Kinetics

Fast or slow, its thermo who says if it (spontaneously) goes.

Chemistry can be divided (somewhat arbitrarily) into the study of structures, equilibria, and rates. Chemical structure is ultimately described by the methods of quantum mechanics; equilibrium phenomena are studied by statistical mechanics and thermodynamics; and the study of rates constitutes the subject of kinetics.

Kinetics can be subdivided into physical kinetics, dealing with physical phenomena such as diffusion and viscosity, and chemical kinetics, which deals with the rates of chemical reactions (including both covalent and noncovalent bond changes).

– Connors, K. (Connors 1990)

Key concepts:

• Kinetics describes the rates of reactions
• Rates can be affected by reactant concentrations as well as environmental factors such as temperature
• Reaction profiles can be drawn as reaction pathways

Basics

Not all reactions are the same

Reactions occur at some “speed” or “rate.” A combustion reaction of gasoline and oxygen is very fast while the oxidation of iron (i.e. rust) is relatively slower. Reaction rates can be expressed in different ways. Imagine a reaction taking place in aqueous solution with a single reactant, A.

\[\mathrm{A}(aq) \longrightarrow \mathrm{products}\]

We can track the molar concentration of A vs. time (t). Here, the initial molar concentration of A is 1.0 M and the consumption of A with respect to time has the form

\[y = e^{-x}\]

or

\[[\mathrm{A}] = e^{-t}\]

Note: All reaction rates are positive.

Average Rate

An average rate of reaction is simply a change in reactant (or product) concentration over an amount of time. We can choose two points in time (t₁ and t₂) and measure the concentration of the reactant at both of those points (c₁ and c₂). We can then determine an average rate of reaction for that time frame via the following expression

\[\mathrm{avg.~rate} = -\dfrac{\Delta c}{\Delta t} = -\dfrac{c_2 - c_1}{t_2-t_1} \] Notice the negative sign. Since the concentration of the reactant is decreasing with increasing time, the numerator will be negative. BUT, all rates are positive so we insert a negative sign to ensure a positive value.

For our reaction, let us choose two points.

Point	t	c
1	0.25	0.7788
2	2.25	0.1050

The plot below shows the two points (in blue).

The slope of the line connecting the two points gives the average rate of reaction across the chosen time interval 0.25 → 2.25.

\[\begin{align*} \mathrm{avg.~rate} &= -\dfrac{c_2 - c_1}{t_2-t_1} \\[1.5ex] &= -\dfrac{0.1050 - 0.7788}{2.25 - 0.25} \\[1.5ex] &= 0.3369~c~t^{-1} \end{align*}\]

where c is simply the unit of concentration and t^–1 is a unit of time.

Let us choose another two points in the reaction and recalculate the average rate.

Point	t	c
3	2	0.1353
4	3	0.0183

Notice the shallow slope of the line. The average rate of the reaction has slowed down as the reaction progressed (i.e. as reactant A is consumed).

We calculate the average rate of reaction here for the time interval 2.00 → 4.00.

\[\begin{align*} \mathrm{avg.~rate} &= -\dfrac{c_2 - c_1}{t_2-t_1} \\[1.5ex] &= -\dfrac{0.0183 - 0.1353}{4.00 - 2.00} \\[1.5ex] &= 0.0585~c~t^{-1} \end{align*}\]

It is important to realize that the rate of a reaction can change as the reaction progresses!

Note: We could track the appearance of product and determine the average rate of reaction. Because the change in product concentration is positive, we would not include the negative sign as our rate will already be positive.

The average rate of reaction can be expressed from a balanced chemical equation. We must take into account the stoichiometric coefficients.

For a general reaction

\[aA + bB \longrightarrow cC + dD\]

the average rate of reaction is written (with molar concentrations) as follows

\[ \require{color} \mathrm{avg.~rate} = \dfrac{1}{a} {\textcolor{red}{\dfrac{-\Delta [A]}{\Delta t}}} = \dfrac{1}{b}\textcolor{blue}{\dfrac{-\Delta [B]}{\Delta t}} = \dfrac{1}{c}\textcolor{green}{\dfrac{\Delta [C]}{\Delta t}} = \dfrac{1}{d}\textcolor{orange}{\dfrac{\Delta [D]}{\Delta t}} \]

The fractions that are colored are as follows:

• red - “rate of consumption of A”
• blue - “rate of consumption of B”
• green - “rate of production of C”
• orange - “rate of production of D”

Similar words for “consumption” (e.g disappearance, expenditure, depletion, etc.) and “production” (e.g. appearance, creation, etc.) can be used.

It is critical to rationalize these terms. They are all rates. Rates are numbers. The fraction, when evaluated, equals a rate which is simply a number.

Practice

What is the average rate expression for the following reaction?

\[2\mathrm{NO}_2(g) \longrightarrow \mathrm{N_2O_4}(g)\]

Answer

\[\mathrm{avg.~rate} = \dfrac{1}{2}\dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} = \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t}\]

Let’s see if we understand what the fractions mean.

Practice

What is the average rate of reaction (in M s^–1) if the rate of disappearance of NO₂ = 0.125 M s^–1?

\[2\mathrm{NO}_2(g) \longrightarrow \mathrm{N_2O_4}(g)\]

Answer

\[\begin{align*} \mathrm{avg.~rate} &= \dfrac{1}{2}\dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} = \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t}\\[5ex] \mathrm{avg.~rate} &= \dfrac{1}{2}\dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} \\[1.5ex] &= \dfrac{1}{2} \left ( 0.125~M~\mathrm{s^{-1}} \right ) \\[1.5ex] &= 0.0625~M~\mathrm{s^{-1}} \end{align*}\]

Let’s try that again!

Practice

What is the average rate of reaction (in M s^–1) if the rate of appearance of N₂O₄ = 0.250 M s^–1?

\[2\mathrm{NO}_2(g) \longrightarrow \mathrm{N_2O_4}(g)\]

Answer

\[\begin{align*} \mathrm{avg.~rate} &= \dfrac{1}{2}\dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} = \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t}\\[5ex] \mathrm{avg.~rate} &= \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t} \\[1.5ex] &= 0.250~M~\mathrm{s^{-1}} \end{align*}\]

Once more.

