 Solution

Write $M=m_A+m_B$ and $\mu =m_A{m}_B/M$. An atom has no rotational or vibrational internal factor under the stated electronic assumption. The transition-state moment of inertia is $I=\mu R^2$, so the high-temperature rigid rotor result isThe translational factors therefore giveSubstituting into the transition state theory expression and simplifying,For distinct atoms, $\sigma =1$. This is a number-density rate coefficient, with units of volume per time. A rate law in molar concentrations uses $N_A{k}_{2nd}$. The derivation assumes a classical high-temperature rotor at the dividing surface and a transmission coefficient of one.