integrated rate laws
DESCRIPTION
Integrated Rate Laws. Finally a use for calculus!. What is a rate?. It’s a “delta/delta”! Rate of reaction = 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!. A rate as a derivative. - PowerPoint PPT PresentationTRANSCRIPT
Text 692019 and Questions to 37607 1
Integrated Rate Laws
Finally a use for calculus!
Text 692019 and Questions to 37607 2
What is a rate?
It’s a “delta/delta”!
Rate of reaction =
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 3
A rate as a derivative
Rate of reaction =
If Δtime is small enough, we have:
Why “-”? Because you are losing reactants and the rate should always be positive.
Text 692019 and Questions to 37607 4
Let’s look at the rate law
Rate = k[A]
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 5
Solving the equation
I’ll show you how to solve it, but it is only the solution that you need to know.
We collect the [A] on one side and get:
Text 692019 and Questions to 37607 6
Solving the equation
Now you can integrate both sides:
Text 692019 and Questions to 37607 7
Solving the equation
This is the only equation we really need. This is called the “integrated rate law”…well, because we integrated the rate law.
Text 692019 and Questions to 37607 8
What it means…
What it means is that the concentration at any time decays logarithmically from the initial concentration. If I rearrange the equation a little:
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 9
ln [𝐴 ] 𝑓𝑖𝑛𝑎𝑙=−𝑘𝑡+ ln [𝐴 ]𝑖𝑛𝑖𝑡𝑖𝑎𝑙If 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 s-
1, I need my time to be in seconds.Plug and chug, baby!
Text 692019 and Questions to 37607 10
Using the integrated rate law
ln[A]final = -0.015s-1*60 s + ln(0.250 M)
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 11
Compare the integrated rate law to the rate law
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/sI 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 12
Other uses of the integrated rate law
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 13
For example, suppose I monitor [A]Time (seconds) [A] (M)
0 0.2510 0.2020 0.1760 0.075
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 14
For example, suppose I monitor [A]Time (seconds) [A] (M) ln[A]
0 0.25 -1.38610 0.20 -1.60920 0.17 -1.77260 0.075 -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 15
LOOK! IT’S A POINT! IT’S A PLANE!NO!!! IT’S A STRAIGHT LINE!
0 10 20 30 40 50 60 70
-2.8
-2.6
-2.4
-2.2
-2
-1.8
-1.6
-1.4
-1.2
-1
f(x) = − 0.02 x − 1.38629436111989R² = 1
Text 692019 and Questions to 37607 16
So, what’s the rate constant?y = -0.02x - 1.3863
ln[A]final = - kt + ln[A]t=0
m= slope=-0.02m=-kk=-(-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 17
Different reaction2 H2
+ O2 → 2 H2OTime (seconds) [H2] (M) ln[H2]
0 0.500 -0.6931510 0.300 -1.2039720 0.200 -1.6094460 0.100 -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 18
NOT a straight line – NOT a 1st order reaction!
0 10 20 30 40 50 60 70
-2.5
-2
-1.5
-1
-0.5
0
f(x) = − 0.0249049998021776 x − 0.891923252029512R² = 0.92882867596042
Text 692019 and Questions to 37607 19
Is it or isn’t it a straight line?
A. It is a straight lineB. It is NOT a straight lineC. I can’t tell without error barsD. I really don’t care it’s MondayE. Your mother!
Text 692019 and Questions to 37607 20
This works for other orders of reaction also.
For a second order reaction:
Rate = k[A]2
You get an integrated rate law
=
Same idea, it’s a straight line (y = mx+b) where:Slope = kIntercept =
Text 692019 and Questions to 37607 21
Hey! It’s 2nd order!
0 10 20 30 40 50 60 700
2
4
6
8
10
12
f(x) = 0.132931726907631 x + 2.09236947791164R² = 0.997741533025526
Time (s)
1/[H
2}
Text 692019 and Questions to 37607 22
Also, there’s the rare zeroth order
If you integrate
[A]t = -kt + [A]
Text 692019 and Questions to 37607 23
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 24
But a lot of reactions fall into those three categories.
0th order
[A]t = -kt + [A]
1st order
2nd order
=
Text 692019 and Questions to 37607 25
How do we use this?
Time [N2(g)] (M)
0 min 0.40
5 min 0.25
10 min 0.17
30 min 0.04
60 min 0.005
N2 (g)+ 3 Cl2(g)→ 3 NCl3(g)
Given the following data, determine the rate law.
Text 692019 and Questions to 37607 26
GRAPH IT!
Text 692019 and Questions to 37607 27
Graph It!
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
N2 (g)+ 3 Cl2(g)→ 3 NCl3(g)
Given the following data, determine the rate law.
Text 692019 and Questions to 37607 28
Try all 3 and see which one fits!
Text 692019 and Questions to 37607 29
Not 0th order – unless it was a sloppy experiment
0 5 10 15 20 25 30 350
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
f(x) = − 0.0106506024096386 x + 0.334819277108434R² = 0.86409072413
time (s)
[H2]
(M)
Text 692019 and Questions to 37607 30
Maybe 2nd order – it’s not a horrible fit.
0 5 10 15 20 25 30 350
5
10
15
20
25
30
f(x) = 0.787030474840539 x + 0.491495393338058R² = 0.966406872923231
Time (s)
1/[H
2}
Text 692019 and Questions to 37607 31
Hey, Goldilocks! It Fits 1st order!0 10 20 30 40 50 60 70
-6
-5
-4
-3
-2
-1
0
f(x) = − 0.0721859170705212 x − 1.00244276678749R² = 0.998887091882838
time (s)
ln([H
2]
Text 692019 and Questions to 37607 32
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 33
What if I don’t want to make a graph?