Practice

What is the rate of disappearance of NO₂ (in M s^–1) if the rate of appearance of N₂O₄ = 0.300 M s^–1?

\[2\mathrm{NO}_2(g) \longrightarrow \mathrm{N_2O_4}(g)\]

Answer

\[\begin{align*} \mathrm{avg.~rate} &= \dfrac{1}{2}\dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} = \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t}\\[5ex] \dfrac{1}{2}\dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} &= \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t} \\[1.5ex] \dfrac{-\Delta[\mathrm{NO_2}]}{\Delta t} &= 2 \left ( \dfrac{\Delta[\mathrm{N_2O_4}]}{\Delta t} \right) \\[1.5ex] &= 2\left ( 0.300~M~\mathrm{s^{-1}} \right )\\[1.5ex] &= 0.600~M~\mathrm{s^{-1}} \end{align*}\]

Practice Some More

Fill in the table for the following reaction:

\[\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\dfrac{1}{2}\mathrm{O}_{2}(g)\]

Time (h)	$$\dfrac{-\Delta [\mathrm{H_2O_2}]}{\Delta t}$$	$$\dfrac{\Delta [\mathrm{H_2O}]}{\Delta t}$$	$$\dfrac{\Delta [\mathrm{O_2}]}{\Delta t}$$	Rate
0.00	…	…	…	…
6.00	0.0833	0.0833	0.04165
12.00	0.0417	0.0417
18.00	0.0208
24.00				0.0103

Solution

Time (h)	$$\dfrac{-\Delta [\mathrm{H_2O_2}]}{\Delta t}$$	$$\dfrac{\Delta [\mathrm{H_2O}]}{\Delta t}$$	$$\dfrac{\Delta [\mathrm{O_2}]}{\Delta t}$$	Rate
0.00	…	…	…	…
6.00	0.0833	0.0833	0.04165	0.0833
12.00	0.0417	0.0417	0.02085	0.0417
18.00	0.0208	0.0208	0.0104	0.0208
24.00	0.0103	0.0103	0.00515	0.0103

Instantaneous Rate

An instantaneous rate of reaction is the rate of a reaction at an instant in time. This is the slope of a line that is tangent to any one point along the reaction curve. For example, let us consider the tangent line to e^–x at x = 1.

The tangent line was given by WolframAlpha. We can solve for y when x = 1.

\[\begin{align*} y &= \dfrac{2}{e^2} - \dfrac{x}{e^2} \\[1.5ex] &= \dfrac{2}{e} - \dfrac{1}{e} \\[1.5ex] &= 0.3679 \end{align*}\]

The blue point shown below (x, y) is (1, 0.3679) and the blue line is the tangent line.

The slope of the line is

\[m = -0.3679\] Therefore, the instantaneous rate of reaction at t = 1 is 0.3679 M s^–1.

Let us determine the instantaneous rate of reaction when t = 2. The tangent line was given by WolframAlpha. We can solve for y when x = 2.

\[\begin{align*} y &= \dfrac{3}{e^2} - \dfrac{x}{e^2} \\[1.5ex] &= \dfrac{3}{e^2} - \dfrac{2}{e^2} \\[1.5ex] &= 0.13534 \end{align*}\]

The blue point shown below (x, y) is (2, 0.1353) and the blue line is the tangent line.

The slope of the line is

\[m = -0.1353\]

Therefore, the instantaneous rate of reaction at t = 2 is 0.1353 M s^–1. Notice how the rate of reaction is slower later in the reaction (t = 2) than it was earlier in the reaction (t = 1).

Rate Law

Rate vs. Concentration

Imagine a reaction with more than one reactant such like the one below.

\[\mathrm{A}(aq) + \mathrm{B}(aq) + \cdots \longrightarrow \mathrm{products}\]

As we have seen, the concentrations of the reactants will decrease with reaction progress. The consumption behavior of each product may be identical or dramatically different. Both will have an effect on the rate of the reaction and the effect is multiplicative. We can express the dependence of the rate on reactant concentrations as follows

\[\mathrm{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n \cdots\]

This is called a rate law and can be written for any reaction. The rate constant, k, is independent of reactant concentrations and takes into account environmental factors such as temperature and effective particle collisions (see Collision Theory and Arrhenius Equation).

It is important to note the exponents in the rate law. The exponents are not related to the stoichiometric coefficients in a balanced chemical equation unless the reaction is an elementary one. We can only determine the value of these exponents from experimental data using a method called the Method of Initial Rates (discussed below).

Order of Reaction Terminology

The exponents in a rate law determine the order of each reactant and are typically integers (e.g. 0, 1, 2 …) though they can be fractions or negative. Imagine the following reaction involving three reactants

\[\mathrm{A}(aq) + \mathrm{B}(aq) + \mathrm{C}(aq) \longrightarrow \mathrm{products}\]

and the corresponding rate law.

\[\mathrm{rate} = k [\mathrm{A}][\mathrm{B}]^2\]

C does not appear in the rate law since [C]⁰ = 1.

We would say that the reaction is first-order in A, second-order in B, and zeroth-order in C. Summing over all the exponents give the overall order of reaction. Here the overall order would be 3, hence, the reaction is third-order.

Method of Initial Rates

To determine the exponents in a rate law, one must perform the Method of Initial Rates using experimentally obtained data. Suppose we wish to determine the order of the following reaction

\[\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\dfrac{1}{2}\mathrm{O}_{2}(g)\]

We write the rate law for the reaction, remembering to write the exponent as a variable.

\[\mathrm{rate} = k[\mathrm{H_2O_2}]^m\] The experiment is run multiple times under identical environmental conditions while the initial concentrations of the reactants are intelligently varied. The following information is tabulated: (1) the initial reactant concentration and (2) the initial rate of reaction. We then perform the Method of Initial Rates. The steps are:

1. Determine exponent for a chosen reactant in the rate law

   1. Select two experiments where only the initial concentration of the chosen reactant changes
   2. Make ratio of each rate law for experiments chosen
   3. Determine exponent
   4. Repeat Step 1 until all exponents are determined

2. Write rate law

3. Use data from any experiment to find rate constant, k

Given the experimental data below (obtained at room temperature), we will determine the order of reaction.

Experiment	[H2O2]i	Initial Rate (M h–1)
1	0.0235	0.166
2	0.0235	0.166
3	0.047	0.333

Step 1:

Choose a reactant whose exponent you wish to determine. Here we only have one reactant so we choose H₂O₂.

Step 1a:

We must now find two experiments where only the initial concentration of the chosen reactant changes. Looking at the table, we see that the initial concentration of H₂O₂ changes between Experiments 2 and 3.

Step 1b:

Now we make a ratio of each rate law for the experiments chosen.

