# Chapter 4  Kinetics

## 4.1  Average Rates

Kinetics is the study of the rate of reactions; how fast they go and why.

The rate of a reaction can be quantified by the rate of change of the concentration of a reactant or product.

The following reaction will be used as an example.

N2 (g) + 3 H2 (g) → 2 NH3 (g)

The average rate of change of N2, rN2, is calculated in terms of the change, Δ, in concentration, [N2], in time, t, and is given by:

rN2 =
 Δ [N2] Δ t
=
 [N2]f − [N2]i tf − ti

If we were to monitor the concentration of N2 in time we could calculate this average rate of change.

The solid line is the curve that passes through the measured points of the [N2]; the dashed line is the curve that connects the first and last points measured which we will calculate the average from:

rN2 =

 0.2 M − 1.0 M 60 s − 0 s
=
 −0.8 M 60 s
= −0.013 M/s N2

rN2 is simply the slope of the straight line from initial to final time; note that it is negative because N2 is a reactant and therefore its concentration should be decreasing as the N2 is being consumed.

We were not monitoring the concentration of the other species in the reaction, but their average rates of change can be related to rN2 by stoichiometry.

rH2 = rN2 ×

 3 mol H2 1 mol N2
= −0.013 M/s N2 ×
 3 mol H2 1 mol N2
= −0.039 M/s H2

One problem encountered when using stoichiometry blindly is that we obtain a negative average rate of change for NH3 which is incorrect because it is a product whose concentration should be increasing!

rNH3 = rN2 ×

 2 mol NH3 1 mol N2
= −0.013 M/s N2 ×
 2 mol NH3 1 mol N2
= −0.026 M/s NH3

Since NH3 is being produced, it should have a positive average rate of change which you can see represented in the plot above:

 rNH3 =  +0.026 M/s NH3

Sometimes positive average rates of consumption are reported for reactants because the word consumption replaces the negative sign.

The next step is to convert these average rates of change of reactants/products into an overall average rate of reaction, rrxn = r.

Again, stoichiometry is used, but you are to convert to mol rxn instead of mol reactant/product.

r = rN2 ×

 1 mol rxn 1 mol N2
= −0.013 M/s N2 ×
 1 mol rxn 1 mol N2
= −0.013 M/s rxn

All rates of reaction are positive values, and we leave off the "rxn" designation:

 r = +0.013  M/s

Notice that r is the same as rN2, except for the sign, because the stoichiometry coefficient of N2 is 1.

This stoichiometric calculation works regardless of which reactant or product you use to start with.

r = rH2 ×

 1 mol rxn 3 mol H2
= −0.039 M/s H2 ×
 1 mol rxn 3 mol H2
= −0.013 M/s rxn → +0.013 M/s

r = rNH3 ×

 1 mol rxn 2 mol NH3
= 0.026 M/s NH3 ×
 1 mol rxn 1 mol NH3
= 0.013 M/s rxn → 0.013 M/s

The calculation of the average rate of reaction, r, from reactant/product average rates of change is often presented as just the average rate of change of a reactant/product divided by its stoichiometric coefficient, which is mathematically equivalent to the stoichiometric version above, but not conceptually equivalent; also notice the negative signs in front of the reactant average rates of change to ensure the average rate of reaction is positive.

r = −
 rN2 1
= −
 rH2 3
=
 rNH3 2

## 4.2  Instantaneous Rates

If the initial and final time points of measured reactant/product concentration are infinitesimally close together, the average rate of reaction, r becomes an instant rate of reaction, r.

The instant rate of change, r, is simply represented by the tangent line of the curve that is tracing the measured reactant/product concentrations in time.

In the plot above, the earlier tangent line has a steeper slope indicating that the reaction is going faster whereas the later tangent line has a shallower slope indicating the reaction is going slower.

The dotted line represents r, the overall average rate of reaction, and is an average of ALL the instantaneous rates, both drawn and not drawn in the plot, between the beginning and end times of the experiment.

Since measurements of reactant/product concentration cannot be taken at infinitesimally small intervals, an approximation to the instantaneous rate is made by using two time points that are as close together as possible.

### 4.2.1  Rate Laws

Many kinetics experiments have been done over the years and a general equation for the instantaneous rate of reaction as been empirically observed; it is called the rate law.

For a generic reaction,

a A + b B → products

where A and B are reactants and a and b are their stoichiometric coefficients, the rate law is

 r = k [A]x [B]y     (1)

where the variables that lend flexibility to the equation are k, the rate constant, and x and y which are individual reaction orders.

Individual reaction orders, x and y, are often whole numbers, but they can be fractional as well, and are sometimes, but not always, the same number as the stoichiometric coefficients, a and b.

The sum of the individual reaction orders is the overall reaction order, x+y.

All types of reaction orders are referred to as first, second, third, etc. unless they turn out to be fractional.

##### Problem

The following reaction is 2nd order overall and 1st order in [H2]. What is the order in [I]? What is the rate law? What are the units of k?

H2 (g) + I (g) → HI (g) + H (g)
 overall order = 2 = x + y = 1 + y; y = 2 − 1 = 1

The reaction is first-order in [I] because y = 1.

The rate law is then:

 r = k [H2]1 [I]1 = k [H2] [I]

The units of k depend on the overall reaction order because the overall reaction order is the exponent to the unit molarity, M, that comes from the number of times a concentration is being multiplied by another concentration on the right hand side of the rate law.

