Chapter 16 Notes Chemical Kinetics

Chapter 16 NotesChemical Kinetics

Chemical kinetics is the study of the rates of chemical reactions, the factors that affect these rates, and the mechanisms by which the reaction occurs.

The Rate of a Reaction

The rate of the reaction describes how fast the reactants are used up and the products form.

The rate of reactions are usually expressed in concentration per time, or in other words, moles per liter per time:

mol or mol L s L min

Getting the measurements of the concentration while a chemical reaction is proceeding can be tricky. Perhaps for a slow reaction involving an acid or base titration could be used. If the reaction involves a color change, then equipment that can measure the intensity of the color by light absorption can be used. If a gas is given off or used up, then a change in pressure can be used.

A balanced chemical reaction also works for the rate of change of the reactants or products.Consider: 2N2O5(g) → 4NO2(g) + O2(g)

If NO2 is forming at the rate of 0.0072 mol L-1 s-1, then what is the rate of change for the other product and reactant? What is the rate of the reaction?

It might be helpful to see in a graph what is happening:

Here’s the math that we are attempting to do, without needing calculus to take the derivative.

H2 + I2 → 2HI

Factors that Affect Reaction Rates

The rate of a chemical reaction depends on several factors, which further adds to the problem of finding ways to measure the reaction rates.

A) CatalystsB) TemperatureC) Nature of the reactantsD) Concentration of the reactants

The effects of catalysts

Catalysts are substances that increase the rate of a chemical reaction without being consumed. There are two basic ways that catalysts can work.

1) Catalysts can lower the activation energy by reacting to form a reaction intermediate of lower potential energy.

2) Catalysts can offer a surface for the reactants to “find” each other on.

A catalysts that increases the rate of a reaction by forming an intermediate takes an active role in the reaction mechanism. Thus it must be in the same phase as the reactants, so it is sometimes called a homogeneous catalyst.

Consider this aqueous redox reaction:

2 Ce4+ + Tl+ → 2Ce3+ + Tl3+

Why is this redox?

Consider this aqueous redox reaction:

2 Ce4+ + Tl+ → 2Ce3+ + Tl3+

What would it take for this reaction to occur as it is written? What would need to collide? What would need to happen when they collide?

Consider this aqueous redox reaction:

2 Ce4+ + Tl+ → 2Ce3+ + Tl3+

Adding manganese (II) ions increases the rate of this reaction. It is thought to proceed like this:

Ce4+ + Mn2+ → Ce3+ + Mn3+

Ce4+ + Mn3+ → Ce3+ + Mn4+

Tl+ + Mn4+ → Tl3+ + Mn2+

2Ce4+ + Tl+ → 2Ce3+ + Tl3+

Adding manganese (II) ions increases the rate of this reaction. It is thought to proceed like this:

Ce4+ + Mn2+ → Ce3+ + Mn3+

Ce4+ + Mn3+ → Ce3+ + Mn4+

Tl+ + Mn4+ → Tl3+ + Mn2+

2Ce4+ + Tl+ → 2Ce3+ + Tl3+

How is the manganese (II) ion involved? How can it be involved yet not consumed? Why is this “easier” than before?

Because a catalyst increases the likelihood of a reaction to proceed, in chemical terms we would say it lowers the activation energy.

2 Ce4+ + Tl+ alone

with Mn2+

Why is this referred to as tunneling?Why is this referred to as an alternative pathway?Why does the activation energy decrease?

Another type of catalyst does not take part in the reaction, but offers a location for the reaction to occur. This catalyst is in a different phase than the reactants (usually the solid phase) so it is called a heterogeneous catalyst (also sometimes a contact catalyst).

Why is the solid phase going to be different than the reactants’ phase?

Enzymes in your body work as catalysts to increase the rate of the reactions in your body. Without them the reactions in our body would happen at the rate of a tortoise.

Consider the catalyst bed in your catalytic converter.

CO + O2

CO + O + O

CO2 + O

The effects of temperature

Temperature is the measure of the average kinetic energy of the particles (relative to absolute zero) in the system.

It takes energy when atoms collide to cause the reactions (Ea), so if they do not collide with enough energy there will not be a reaction.

So roughly temperature can measure the relative ability of the atoms to react. At low temperatures only a few atoms will react even if a lot collide and at high temperatures a lot more will react when they collide.

Where is the Ea?Why are the crests in different spots?What is this really showing?

Svante Arrhenius calculated the relationship between activation energy and temperature and the amount of possibly productive collisions:

k = the specific rate constant of the reactionA = a constant that represents the fraction of the collisions with the proper orientation when all concentrations are 1 MR and T are as before

ln k = ln A - Ea

RT

Notice, the larger the Ea the smaller the value of k, meaning the slower the reaction takes. Why?

