Lecture 2 Kinetics

Kinetics

The speed with which reactions proceed.

Chemists make things. Chemistry is the study of how to make things, how to measure the things we make, and the principles that govern these syntheses.

If a reaction can go, then kinetics is the study of the speed with which it goes and the construction of a plausible mechanism of the reaction.

Overarching Goal: the design of a mechanism for a reaction.

In Chem 104 we learn pieces of this big goal.

STEP ONEThe Rate

Need to be able to express the speed of a reaction mathematically and relate it to the stoichiometric equation. We need to understand what the rate of reaction means conceptually.

Consider the reaction:

2NO2 2NO(g) + O2(g)Balanced?

Look at these data. Now we are really thinking about how long (time) it takes for a reaction to take place – this differs from CHEM 102 when we computed endpoints.

Spend a moment understanding these data. Fill in the blanks.

Reaction:

2NO2(g)

2NO(g) + O2(g)

Seconds MOLES

Time NO2 NO O2

0 0.02 ___ ___100 0.013 ___ ___200 0.0096 ___ ___

Go back to the original data and now let’s graph the concentrations:

Huge amount of mathematical information in these plots.

Notice1) The concentration of

reactants and products change with time consistent with the stoichiometry of the reaction.

2) The RATE of the reaction is defined as the change in concentration with change in time.

3) The instantaneous rate is the slope of the tangent to the curve at a given time – this is the derivative from calculus.

Notice4) The slope of the tangent

line changes with time!5) This means the rate of the

reaction changes with time.

6) Reactants --Rate is NEGATIVE

7) PRODUCTS – Rate is POSITIVE

8) Magnitude of the rate DECREASES with time – that is reactions slow down

Mathematically:

The rate of the reaction can be written as

D[species]/Dt

In this example we have:D[NO2]/DtD[NO]/DtD[O2]/Dt

Here’s the rub: these rates are not equal

However, not so grim:

For: 2NO2 2NO(g) + O2(g)

We have

t

O

t

NO

t

NOR

][][

2

1][

2

1 22

Where R means “rate of reaction”. Notice that by convention the rate of the reaction is taken to be POSITIVE.

For: 2NO2 2NO(g) + O2(g)

t

O

t

NO

t

NOR

][][

2

1][

2

1 22

If the rate of this reaction at 100 s was measured to be 0.1M/s, then what is the rate of loss of nitrogen dioxide, the rate of production of nitrogen monoxide and the rate of production of oxygen at this time?

Consider the Reaction:

2N2O5 (soln)→4NO2(soln) + O2(g)

If the rate of the reaction is 10M/min, what is ? t

NO

][ 2

A) 10M/min B) 40M/min C) 2.5M/min D) 5M/min



Which rate (absolute value) has the largest number? A)Rate of loss of N2O5

B) Rate of production of NO2

C) Rate of production of O2



If = -5M/s, then what is

A) -10M/s B) 10M/s C) -5M/s D) 5M/s

t

ON

][ 52

t

NO

][ 2

STEP TWOThe Rate Law

What influences the rate of a reaction?

Following the logic flow (which is all it is, this is not a derivation)

1) For a reaction to go, the reacting species must collide. That means that the particles have to be in the same place at the same time (or near enough) to react. Thus the rate of the reaction must be proportional to the probability of the species being at the specific place.

2) The probability that the reacting species are present at a particular place, will be proportional to the concentration of the reacting species.

3) All the reacting species have to be present at the same place (or near enough) at the same time. This is like the following probability problem: suppose you have 3 dice. What is the probability that you will throw a 3 with one dice and then another 3 and then another 3? 1/6 x 1/6 x 1/6 (AND probability)

Putting this altogether

Rate [Reactant 1]x[Reactant 2]y [Reactant 3]z …

Or, creating an equation,

Rate = k(T) [Reactant 1]x[Reactant 2]y [Reactant 3]z

Where k is called the “rate constant” and is dependent upon T butindependent upon concentration

Here’s where you need to think about the form of this equation. It actually doesn’t give us a lot of information – yet.

The details have to be determined – and the details mean

Determine:

x, y, z, and k

Go over some language

1) Rate law – you can be asked to determine the rate law. This means find the exponents x, y, z, etc and find k, the rate constant.

2) The rate law will give us the “order” of the reaction. Be careful with the language

1) The overall order of the reaction is the sum of the exponents OR

2) You can talk about the order of a particular reactant

3) If the exponent is 1 the reaction is first order; if 2 second order, etc.

