Integrated Rate Laws

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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

Rate of reaction =

If Δtime is small enough, we have:

Why “-”? Because you are losing reactants and the rate should always be positive.


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!]


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:



Now you can integrate both sides:



This is the only equation we really need. This is called the “integrated rate law”…well, because we integrated the rate law.


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)!


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!


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!)


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.


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.


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.


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.


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


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.


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.


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


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!


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 =


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}


Also, there’s the rare zeroth order

If you integrate

[A]t = -kt + [A]


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.


But a lot of reactions fall into those three categories.

0th order

[A]t = -kt + [A]

1st order

2nd order

=


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.


GRAPH IT!


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)



Try all 3 and see which one fits!


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)


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}


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]


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.


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)



3 possibilities

Rate = kRate = k[N2]Rate = k[N2]2


k is the rate CONSTANT and it’s the slope of the line

0th order

[A]t = -kt + [A]

1st order

Or

2nd order

=


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.


Problem recognition

What’s the tell?

How do I know how to handle the problem?


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


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.


Does that make sense?

A. YesB. NoC. Maybe


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.


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?


Do I know the rate constant?

A. YesB. NoC. Not directly but implicitlyD. I have no clueE. You look beautiful today


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



Rate=k[Cl2]Once I know that, the I.R.L. is automatic:

Ln[Cl2]time = - kt + ln[Cl2]time=0


Rate=k[Cl2]

Does this help me? What do I need to know?





Rate=k[Cl2]

Time=10 minutes[H2]initial = 2M[Cl2]initial = 2MRateinitial = 3.82x10-3 M/s


3.82x10-3 M/s = k[2M]k=1.91x10-3 s-1

This allows me to use the I.R.L.


[Cl2]10 min = 0.632 M


A. YesB. NoC. MaybeD. You look like crapE. You look beautiful





In a “word”…

A. ExploitationB. DeathC. LifeD. StoichiometryE. Integration


Rate = k[Cl2]Rate = 1.92x10-3 s-1 (0.632 M) = 1.2135x10-3 M/s


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!


Cl2 (g) + H2 (g) = 2 HCl (g)

6.84𝑚𝑜𝑙𝐶𝑙2𝑟𝑒𝑎𝑐𝑡𝑒𝑑2𝑚𝑜𝑙𝐻𝐶𝑙1𝑚𝑜𝑙𝐶𝑙2

=13.68𝑚𝑜𝑙𝐻𝐶𝑙


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…


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)


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]


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”.


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.


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.


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.


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!


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!


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….


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.

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Integrated Rate Laws