For Experiment 2, the rate law is as follows

\[\mathrm{rate}_{e2} = k[\mathrm{H_2O_2}]_{e2}^m\]

and the rate law for Experiment 3 is

\[\mathrm{rate}_{e3} = k[\mathrm{H_2O_2}]_{e3}^m\]

e2 = Experiment 2 and e3 = Experiment 3 and is simply written for bookkeeping.

We now make a ratio of the two rate laws. This means we will write a fraction with one rate law in the numerator and one in the denominator. It does not matter which rate law is placed where. Here I’ve chosen to place the rate law for Experiment 3 in the numerator and the rate law for Experiment 2 in the denominator.

\[\dfrac{\text{rate}_{e3}}{\text{rate}_{e2}} = \dfrac{k[\mathrm{H_2O_2}]_{e3}^{m}}{k[\mathrm{H_2O_2}]_{e2}^{m}}\]

We note that the rate constant, k is identical between the experiments since the environmental factors were kept the same among the experiments performed. Therefore, the rate constants cancel leaving us with

\[\dfrac{\text{rate}_{e3}}{\text{rate}_{e2}} = \dfrac{[\mathrm{H_2O_2}]_{e3}^{m}}{[\mathrm{H_2O_2}]_{e2}^{m}}\] Step 1c:

Next, plug in the rate and concentration values into the equation and solve for the exponent i.

\[\begin{align*} \dfrac{0.333~M~\mathrm{h}^{-1}}{0.166~M~\mathrm{h}^{-1}} &= \dfrac{(0.0470~M)^{m}}{(0.0235~M)^{m}} \\[1.5ex] 2 &= 2^m \\[1.5ex] \ln(2) &= \ln(2^m) \\[1.5ex] \ln(2) &= m\ln(2) \\[1.5ex] m &= \dfrac{\ln(2)}{\ln(2)} \\[1.5ex] &= 1 \end{align*}\]

Log identity: ln(x^y) = yln(x)

For this class, exponents will be integers. The experimental data given to you should resolve to a number that is very close to an integer. Just round the number to the nearest integer if necessary.

If there were a second reactant, we would repeat the process and solve for the next exponent. Given that this reaction has one reactant, we are now done with Step 1.

Step 2:

We can now write the rate law with our determined exponent.

\[\begin{align*} \mathrm{rate} &= k[\mathrm{H_2O_2}]^1 \\[1.5ex] \mathrm{rate} &= k[\mathrm{H_2O_2}] \end{align*}\]

Since we have a power of 1, we don’t explicitly write it in the rate law (as it is assumed).

Step 3:

If you are tasked with determining the rate constant, k, we can do that now. Simply choose any experiment and plug the data into the rate law and solve for k. Determine the rate constant in M s^–1 Here I choose to use data from Experiment 1.

\[\begin{align*} \mathrm{rate} &= k[\mathrm{H_2O_2}] \\[1.5ex] k &= \dfrac{\mathrm{rate}}{[\mathrm{H_2O_2}]}\\[1.5ex] &= \dfrac{0.166~M~\mathrm{h}^{-1}}{0.0235~M} \\[1.5ex] &= 7.064~\mathrm{h}^{-1} \end{align*}\]

Convert hours to seconds.

\[\begin{align*} k &= 7.064~\mathrm{h}^{-1} \left ( \dfrac{\mathrm{h}}{60~\mathrm{min}} \right ) \left ( \dfrac{\mathrm{min}}{60~\mathrm{s}} \right )\\[1.5ex] &= 0.00196~\mathrm{s}^{-1} \end{align*}\]

We could write out the rate law once again with the rate constant explicitly written.

\[\mathrm{rate} = 0.00196~\mathrm{s}^{-1}~[\mathrm{H_2O_2}]\] Order of Reaction

The order of this reaction is first-order and that the reaction is first-order in H₂O₂.

Practice a two-reactant problem here.

Visualize the first-order Rate Law

Now that we have our rate law, we can plot the rate of the reaction vs. the reactant concentration. Let us choose an initial concentration of 1 M H₂O₂ and run it to zero in increments of 0.1 M.

Figure 1: Rate plot for the first-order hydrogen peroxide decomposition reaction.

Notice how the rate of the reaction decreases linearly with reactant consumption. The change in rate remains constant for a first-order reaction.

Rate Constant, k

The rate constant is determined using the Arrhenius equation.

\[k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}\]

We will explore the Arrhenius equation in detail later. For now, notice the location of temperature, T. As temperature increases, the rate constant, k, increases. Let us visualize this. Below are three values for k obtained at various temperatures.

T (°C)	T (K)	k
25.0	298.15	0.00196
37.1	310.26	0.00250
77.7	350.85	0.00500

Obtaining k at various temperatures

I obtained k at two elevated temperatures by using the Arrhenius equation and making a few assumptions.

First, we begin by understanding that the rate constant, k, that we calculated previously (k = 0.00196 s^–1) was determined from experimental values obtained at 25 °C.

We now use the Arrhenius equation to determine the activation energy, E_a for the reaction. Let us assume that the frequency factor, A, is 1 to make this exercise simple. We then solve for E_a in kJ.

\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\[1.5ex] E_{\mathrm{a}} &= -RT\ln \left (\dfrac{k}{A} \right ) \\[1.5ex] &= - \left (\dfrac{8.314~\mathrm{J}}{\mathrm{mol~K}} \right ) \left (\dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left( 298.15~\mathrm{K} \right ) \ln \left (\dfrac{0.00196~\mathrm{s}^{-1}}{1} \right ) \\[1.5ex] &= 15.455~\mathrm{kJ~mol} \end{align*}\]

We can now use the energy of activation to find k at other temperatures. Here I solve for the rate constant at 310.26 K (or about 37 °C). Remember, A is equal to 1.

\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\[1.5ex] &= e^{\frac{-(15.455~\mathrm{kJ})}{\frac{8.314~\mathrm{J}}{\mathrm{mol~K}} \left (\frac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 310.26~\mathrm{K} \right ) }}\\[1.5ex] &= 0.00250~\mathrm{s^{-1}} \end{align*}\]

Repeat this again for T = 350.85 K (or 77.7 °C) and k is found to be 0.00500 s^–1.

Let us plot again the rate of reaction vs. reactant concentration but at each temperature for this first-order reaction.

Figure 2: Rate plot for a first-order reaction at various temperatures.

Rationalize this plot. At higher temperatures, the rate of the reaction is faster than at lower temperatures. The plot doesn’t show this but ask yourself the following

Assuming the three reactions start with the same reactant concentration (e.g. 1.0 M), which reaction will go to completion first?

