Integrated Rate Laws
Finally a use for calculus!
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What is a rate?
It’s a “delta/delta”!
Rate of reaction =
Δ𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
Δ𝑡𝑖𝑚𝑒
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 =
Δ𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
Δ𝑡𝑖𝑚𝑒
If Δtime is small enough, we have:
−𝑑[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =
𝑑𝑡
Why “-”? Because you are losing reactants and the rate
should always be positive.
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Let’s look at the rate law
Rate = k[A]
−𝑑[𝐴]
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =
= 𝑘[𝐴]
𝑑𝑡
This is actually an integrable equation.
[Don’t worry, this isn’t a math class…it’s just
masquerading as one!]
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Solving the equation
I’ll show you how to solve it, but it is only the
solution that you need to know.
−𝑑[𝐴]
= 𝑘[𝐴]
𝑑𝑡
We collect the [A] on one side and get:
𝑑[𝐴]
= −𝑘𝑑𝑡
[𝐴]
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Solving the equation
𝑑[𝐴]
= −𝑘𝑑𝑡
[𝐴]
Now you can integrate both sides:
𝐴 𝑓𝑖𝑛𝑎𝑙
𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑑[𝐴]
=−
[𝐴]
𝑓𝑖𝑛𝑎𝑙 𝑡𝑖𝑚𝑒
𝑘𝑑𝑡
𝑡𝑖𝑚𝑒=0
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Solving the equation
𝐴 𝑓𝑖𝑛𝑎𝑙
𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
ln 𝐴
𝑑[𝐴]
=−
[𝐴]
𝑓𝑖𝑛𝑎𝑙
−ln 𝐴
𝑓𝑖𝑛𝑎𝑙 𝑡𝑖𝑚𝑒
𝑘𝑑𝑡
𝑡𝑖𝑚𝑒=0
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −𝑘𝑡
This is the only equation we really need. This is
called the “integrated rate law”…well, because
we integrated the rate law. 
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What it means…
ln 𝐴
𝑓𝑖𝑛𝑎𝑙
− ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −𝑘𝑡
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 𝑓𝑖𝑛𝑎𝑙 = −𝑘𝑡 + ln 𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
What does this look like to you?
Yes, it is the equation of a straight line (y=mx+b)!
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ln 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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 𝐴 𝑓𝑖𝑛𝑎𝑙 = −𝑘𝑡 + ln 𝐴
ln[A]final = -0.015s-1*60 s + ln(0.250 M)
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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.
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Other uses of the integrated rate law
ln 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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.
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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! 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
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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.
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Different reaction
2 H2 + O2 → 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 – 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 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!
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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
𝐴
= 𝑘𝑡 +
𝑓𝑖𝑛𝑎𝑙
1
𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
Same idea, it’s a straight line (y = mx+b) where:
Slope = k
Intercept =
1
𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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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)
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Also, there’s the rare zeroth order
−𝑑[𝐴]
𝑅𝑎𝑡𝑒 =
= 𝑘
𝑑[𝑡]
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 we’ll
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 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
2nd order
1
𝐴
𝑓𝑖𝑛𝑎𝑙
= 𝑘𝑡 +
1
𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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How do we use this?
N2 (g)+ 3 Cl2(g)→ 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)→ 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 – 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 – 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)
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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 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.
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What if I don’t want to make a graph?
N2 (g)+ 3 Cl2(g)→ 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
−∆[𝑁2 ]
𝑅𝑎𝑡𝑒 =
∆𝑡
−{ 𝑁2 𝑙𝑎𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 − 𝑁2 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑡𝑖𝑚𝑒}
=
𝑙𝑎𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 − 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑡𝑖𝑚𝑒
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k is the rate CONSTANT and it’s the
slope of the line
0th order
[A]t = -kt + [A]
1st order
ln 𝐴
𝑓𝑖𝑛𝑎𝑙 = − 𝑘𝑡
Or 𝑙𝑛
[𝐴]𝑓𝑖𝑛𝑎𝑙
[𝐴]𝑖𝑛𝑖𝑡𝑎𝑙
+ ln[𝐴] 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −𝑘𝑡
2nd order
1
𝐴
𝑓𝑖𝑛𝑎𝑙
= 𝑘𝑡 +
1
𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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Slope is all over the place except 1st
order
N2 (g)+ 3 Cl2(g)→ 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
=
−(0.25 − 0.40)
5 min − 0 𝑚𝑖𝑛
= 0.03 𝑀/𝑚𝑖𝑛
1
1
−[ln 0.25 − ln(0.40)]
−
0.25 0.40 = 0.30
5 min − 0 𝑚𝑖𝑛
5 𝑚𝑖𝑛
= 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 – 2nd order
36
Problem recognition
What’s the tell?
