Chapter 18, Problem 18A.1ST

Interpretation Introduction

Interpretation:

Value of P has to be estimated for the harpoon mechanism between $\text{Na}$ and ${\text{Cl}}_{\text{2}}$.

Explanation of Solution

Electron affinity ($E_{ea}$) for chlorine is given as $\text{230}{\text{kJmol}}^{\text{-1}}$.  Ionization energy ($I$) of sodium is $\text{496}{\text{kJmol}}^{\text{-1}}$.  Sum of radii ($d$) is given as $\text{350}\text{pm}$.

Steric factor can be calculated using the equation given below,

    $P=\frac{{\sigma }^{\text{*}}}{\sigma }------------(1)$

Where,

    ${\sigma }^{*}$ is the reactive cross section.

    $\sigma$ is the non-reactive collision cross section.

Non reactive collision cross section can be given as,

    $\sigma =\pi d^2$

Reactive cross section can be given as,

    $\sigma =\pi R^{*}{}^2$

Net change in energy is given as,

    $E=I-E_{ea}-\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}R}$

Where,

    $-\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}R}$ is the Coulombic attraction energy.

When $R$ is equal to some critical value $R^{*}$, the energy is zero.  Therefore, the above equation can be written as,

    $\begin{array}{l}I-E_{ea}-\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}{R}^{*}}=0\\ R^{*}=\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}(I-E_{ea})}----------(2)\end{array}$

Steric factor can be calculated as,

    $\begin{array}{l}P=\frac{{\sigma }^{\text{*}}}{\sigma }\\ =\frac{\pi R^{\text{*2}}}{\pi d^2}------------(3)\end{array}$

Substituting for $R^{*}$ in equation (3),

    $P={\left(\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}d(I-E_{ea})}\right)}^2------------(4)$

Substitution of the values in equation (4), the steric factor can be calculated as,

    $P={\left(\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}d(I-E_{ea})}\right)}^2$

Where,

    $e$ is the charge of electron which is $1.602\times {10}^{-19}\text{C}$.

    ${\epsilon }_0$ is the permitivity of vacuum which is $8.8\times {10}^{-12}\text{F/m}$.

Ionization energy of sodium in attojoules is $\text{0}\text{.83}\text{aJ}$ and electron affinity of chlorine in attojoules is $0.39aJ$.

    $\begin{array}{l}P={\left(\frac{e^2}{\text{4}\pi {\epsilon }_{\text{0}}d(I-E_{ea})}\right)}^2\\ ={\left(\frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^2}{\text{4}\times \text{3}\text{.14}\times 8.8\times {10}^{-12}\text{F/m}\times 350\times {10}^{-12}m\times (0.83-0.39)\text{aJ}}\right)}^2\\ ={\left(\frac{2.566\times {10}^{-38}{\text{C}}^{\text{2}}}{17021.312\times {10}^{-24}{\text{aJFm}}^{\text{-2}}}\right)}^2\\ =\left(\frac{6.5843\times {10}^{-76}{\text{C}}^{\text{4}}}{289725062.2\times {10}^{-48}{\text{aJ}}^{\text{2}}{\text{mol}}^{\text{-2}}{\text{F}}^{\text{2}}{\text{m}}^{\text{-4}}}\right)\\ =2.2\end{array}$

Therefore, the steric factor is calculated to be $2.2$.