The unit of rate is always M/s so the units of k must convert Moverall order to M/s as indicated by the following unit analysis.

 M s
= k Moverall order

k =

 M s · Moverall order
=
 1 s · Moverall order − 1

For our problem above

 M s
= k M2

k =

 1 s · M2−1
=
 1 M · s

———————————————————————–

It’s pretty easy to see a trend in the units of k as the overall order of reactions change.

 overall reaction order 0 1 2 3 units of k M/s 1/s 1/Ms 1/M2s

### 4.2.2  The Method of Initial Rates

The method of initial rates is an experimental method that uses the instantaneous rate of reaction, measured approximately using two very closely spaced time points, to determine the individual reaction orders; once those are determined the value and units of k can be determined and the full rate law can be known.

The experiment requires several iterations of running the reaction with varying initial concentrations of the reactants to probe their effect on the initial instantaneous rate of reaction so that their individual reaction orders can be determined.

Initial instantaneous reaction rates are measured because it is often only at initial time that we know what the concentrations of the all the reactants are.

##### Problem

The method of initial rates experiment was conducted on the following (generic) reaction. What are the individual orders, overall order, and value and units of k?

2 A + B → products
 Trial [A]0 (M) [B]0 (M) initial rate, r r = k [A]x [B]y 1 0.1 0.05 2 2 = k (0.1)x (0.05)y 2 0.2 0.05 8 8 = k (0.2)x (0.05)y 3 0.1 0.1 4 4 = k (0.1)x (0.1)y

Notice that only 1 concentration was changed at a time, otherwise the effect of the concentration of one reactant can’t be seen clearly in the reaction rate.

We’ll compare the two trials where the [A]0 changed by dividing Trial 2 by Trial 1, since Trial 2 has the higher rate.

 Trial 2 Trial 1

=
 8 2
=
 k (0.2)x (0.05)y k (0.1)x (0.05)y

4 =
 (0.2)x (0.1)x
=

 0.2 0.1

 x
= 2x

To solve 4 = 2x we use logs:

 log(4) = log(2x) = x log(2)
x =
 log(4) log(2)
= 2

In this case it was plain to see that x = 2.

The reaction is 2nd order in [A].

Do the same from Trials 3 and 1 to find y.

 Trial 3 Trial 1

=
 4 2
=
 k (0.1)x (0.1)y k (0.1)x (0.05)y

2 =
 (0.1)y (0.05)y
=

 0.1 0.05

 y
= 2y

It is plain to see that 2 = 2y yields y = 1.

The reaction is 1st order in [B].

The rate law is then r = k [A]2 [B]1 = k [A]2 [B].

The overall reaction order is 3rd order meaning the units of k are 1/M2s.

In this case, that the individual reaction orders were the same as the stoichiometric coefficients is coincidental.

The value of k is determined by picking one Trial, or averaging the results of all three, and solving for the only remaining variable, k.

Using Trial 2:

8 = k (0.2)2 (0.05); k = 4000
 1 M2s
= 4000 M−2 s−1

———————————————————————–

## 4.3  The 3 Common Rate Laws

We are going to limit ourselves to the following reaction:

a A → products

We can arrive at three different rate laws depending upon the reaction mechanism (the sequence of reaction steps):

 r = k [A]x  where x = 0, 1, 2     (2)

### 4.3.1  Second-order reactions

Second-order reactions, x = 2, are the easiest to understand conceptually.

The rate law is:

 r = k [A]2

This rate law means that the rate of the reaction increases to the square of the reactant A concentration.

That the rate increases to the square of [A] makes sense if the reaction requires one A to collide with another A.

By increasing [A] we’re making it more and more likely that the collision will occur.

A plot of this rate law looks like half of a parabola:

### 4.3.2  First-order reactions

The rate of a first-order reaction increases linearly with increasing [A] as given by its rate law:

 r = k [A]1 = k [A]

In this case, the linear rate increase with [A] indicates that the rate is only going faster because there is more A present, but not that we’re waiting for the A reactants to collide with each other.

Radioactive nuclear decay follows first-order kinetics; imagine radioactive nuclei sitting in a rock waiting to decay, the more nuclei you have the more often you’ll see one decay.

The slope of the plot of rate versus [A] turns out to be equal to the rate constant, k; the y-intercept is obviously zero because with any A you can have no reaction.

### 4.3.3  Zero-order reactions

The simplest rate law is one that is zero-order in [A].

 r = k [A]0 = k

This rate indicates that the rate of the reaction has no dependence on how much of reactant A is present.

This is a very odd rate law that makes no sense as it seems to imply the reaction would go even without any A present!

r = k and that’s it.

However, in some cases it is possible to observe a reaction that is occurring at a rate independent of the reactant concentration.

Such reactions generally involve very high concentrations of A with some other barrier to reaction that does not involve how much A is present; the high concentration of A ensuring the reaction barrier is always being crossed thereby giving a constant rate.

The plot of this rate law indicates that there is no slope with only a y-intercept of r = k:

## 4.4  Integrated Rate Laws

If we can measure the average reaction rate, r, over infinitesimally small time steps it effectively becomes an instantaneous reaction rate, r.

This is a trick of calculus and is denoted mathematically by replacing the Δ with a lowercase d:

r = −
 Δ [A] a Δ t
→calculus→ r = −
 d[A] a dt

We’ll assume the stoichiometric coefficient is a = 1, for convenience.

There are now 2 equations that are equal to the instantaneous rate, r.