Examples:

Consider reaction NH4+(aq)+NO2-(aq) → N2(g)+2H2O(l)

The rate law for this reaction was determined to be:

R = k[NH4+][NO2

-]

A. zeroeth, B. first C. second D. third

The rate is 1st order in ammonium, 1st order in nitrite, and 2nd order overall.

Questions:

The reaction 2N2O5 (soln)→4NO2(soln) + O2(g) is first order in N2O5 .

The rate law for the reaction is

A) R=k[N2O5] B) R=k[N2O5]2 C) R=k[N2O5]0

Questions:

The reaction 2N2O5 (soln)→4NO2(soln) + O2(g) is first order in N2O5 .

The units of the rate constant are

A) M/sec B) sec-1 C) M2 sec-1

Questions:

The reaction 2N2O5 (soln)→4NO2(soln) + O2(g) is first order in N2O5 . At 300 K the initial rate of the reaction was found to be 2.5 millimolar (mM)/min when the dinitrogen pentoxide concentration was 0.10 millimolar. The value of k is

The units of the rate constant are

A) 0.25 min-1 B) 2.5 min-1 C) 25 min-1

One more question:

The reaction 2N2O5 (soln)→4NO2(soln) + O2(g) is first order in N2O5 . At 300 K the initial rate of the reaction was found to be 2.5 millimolar (mM)/min when the dinitrogen pentoxide concentration was 0.10 millimolar. If the initial concentration was quadrupoled in another experiment, the rate and the rate constant are expected to

A) Remain the same

B) Quadruple

C) Double

D) The rate will quadruple and the rate constant will remain the same

STEP THREEDetermining the Rate Law by

the Method of Initial Rates

So we now know what the rate law is, but we need to figure out (measure) the exponents.

Since the rate of the reaction depends on the concentration of reacting species, we must pick a time to measure the rates. We will pick the INITIAL TIME and measure the initial rates of the reactions.

We do several experiments with different initial conditions. I am going to obtain the exponents by inspection.

First, if I compare expts 1 and 2 I see that I don’t change the ammonium conc and I double the nitrite concentration. When I double the nitrite, the rate doubles.

THEREFORE: the rate law is FIRST ORDER in nitrite.

You must understand this logic.

Note the math:

Suppose I have an equation R = kx, where k is a constant and x is the variable. If I double x, what happens to R?

If the exponent is 1 and I double the concentration then R will double.

So if I double the concentration and R doubles, then the exponent must have been 1.

Let’s look at the case when the exponent is 2.

R=kx2.

If I double x, note that R quadruples (22=4).

So if I double x and R quadruples then the exponent must be 2.

And one more case – suppose I double the concentration and R doesn’t change?

Then the exponent has to be 0 because the rate doesn’t dependent on the concentration. Recall that anything to the power of 0 = 1.

So now, by judicious choice of experiments we can determine the rate law.

Determine the rate law for the following reaction:

A) R=k[BrO3-][Br-][H+]

B) R=k[BrO3-][Br-]

C) R=k[BrO3-][Br-][H+]2

The rate law was found to be R=k[BrO3-][Br-][H+]2

The value of the rate constant is

A) 8.0 Ms-1

B) 8.0 M-2s-1

C) 8.0 M-3s-1

Determining the Rate Law for Special Cases

First Order Reactions:

R = k[A] where A is the relevant species

First order reactions are very important. Nuclear reactions (radiochemistry) are ALL first order reactions.

First order reactions produce a very specific outcome – the half lives of the reactant are independent of the amount of reactant present.

Calculus Interlude (if you don’t know calc, don’t worry, just realize that there is a mathematical way to arrive at the answer)

See next page

ktAA

AC

CktA

dtk

kdtA

Akdt

Ad

AkR

0

0

0

]ln[]ln[

togetherallit Putting

]ln[ Therefore

[A] written A, ofion concentrat initial[A] 0,At t

constant.n integratio theis C where]ln[

[A]

d[A]

gintegratin Now

right on the are t involving thoseand

left on the areA involving terms the where][

d[A]

Rfor expression thedsubstitute have we where][][

A""reactant theofion concentrat theis [A] where][

The final outcome is what matters and can be expressed in several ways (called the integrated rate eqn):

ln[A] =ln[A]0-kt

ln[A]/[A]0 = -kt

[A]/[A] 0 = e-kt

Graphs of ln[A] vs t are linear and the slope = -k.

Let’s use excel to graph the decay of Mo-99. The half life of Mo-99 is 66 hours.