The reaction which consumes all reactant first is the reaction carried out at 350.85 K (77.7 °C). The rate of this reaction is much faster than the others at lower temperature.

So how do we visualize the reaction progress with time?

We’ll get to that shortly (see Integrated Rate Laws). For now, let us visualize some more rate laws. Above, we looked at a rate law for a first order reaction. Let us see what the others (zeroth- and second-order) look like.

Consider the following rate law for a zeroth-order reaction.

\[\begin{align*} \mathrm{rate} &= k[\mathrm{A}]^0 \\[1.5ex] &= k \end{align*}\]

Anything to the zeroth power is 1 so the concentration of A simply disappears (there is an assumed 1).

Let us stick with our three rate constants and temperatures from above for this exercise. We should be able to rationalize the dependence of the rate on the reactant concentration before even looking at the resulting plot. Here, rate is only dependent on k so the rate should not change with reactant concentration!

Figure 3: Rate plot for a zeroth-order reaction at various temperatures.

Sure enough, the rate does not change with reactant concentration, a characteristic of a zeroth-order reaction. We see again that the reaction performed at higher temperatures has a faster rate than those carried out at lower temperatures.

Now let us consider the plot again for a second-order reaction.

\[\mathrm{rate} = k[\mathrm{A}]^2\]

Figure 4: Rate plot for a second-order reaction at various temperatures.

Here we see an exponential decrease in the rate of reaction as the reactant is consumed. The reaction is slowing down exponentially as the reaction progresses for a second-order reaction.

Visualizing a negative order reaction

What does a rate plot look like for a negative order reaction? Consider the following rate law:

\[\mathrm{rate} = k[\mathrm{A}]^{-1}\]

The resulting rate plot (at various T) is given below.

Figure 5: Rate plot for an negative order reaction at various temperatures.

Here we see that the rate of the reaction increases as reactant is consumed. A common example of this type of behavior is the formation of oxygen from ozone. The overall reaction for this is given below.

\[2\mathrm{O_3}(g) \longrightarrow 3\mathrm{O_2}(g)\]

This is a multi-step reaction that can be broken down into two elementary steps.

\[\begin{align*} \mathrm{O_3}(g) &\overset{h\nu}\rightleftharpoons \mathrm{O_2}(g) + \mathrm{O}(g) &&\mathrm{rate}_1 = k_1[\mathrm{O_3}]\\[1.5ex] \mathrm{O}(g) + \mathrm{O_3}(g) &\longrightarrow \mathrm{2O_2}(g) &&\mathrm{rate}_2 = k_2[\mathrm{O}][\mathrm{O_3}] \end{align*}\]

and the resulting rate law for the overall reaction is

\[\mathrm{rate} = k[\mathrm{O}_3]^2[\mathrm{O}_2]^{-1}\]

In the first step, ozone is split into oxygen gas and free oxygen when hit with UV light from the sun. This step is the faster of the two steps and can reach equilibrium.

In the second, slower (i.e. rate-limiting) step, a free oxygen atom can collide with an ozone molecule to form oxygen gas. However, if a free oxygen atom reacts with another free oxygen atom, oxygen gas is formed. This reaction readily occurs.

\[\mathrm{O}(g) + \mathrm{O}(g) \longrightarrow \mathrm{O_2}(g)\] The more O₂ that is present, the less free oxygen atoms that are present to become ozone (as seen in reaction step 2 above). Therefore, the higher [O₂], the lower [O], and ultimately the lower the rate of reaction giving us the rate law

\[\mathrm{rate} = k[\mathrm{O_3}]^2[\mathrm{O}_2]^{-1}\]

or

\[\mathrm{rate} = k\dfrac{[\mathrm{O_3}]^2}{[\mathrm{O}_2]}\]

Rate Constant Units and Order

Our example of the Method of Initial Rates (above) analyzed a first-order reaction. The resulting rate constant had units of s^–1. The rate constant for all first-order reactions will have units of inverse time. Therefore, if you were given a rate constant of k = 0.234 h^–1, you could immediately conclude that the reaction was first order!

The units of the rate constant will tell you what order the reaction is! See the table below showing rate constant units and what reaction order they belong to. The units are generalized (c = concentration and t = time) but are usually reported as M s^–1 or similar.

Order	k
Zeroth	c t–1
First	t–1
Second	c–1 t–1

Practice

A reaction was found to have a rate constant of 0.132 M^–1 s^–1. What is the order of the reaction?

Answer

second-order

Practice

The following reaction has a rate constant of 0.3387 M^–1 s^–1 at 326.86 °C (or 600 K).

\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\] If the initial concentration of NO₂ = 5.00 M, what is the initial rate of the reaction (in M s^–1)?

Visualize This

A 5 mol sample of NO₂ at 25 °C and 1 atm of pressure occupies 122.33 L of space if treated as an ideal gas.

\[\begin{align*} PV &= nRT \\[1.5ex] V &= \dfrac{nRT}{P} = 122.33~\mathrm{L} \end{align*}\]

If we compress 5 mol of NO₂ into 1 L (to give a 5 molar concentration), the pressure would be an astounding 122.33 atm of pressure! If the reaction was carried out in a closed container, the pressure would only increase as the gas decomposed. That seems dangerous.

Data

The rate constant was published by the National Bureau of Standards (Westley 1980) and can be found on pg. 40.

Answer

Realize that the reaction is second-order by analyzing the rate constant units then write out the rate law. Plug in your givens and solve.

\[\begin{align*} \mathrm{rate} &= k[\mathrm{NO_2}]^2 \\[1.5ex] &= (0.3387~M^{-1}~\mathrm{s}^{-1}) ( 5.00~M )^2 \\[1.5ex] &= 8.468~M~\mathrm{s}^{-1} \end{align*}\]

Remember, this is the initial rate of the reaction since the reaction initially began with 5.0 M NO₂.

Extend

What is the rate of reaction when the reactant concentration reaches 2.0 M?

Rationalize:

Given that this is a second-order reaction, the rate should be lower!

Solve

\[\begin{align*} \mathrm{rate} &= k[\mathrm{NO_2}]^2 \\[1.5ex] &= (0.3387~M^{-1}~\mathrm{s}^{-1}) ( 2.0~M )^2 \\[1.5ex] &= 1.3548~M~\mathrm{s}^{-1} \end{align*}\]

We see that the reaction is dramatically slower when the concentration of reactant is 2.0 M!