1. r = −d[A]/a dt
2. r = k [A]x

Setting these two equal to each other gives the differential rate law:

 d[A] dt
= k[A]x     (3)

The differential rate law can be reorganized and solved using methods of differential equations, like simple integration:

 d[A] [A]x
= −kdt
 d[A] [A]x
= −k dt

### 4.4.1  Zero-order Integrated Rate Law

The solution of the zero-order differential rate law is easiest to understand since the rate of reaction is constant, r = k.

 d[A] [A]0
= d[A] = −k dt

The solution to the above equation is called an integrated rate law which is a "time machine" that allows the calculation of [A] at various times during the reaction.

 [A]f − [A]0 = − k(tf − t0)

Solving this equation for the rate constant, k, gives the same equation as the average reaction rate.

k = −
 [A]f − [A]0 tf − t0
= −
 Δ [A] Δ t
= r

Since the instantaneous rate of reaction is constant for zero-order reactions (because r = k), it makes sense that the instantaneous rate is always equal to the average rate, r = r.

Assuming t0 = 0 s, making tf = t, and solving for [A]f gives the standard way this equation is presented:

 [A]f = [A]0 − kt     (4)

The units of k for a zero-order reaction, M/s, remind us it has the exact same units as the rate because r = k for zero-order reactions, allowing us to write the zero-order rate law as:

 [A]f = [A]0 − rt

A unit analysis of equation 4.4 reveals:

Mf = M0 −

 M s

(s

This equation shows us that the final concentration is the difference between the initial concentration and how much of A is consumed in some amount of time, t, at a reaction rate, r.

This is similar to a one dimensional position problem one might encounter in a physics course.

 xf = x0 + vx t

Your final position, xf, is equal to your initial position, x0, plus how fast you were going, vx (the velocity in the x direction), times how long you were going that fast, t.

vx could be negative or positive depending on whether you were traveling toward more negative or more positive values of x, respectively.

A plot of the zero-order integrated rate law, equation 4.4, shows how [A]0 acts as the y-intercept and k acts as the slope along the x-axis of time, t.

Again, the plot makes it plain to see that any instantaneous reaction rate (as given by the tangent to the curve at any point) is equal to and the same as the average reaction rate between the first and last measured points.

### 4.4.2  First-order Integrated Rate Law

Solution of the reorganized differential rate law:

 d[A] [A]1
=
 d[A] [A]
= −k dt

gives the first-order integrated rate law

 ln[A]f − ln[A]0 = − kt

This first-order integrated rate law can be rewritten in a convenient form that allows one to work in fractions (or percentages) of reactant left, [A]f/[A]0, when the initial concentration is unknown or irrelevant.

ln

 [A]f [A]0

= − kt

This form is also the most mathematically correct since you cannot, formally speaking, take the natural log of a number that has a unit (e.g. molarity); in this case the ln(M) doesn’t mean anything.

By using the ratio of concentrations the molarity unit is canceled before the ln is taken.

To write this integrated rate law in a way that solves for [A]f gives almost the exact same form as for the zero-order integrated rate law except it uses the ln[A] instead of plain [A].

 ln[A]f = ln[A]0 − kt     (5)

A similar interpretation can be had as before, the final amount, ln[A]f, is the difference between the initial amount, ln[A]0, and how much of A is consumed in some amount of time, t, at a rate of k.

However, k is not a true rate because it only has units of s−1 so when it is multiplied by time the result has no units which is consistent with the ln[A] not having any units.

To truly solve for [A]f, one obtains

 [A]f = eln[A]0 − kt = eln[A]0 e− kt = [A]0 e− kt

to finally obtain

 [A]f = [A]0 e− kt     (6)

If you study economics, this equation should look familiar because it is the same as that used for continuously compounded interest.

 \$f = \$0 ert

Where r = the annual percentage rate (APR) and t is expressed in years.

Since nuclear radioactive decay is a good example of a first-order reaction, it is pertinent to say that the unit of time does not have to be seconds; it should be whatever unit is most convenient and natural for the reaction being considered, be it seconds, minutes, hours, days, or years.

Just be sure that the denominator of your rate constant, k, uses the same unit for time, be it s−1, min−1, h−1, d−1, or y−1 (sometimes a−1), respectively.

The plot of [A] versus time for the first-order integrated rate law shows that the average reaction rate is NOT equal to the instantaneous reaction rate except at one specific time where the tangent of the curve has the same slope as the average reaction rate.

As can be seen, it is common to try and obtain a straight line when relating [A] to t, so the y-axis must be altered from [A] to ln[A] to give a straight line when working with first-order reactions.

A straight line is useful because its (constant) slope is the rate constant, k, with y-intercept, ln[A]0.

slope, m = k =

 rise run

=
 ln[A]f − ln[A]0 tf − t0

Of course, you get the exact same equation solving the first-order integrated rate law, equation 4.5, for k,

k =
 ln[A]f − ln[A]0 t

If you can monitor [A] during a reaction, whether you have to plot [A] or ln[A] versus time to get a straight line can tell you whether the reaction is zero or first-order, respectively, without the need to use the method of initial rates.

### 4.4.3  Second-order Integrated rate law

Solution of the reorganized differential rate law:

 d[A] [A]2
= −k dt

gives the second-order integrated rate law

−
 1 [A]f
− −
 1 [A]0
= − kt

Rearranging this gives

 1 [A]f
=
 1 [A]0
+ kt     (7)

As with the first-order integrated rate law, you must plot a modified concentration of A, the reciprocal as it appears in the integrated rate law, 1/[A], versus time to obtain a straight line.

