays logarithmically from the initial concentration. If I
rearrange the equation a little:
ln π΄ πππππ = βππ‘ + ln π΄ ππππ‘πππ
What does this look like to you?
Yes, it is the equation of a straight line (y=mx+b)!
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ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
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!
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Using the integrated rate law
ln π΄ πππππ = βππ‘ + ln π΄
ln[A]final = -0.015s-1*60 s + ln(0.250 M)
ππππ‘πππ
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!)
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Compare the integrated rate law to the
rate law
ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
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 canβt know how long it
takes me to get to that new concentration.
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Other uses of the integrated rate law
ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
Itβs 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 youβll get a straight line where the slope=-k. So you
could use the equation to find the rate constant.
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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.
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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.
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LOOK! ITβS A POINT! ITβS A PLANE!
NO!!! ITβS 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
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So, whatβs 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 itβs 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.
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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.
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NOT a straight line β 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
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Is it or isnβt it a straight line?
A.
B.
C.
D.
E.
It is a straight line
It is NOT a straight line
I canβt tell without error bars
I really donβt care itβs Monday
Your mother!
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This works for other orders of reaction
also.
For a second order reaction:
Rate = k[A]2
You get an integrated rate law
1
π΄
= ππ‘ +
πππππ
1
π΄
ππππ‘πππ
Same idea, itβs a straight line (y = mx+b) where:
Slope = k
Intercept =
1
π΄
ππππ‘πππ
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Hey! Itβs 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)
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Also, thereβs the rare zeroth order
βπ[π΄]
π
ππ‘π =
= π
π[π‘]
If you integrate
[A]t = -kt + [A]
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Those are the easy ones
For more complicated mixed orders like:
Rate = k[A][B]
The math gets much more complicated, so weβll
ignore them until you become a chemistry
major. But you can do a similar thing.
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But a lot of reactions fall into those
three categories.
0th order
[A]t = -kt + [A]
1st order
ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
2nd order
1
π΄
πππππ
= ππ‘ +
1
π΄
ππππ‘πππ
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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
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GRAPH IT!
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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
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Try all 3 and see which one fits!
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Not 0th order β 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)
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Maybe 2nd order β itβs 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)
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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)
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What if I donβt want to or canβt 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.
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What if I donβt 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
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3 possibilities
Rate = k
Rate = k[N2]
Rate = k[N2]2
ββ[π2 ]
π
ππ‘π =
βπ‘
β{ π2 πππ‘ππ π‘πππ β π2 πππππππ π‘πππ}
=
πππ‘ππ π‘πππ β πππππππ π‘πππ
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k is the rate CONSTANT and itβs the
slope of the line
0th order
[A]t = -kt + [A]
1st order
ln π΄
πππππ = β ππ‘
Or ππ
[π΄]πππππ
[π΄]ππππ‘ππ
+ ln[π΄] ππππ‘πππ
= βππ‘
2nd order
1
π΄
πππππ
= ππ‘ +
1
π΄
ππππ‘πππ
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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)
slopeβ 0th order
slope - 1st order
0
0.40
=
β(0.25 β 0.40)
5 min β 0 πππ
= 0.03 π/πππ
1
1
β[ln 0.25 β ln(0.40)]
β
0.25 0.40 = 0.30
5 min β 0 πππ
5 πππ
= 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
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K β 2nd order
36
Problem recognition
Whatβs the tell?
How do I know how to handle the problem?
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Method of initial rates β 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
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Integrated rate law β 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
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Does that make sense?
A. Yes
B. No
C. Maybe
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Once I know the order, howβs it
work�
Once I know the order of the reaction, I can use
the integrated rate law to determine the
concentration at any time.
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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?
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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
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As soon as Iβm talking about TIME, itβs 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
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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:
[πΆπ2 ]π‘πππ=π‘
ππ
= βππ‘
[πΆπ2 ]π‘πππ=0
Ln[Cl2]time = - kt + ln[Cl2]time=0
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Rate=k[Cl2]
[πΆπ2 ]π‘πππ=π‘
ππ
= βππ‘
[πΆπ2 ]π‘πππ=0
Does this help me? What do I need to know?
Text 692019 and Questions to 37607
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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?
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Rate=k[Cl2]
[πΆπ2 ]π‘πππ=π‘
ππ
= βππ‘
[πΆπ2 ]π‘πππ=0
Time=10 minutes
[H2]initial = 2M
[Cl2]initial = 2M
Rateinitial = 3.82x10-3 M/s
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3.82x10-3 M/s = k[2M]
k=1.91x10-3 s-1
This allows me to use the I.R.L.
