# Rate Law

A rate law is an expression that relates the concentrations of reactants to the rate of a reaction. For an **elementary reaction** such as:

\[aA + bB \longrightarrow cC\]

the corresponding rate law can be written as

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

where *k* is the **rate constant**.

The powers, *a* and *b*, are the **partial orders of reaction** for their given components and can be positive or negative integers or fractions as well as zero. For this course, powers will be positive whole numbers or zero. However, these powers can be fractional and even negative.

For a **complex reaction** such as

\[aA + bB \longrightarrow cC\]

the corresponding rate law can be written as

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

Note that the powers *i* and *j* cannot be directly determined from the balanced chemical equation for a complex reaction. You would need (a) to be given the power information or (b) to determine the exponents from *experimental data* using a method called the **method of initial rates**.

## Order of Reaction

The exponents in a rate law always sum to the order of reaction. For example, if the reaction above was second-order, then

\[i + j = 2\]

If the order of reaction was first-order, then

\[i + j = 1\]

If the reaction

\[aA + bB \longrightarrow cC\]

was first-order with respect to *A* and first-order with respect to *B*, you could then write the following rate law:

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

If the reaction was first-order with respect to *A* and third-order overall, you could write the following rate law:

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

since \(i + j = 3\) and \(i = 1\).

## Method of Initial Rates

The *method of initial rates* is a process by which you take experimental data and mathematically determine the exponents in a rate law.

**Procedure:**

- 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 (1) until all exponents are determined

- Write rate law
- Use data from any experiment to find rate constant,
*k*

### Example

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

\[ \mathrm{NO} + \mathrm{O_2} \longrightarrow \mathrm{products}\]

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

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

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

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

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

**1. Determine exponent for NO**

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

\[\begin{align*} \frac{\text { rate }_{e 4}}{\text { rate }_{e 2}} &= \frac{k[\mathrm{NO}]_{e 4}^{i}\left[\mathrm{O}_{2}\right]_{e 4}^{j}}{k[\mathrm{NO}]_{e 2}^{i}\left[\mathrm{O}_{2}\right]_{e 2}^{j}} \end{align*}\]

- Determine exponent

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

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

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

- Determine exponent

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

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

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

## Practice Problems

### Problem 1

What is the rate law for the following reaction?

2C_{2}H_{6} (*g*) + 7O_{2} (*g*) ⟶ 4CO_{2} (*g*) + 6H_{2}O (*l*)

##
**Solution**

\[\begin{align*} \mathrm{rate} = k\lbrack \mathrm{C_2H_6} \rbrack ^i \lbrack \mathrm{O_2}\rbrack ^j \end{align*}\]

### Problem 2

What is the rate law for the following reaction? Assume that the reaction is 0th-order with respect to [B] and second-order overall.

A + B ⟶ C

##
**Solution**

\[\begin{align*} \mathrm{rate} = k[\mathrm{A}]^2 \end{align*}\]

### Problem 3

Both reactants in **Problem 2** above are increased by the same amount. Which reactant will increase the reaction rate the most?

##
**Solution**

\[\begin{align*} \mathrm{A} \end{align*}\]

### Problem 4

The rate constant for a first-order decomposition at 25 °C of dinitrogen pentoxide dissolved in
chloroform is 6.2 × 10^{–4} min^{–1}. What is the rate of reaction
(in *M* min^{–1}) when [N_{2}O_{5}] = 0.40 *M*?

2N\(_2\)O\(_5\) \(\longrightarrow\) 4NO\(_2\) + 2O\(_2\)

##
**Solution**

\[\begin{align*} \mathrm{rate} &= k[\mathrm{N_2O_5}] \\ &= 6.2\times 10^{-4}~\mathrm{min^{-1}} \left ( 0.40~M \right ) \\ &= 2.5\times 10^{-4}~M\mathrm{~min^{-1}} \end{align*}\]

## Timed Assessments

### Method of Initial Rates

##
**Answers**

- rate = [A][B]
^{2}; 3rd order;*k*= 1.03 × 10^{4} - rate = [A][B]; 2nd order;
*k*= 1.97 × 10^{4} - rate = [A][B]; 2nd order;
*k*= 7.97