Plot

Above we’ve solved for two data points on the rate plot. Let’s plot the rate vs. reactant concentration from what was initially 5.0 M all the way down to zero!

Figure 6: Rate vs. concentration plot for this second-order reaction.

Integrated Rate Law

Concentration vs. Time

At the beginning of this page, we saw some concentration vs. time plots and then transitioned into rate vs. concentration plots when analyzing rate laws. We now transition back to concentration vs. time plots for different orders of reactions. These plots are a product of integrated rate laws which show the linear relationship of reactant concentration vs. time and have the form of y=mx+b. There is an integrated rate law for each order (zeroth- first- and second-) of reaction. These are derived from the corresponding rate laws (derivations not shown).

First-order Integrated Rate Law

Consider the following concentration vs. time plot for a first-order reaction.

Figure 7: Concentration vs. time plot for a first-order reaction.

This plot uses, once again, the numerical values of the rate constants and corresponding temperatures found above. The concentration of A decreases exponentially with time.

We can obtain a linear relationship by transforming the data in order to extract the rate constant from the slope. To do this for a first-order reaction, take the natural log of the concentration!

The first-order integrated rate law is

\[\ln\mathrm{[A]}_t = -kt + \ln\mathrm{[A]_0}\] and gives a linear correlation between the natural log of the concentration and time (e.g. reaction progress). The plot below shows the transformed data ([A] → ln[A]).

Figure 8: Linear integrated rate law plot for a first-order reaction.

The slope of each line is the is related to the rate constant.

\[\mathrm{slope} = -k\]

First-order reactions give linear relationships when the natural log of the reactant concentration is plotted vs. time.

Zeroth-order Integrated Rate Law

Consider the following concentration vs. time plot for a zeroth-order reaction. Again, I use the numerical values of the rate constants and corresponding temperatures found above.

Figure 9: Integrated rate law plot for a zeroth-order reaction.

The concentration of A decreases linearly with time. Since the relationship is already in a linear form, there is no need to transform the data! The rate constant can be obtained directly from the slope.

\[\mathrm{slope} = -k\]

The zeroth-order integrated rate law is

\[\mathrm{[A]} = -kt + \mathrm{[A]_0}\]

Zeroth-order reactions give linear relationships when the reactant concentration is plotted vs. time.

Second-order Integrated Rate Law

Consider the following concentration vs. time plot for a second-order reaction. Again, I use the numerical values of the rate constants and corresponding temperatures found above.

Figure 10: Concentration vs. time plot for a second-order reaction.

Once again, we see an exponentially decreasing relationship between reactant concentration and time. We can transform this data, once again, to obtain a linear relationship allowing us to extract the rate constant from the slope. We do this by taking the inverse of the reactant concentration ([A] → 1/[A]).

The second-order integrated rate law is

\[\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0}\] and gives a linear correlation between the inverse of the concentration and time (e.g. reaction progress). The plot below shows the transformed data.

Figure 11: Integrated rate law plot for a second-order reaction.

Second-order reactions give linear relationships when the inverse of the reactant concentration is plotted vs. time.

Practice

Let us revisit our reaction from above. The reaction has a rate constant of 0.3387 M^–1 s^–1 (at 326.86 °C or 600 K) and begins with 5.0 M NO₂.

\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\] How much time (in s) will it take for 1.0 M of the reactant to be consumed?

Answer

Since the problem is asking for time, we realize that we must solve the integrated rate law for this second-order reaction (remember, we know the reaction order by analyzing the units of the given rate constant).

We initially begin with 5.0 M NO₂ and we are finding the time required for 1.0 M of reactant to be consumed. Therefore, [NO₂]₀ = 5.0 and [NO₂]_t = 4.0

Rearrange the equation for t and solve.

\[\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} ~~\longrightarrow \\[1.5ex] t &= \dfrac{\left ( \frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} \right )}{k} \\[1.5ex] &= \dfrac{\left ( \frac{1}{4.0~M} - \frac{1}{5.0~M} \right )}{0.3387~M^{-1}~\mathrm{s}~^{-1}} \\[1.5ex] &= 0.14762~\mathrm{s} \end{align*}\]

In about one-fifteenth of a second, 1 M NO₂ is transformed to products. This is a fast reaction!

Extend

How much time will it take for the reaction to go from 2.0 M NO₂ to 1.0 M NO₂?

Remember, the reaction is second-order and, as we determined from the practice problem above, the reaction is slowing down. We would expect that the time required for this will be longer than 0.15 seconds!

Solve

\[\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} ~~\longrightarrow \\[1.5ex] t &= \dfrac{\left ( \frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} \right )}{k} \\[1.5ex] &= \dfrac{\left ( \frac{1}{1.0~M} - \frac{1}{2.0~M} \right )}{0.3387~M^{-1}~\mathrm{s}~^{-1}} \\[1.5ex] &= 1.476~\mathrm{s} \end{align*}\]

We see that the reaction takes much longer (about twice as long) at this stage in the reaction!

Plot

We’ve determined two points along this reaction numerically. Let’s go ahead and plot the data from 5.0 M down to 0 M and see what it looks like!

Figure 12: Concentration vs. time plot for this second-order reaction.

Here we see the second-order reaction starts out really fast (reactant is being quickly consumed) but as the NO₂ concentration decreases, the reaction rate increases significantly.

Extend Again

Let’s make the concentration vs. time relationship linear! We can get the rate constant from the slope of the line! For a second-order reaction, we simply plot the inverse concentration (i.e. 1/[NO₂]) vs. time.

Figure 13: Integrated rate plot for this second-order reaction.

The equation for the resulting line is

\[y = 0.3387 + 0.2\]

The given rate constant of reaction is 0.3387 s^–1 which exactly matches the slope of our integrated rate plot for a second-order reaction!

Half-Life

The half-life (t_1/2) for any reaction is simply the amount of time required for exactly one half of the reactant to be consumed. Given that each order of reaction responds differently to reactant concentration, the half-life for these reactions vary. We can easily derive the half-life equations from their integrated rate law counterparts.

To perform the derivation, simply start with an appropriate integrated rate law. Make the following modifications to the equation

\[t ~\longrightarrow~ t_{1/2}\] and

\[[\mathrm{A}]_t ~\longrightarrow~ \dfrac{1}{2}[\mathrm{A}]_0\] since, by definition, the half-life of a reaction is when exactly one half of the reactant concentration is consumed. Once that is done, solve for t_1/2.

