# 27.7: The Rate of a Gas-Phase Chemical Reactions

## Rate Law and Collision Theory

Consider the elementary reaction

$\text{A} + \text{B} \longrightarrow \text{C}$

with the associated second-order rate law

$r = k \left[ \text{A} \right] \left[ \text{B} \right] \label{20.1}$

as empirical. As it happens, we can actually derive this using the collision theory. Recall, that the collision rate between two atoms or molecules in a system is

$\gamma = \rho \sigma \left< \left| \textbf{v} \right| \right> \label{20.2}$

where $$\rho$$ is the number density, $$\sigma$$ is the collision cross section, and $$\left< \left| \textbf{v} \right| \right>$$ is the relative velocity between the two atoms or molecules. Now, if the two colliding atoms or molecules are different, and we are interested in the rate of collisions of atoms/molecules of type $$\text{A}$$ with those of type $$\text{B}$$, then the collision rate must be written as

$\gamma_\text{AB} = \sigma_\text{AB} \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{B} \label{20.3}$

Here $$\rho_\text{B}$$ is the density of atoms/molecules of type $$\text{B}$$, $$\left| \textbf{v}_\text{AB} \right| = \left| \textbf{v}_\text{B} - \textbf{v}_\text{A} \right|$$ is the relative speed between $$\text{A}$$ and $$\text{B}$$, and $$\sigma_\text{AB}$$ is the cross section between $$\text{A}$$ and $$\text{B}$$, which, is given the average of arithmetic and geometric averages:

$\sigma_\text{AB} = \dfrac{1}{2} \left[ \left( \dfrac{\sigma_\text{A} + \sigma_\text{B}}{2} \right) + \sqrt{\sigma_\text{A} \sigma_\text{B}} \right] \label{20.4}$

From the Maxwell-Boltzmann distribution,

$\left< \left| \textbf{v}_\text{AB} \right| \right> = \sqrt{\dfrac{8 k_B T}{\pi \mu}} \label{20.5}$

where the reduced mass

$\mu = \dfrac{m_\text{A} m_\text{B}}{m_\text{A} + m_\text{B}} \label{20.6}$

The collision rate $$\gamma_\text{AB}$$ is the rate for the collision of one atom/molecule of $$\text{A}$$. If there are $$N_\text{A}$$ atoms/molecules of $$\text{A}$$, then the total collision rate for $$\text{A}$$ with $$\text{B}$$ is

$\gamma_\text{tot} = N_\text{A} \gamma_\text{AB} = N_\text{A} \sigma_\text{AB} \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{B} \label{20.7}$

However, the number density of $$\text{A}$$ is $$\rho_\text{A} = N_\text{A}/V$$, so we can write the total collision rate as

$\gamma_\text{tot} = \sigma_\text{AB} V \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{A} \rho_\text{B} \label{20.8}$

In a time interval $$dt$$, the number of collisions $$dN_\text{coll}$$ is

$dN_\text{coll} = \gamma_\text{tot} dt = \sigma_\text{AB} V \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{A} \rho_\text{B} dt$

Let $$P_\text{rxn}$$ denote the probability that a collision between $$\text{A}$$ and $$\text{B}$$ leads to product $$\text{C}$$. The rate of decrease of $$N_\text{A}$$ must then be

$dN_\text{A} = -P_\text{rxn} dN_\text{coll} = -\sigma_\text{AB} P_\text{rxn} V \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{A} \rho_\text{B} dt \label{20.9}$

so that

$\dfrac{dN_\text{A}}{dt} = -\sigma_\text{AB} P_\text{rxn} V \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{A} \rho_\text{B} \label{20.10}$

Note that the rate is

$r = -\dfrac{d \left[ \text{A} \right]}{dt} \label{20.11}$

However, $$\left[ \text{A} \right]$$ is in units of moles/liter. The ratio $$N_\text{A}/\left(N_0V \right)$$, where $$N_0$$ is Avogadro’s number, has the proper units of moles/liter, if $$V$$ is in liters. Thus,

\begin{align} \dfrac{d \left[ \text{A} \right]}{dt} &= \dfrac{d \left( N_\text{A}/\left( N_0 V \right) \right)}{dt} \\ &= -\dfrac{1}{N_0 V} \sigma_\text{AB} P_\text{rxn} V \left< \left| \text{v}_\text{AB} \right| \right> \rho_\text{A} \rho_\text{B} \\ &= -\sigma_\text{AB} P_\text{rxn} N_0^{-1} \left< \left| \textbf{v}_\text{AB} \right| \right> \rho_\text{A} \rho_\text{B} \end{align} \label{20.12}