Due to the reciprocal, high concentrations are at the bottom of the y-axis and low concentrations are at the top of the y-axis.

Hence, why the straight line goes up instead of down as time progresses, as the reactant concentration must become lower as the reaction proceeds.

This is also why we see kt added to 1/[A]0 instead of subtracted from it.

Once you take the reciprocal of everything on the right hand side of equation 4.7 to solve for [A]f, anything that has been added to 1/[A]0 will only make [A]f smaller.

Solving equation 4.7 for [A]f leads to a rather ugly equation, so the following form of the integrated rate law is rarely shown, but it does allow you to explicityly see how kt does decrease the value of [A]f despite it being added to 1/[A]0.

[A]f =
 [A]0 1+[A]0 kt

##### Problem

The following reaction has a rate constant of k = 2.80 × 10−3 min−1. If [SO2Cl2]0 = 4.31 × 10−3 M, what is [SO2Cl2]f at t = 520 min?

SO2Cl2 (g) → SO2 (g) + Cl2 (g)

We first recognize that the units of k are min−1, which immediately indicates this is a first order reaction, so we will need the first-order integrated rate law to solve for the [SO2Cl2] at a later time.

 ln[SO2Cl2]f = ln[SO2Cl2]0 − kt
 ln[SO2Cl2]f = ln(4.31 × 10−3  M) − (2.80 × 10−3  min−1)(520  min)
 ln[SO2Cl2]f = −6.9028
 [SO2Cl2]f = e−6.9028
 [SO2Cl2]f = 1.00 × 10−3  M

———————————————————————–

## 4.5  Half-Life

To augment the already long discussion on the first-order integrated rate law, there is something very special about the equation in the following form:

ln

 [A]f [A]0

= − kt

You’ll recall it was stated that this form allows one to work with the fractional amounts of reactant left after time t, e.g. the amount of time it takes for half, 1/2, of the initial amount of A to be consumed, t1/2, doesn’t require you to know what the initial amount actually was because [A]f = 1/2[A]0.

ln

 1/2[A]0 [A]0

= ln
1/2
= − kt1/2

Rearranging to clean up the result gives:

 ln(1) − ln(2) = −kt1/2
 ln(2) − ln(1) = +kt1/2
 ln(2) = kt1/2

Finally, the amount of time it takes to consume half of the initial amount of reactant is called the half-life, t1/2,

t1/2 =
 ln(2) k
(8)

What is so cool about this equation is it does NOT depend on the initial concentration of reactant (for first-order reactions only)!

Also, if you know the half-life, you can derive an equation to calculate the amount of reactant left after some amount of time, tf, that is easier to use as it does not involve natural logs.

ln

 1 2

= − kt1/2
ln

 [A]f [A]0

= − ktf

Both equations above can be written to have −k on one side of the equation and can therefore be set equal to each other having done so:

k =
ln

 1 2

t1/2
=
ln

 [A]f [A]0

tf

Gathering terms gives:

 tf t1/2
ln

 1 2

= ln

 [A]f [A]0

Which, using a property of logs, can be rewritten:

ln

 1 2

 tf t1/2

= ln

 [A]f [A]0

Then raise both sides as powers of e:

e
ln

 1 2

 tf t1/2

= e
ln

 [A]f [A]0

The e and natural logs cancel each other out leaving behind:

 1 2

 tf t1/2

=

 [A]f [A]0

Which can be solved for [A]f to give:

[A]f = [A]0

 1 2

 tf t1/2

(9)

This form of the first-order integrated rate law is most often encountered when studying the radioactive decay of some of number, N, of radioactive atoms:

Nf = N0

 1 2

 tf t1/2

##### Problem

The first-order decay of 15O, oxygen-15, has a rate constant of k = 0.00568 s−1. If [15O]0 = 1.56 × 10−2 M, how long does it take to consume half of the initial concentration? How long to consume 75%? How long to consume 90%?

Calculate the half-life from k:

t1/2 =
 ln(2) 0.00598  s−1
= 122  s

To consume half of 1.56 × 10−2 M takes 122 s.

To consume half of any initial amount takes only 122 s; it does NOT matter how much you start with.

The following table shows the progression of how much is consumed and how much remains with each successive half-life.

Note that half of 50% at the start of the 2nd half-life period is not 0% but 25%, and so on.

 number of half-lives %[15O] remaining %[15O] consumed t (s) 0 100 0 0 1 50 50 122 2 25 75 244 3 12.5 87.5 366 4 6.25 93.75 488

From the table we can see that it takes 2 half-lives, or 244 s, to consume 75% of [15O]0.

The consumption of 90% does not fall on a half-life; but that’s okay, we can calculate the tenth-life, t1/10; though we have to be careful that we understand that this is the amount of time it takes to be left with 10%, NOT the amount of time to consume 10% (such a distinction isn’t necessary for half-life because the amount consumed and the amount remaining after each half-life is the same, half was consumed and half remains):

t1/10 =
 ln(10) k
=
 ln(10) 0.00598  s−1
= 405  s

To have only 10% remaining (90% consumed) takes 405 s, which is equivalent to the passage of 405/122 = 3.32 half-lives.