First-order Half-Life

Derivation

\[\begin{align*} \ln\mathrm{[A]}_t &= -kt + \ln\mathrm{[A]_0} \\[1.5ex] \ln \left (\frac{1}{2}[\mathrm{A}]_0 \right ) &= -kt_{1/2} + \ln\mathrm{[A]_0}\\[1.5ex] \ln \left (\frac{1}{2}[\mathrm{A}]_0 \right ) - \ln\mathrm{[A]_0} &= -kt_{1/2} \\[1.5ex] \ln\left ( \dfrac{\frac{1}{2}[\mathrm{A}]_0}{[\mathrm{A}]_0} \right ) &= -kt_{1/2} \\[1.5ex] \ln\dfrac{1}{2} &= -kt_{1/2} \\[1.5ex] t_{1/2} &= -\dfrac{\ln\frac{1}{2}}{k} \\[1.5ex] &= \dfrac{\ln 2}{k}\\[1.5ex] & = \dfrac{0.693}{k} \end{align*}\]

\[\begin{align*} t_{1/2} &= \dfrac{\ln 2}{k}\\[1.5ex] &= \dfrac{0.693}{k} \end{align*}\]

Zeroth-order Half-Life

Derivation

\[\begin{align*} [\mathrm{A}]_t &= -kt + [\mathrm{A}]_0 \\ \frac{1}{2}[\mathrm{A}]_0 &= -kt_{1/2} + [\mathrm{A}]_0 \\ kt_{1/2} &= [\mathrm{A}]_0 - \frac{1}{2}[\mathrm{A}]_0 \\ t_{1/2} &= \dfrac{[\mathrm{A}]_0}{2k} \end{align*}\]

\[t_{1/2} = \dfrac{[\mathrm{A}]_0}{2k}\]

Second-order Half-Life

Derivation

\[\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} \\ \dfrac{1}{\frac{1}{2}[\mathrm{A}]_0} &= kt_{1/2} + \dfrac{1}{[\mathrm{A}]_0} \\ \dfrac{2}{[\mathrm{A}]_0} - \dfrac{1}{[\mathrm{A}]_0}&= kt_{1/2} \\ \dfrac{1}{[\mathrm{A}]_0}&= kt_{1/2} \\ \dfrac{1}{k[\mathrm{A}]_0}&= t_{1/2} \\ t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \end{align*}\]

\[t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}\]

Practice

Let us continue using our reaction from above. The reaction has a rate constant of 0.3387 M^–1 s^–1 (at 326.86 °C or 600 K) and begins with 5.0 M NO₂.

\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\]

What is the half-life (in s) of the reaction?

Answer

We resort to the second-order half-life equation to solve this. Solve for t_1/2 (in s).

\[\begin{align*} t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \\[1.5ex] &= \dfrac{1}{0.3387~M^{-1}~\mathrm{s}^{-1} \left ( 5.0~M \right )} \\[1.5ex] &= 0.5905~\mathrm{s} \end{align*}\]

Therefore, half the NO₂ concentration gets consumed about every sixth of a second.

Extend

Does the half-life change with a different initial concentration?

For a second-order reaction it does! Let’s determine the half-life of the reaction if the initial concentration was 1.0 M.

Solve

\[\begin{align*} t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \\[1.5ex] &= \dfrac{1}{0.3387~M^{-1}~\mathrm{s}^{-1} \left ( 1.0~M \right )} \\[1.5ex] &= 2.952~\mathrm{s} \end{align*}\]

Because a second-order reaction has a rate that slows exponentially with reactant consumption, the half-life increases with decreasing rate!

Plot

Let’s visualize the change in half-life as reactant gets consumed.

Figure 14: Half-life plot for this second-order reaction.

As expected, the half-life increases with reaction progress (e.g. as NO₂ gets consumed).

Follow Through

So we see that the half-life changes as a second-order reaction progresses. Will the half-life change as a zeroth-order reaction progresses? What about a first-order reaction?

Summary of Reaction Profiles

Reaction	Order	Rate Law	Integrated Rate Law	Half-Life	k
$$\mathrm{A \rightarrow P}$$	Zeroth	$$\mathrm{rate} = k$$	$$\mathrm{[A]}_t = -kt + \mathrm{[A]_0}$$	$$t_{1/2} = \dfrac{[\mathrm{A}]_0}{2k}$$	M s–1
$$\mathrm{A \rightarrow P}$$	First	$$\mathrm{rate} = k[\mathrm{A}]$$	$$\ln[\mathrm{A}]_{t} = -kt + \ln[\mathrm{A}]_0$$	$$t_{1/2} = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k}$$	s–1
$$\mathrm{A + A \rightarrow P}$$	Second	$$\mathrm{rate} = k[\mathrm{A}]^2$$	$$\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0}$$	$$t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}$$	M–1 s–1
$$\mathrm{A + B \rightarrow P}$$	Second	$$\mathrm{rate} = k[\mathrm{A}][\mathrm{B}]$$	$$\dfrac{1}{[\mathrm{B}]_0 - [\mathrm{A}]_0}\ln\dfrac{[\mathrm{B}][\mathrm{A}]_0}{[\mathrm{A}][\mathrm{B}]_0} = kt$$	$$\cdots$$	M–1 s–1

Arrhenius Equation

We have already seen how temperature affects the rate of reaction via the rate constant. The relationship between temperature, activation energy and rate constant is given by the Arrhenius equation.

Exponential Form

\[k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}\]

As temperature increases, the rate constant increases and therefore the rate of reaction (as seen with the rate law).

The Arrhenius equation has a similar form to the Clausius-Clapeyron equation and can therefore be written in a linear and two-point form.

Linear Form

\[\ln k = \dfrac{-E_{\mathrm{a}}}{R} \left (\dfrac{1}{T}\right) + \ln A\]

Two-point Form

\[\ln\dfrac{k_2}{k_1} = \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\]

The explanations for these equations are similar to the Clausius-Clapeyron equations except here we relate rate constant with temperature.

We can numerically determine a set of rate constants for a range of temperatures and then plot these via the linear form (ln(k) vs 1/T). The resulting slope allows us to determine the energy of activation. The pre-exponential factor, A, is discussed below.

\[\mathrm{slope} = \dfrac{-E_{\mathrm{a}}}{R}\]

Additionally, we can determine the energy of activation by evaluating the two-point Arrhenius equation. The activation energy does not change with temperature.

Practice

As seen previously, the following reaction has a rate constant of 0.3387 M^–1 s^–1 at 326.86 °C (or 600 K).

\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\]

At 526.85 °C (800 K), the rate constant is 93.823 M^–1 s^–1. What is the energy of activation (in kJ mol^–1)?