Since $$\rho_\text{A}$$ has units of (molecules of $$\text{A}$$/liters), we can write $$\rho_\text{A} = N_0 \left[ \text{A} \right]$$, and similarly, $$\rho_\text{B} = N_0 \left[ \text{B} \right]$$. This gives

$\dfrac{d \left[ \text{A} \right]}{dt} = -\sigma_\text{AB} P_\text{rxn} N_0 \left< \left| \textbf{v}_\text{AB} \right| \right> \left[ \text{A} \right] \left[ \text{B} \right] = -k \left[ \text{A} \right] \left[ \text{B} \right] \label{20.13}$

where the rate constant is

$k = \sigma_\text{AB} P_\text{rxn} N_0 \left< \left| \textbf{v}_\text{AB} \right| \right> \label{20.14}$

To determine the reaction probability $$P_\text{rxn}$$, consider the energy profile for the reaction in Figure $$\PageIndex{1}$$. In the gas phase, the activation “energy”, denote $$E_a$$ in the figure is the potential energy at the top of the hill, which we denote as $$\mathcal{E}^\ddagger$$. If the reaction takes place in a condensed phase, such as in solution, then the activation “energy” is the free energy $$\Delta G^\ddagger$$. Figure $$\PageIndex{1}$$: Illustration of a reaction energy profile.

If $$\text{A}$$ and $$\text{B}$$ are atoms, then $$P_\text{rxn}$$ is the probability that the energy $$E_\text{AB}$$ between $$\text{A}$$ and $$\text{B}$$ must be larger than this energy $$E_a$$ in order for the collision to yield product $$\text{C}$$. If $$\text{A}$$ and $$\text{B}$$ are molecules, then $$\text{A}$$ and $$\text{B}$$ must also have the right orientation in addition to a sufficiently high energy. The probability that they have the right orientation is a fraction $$f < 1$$, which we call the steric factor. When $$\text{A}$$ and $$\text{B}$$ are atoms, $$f = 1$$. Generally, we can write

$P_\text{rxn} = f P \left( E_\text{AB} > E_a \right) \label{20.15}$

The general probability distribution $$p \left( E_\text{AB} \right)$$ is just given by the Boltzmann distribution

$p \left( E_\text{AB} \right) Ce^{-\beta E_\text{AB}} \label{20.16}$

where $$C$$ is a normalization constant. The normalization condition is

$\int_0^\infty p \left( E_\text{AB} \right) d E_\text{AB} = C \int_0^\infty e^{-\beta E_\text{AB}} = \left. -\dfrac{C}{\beta} e^{-\beta E_\text{AB}} \right|_0^\infty = 1 \label{20.17}$

which gives $$C = \beta = 1/\left( k_B T \right)$$. Thus,

$p \left( E_\text{AB} \right) = \beta e^{-\beta E_\text{AB}} \label{20.18}$

Now, the probability $$P \left( E_\text{AB} > E_a \right)$$ that $$E_\text{AB} > E_a$$ is

$P \left( E_\text{AB} > E_a \right) = \beta \int_{E_a}^\infty e^{-\beta E_\text{AB}} d E_\text{AB} = e^{-\beta E_a} \label{20.19}$

which gives the rate constant as

$k = \sigma_\text{AB} N_0 f e^{-\beta E_a} \left< \left| \textbf{v}_\text{AB} \right| \right> = \sigma_\text{AB} N_0 f e^{-\beta E_a} \left( \dfrac{8}{\beta \pi \mu} \right)^{1/2} \label{20.20}$

We see, generally, that

$k = \text{A} e^{-\beta E_a} \label{20.21}$

where $$E_a$$ is the activation potential $$\mathcal{E}^\ddagger$$ in the gas phase and the activation free energy $$\Delta G^\ddagger$$ in condensed phases. This is known as the Arrhenius law.

Note that if we plot $$\text{ln} \: k$$ vs. $$1/T$$, which is given by

$\text{ln} \: k = \text{ln} \: \text{A} - \dfrac{E_a}{k_B T} \label{20.22}$

the plot will be a line with slope $$-E_a/k_B$$. Such a plot is called an Arrhenius plot. Note, moreover, that if $$\text{A}$$ and $$\text{B}$$ are the same atom or molecule type, then the rate law we derived, would take the form of a second-order rate law

$\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right]^2 \label{20.23}$