[πΆπ2 ]π‘πππ=π‘
ππ
= βππ‘
[πΆπ2 ]π‘πππ=0
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[πΆπ2 ]π‘πππ=π‘
ππ
= βππ‘
[πΆπ2 ]π‘πππ=0
[πΆπ2 ]π‘πππ=π‘
ππ
= β(1.92 × 10β3 π β1 ) (600 π )
2π
([πΆπ2 ]10 πππ )
ln
= β1.152
2π
[πΆπ2 ]10 πππ
= π β1.152 = 0.316
2π
[Cl2]10 min = 0.632 M
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A.
B.
C.
D.
E.
Yes
No
Maybe
You look like crap
You look beautiful
Text 692019 and Questions to 37607
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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?
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In a βwordββ¦
A.
B.
C.
D.
E.
Exploitation
Death
Life
Stoichiometry
Integration
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Rate = k[Cl2]
Rate = 1.92x10-3 s-1 (0.632 M) = 1.2135x10-3 M/s
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How much HCl?
Just stoichiometry folksβ¦
I started with 10 moles Cl2 :
2 πππ
πΏ
× 5 πΏ = 10 πππππ ππππ‘πππ
I end up with:
0.632 πππ
× 5πΏ = 3.16 πππ ππππ‘
πΏ
Soβ¦
10 moles initial β 3.16 mol left = 6.84 mol reacted!
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Cl2 (g) + H2 (g) = 2 HCl (g)
2 πππ π»πΆπ
6.84 πππ πΆπ2 πππππ‘ππ
1 πππ πΆπ2
= 13.68 πππ π»πΆπ
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Another fun little rate thingβ¦
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.
Letβs look at our earlier exampleβ¦
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Not 0th order β 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)
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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)
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ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
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% leftβ¦eventually 50% left.
The time it takes for ½ (50%) of the A to react is
called the βhalf-lifeβ.
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ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
Letβs do a little algebra.
I start with [A]initial.
I end up with ½ [A]initial.
1
ln π΄
2
1
ln π΄
2
ππππ‘πππ
ππππ‘πππ
β ln π΄
= βππ‘1/2 + ln π΄
ππππ‘πππ
ππππ‘πππ
1
π΄ ππππ‘πππ
1
2
= ln
= ln
= βππ‘1/2
π΄ ππππ‘πππ
2
Or:
β0.693 = βππ‘1/2
0.693
π‘1/2 =
π
The half-life of a reaction (in this case 1st order) is just another way of
specifying k.
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ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
I start with [A]initial. I end up with ½ [A]initial.
NOTICE, I DIDNβT USE ANY PARTICULAR AMOUNT
1
ln π΄ ππππ‘πππ = βππ‘1/2 + ln π΄ ππππ‘πππ
2
1
ln π΄
2
ππππ‘πππ
β ln π΄
ππππ‘πππ
1
π΄ ππππ‘πππ
1
2
= ln
= ln
= βππ‘1/2
π΄ ππππ‘πππ
2
Or:
β0.693 = βππ‘1/2
0.693
π‘1/2 =
π
The half-life is always the same (for a given k) no matter how
much you start with.
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t1/2 = 2 hours
So, letβs 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 donβt know enough to calculate it.
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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!
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1st order is specialβ¦
Radioactive decays show 1st order kinetics.
Thatβs why you hear βhalf-lifeβ when people are
talking about reactivity. But βhalf-lifeβ actually
applies to any reaction: itβs the time it takes for
½ the reactants to react!
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It also doesnβt have to be ½ life.
Suppose Iβm arrogant, obstinate, and just a
general pain in the patootieβ¦
I insist on using t9/10 β the time it takes for 90%
of the reactants to react.
Again, if it is first orderβ¦.
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ln π΄
πππππ
= βππ‘ + ln π΄
ππππ‘πππ
Letβs do a little algebra.
I start with [A]initial.
I end up with 1/10 [A]initial.
1
ln
π΄
10
1
ln
π΄
10
ππππ‘πππ
ππππ‘πππ
β ln π΄
ππππ‘πππ
= βππ‘1/2 + ln π΄
ππππ‘πππ
1
π΄ ππππ‘πππ
1
10
= ln
= ln
= βππ‘9/10
π΄ ππππ‘πππ
10
Or:
β2.303 = βππ‘9/10
3.3033
π‘9/10 =
π
The 9/10th life of a reaction (in this case 1st order) is just another way of
specifying k.
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