Of course, we could have just used the first-order integrated rate law directly, though that’s basically what we already did.

ln

 0.10 [15O]0 [15O]0

= −(0.00568 s−1)t

t =
 ln(0.1) −0.00568  s−1
= 405  s

———————————————————————–

There are half-life equations for zero and second-order reactions as well, however they depend on the amount you start with and so there’s nothing special about them.

The zero-order half-life depends on the amount you start with because the rate never changes, so the more you start with the longer it will take to consume half of it.

t1/2 =
 1/2[A]0 k
=
 1/2[A]0 r

The second-order half-life depends inversely on the amount you start with since the rate depends on how frequently 2 reactants collide.

t1/2 =
 1 k[A]0

The higher [A]0 is, the more frequently reactants will collide causing a higher reaction rate which then requires less time to consume half of [A]0.

Opposite of the zero-order half-life, the more reactant you start with in a second-order reaction will give rise to a shorter half-life.

### 4.5.1  Mean-life

One last curiosity of the first-order reaction is the mean-life, t, which is how long, on average, does it take one reaction to occur.

The mean-life is simply the reciprocal of the rate constant.

t =
 1 k

We can imagine that the mean-life is slightly longer than the half-life because the first half of [A]0 has been consumed by the first half-life and the other half of [A]0 will be consumed over the course of the next few half-lives.

For 15O, the mean-life is

t =
 1 0.00568  s−1
= 176  s

## 4.6  The Rate Constant, k

We have already explored one of the primary factors in the rate of a reaction, reactant concentration.

What is left is to understand is the role and meaning of the rate constant, k.

The rate constant is primarily responsible for accounting for the temperature of the reaction conditions, but it also accounts for orientational requirements in order for collision events to become reaction events.

In the spirit of the method of initial rates, the same experiment can be run at various temperatures instead of various concentrations to determine how the rate constant, k, changes with temperature; usually under the assumption that the individual reaction orders are independent of temperature.

This has been done for many reactions and a general trend as emerged; there is a linear relationship between ln(k) (y-axis) and 1/T (x-axis):

 ln(k) = m ⎛ ⎝ 1/T ⎞ ⎠ + b

This linear trend is observed in the Arrhenius plot:

Since the reciprocal of T is on the x-axis, the higher temperatures are on the left and the lower temperatures are on the right.

In the limit of very high temperature, meaning the temperatures are so high that k becomes constant, the only part of this relationship that matters is the y-intercept, b.

The y-intercept, b, is related to the frequency factor, A:

 b = ln(A)

A has the same units as k which is consistent with its name as the frequency factor because frequency has units of s−1; of course we can’t forget there may be some factor of molarity in the unit of k and therefore A as well.

The frequency factor, A, is broken into the product of 2 pieces, p and z.

 A = pz

The collision frequency, z, is ultimately where the units of s−1 come from as it accounts for the frequency of collisions, the number of collisions that occur in a given time.

Not every collision accounted for in the collision frequency results in a reaction because perhaps the reactants are picky about what parts of their molecules are involved in the collision in order for reaction to occur.

The orientation factor, p, is a number between 0 and 1 that reflects the probability that any given collision will collide with the correct orientation for reaction.

Therefore, the number of reactive collisions is less than or equal to the number of collisions, pzz.

• A-A + B → A-A-B probably has an orientation factor, p ≈ 1, because regardless of which side of the A-A molecule B collides with, the product A-A-B will still be made
• C-A + B → C-A-B probably has an orientation factor, p ≈ 0.5, because only half of the C-A molecule has an A for the B to collide and then react with; collision with the C side of the C-A molecule results in no reaction

The temperature dependence of the rate constant, k, is more directly related to the slope, m, of the plot above at lower temperatures where the value of k does change with changing temperature.

Specifically, the slope is related to the activation energy, Ea, the amount of kinetic energy needed to surmount any energetic barriers to reaction.

It is not enough that reactants collide with the correct orientation, they must also collide with enough kinetic energy to force the rearrangement of atoms in the molecules; they must overcome that activation energy, Ea.

The slope, m, of ln(k) versus 1/T must have units of temperature in kelvin so as to cancel with the 1/K in the x-axis term, 1/T.

The activation energy, Ea, has units of kJ/mol so we divide it by the universal gas constant, R, expressed in units of kJ/mol·K (8.314 × 10−3 kJ/mol· K) in order to achieve the necessary unit of kelvin.

We also insert a negative sign so that Ea is a positive number in relation to the slope, m, so that Ea represents a conceptually consistent positive energetic barrier to reaction.

m = −
 Ea R

##### Problem

The k over a range of temperatures are measured and the results graphed in an Arrhenius plot whose slope turns out to be m = −3.27 × 104 K. What is Ea for this reaction?

−3.27 × 104  K = −
 Ea 8.314 × 10−3  kJ/mol · K

 Ea = −(−3.27 × 104  K)(8.314 × 10−3  kJ/mol · K) = 272  kJ/mol

———————————————————————–

A plot of the potential energy along the reaction coordinate (the collision path with correct orientation) elucidates the meaning of Ea which appears as an energetic "hill" that must be overcome in order to convert reactants into products.

The arrangement of atoms present at the top of the activation energy hill is called the activated complex or alternatively, the transition state

Substituting these quantities in for the y-intercept, b, and slope, m, gives the linear form of the Arrhenius equation.

ln(k) = −
 Ea R

 1 T

+ ln(A
 y = m(x) + b

The more common form of the Arrhenius equation, known as the exponential form, is written in terms of k.

 k = Ae−Ea/RT = pze−Ea/RT     (10)

What we see is a unitless exponential factor known as a Boltzmann distribution which gives the probability that colliding reactants will have enough energy to surmount Ea at the current reaction temperature, T.

