Skip to main content
Chemistry LibreTexts

13.3: The Rate of a Gas-Phase Chemical Reactions

  • Page ID
  • 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

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

    Contributors and Attributions

    • Was this article helpful?