Answer

We have two rate constants and two temperatures. We can solve the Arrhenius two-point form.

\[\begin{align} \ln \left ( \dfrac{k_2}{k_1} \right ) &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow\\[4ex] E_{\mathrm{a}} &= R\left (\dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[2ex] &= 8.314~\mathrm{J~mol^{-1}~K^{-1}} \left (\dfrac{\ln\left (\dfrac{93.82~M^{-1}~\mathrm{s}^{-1}}{0.3387~M^{-1}~\mathrm{s}^{-1}}\right )} {\dfrac{1}{600~\mathrm{K}} - \dfrac{1}{800~\mathrm{K}}} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 112.22~\mathrm{kJ~mol^{-1}} \end{align}\]

Rationalize

The energy of activation for this reaction is quite high. Many reactions that take place at around temperature have activation energies of less than 80 kJ mol^–1. This reaction is very fast at high temperatures (300+ °C) and is very slow at low temperatures (such as room temperature).

Extend

Estimate the rate constant for this reaction at room temperature (25 °C). The pre-exponential factor for the reaction is 2 × 10⁹.

Solve

\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\[1.5ex] &= (2\times 10^{9}) e^{\frac{-(112.23~\mathrm{kJ})}{\frac{8.314~\mathrm{J}}{\mathrm{mol~K}} \left (\frac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( 298.15~\mathrm{K} \right ) }}\\[1.5ex] &= 4.346\times 10^{-11}~M^{-1}~\mathrm{s^{-1}} \end{align*}\]

The reaction is ten orders of magnitude slower at room temperature!

Plot

Above we’ve solved for two data points on a k vs. T plot. Let’s plot the rate constant vs. T from 600 K to 2000 K (326.85 °C to 1,726.85 °C).

Figure 15: Rate constant vs temperature for this second-order reaction.

This reaction is FAST at higher temperatures.

Let’s transform the data to give a straight line as the linear-form of the Arrhenius equation is set up to do. We simply plot ln(k) vs. 1/T.

Figure 16: Linear-form of the Arrhenius equation.

The resulting slope is

\[\mathrm{slope} = -13,498\]

The energy of activation (in kJ mol^–1) can be determined.

\[\begin{align*} \dfrac{-E_{\mathrm{a}}}{R} &= \mathrm{slope} \\[1.5ex] E_{\mathrm{a}} &= -\mathrm{slope} \times R \\[1.5ex] E_{\mathrm{a}} &= - (-13,498) \left ( \dfrac{8.314~\mathrm{J}}{\mathrm{mol~K}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= 112.22~\mathrm{kJ~mol^{-1}} \end{align*}\]

Collision Theory

Collision theory aims to explain how reactions occur using a collision model. One fundamental principle of collision theory is that two particles must interact (i.e. collide) in order for those two particles to undergo a chemical change. Consider the following reaction:

\[2\mathrm{HI}(g) \longrightarrow \mathrm{H_2}(g) + \mathrm{I_2}(g)\]

Here, two hydrogen iodide molecules must collide to undergo a chemical change that results in the breaking and formation of covalent bonds to produce the products.

A collision of the two molecules on its own does not guarantee that the reactants can become products. The orientation of the two molecules also matter. They must collide in the proper orientation! If the two molecules collide in the orientation shown below, the hydrogen atoms are out of position to potentially form a covalent bond.

However, if the orientation was like this (below), the molecules can react to proceed to product. The transition state is referred to as an “activated complex” and is the geometry the system must have to connect reactants and products together. It has a particular energy just as the reactants and products do. Transition states are higher in energy than the structures they connect.

In a sample of particles, collisions occur at a certain rate (or frequency) and this rate will increase with increasing temperature. A fraction of the collisions will have the proper orientation and a fraction of those will collide with sufficient energy to move on to products (they must collide with energy equal to or greater than the activation energy of reaction). Collisions that meet all the criteria to facilitate formation of product are referred to as effective collisions. The pre-exponential factor (i.e. Arrhenius factor), A, is correlated with the number of these effective collisions and has the same units as the rate constant.

Reaction Mechanisms

Chemical reactions can be categorized as complex reactions or elementary reactions. Consider the reaction we have been working with.

\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\]

This reaction looks seemingly simple. We might look at this and interpret this reaction as two NO₂ molecules colliding to undergo a chemical change to make products. However, this reaction is actually a multi-step reaction.

First Proposed Mechanism, slow second-step

One proposed mechanism is the following two-step mechanism.

\[\begin{align*} 2\mathrm{NO_2}(g) &\underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{(NO_2)_2}(g) &&\mathrm{(fast)} \\[1.5ex] \mathrm{(NO_2)_2}(g) &\overset{k_2}\longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g) &&\mathrm{(slow)} \end{align*}\]

Notice that adding both elementary steps together sums to the overall reaction.

The proposed set of steps for the overall reaction is called a reaction mechanism. The molecularity of the first step is bimolecular (involving two reactant particles) whereas the molecularity of the second step is unimolecular (involving one reactant particle). The first step produces an intermediate of reaction, a particle that a reaction produces and then later consumes. Here, the intermediate is (NO₂)₂ (e.g. the NO₂ dimer) and this step is fast. The second step is the slower of the two. Because the second step is the slowest step in the mechanism, it is referred to as the rate determining step and this is the step that governs the overall rate of the reaction.

Elementary reactions involving the collision of 3 particles are called termolecular but these types of reactions are very rare.

We can write a rate law for each elementary step.

\[\begin{align*} \mathrm{rate}_1 &= k_1[\mathrm{NO_2}]^2 \\[1.5ex] \mathrm{rate}_2 &= k_2[(\mathrm{NO_2})_2] \end{align*}\]

where rate₁ > rate₂. The first step is second-order and the second step is first-order.

NOTE: Here that the exponents in the written rate laws come from the stoichiometric coefficients of the balanced elementary chemical equations. We can only do this with elementary reactions. With complex reactions, we must use the method of initial rates to determine the orders.

Let us reiterate the fact that the rate of this reaction is dictated by the slowest step of the reaction. Therefore, the rate law might be predicted to be

\[\mathrm{rate}_2 = k_2[(\mathrm{NO_2})_2] \qquad \longleftarrow \mathrm{this~is~wrong!}\] The (NO₂)₂ complex is an intermediate of reaction and not a reactant. This cannot be the appropriate rate law for the overall reaction since rate laws are expressed in terms of the reactants! What do we do here?