Time [N2(g)] (M)
0 min 0.40
5 min 0.25
10 min 0.17
30 min 0.04
60 min 0.005
N2 (g)+ 3 Cl2(g)→ 3 NCl3(g)
Given the following data, determine the rate law.
Text 692019 and Questions to 37607 34
3 possibilities
Rate = kRate = k[N2]Rate = k[N2]2
Text 692019 and Questions to 37607 35
k is the rate CONSTANT and it’s the slope of the line
0th order
[A]t = -kt + [A]
1st order
Or
2nd order
=
Text 692019 and Questions to 37607 36
Slope is all over the place except 1st order
Time (min)
[N2(g)] (M)
slope– 0th order slope - 1st order K – 2nd order
0 0.40
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
N2 (g)+ 3 Cl2(g)→ 3 NCl3(g)
Given the following date, determine the rate law.
Text 692019 and Questions to 37607 37
Problem recognition
What’s the tell?
How do I know how to handle the problem?
Text 692019 and Questions to 37607 38
Method of initial rates – Rates measured for different initial mixes
The reaction:
2 I-(aq) + S2O8
2-(aq) → 6 I2 (aq) + 2 SO4
2-(aq)
was studied at 25° C. The following results were obtained for the rate of disappearance of S2O8
2-
[I-]0 (M) [S2O82-]0 (M) Initial rate (M/s)
0.080 0.040 12.5x10-60.040 0.040 6.25x10-60.080 0.020 6.25x10-60.032 0.040 5.00x10-60.060 0.030 7.00x10-6
Text 692019 and Questions to 37607 39
Integrated rate law – concentration at different times
Time [N2(g)] (M)
0 min 0.40
5 min 0.25
10 min 0.17
30 min 0.04
60 min 0.005
N2 (g)+ 3 Cl2(g)→ 3 NCl3(g)
Given the following date, determine the rate law.
Text 692019 and Questions to 37607 40
Does that make sense?
A. YesB. NoC. Maybe
Text 692019 and Questions to 37607 41
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 42
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?
Text 692019 and Questions to 37607 43
Do I know the rate constant?
A. YesB. NoC. Not directly but implicitlyD. I have no clueE. You look beautiful today
Text 692019 and Questions to 37607 44
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 45
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:
Ln[Cl2]time = - kt + ln[Cl2]time=0
Text 692019 and Questions to 37607 46
Rate=k[Cl2]
Does this help me? What do I need to know?
Text 692019 and Questions to 37607 47
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?
Text 692019 and Questions to 37607 48
Rate=k[Cl2]
Time=10 minutes[H2]initial = 2M[Cl2]initial = 2MRateinitial = 3.82x10-3 M/s
Text 692019 and Questions to 37607 49
3.82x10-3 M/s = k[2M]k=1.91x10-3 s-1
This allows me to use the I.R.L.
Text 692019 and Questions to 37607 50
[Cl2]10 min = 0.632 M
Text 692019 and Questions to 37607 51
A. YesB. NoC. MaybeD. You look like crapE. You look beautiful
Text 692019 and Questions to 37607 52
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?
Text 692019 and Questions to 37607 53
In a “word”…
A. ExploitationB. DeathC. LifeD. StoichiometryE. Integration
Text 692019 and Questions to 37607 54
Rate = k[Cl2]Rate = 1.92x10-3 s-1 (0.632 M) = 1.2135x10-3 M/s
Text 692019 and Questions to 37607 55
How much HCl?
Just stoichiometry folks…
I started with 10 moles Cl2 :
I end up with:
So…10 moles initial – 3.16 mol left = 6.84 mol reacted!
Text 692019 and Questions to 37607 56
Cl2 (g) + H2 (g) = 2 HCl (g)
6.84𝑚𝑜𝑙𝐶𝑙2𝑟𝑒𝑎𝑐𝑡𝑒𝑑2𝑚𝑜𝑙𝐻𝐶𝑙1𝑚𝑜𝑙𝐶𝑙2
=13.68𝑚𝑜𝑙𝐻𝐶𝑙
Text 692019 and Questions to 37607 57
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 58
Not 0th order – unless it was a sloppy experiment
0 5 10 15 20 25 30 350
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
f(x) = − 0.0106506024096386 x + 0.334819277108434R² = 0.86409072413
time (s)
[H2]
(M)
Text 692019 and Questions to 37607 59
Hey, Goldilocks! It Fits 1st order!0 10 20 30 40 50 60 70
-6
-5
-4
-3
-2
-1
0
f(x) = − 0.0721859170705212 x − 1.00244276678749R² = 0.998887091882838
time (s)
ln([H
2]
Text 692019 and Questions to 37607 60
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”.
Text 692019 and Questions to 37607 61
Let’s do a little algebra. I start with [A]initial.
I end up with ½ [A]initial.
Or:
The half-life of a reaction (in this case 1st order) is just another way of specifying k.
Text 692019 and Questions to 37607 62
I start with [A]initial. I end up with ½ [A]initial.NOTICE, I DIDN’T USE ANY PARTICULAR AMOUNT
Or:
The half-life is always the same (for a given k) no matter how much you start with.
Text 692019 and Questions to 37607 63
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 molB. 0.75 molC. 0.50 molD. 0.25 molE. I don’t know enough to calculate it.
Text 692019 and Questions to 37607 64
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 molB. 0.008 molC. 0.004 molD. 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 65
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!
Text 692019 and Questions to 37607 66
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 67
Let’s do a little algebra. I start with [A]initial.
I end up with 1/10 [A]initial.
Or:
The 9/10th life of a reaction (in this case 1st order) is just another way of specifying k.