What happens to Ea when a catalyst is present?What does this mean to the value of k and the rate of the reaction.

ln k = ln A - Ea

RT

It is possible to compare the rate at two different temperatures. We just calculate k at each temperature and just subtract, or:

Try this: what is the change in rate if the activation energy is 50 kJ mol-1 and the reaction is raised from 300K to 310K?

lnk2 =

Ea (1

-1

)k1 R T1 T2

For a given reaction the rate constant, k, is 9.16 x 10-

3 s-1 at 0.0°C and the activation energy is 88.0 kJ mol-

1. What is the rate constant at 2.0°C?

At 600. K the value of k is 1.60 x 10-5 s-1. When the temperature is raised to 700. K, the value of k increases to 6.36 x 10-3 s-1. What is the Ea?

Determining the activation energy based on just two temperatures is not necessarily accurate.But if the equation is rearranged a little, something interesting happens:

which can be: y = m x + b

ln k = - (Ea )(

1)+ ln A

R T

which can be: y = m x + bHow would this look to sketch it?

ln k = - (Ea )(

1)+ ln A

R T

The effects of the nature of the reactants.

What does it take on the atomic level for reactants to react?

Check this out!

To facilitate a reaction atoms need to be mobile and as the reactions occur where different atoms can touch each other it helps to have a high surface area.

What are the situations in which these conditions exist?

Also most chemical reactions are redox reactions, so having low ionization energies would help. Why?

What, then, is the “nature” part of the nature of reactants?

The effects of the concentration of the reactants.

Why does the concentration of the reactants matter to the rate of a reaction?

Check this out!

Because the concentration of the reactants has a HUGE effect on the rate of the reaction, it is common to describe the rate of a reaction in terms of the concentrations and a rate constant, k.

rate = k [A]x[B]y[C]z…

What is the concentration measured in?

rate = k [A]x[B]y[C]z…

The superscripts are called the “order” of the substances, and the sum of the superscripts are called the “order of the reaction” or “order overall”.

The superscripts DO NOT come from the coefficients in front of the substance. If they happen to match, it is purely coincidental.

Let’s look as some examples:

3NO(g) → N2O(g) + NO2(g)the rate was determined to be = k[NO]2

This reaction is second order with respect to [NO] and second order overall. This means if we doubled [NO] the reaction would increase by 4 times. This relationship is determined experimentally at first.

Why don’t the products appear in the rate law?

Let’s look as some examples:

2NO2(g) + F2(g) → 2NO2F(g)

rate = k[NO2][F2]

This reaction is first order with respect to [NO2] and first order with respect to [F2], and second order overall.

What would happen to the rate if [NO2] was doubled? What about both [NO2] and [F2]?

Let’s look as some examples:

H2O2(aq) + 3I-(aq) + 2H+(aq) → 2H2O(l) + I3-(aq)

rate = k[H2O2][I-]2

What is the order of each reactant and the order of the reaction?

What does it mean to be zero in order?

What would happen to the rate if the [] of everything were doubled?

rate = k [A]x[B]y[C]z…

Things you need to know about k:1. k can only be determined experimentally, the

balanced equation will not be useful2. the units of k will change depending on the

overall order of the reaction3. k does NOT change if the [] changes or as time

changes4. k does depend on temperature, so often the

temperature of the reaction is specified (or assumed 298 k)

5. k does change if a catalyst is added

Consider a simple reaction : A + 2B → CFrom experiments we get:

let’s find the k and the orders: rate = k [A]x[B]y

Look at experiments 1 and 2. What stays the same, and what changes?

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

Consider a simple reaction : A + 2B → CFrom experiments we get:

What is the result of doubling [B]? How would you say this on the AP test?Between experiment 1 and 2 the [A] is held constant and the [B] is doubled, and the rate of formation of C doubles. So 1 x 2y = 2. So y = 1, therefore the reaction is first order with respect to [B].

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

Consider a simple reaction : A + 2B → CFrom experiments we get:

Look at experiments 1 and 3. What stays the same, and what changes?

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

Consider a simple reaction : A + 2B → CFrom experiments we get:

What is the result of doubling [A]? How would you say this on the AP test?Between experiment 1 and 3 the [B] is held constant and the [A] is doubled, and the rate of formation of C increases by 4. So 2x x 1 = 4. So x = 2, therefore the reaction is second order with respect to [A].

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

Consider a simple reaction : A + 2B → CFrom experiments we get:

What we know so far: rate = k [A]2[B]Why did we not compare 2 and 3?

How can we determine k?

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

Consider a simple reaction : A + 2B → CFrom experiments we get:

rate = k [A]2[B]from experiment 1: 1.5 x 10-6 M s-1 = k (1.0 x 10-2 M)2(1.0 x 10-2 M)

What determines the label for k?

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

k = 1.5 x 10-6 M s-1

= 1.5 M-2 s-1

1.0 x 10-4 M3

Consider a simple reaction : A + 2B → CFrom experiments we get:

so, rate = 1.5 M-2 s-1 [A]2[B]

What is another way to write the label of k?