Take a look at that first step. It is a reversible step (see the double harpoons)? The first step is fast and when two NO₂ molecules reacts to produce (NO₂)₂, we will see a rapid build up in intermediate. The second step of the reaction (the rate determining step) is chugging along slowly, consuming intermediate but it will not be able to keep up with how fast step 1 is producing it. The intermediate is hanging around in large quantities and is highly reactive. It readily dissociates into 2NO₂.

Here’s the key. The first step will reach equilibrium. The rate at which (NO₂)₂ is produced will eventually equal the rate at which it is consumed (whether in step 2 or by dissociating and returning to 2NO₂). We say that the rate of the forward reaction is equal to the rate of the reverse reaction.

Let us right out the rate laws for the forward and reverse reaction of step 1.

\[\begin{align*} \mathrm{rate}_{1,\mathrm{f}} &= k_1[\mathrm{NO_2}]^2 \\[1.5ex] \mathrm{rate}_{1,\mathrm{r}} &= k_{-1}[\mathrm{(NO_2)_2}] \end{align*}\]

At equilibrium, these two rates are equal. Follow the math here.

\[\begin{align*} \mathrm{rate}_{1,\mathrm{f}} &= \mathrm{rate}_{1,\mathrm{r}} \quad \therefore \\[1.5ex] k_1[\mathrm{NO_2}]^2 &= k_{-1}[\mathrm{(NO_2)_2}] \end{align*}\]

Remember how we said that the rate law for the overall reaction should only be expressed in terms of reactants? We can eliminate the intermediate term from the rate law by simply expressing it in terms of NO₂. Watch this.

\[\begin{align*} k_1[\mathrm{NO_2}]^2 &= k_{-1}[\mathrm{(NO_2)_2}] \\[1.5ex] [\mathrm{(NO_2)_2}] &= \dfrac{k_1}{k_{-1}}[\mathrm{NO_2}]^2 \end{align*}\]

Here we “solved” for (NO₂)₂. Why? Take the right side of the expression and plug it into our “good guess but ultimately wrong” rate law!

\[\begin{align*} \mathrm{rate} &= k_2[(\mathrm{NO_2})_2] \qquad &&\longleftarrow \mathrm{this~is~wrong!} \\[1.5ex] &= k_2\dfrac{k_1}{k_{-1}}[\mathrm{NO_2}]^2 \\[1.5ex] &= k_2K_1[\mathrm{NO_2}]^2 \\[1.5ex] \mathrm{rate} &= k[\mathrm{NO_2}]^2 &&\longleftarrow \mathrm{this~is~right!} \end{align*}\]

The k₁/k_–1 term becomes K₁, an equilibrium constant for the fast, reversible step and K₁ × k₂ becomes another rate constant. We will see equilibrium constants in the next chapter.

We can draw a reaction diagram for this reaction. A reaction diagram places the structures (reactants, products, intermediates, and transition states) on an energy diagram in the order that the reaction takes place.

Plot was created using mechaSVG (Angnes 2020).

Let us break down this diagram. First, there are five stationary points shown (reactants, two transition-states, an intermediate, and products) and their corresponding relative energies. Transition-states are generally higher in energy than the points they connect. The reaction is exothermic since the products of reaction is lower in energy than the reactants.

The plot indicates a two-step reaction. The first step shows reactants proceeding through a transition state (TS2) and onward to the intermediate of reaction. The second step shows the intermediate proceeding to products through TS2. While each individual step has an activation energy, the activation energy of the overall reaction is given by step 2 as this is the largest energy change between reactants and the highest energy stationary point on the diagram. Step 1 is fast whereas step two is slow. This means that step 2 is the rate determining step (it contains the highest energy barrier in the entire reaction). If the energies were enthalpies, the reaction is endothermic (the products are higher in energy than the reactants).

Second Proposed Mechanism, slow first-step

The second proposed mechanism is the following two-step mechanism.

\[\begin{align*} 2\mathrm{NO_2}(g) &\overset{k_1}\longrightarrow \mathrm{NO_3}(g) + \mathrm{NO}(g) &&\mathrm{(slow)} \\[1.5ex] \mathrm{NO_3}(g) + \mathrm{NO}(g) &\underset{k_{-2}}{\stackrel{k_2}{\rightleftharpoons}} \mathrm{2NO}(g) + \mathrm{O_2}(g) &&\mathrm{(fast)} \end{align*}\]

Here, the first step is slow while the second step is fast. The predicted rate law (which should depend on the slowest step of the reaction) is predicted to be

\[\mathrm{rate} = k[\mathrm{NO_2}]^2\] which matches our predicted rate law for the first proposed mechanism! In fact, both mechanisms are plausible. The rate laws between the two are in agreement, both reaction mechanisms are bimolecular, and both reaction diagrams indicate an endothermic reaction!

Here is the reaction diagram for this mechanism.

Practice

Determine the following for the given reaction diagram:

1. Endothermic or Exothermic
2. Number of transition states
3. Number of intermediates
4. Number of steps
5. Slowest step

Answer

1. Endothermic
2. Transition states: 2
3. Intermediates: 1
4. Steps: 2
5. Slowest step: Step 1

Practice

Determine the following for the given reaction diagram:

1. Endothermic or Exothermic
2. Number of transition states
3. Number of intermediates
4. Number of steps
5. Slowest step

Answer

1. Exothermic
2. Transition states: 4
3. Intermediates: 3
4. Steps: 4
5. Slowest step: Step 4

Catalysis

Many reactions can be catalyzed (i.e. sped up; rate increased) by the addition of a catalyst. There is a huge body of work devoted to this. A catalyst is a substance that is added to a reaction and is not consumed by the reaction.

\[\mathrm{A} + \mathrm{B} + \mathrm{catalyst} \longrightarrow \mathrm{products} + \mathrm{catalyst}\]

Here we see a catalyst going into a reaction and coming back out. It would not be seen in a typical chemical equation since it would cancel out. The catalyst speeds up the reaction by lowering the energy of activation. Consider the reaction pathway given below.

The uncatalyzed reaction pathway (black line) has a higher energy of activation than the catalyzed reaction pathway (red line). The catalyzed reaction is faster because the energy of activation is lowered.

There are many types of catalysts and they can be categorized in many different ways. For example, homogeneous catalysts are substances that adopt the same phase as the reactants whereas heterogeneous catalysts do not.

Catalysts allow us to facilitate the formation of products on a more convenient timescale, particularly for reactions that would be too slow without the presence of a catalyst.