HALF LIVES

When ½ of the amount of initial material has decayed or reacted from its initial amount, then

[A] ½ = ½[A]0. Substituting this into the integrated rate equations:

ln (½[A]0/[A]0) = -kt1/2

ln(0.5) = -kt1/2

ln2 = -k - -t kT, where T=t=t1/2

KEY EQUATIONS

R=k[A]x[B]y[C]z ….

ln[A]=ln[A]0 – kt

ln2=kt

In the context of radioactive decay the symbol t stands for:

a) The time for the substance to decay

b) The time for ½ of the substance to decay

c) The amount of substance present at the half life

The decay of P-31, like all radioactive decay, exhibits first order kinetics. It takes 14 days for 10 mg of P-31 to decay to 5 mg. What is the rate constant for this reaction?

A) 0.05 days-1

B) 0.02 days-1

C) 5 days

D) 2 days

The following data were obtained for the reaction of NO with O2. Concentrations are in molecules/cm3 and rates are in molecules/cm3×s.

What is the rate law?

a) Rate = k[NO][O2]

b) Rate = k[NO][O2]2

c) Rate = k[NO]2[O2]

d) Rate = k[NO]2

e) Rate = k[NO]2[O2]2

[NO]0

1 1018

2 1018

3 1018

1 1018

1 1018

[O2]0

1 1018

1 1018

1 1018

2 1018

3 1018

Initial Rate2.0 1016

8.0 1016

18.0 1016

4.0 1016

6.0 1016

For a reaction in which A and B react to form C, the following initial rate data were obtained:

What is the rate law?

a) Rate = k[A][B]

b) Rate = k[A]2[B]

c) Rate = k[A][B]2

d) Rate = k[A]2[B]2

e) Rate = k[A]3

[A](mol/L)

0.100.100.20

[B](mol/L)

0.100.200.40

Initial Rate of Formation of C(mol/L·s)1.001.008.00

For the reaction 2N2O5(g) 4NO2(g) + O2(g), the following data were collected:

t (minutes) [N2O5] (mol/L)

0 1.24 x 10–2

10. 0.92 x 10–2

20. 0.68 x 10–2

30. 0.50 x 10–2

40. 0.37 x 10–2

50. 0.28 x 10–2

70. 0.15 x 10–2

The order of this reaction in N2O5 isa) 0 b) 1 c) 2d) 3 e) none of these

2N2O5(g) 4NO2(g) + O2(g)


0 1.24 x 10–2

10. 0.92 x 10–2

20. 0.68 x 10–2

30. 0.50 x 10–2

40. 0.37 x 10–2

50. 0.28 x 10–2

70. 0.15 x 10–2

The concentration of O2 at t = 10. minutes isa) 2.0 x 10–4 mol/L b) 0.32 x 10–2 mol/Lc) 0.16 x 10–2 mol/L d) 0.64 ´x10–2 mol/Le) none of these

2N2O5(g) 4NO2(g) + O2(g)


0 1.24 x 10–2

10. 0.92 x 10–2

20. 0.68 x 10–2

30. 0.50 x 10–2

40. 0.37 x 10–2

50. 0.28 x 10–2

70. 0.15 x 10–2

The half-life of this reaction is approximatelya) 15 minutes b) 18 minutesc) 23 minutes d) 36 minutese) 45 minutes

2N2O5(g) 4NO2(g) + O2(g)


0 1.24 x 10–2

10. 0.92 x 10–2

20. 0.68 x 10–2

30. 0.50 x 10–2

40. 0.37 x 10–2

50. 0.28 x 10–2

70. 0.15 x 10–2

The initial rate of production of NO2 for this reaction is approximatelya) 7.4 x 10–4 mol/L·min b) 3.2 x 10–4 mol/L·minc) 1.24 x 10–2 mol/L·min d) 1.6 x 10–4 mol/L·mine) none of these

Technetium-99 is used as a radiographic agent in medicine. If Tc-99 has a half-life of 6.0 hours, what fraction of an administered dose of 100 mg remains in a patient’s body after 2.0 days.

a) b) c) d)

22

142

182

1162

1

Here is a challenging radioactive decay problem. 99mTc is a common radioisotope used to image soft tissue and bones. 99mTc has a half life of 6 hours. Therefore, the isotope is generated just before administration to the patient.

A. How many millicuries of radiation is delivered to the patient if 2ml of 10-8M of 99mTc is initially injected? 1 curie of radiation is defined as 3.7x1010 disintegrations/sec.

B. 99mTc decays to 99Tc with the production of a g ray. It is the number of g photons that are counted in the medical imaging. After 2 days, how many millicuries of radiation are present from the 99mTc? Assume none of the 99mTc has been excreted.

C. 99Tc has a half life of 2x105 years and decays by emitting a beta particle. How many curies of radiation are present in the body 2 days after initial administration? Assume that all of the 99mTc has decayed to 99Tc.

Lecture 2 Kinetics

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