Question 17.1: The pre-exponential factor for the elementary reaction O + H...
The pre-exponential factor for the elementary reaction O + H_{2} → OH + H has been determined experimentally to be 8.04 × 10^{11} cm³/mol·s at 500 K. Calculate the pre-exponential factor for this elementary chemical reaction at the same temperature using collision theory and compare your result with the given experimental value.
From Eqs. (17.6) and (17.19), the pre-exponential factor provided by kinetic theory is
A = BT^{n}, (17.6)
B=2 N_{A} \sigma_{A B}^{2}\left(\frac{2 \pi k T}{\mu}\right)^{1 / 2}, (17.19)
A=2 N_{A} \sigma_{A B}^{2}\left(\frac{2 \pi k T}{\mu}\right)^{1 / 2}.
From Appendix O, we may take the rigid-sphere diameter of H_{2} to be 2.92 Å and that for atomic oxygen to be half of its value for O_{2} or 1.72 Å. On this basis, we obtain, from Eq. (16.55),
\sigma_{A B}=\frac{1}{2}\left(\sigma_{A}+\sigma_{B}\right) (16.55)
\sigma_{AB}=\frac{\sigma_{A}+ \sigma_{B}}{2}=\frac{2.92+1.72}{2} = 2.32 Å.
From Eq. (16.6), the reduced mass for this binary reaction is
\mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}. (16.6)
\mu=\frac{m_{A} m_{B}}{m_{A}+m_{B}}=\frac{(2)(16)}{2+16}\left(1.6605 \times 10^{-24} g \right)=2.952 \times 10^{-24} g.
Therefore, the pre-exponential factor from kinetic theory becomes
A=2\left(6.022 \times 10^{23} mol ^{-1}\right)\left(2.32 \times 10^{-8} cm \right)^{2}\times \sqrt{\frac{2 \pi\left(1.381 \times 10^{-16} g \cdot cm ^{2} / s ^{2} \cdot K \right)(500 K )}{2.952 \times 10^{-24} g }}
= 2.48 × 10^{14} cm³/mol · s.
The calculated pre-exponential factor from kinetic theory is over 300 times greater than that provided by experiment. This huge discrepancy arises from the internal energy modes of molecular hydrogen, which are not accounted for in a purely kinetic model.