Experiment Initial [A] Initial [B] Initial Rate of Formation of C

1 1.0 x 10-2 M 1.0 x 10-2 M 1.5 x 10-6 M s-1

2 1.0 x 10-2 M 2.0 x 10-2 M 3.0 x 10-6 M s-1

3 2.0 x 10-2 M 1.0 x 10-2 M 6.0 x 10-6 M s-1

Your turn: 2A + B + C → D + EFrom experiments we get:

Determine the rate law for this reaction.

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

Your turn: 2A + B + C → D + EWhat is the general rate law form?

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

Your turn: 2A + B + C → D + EWhat is the order in respect to [A]?

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

Your turn: 2A + B + C → D + EWhat is the order in respect to [B]?

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

Your turn: 2A + B + C → D + EWhat is the order in respect to [C]?

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

Your turn: 2A + B + C → D + EWhat is the value of k?

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

Your turn: 2A + B + C → D + EWhat is the rate law?

Experiment Initial [A] Initial [B] Initial [C] Initial Rate of Formation of E

1 0.20 M 0.20 M 0.20 M 2.4 x 10-6 M min-1

2 0.40 M 0.30 M 0.20 M 9.6 x 10-6 M min-1

3 0.20 M 0.30 M 0.20 M 2.4 x 10-6 M min-1

4 0.20 M 0.40 M 0.60 M 7.2 x 10-6 M min-1

What if you started a reaction and wanted to know the concentration of a reactant after some time, like “when would half the reactants be left?”

To do this we use a different rate expression, called the integrated rate law (who loves calculus?), but the form of the integrated rate law changes depending on the order of the reaction.

For our use, we will use the hypothetical equation of aA → products

aA → products: first order in respect to A and overall

ln ([A]0/[A]) = a k t

where [A]0 = initial [A] [A] = [A] at the time of interest a = coefficient of A k = rate constant t = time of interest

for example: a radioactive atom decays to form less energetic particles, the reaction is first order overall and has a rate constant of 0.0450 s-1

What is the half life of this atom?ln ([A]0/[A]) = a k t

aA → products: second order in respect to A and overall

1/[A] - 1/[A]0 = a k t

where [A]0 = initial [A] [A] = [A] at the time of interest a = coefficient of A k = rate constant t = time of interest

for example: a reaction of A and B is second order in respect to A and second order overall, and has a rate constant of 0.622 M-1 s-1

What is the half life of 4.10x 10-2 M of A was initially present?1/[A] - 1/[A]0 = a k t

aA → products: zero order in respect to A and overall

[A] = [A]0 - a k t

where [A]0 = initial [A] [A] = [A] at the time of interest a = coefficient of A k = rate constant t = time of interest

Now what would the units of k be?

Reaction Mechanisms

Chemists describe reactions occurring when reactants collide into each other. This is called collision theory. However, not every collision guarantees a reaction. Kinetic energies, ionization energies, and orientation are involved. Some collisions are perfectly elastic and no reaction occurs, just deflection of the particles. Sometimes an inelastic collision occurs long enough for a reaction to actually take place.Consider CO2: What shape is this molecule? How would all three atoms need to collide? Is that likely?

Consider this simple reaction: S + FeSe → Se + FeSConsider this mechanism:

S + Fe - Se → S - Fe - Se → S - Fe + Se

reactants → transition state → products

What orientations would not work?S + Fe - Se → S - Fe - Se → S - Fe + Se

Consider I- + CH3Cl → CH3I + Cl-

What orientations would not work? Would this be more or less likely to occur than the first equation?

The preceding examples are considered very simple because they only involve 1 step (1 collision between two particles). Most reactions occur in many steps. These steps are called elementary steps and comprise the entire reaction mechanism.

A reaction mechanism cannot occur faster than the slowest elementary step. This is called the rate determining step.

The reaction order is equal to the coefficients of the rate determining elementary step.

if rate = k[A]x[B]y, then the rate determining step is xA + yB → products

This can help us determine a mechanism by giving a guide to the rate determining step.

Consider NO2(g) + CO(g) → NO(g) + CO2(g)and rate = k [NO2]2

So the rate determining step is 2 NO2 → N2O4 (probably)so what’s left? 2 NO2 → N2O4

N2O4 + CO → NO + CO2 + NO2

------------------------------------------ NO2 + CO → NO + CO2

N2O4 is called a reaction intermediate. What is characteristic of a reaction intermediate?

Consider NO2(g) + CO(g) → NO(g) + CO2(g)and rate = k [NO2]2

Careful study has not detected N2O4, but has found NO3 in this reaction. How does that change things?

Consider 2NO(g) + Br2(g) → 2NOBr(g)and rate = k [NO]2[Br2]

A collision involving only two particles seems more likely than one that involves three. This time the rate determining step is the second.

NO + Br2 → NOBr2

NOBr2 + NO → 2NOBr (rate step)2NO + Br2 → 2NOBr

How does this match the rate law?Why does this seem better than only one step?

The rate determine step was rate = k2[NOBr2][NO]How does this become rate = k[NO]2[Br2]?

In what way could a catalyst be involved?