 Boltzmann distribution = e−Ea/RT

Thus, the rate constant, k, is the collision frequency, z, multiplied by the probability of colliding with the correct orientation, p, and the probability of colliding with enough energy to surmount Ea, eEa/RT.

At very high temperatures, we see the Boltzmann exponential factor effectively become the number 1 since dividing the exponent by a very large T makes the whole exponent approach 0, and e0 = 1.

k =
 lim T → ∞
pzeEa/RT = pze0 = pz

This makes sense because at very high temperatures, the probability that colliding molecules have enough energy to surmount Ea is 1, complete certainty, and e0 = 1.

We stated earlier that at higher temperatures, the only thing that matters is the y-intercept, b, which turned out to be related to A=pz which is just the collision frequency and orientation factor.

k is inversely proportional to Ea when you consider the exponential form of the Arrhenius equation without the negative exponent.

k =
 A eEa/RT

k ∝
 1 Ea

Thus, the larger the Ea for a reaction, the smaller k will be leading to a slower reaction because fewer colliding molecules will have enough kinetic energy to surmount the larger Ea.

A smaller Ea will lead to a relatively larger k and a faster reaction because more colliding molecules will have enough kinetic energy to surmount the smaller Ea.

k is inversely proportional to the reciprocal of T making it directly proportional to T, so when T increases so does k, and so does the reaction rate.

k ∝
 1 1/T
∝ T

Of course, we’ve already seen this direct proportionality between k and T in the Arrhenius plot above when we plotted ln(k) versus 1/T and saw a negative slope.

### 4.6.1  Two-point form of the Arrhenius Equation

It only takes 2 points to make a straight line, so one could measure k at 2 different temperatures in order to determine Ea.

The slope of a line is often stated as rise over run, and considering our y-axis is ln(k) and our x-axis is 1/T we get

m =

 rise run

=
 y2 − y1 x2 − x1
=
ln(k2) − ln(k1)

 1 T2
−
 1 T1

= −
 Ea R

The right hand side of the equation above is the two-point form of the Arrhenius equation, often written as in equation 4.11.

ln

 k2 k1

= −
 Ea R

 1 T2
−
 1 T1

(11)
##### Problem

If k1 = 0.125 s−1 at 200 K and k2 = 0.347 s−1 at 350 K, what is Ea and what is the pre-exponential factor, A? What is the probability that colliding molecules will have enough kinetic energy to overcome Ea at 200 K and 350 K?

Solving for the activation energy:

Ea = −R ×
ln

 k2 k1

 1 T2
−
 1 T1

= −8.314 × 10−3  kJ/mol · K ×
ln

 0.347  s−1 0.125  s−1

 1 350  K
−
 1 200  K

= 3.96  kJ/mol

Pick either conditions 1 or 2 to calculate A:

 k2 = Ae−Ea/RT2
 0.347  s−1 = Ae−3.96  kJ/mol/R × 350  K
A =
 0.347  s−1 e−3.96  kJ/mol/R × 350  K
= 1.35  s−1

Since A = pz, there are apparently 1.35 collisions with the correct orientation for reaction every second.

The probability of colliding reactants having enough kinetic energy to surmount Ea is just the Boltzmann distribution:

 Boltzmann distribution at 200 K = e−3.96  kJ/mol/R × 200  K = 0.0924
 Boltzmann distribution at 350 K = e−3.96  kJ/mol/R × 350  K = 0.256
 0.256 0.0924
= 2.77

Colliding molecules at 350 K have a 2.77 times higher chance of reacting than those at 200 K, assuming A is constant and temperature independent (which is something we’ve assumed so far but isn’t necessarily true).

———————————————————————–

### 4.6.2  The Universal Gas Constant, R

Lastly, the universal gas constant, R, is often thought of as just a number that allows the conversion of L, atm, mol, and K in the ideal gas equation, PV=nRT, where it takes on the form R = 0.08206 L·atm/mol·K.

A closer look at PV=nRT reveals that PV is a unit of energy (e.g. remember enthalpy, H = E + PV), nT is a representation of how much energy n moles of gas have at temperature T, and R just converts one unit of energy, L·atm, to another, K (kelvin).

In the context of the Arrhenius equation, R is expressed in units of kJ/mol·K since kJ is a more natural unit of energy than L·atm.

In the case of the Arrhenius equation, R’s job is to convert the temperature, T (a measure of the average kinetic energy of the reactants), into units of kJ/mol so it can be compared to Ea, to arrive at the probability of the reactants having enough kinetic energy to surmount Ea as given by the Boltzmann distribution.
It is very important to use the correct R in your equation depending on the application.

One is 8.206 × 10−2 L·atm/mol·K and the other is 8.314 × 10−3 kJ/mol·K; they’re almost the same number being different by only a factor of ≈ 10, so if you use the wrong one you might end up with semi-correct looking numbers but with the decimal point in the wrong spot.

We can use these two versions of R to create a conversion factor that allows one to convert energy units of L·atm to kJ.

 8.314 × 10−3 kJ/mol · K 0.08206 L · atm/mol · K
= 0.1013 kJ/L · atm

That conversion factor may be familiar because it is the same conversion factor (within a few powers of ten) that converts pressure units of Pascals (Pa) to atm.

