## 3.3 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** (or *differential 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).

### 3.3.1 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.

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

We would say that the reaction is first-order in A, second-order in B, and zeroth-order in C.

Reactant C does not appear in the rate law since [C]^{0} = 1.

Summing over all the exponents give the overall order of reaction. Here the overall order would be 3, hence, the reaction is third-order.

### 3.3.2 Method of Initial Rates

To determine the exponents in a rate law, one must perform the **Method of Initial
Rates** that requires *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:

- Initial reactant concentration
- Initial rate of reaction

We then perform the Method of Initial Rates following the steps below:

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

- Write rate law
- 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 | [H_{2}O_{2}]_{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 H_{2}O_{2}.

**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 H_{2}O_{2} 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

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

The e2 (Experiment 2) and e3 (Experiment 3) labels are 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*.

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

Note the following log identity used: ln(x^{y}) = yln(x)

Exponents in rate laws can be whole numbers, fractions, zero, or even negative. 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 H_{2}O_{2}.

**Practice**

Given the following table of experimental data for the given reaction, determine the full rate law.

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

Experiment | [NO]_{0} (M) |
[O_{2}] (M) |
Initial Rate (M s^{–1}) |
---|---|---|---|

1 | 0.0235 | 0.0125 | 7.98 × 10^{–3} |

2 | 0.0235 | 0.0250 | 15.9 × 10^{–3} |

3 | 0.0470 | 0.0125 | 32.0 × 10^{–3} |

4 | 0.0470 | 0.0250 | 63.5 × 10^{–3} |

##
**Solution**

**1. Determine exponent for NO**

- Select experiments 2 and 4
- Make ratio of rate laws for experiments 2 and 4

\[\frac{\text { rate }_{e 4}}{\text { rate }_{e 2}} = \frac{k[\mathrm{NO}]_{e 4}^{m}\left[\mathrm{O}_{2}\right]_{e 4}^{n}}{k[\mathrm{NO}]_{e 2}^{m}\left[\mathrm{O}_{2}\right]_{e 2}^{n}}\]

- Determine exponent

\[\begin{align*} \frac{63.5 \times 10^{-3}}{15.9 \times 10^{-3}}&=\frac{k(0.0470)^{m}(0.0250)^{n}}{k(0.0235)^{m}(0.0250)^{n}}\\[2ex] 3.994 &= 2^m \\ \log (3.994) &= \log(2^m) \\ \log (3.994) &= m\log(2)\\ m &= \dfrac{\log(3.994)}{\log (2)}\\ &= 1.998 \approx 2 \end{align*}\]

The exponent for [NO] in the rate law is 2.

- Repeat step 1 for [O
_{2}]

**1. Determine exponent for O _{2}**

Select experiments 1 and 2

Make ratio of rate laws for experiments 1 and 2

\[\begin{align*} \frac{\operatorname{rate}_{e 2}}{\operatorname{rate}_{e 1}}=\frac{k[\mathrm{NO}]_{e 2}^{m}\left[\mathrm{O}_{2}\right]_{e 2}^{n}}{k[\mathrm{NO}]_{e 1}^{m}\left[\mathrm{O}_{2}\right]_{e 1}^{n}} \end{align*}\]

- Determine exponent

\[\begin{align*} \frac{15.9 \times 10^{-3}}{7.98 \times 10^{-3}} &= \frac{k(0.235)^{m}(0.0250)^{n}}{k(0.0235)^{m}(0.0125)^{n}}\\ 1.990 &= 2^n \\ \log (1.990) &= \log(2^n) \\ \log (1.990) &= n\log(2)\\ n &= \dfrac{\log(1.990)}{\log (2)}\\ &= 0.994 \approx 1 \end{align*}\]

The exponent for [O_{2}] in the rate law is 1.

**2. Write rate law**

\[\mathrm{rate} = k[\mathrm{NO}]^2[\mathrm{O_2}]\]

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

Here I’ve chosen data from experiment 1

\[\begin{align*} k &= \dfrac{7.98\times 10^{-3}~M~\mathrm{s^{-1}}}{(0.0235~M)^2(0.0125~M)} \\ &= 1.16\times 10^{3}~M^{-2}~\mathrm{s^{-1}} \end{align*}\]

The complete rate law is now

\[\mathrm{rate} = 1.16\times 10^{3}~M^{-2}~\mathrm{s^{-1}}[\mathrm{NO}]^2[\mathrm{O_2}]\]

The order of reaction is third-order.

**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*
H_{2}O_{2} and run it to zero in increments of 0.1 *M*.

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

### 3.3.3 Rate Constant, *k*

The rate constant is determined using the **Arrhenius equation**.

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

(See the rate constants for some reactions here). 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

*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*, 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.

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

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**.

Rationalize this plot. At higher temperatures, the rate of the reaction is faster than at lower temperatures.

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 [A] term is omitted.

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!

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**.

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

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.

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 O_{2} that is present, the less free oxygen atoms that are present to
become ozone (as seen in reaction step 2 above).
Therefore, the higher [O_{2}], 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

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

### 3.3.4 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 here (c = concentration and t = time).

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?

##
**Solution**

second-order

**Practice**

The following reaction has a rate constant of 0.3387 *M*^{–1} s^{–1}
at 326.85 °C (600 K).

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

**Visualize This**

A 5 mol sample of NO_{2} 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 NO_{2} 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^{6}
and can be found on pg. 40.

##
**Solution**

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

\[\begin{align*} \mathrm{rate} &= k[\mathrm{NO_2}]^2 \\[1.5ex] &= (0.3387~M^{-1}~\mathrm{s}^{-1}) ( 5.00~M )^2 \\[1.5ex] &= 8.467~M~\mathrm{s}^{-1} \end{align*}\]

Remember, this is the initial rate of the reaction since the reaction initially
began with 5.0 *M* NO_{2}.

**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!

### References

*Table of Recommended Rate Constants for Chemical Reactions Occurring in Combustion*; Westley, F., Ed.; National Bureau of Standards, 1980.