How do I know how to handle the problem?
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Method of initial rates – Rates
measured for different initial mixes
The reaction:
2 I-(aq) + S2O82-(aq) → 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 – concentration at
different times
N2 (g)+ 3 Cl2(g)→ 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, 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.
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The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) → 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 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
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The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) → 2 HCl(g)
Rate=k[Cl2]
Once I know that, the I.R.L. is automatic:
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=𝑡
𝑙𝑛
= −𝑘𝑡
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=0
Ln[Cl2]time = - kt + ln[Cl2]time=0
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Rate=k[Cl2]
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=𝑡
𝑙𝑛
= −𝑘𝑡
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=0
Does this help me? What do I need to know?
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The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) → 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]
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=𝑡
𝑙𝑛
= −𝑘𝑡
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=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.
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=𝑡
𝑙𝑛
= −𝑘𝑡
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=0
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[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=𝑡
𝑙𝑛
= −𝑘𝑡
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=0
[𝐶𝑙2 ]𝑡𝑖𝑚𝑒=𝑡
𝑙𝑛
= −(1.92 × 10−3 𝑠 −1 ) (600 𝑠)
2𝑀
([𝐶𝑙2 ]10 𝑚𝑖𝑛 )
ln
= −1.152
2𝑀
[𝐶𝑙2 ]10 𝑚𝑖𝑛
= 𝑒 −1.152 = 0.316
2𝑀
[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
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The following reaction is 1st order in Cl2 and 1st
order overall.
H2 (g) + Cl2 (g) → 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 “word”…
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 folks…
I started with 10 moles Cl2 :
2 𝑚𝑜𝑙
𝐿
× 5 𝐿 = 10 𝑚𝑜𝑙𝑒𝑠 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
I end up with:
0.632 𝑚𝑜𝑙
× 5𝐿 = 3.16 𝑚𝑜𝑙 𝑙𝑒𝑓𝑡
𝐿
So…
10 moles initial – 3.16 mol left = 6.84 mol reacted!
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Cl2 (g) + H2 (g) = 2 HCl (g)
2 𝑚𝑜𝑙 𝐻𝐶𝑙
6.84 𝑚𝑜𝑙 𝐶𝑙2 𝑟𝑒𝑎𝑐𝑡𝑒𝑑
1 𝑚𝑜𝑙 𝐶𝑙2
= 13.68 𝑚𝑜𝑙 𝐻𝐶𝑙
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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…
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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)
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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 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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 “half-life”.
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ln 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
− ln 𝐴
= −𝑘𝑡1/2 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
2
= ln
= ln
= −𝑘𝑡1/2
𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
2
Or:
−0.693 = −𝑘𝑡1/2
0.693
𝑡1/2 =
𝑘
The half-life of a reaction (in this case 1st order) is just another way of
specifying k.
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ln 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
I start with [A]initial. I end up with ½ [A]initial.
NOTICE, I DIDN’T USE ANY PARTICULAR AMOUNT
1
ln 𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = −𝑘𝑡1/2 + ln 𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
2
1
ln 𝐴
2
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
− ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
2
= ln
= ln
= −𝑘𝑡1/2
𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
2
Or:
−0.693 = −𝑘𝑡1/2
0.693
𝑡1/2 =
𝑘
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, 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.
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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 special…
Radioactive decays show 1st order kinetics.
That’s why you hear “half-life” when people are
talking about reactivity. But “half-life” actually
applies to any reaction: it’s the time it takes for
½ the reactants to react!
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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….
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ln 𝐴
𝑓𝑖𝑛𝑎𝑙
= −𝑘𝑡 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
− ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −𝑘𝑡1/2 + ln 𝐴
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
10
= ln
= ln
= −𝑘𝑡9/10
𝐴 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
10
Or:
−2.303 = −𝑘𝑡9/10
3.3033
𝑡9/10 =
𝑘
The 9/10th life of a reaction (in this case 1st order) is just another way of
specifying k.
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