 1 atm = 101325 Pa

Remember that pressure is force per unit area, F/A, yielding the SI unit of pressure, 1 Pa = 1 N/m2, where the unit of force is a newton, N.

F = ma =
kg × m/s2 =
 kg · m s2
= N

In a proof that pressure times volume, PV, is an energy, the product of their SI units yields the units of force times distance, Fx, which is the definition of work, w, which is a means to exchange energy, which has units of joules, J:

 PV = Fx = w ( N/m2 ) × m3 = N × m = ( kg · m/s2 ) × m = kg · m2/s2 = J

## 4.7  Reaction Mechanisms

Reaction mechanisms are the sequence of steps, or collisions, that covert reactants into products.

Chemical equations can represent one of two different reaction types which differ in the number of steps in the reaction mechansim.

1. Elementary reactions have a 1 step reaction mechanism.

The number of molecules involved in that 1 step is indicated by the elementary reaction’s molecularity which is intimately tied to its rate law.

Elementary reactions have individual reaction orders equal to stoichiometric coefficients.

• unimolecular
• A → products; r = k [A]
• bimolecular
• 2A → products; r = k [A]2
• A + B → products; r = k [A][B]
• termolecular (this type is an unlikely reaction mechanism because 3 molecules colliding simultaneously with the correct orientation is rare)
• 3A → products; r = k [A]3
• 2A + B → products; r = k [A]2[B]
• A + B + C → products; r = k [A][B][C]
2. Composite reactions have a multi-step reaction mechanism; therefore they are made of a series of elementary reactions.

The individual reaction orders may or may not equal stoichiometric coefficients.

Because of the strong link between the rate law (i.e. individual reaction orders) and the molecularity (i.e. stoichiometry) of elementary reactions (which are the components of composite reactions), the experimentally determined rate law can provide clues as to what the reaction mechanism is.

##### Problem

Propose a reaction mechanism for the following reaction with an experimentally determined rate law of r = k [NO2] [F2].

2 NO2 (g) + F2 (g) → 2 NO2F (g)

The experimental rate law indicates a rate-limiting step of a bimolecular elementary collision between one NO2 and one F2.

At some point the F2 molecule has to break apart, so might as well accomplish that in the collision between the one NO2 and F2 and attach one of the resulting F atoms to the NO2 to make one of the two NO2F products.

NO2 (g) + F2 (g) → NO2F (g) + F (g)

In another step, the produced F atom needs to react with a second NO2 to form another NO2F in order for the reaction mechanism to respect the stoichiometry of the chemical equation; an F atom is also not a product so it must be consumed at some point in the reaction mechanism.

NO2 (g) + F (g) → NO2F (g)

The consumption of the F atom made in the first step by a second NO2 must proceed very quickly since we don’t see evidence of this reaction in the experimentally determined rate law.

We may now propose a reaction mechanism which in addition to listing the sequence of elementary reaction steps, must also specify the relative speed of the steps to each other.

 Step Elementary Reaction Speed Rate Law 1 NO2 (g) + F2 (g) → NO2F (g) + F (g) slow r1 = k1 [NO2] [F2] 2 NO2 (g) + F (g) → NO2F (g) fast r2 = k2 [NO2] [F] sum 2 NO2 (g) + F2 (g) + F (g) → 2 NO2F (g) + F (g)

The relatively slow step will always dictate what the predicted rate law will be so it must be chosen to coincide with the experimentally observed rate law.

Since the rate law of the bimolecular first step matches the experimentally observed rate law, the first step must be the slow, rate-limiting step.

It is important that the sum of the steps returns the original chemical equation.

The F atom produced in the first step and consumed in the second is called a reaction intermediate.

A possible potential energy profile for a two step reaction could look like this with the first Ea larger than the second since the first step is slower:

Having established a reasonable reaction mechanism, let’s look at other potential reaction mechanisms and see how they are not consistent with the experimentally observed rate law.
Experimentally Inconsistent Reaction Mechanism #1

This mechanism will require that the F2 molecule break apart before it is capable of reacting with the NO2.

 Step Elementary Reaction Speed Rate Law 1 F2 (g) → 2 F (g) slow r1 = k1 [F2] 2 (×2) NO2 (g) + F (g) → NO2F (g) fast r2 = k2 [NO2] [F]

The above reaction mechanism is implausible because it predicts the rate law, r = k [F2], due to the slow first step, which is not the same as the actual experimentally stated rate law given in the problem.

Experimentally Inconsistent Reaction Mechanism #2

If we switch which of those steps fastest, the mechanism is still implausible.

 Step Elementary Reaction Speed Rate Law 1 F2 (g) → 2 F (g) fast r1 = k1 [F2] 2 (×2) NO2 (g) + F (g) → NO2F (g) slow r2 = k2 [NO2] [F]

In this case, the predicted rate law would be, r = k [NO2] [F], consistent with the slow second step.

The slowest step in any proposed reaction mechanism, regardless of the number of steps, is the one that should reflect in the experimentally observed rate law.

It is not immediately obvious why this reaction mechanism is implausible because the rate law it predicts relies on the concentration of an intermediate, the F atom.

This is a problem because those doing the experiments may not even know that there is an F atom in the reaction anywhere, and as such will not account for it when they discover the experimentally determined rate law, which in this problem was, r = k [NO2] [F2].

In the method of initial rates, the experimenters are only aware of the reactants they purposefully put in the reactor so those are the only chemicals that will ever appear in the experimentally determined rate law.

Before we can see that this reaction mechanism is also implausible, we need to write the predicted rate law in terms of concentrations of known reactants only, without intermediates.

### 4.7.1  Dynamic Equilibrium

In order to do this, it needs to be understood that every reaction is capable of going backwards.

Specifically, it is important to understand that the 1 step will start going backwards with F (g) recombining to make F2 (g), while F2 (g) is still simultaneously producing F(g).

Because the first step is fastest, a mixture of F2 (g) and F (g) is formed before the second, slow step reaction ever gets started.

This mixture of F2 (g) and F (g) reaches dynamic equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction effectively canceling each other so that the concentrations of F2 and F in the mixture no longer change despite the fact that the forward and reverse reactions are still occurring.

The reverse reaction has its own rate law, r−1 = k−1 [F]2, with the −1 subscripts indicating it is the rate law of the reaction that is the reverse of the first step.

Because the forward and reverse reactions are canceling each other out by going the same speed, r1 = r−1 (this is the mathematical definition of dynamic equilibrium).

Setting the rest of those two rate laws equal to one another gives a relationship between the concentration of F2 and F which can be solved for [F], the intermediate we’re trying to get rid of in our predicted rate law.

 k1 [F2] = k−1 [F]2
[F] =

 k1 k−1
[F2

 1/2

We can now prove why Experimentally Inconsistent Reaction Mechanism #2 is implausible by substituting the above expression for [F] into the predicted rate law for this mechanism, r = k [NO2] [F].

r = k [NO2] [F] = k [NO2

 k1 k−1
[F2

 1/2
= k [NO2] [F2]1/2

Thus, instead of being first-order in [F2], Experimentally Inconsistent Reaction Mechanism #2 would predict the reaction would be half-order in [F2], which is not what was observed, so this mechanism is implausible.

Note that k1 and k−1 are just all lumped together into k, since they cannot be differentiated from each other.

———————————————————————–

##### Problem

What is the predicted rate law for the following reaction mechanism?

 Step Elementary Reaction Speed Rate Law 1 2 NO (g) → N2O2 (g) slow r1 = k1 [NO]2 2 N2O2 (g) + O2 (g) → 2 NO2 (g) slower r2 = k2 [N2O2] [O2] sum 2 NO (g) + N2O2 (g) + O2 (g) → N2O2 (g) + 2 NO2 (g)

The predicted rate law is that of the slower step, r = k [N2O2] [O2].

As before, this rate law must be reworked to only include the concentrations of reactants without intermediates (i.e. without N2O2).

Appealing to the dynamic equilibrium mixture of NO (g) and N2O2 (g) achieved by the first step because it is faster, the rate law of the reverse of the first reaction, r−1 = k−1 [N2O2], is set equal to the rate law of the forward reaction as allowed by the dynamic equilibrium condition that r1 = r−1.

 k1 [NO]2 = k−1 [N2O2]

Solving this for [N2O2] gives

[N2O2] =
 k1 k−1
[NO]2

Substituting this into the rate law for the second slow step gives

r2 = k2 [N2O2] [O2] = k2

 k1 k−1
[NO]2

[O2] = k [NO]2 [O2

So the experimentally observed rate law, if this mechanism is representative of what is really happening, should be r = k [NO]2 [O2], where again all the various k are just all lumped together into one.
What is interesting is that this is the exact same rate law that would be observed if we assumed the reaction was a termolecular elementary reaction with the following reaction mechanism.

 Step Elementary Reaction Speed Rate Law 1 2 NO (g) + O2 (g) → 2 NO2 (g) slow r1 = k1 [NO]2 [O2]

It is the unlikelihood of a termolecular collision that convinces us that the overall reaction is more likely to be a 2 step composite reaction as proposed at the beginning of the problem.

There are other unlikely mechanisms as well:

 Step Elementary Reaction Speed Rate Law 1 NO (g) + O2 (g) → NO2 (g) + O (g) slow r1 = k1 [NO][O2] 2 NO (g) + O (g) → NO2 (g) slower r2 = k2 [NO][O] sum 2 NO (g) + O (g) + O2 (g) → O (g) + 2 NO2 (g)

———————————————————————–

## 4.8  Catalysis

Catalyst increase the rate of conversion of reactants to products by providing an alternate reaction mechanism that has a smaller Ea for the rate-limiting step.

• Eacatalyzed < Eauncatalyzed
• The rate law will probably change because the mechanism is different.

The concentration of the catalyst might become part of the rate law, e.g. r = k [A] [catalyst]

• The overall reaction is exactly the same as before.
• Catalysts are consumed in an early step of the reaction, but are then regenerated in a later step.

Hence why the overall reaction is exactly the same as the uncatalyzed reaction, the catalyst cancels itself out as an initial reactant and a final product.

One step uncatalyzed reaction mechanism

A + C → D

———————————————————————–

Two step catalyzed reaction mechansim

 A + catalyst → B B + C —→ D + catalyst A + catalyst + B + C → B + D + catalyst

Common notation for the chemical equation for a catalyzed reaction

A + C →catalyst→ D

• Because the overall reaction is the same, Δ Hcatalyzed = Δ Huncatalyzed
• homogeneous catalysts are in the same phase as the reaction

heterogeneous catalysts are in a different phase (e.g. solid catalyst for a liquid or gas phase reaction)

Catalysts are useful because they increase the reaction rate without increasing the temperature.

There are many biological catalysts (i.e. enzymes) since most organisms need a fairly constant, set temperature.