Chapter 18, Problem 18C.4AE

Interpretation Introduction

Interpretation:

Entropy for activation at $\text{300}\text{K}$ has to be calculated for the collision between two structureless particles considering $M=65{\text{gmol}}^{\text{-1}}$ and $\sigma =0.35{\text{nm}}^{\text{2}}$.

Explanation of Solution

Given information:

    $\begin{array}{l}M=\text{65}\text{g/mol}\\ \text{T}\text{=}\text{300}\text{K}\\ \sigma =\text{0}\text{.35}{\text{nm}}^{\text{2}}\end{array}$

Entropy of collisions between two particles can be calculated using the formula,

    ${\Delta }^{‡}S=R\left(\ln \frac{A}{B}-2\right)-------------------(1)$

Where,

    $A$ is the activity coefficient.

    ${\Delta }^{‡}S$ is the entropy of collision.

    $R$ is the gas constant.

Activity coefficient can be expressed as given below,

  $A=\sigma {\left(\frac{\text{8}k\text{T}}{\pi \mu }\right)}^{1/2}{N}_A------------------(2)$

Where,

    $\mu$ is reduced mass

    $\sigma$ is collision cross section area

    $N_A$ is Avogadro number

    $k$ is the Boltzmann constant

Calculation for identification of activity coefficient:

As the collision is between two identical particles, the reduced mass is $m/2$.

Boltzmann constant and gas constant is related as shown below,

    $\frac{k}{\text{m}}=\frac{R}{M}$

Substituting the values in equation (2), activity coefficient can be obtained as shown below,

    $\begin{array}{l}A=\sigma {\left(\frac{\text{8}k\text{T}}{\pi \mu }\right)}^{1/2}{N}_A\\ =\text{0}\text{.35}\text{x}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}\text{x}{\left(\frac{\text{8}\text{x}k\text{x}\text{300}\text{K}}{3.14\text{x}\frac{m}{\text{2}}}\right)}^{1/2}\text{x}\text{6}\text{.023}\text{x}{\text{10}}^{\text{23}}{\text{mol}}^{\text{-1}}\\ =\text{0}\text{.35}\text{x}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}\text{x}{\left(\text{1528}\text{K}\text{x}\frac{R}{M}\right)}^{1/2}\text{x}\text{6}\text{.023}\text{x}{\text{10}}^{\text{23}}{\text{mol}}^{\text{-1}}\\ =\text{2}\text{.10805}\text{x}{\text{10}}^{\text{5}}{\text{m}}^{\text{2}}{\text{mol}}^{\text{-1}}\text{x}{\left(\text{1528}\text{K}\text{x}\frac{8.314{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}}{\text{65}\text{x}{\text{10}}^{\text{-3}}{\text{kgmol}}^{\text{-1}}}\right)}^{1/2}\\ =\text{2}\text{.10805}\text{x}{\text{10}}^{\text{5}}{\text{m}}^{\text{2}}{\text{mol}}^{\text{-1}}\text{x}{\left(\text{195}\text{.44}\text{x}{\text{10}}^{\text{3}}{\text{Jkg}}^{\text{-1}}\right)}^{1/2}\end{array}$

    $\begin{array}{l}\because \text{1J}\text{=}{\text{1kgm}}^{\text{2}}{\text{s}}^{\text{-2}}\\ \text{2}\text{.10805}\text{x}{\text{10}}^{\text{5}}{\text{m}}^{\text{2}}{\text{mol}}^{\text{-1}}\text{x}{\left(\text{195}\text{.44}\text{x}{\text{10}}^{\text{3}}{\text{Jkg}}^{\text{-1}}\right)}^{1/2}\\ =2.\text{10805}\text{x}{\text{10}}^{\text{5}}{\text{m}}^{\text{2}}{\text{mol}}^{\text{-1}}\text{x}{\left(\text{195}\text{.44}\text{x}{\text{10}}^{\text{3}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right)}^{1/2}\\ =2.\text{10805}\text{x}{\text{10}}^{\text{5}}{\text{m}}^{\text{2}}{\text{mol}}^{\text{-1}}\text{x}\left(\text{442}\text{.085}{\text{ms}}^{\text{-1}}\right)\\ =931.9505\text{x}{\text{10}}^{\text{5}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ =0.93\text{x}{\text{10}}^{\text{8}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\end{array}$

Activity coefficient is calculated as $0.93\text{x}{\text{10}}^{\text{8}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}$.

Entropy of activation:

The rate constant equation can also be given as shown below by use of Eyring equation.

    $\begin{array}{l}{k}_r=B{e}^{{\Delta }^{‡}S/\mathrm{R}}{e}^{-{\Delta }^{‡}H/RT}\\ Where,\\ B=\left(\frac{kT}{h}\right)\left(\frac{RT}{p^{\sout{\text{O}}}}\right)=\frac{kR{T}^2}{h{p}^{\sout{\text{O}}}}\end{array}$

$B$ can be calculated as shown below,

$\begin{array}{l}B=\frac{kR{T}^2}{h{p}^{\sout{\text{O}}}}\\ =\frac{\left(\text{1}\text{.381}\text{x}{\text{10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\text{x}\left(\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{x}{\left(\text{300}\text{K}\right)}^2}{\text{6}\text{.626}\text{x}{\text{10}}^{\text{-34}}\text{Js}\text{x}{\text{10}}^{\text{5}}\text{Pa}}\\ =\frac{\left(\text{1}\text{.381}\text{x}{\text{10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\text{x}\text{300}\text{K}\right)}{\text{6}\text{.626}\text{x}{\text{10}}^{\text{-34}}\text{Js}}\text{x}\frac{\left(\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{x}\frac{\text{1}{\text{kgm}}^{\text{2}}{\text{s}}^{\text{-2}}}{\text{1}\text{J}}\right)\text{x}\left(\text{300}\text{K}\right)}{{\text{10}}^{\text{5}}\text{Pa}\text{x}\frac{\text{1}{\text{kgm}}^{\text{-1}}{\text{s}}^{\text{-2}}}{\text{1}\text{Pa}}}\\ =(62.52\text{x}{\text{10}}^{\text{11}}{\text{s}}^{\text{-1}})(2494.2\text{x}{\text{10}}^{\text{-5}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}\text{)}\\ \text{=}(\text{62}\text{.52}\text{x}{\text{10}}^{\text{11}}{\text{s}}^{\text{-1}})(2.494\text{x}{\text{10}}^{\text{-2}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}\text{)}\\ \text{=}\text{155}\text{.92}\text{x}{\text{10}}^{\text{9}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ =\text{1}\text{.56}\text{x}{\text{10}}^{\text{11}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\end{array}$

Substitution of the obtained values in the equation (1), entropy of activation can be calculated as shown below,

  $\begin{array}{l}{\Delta }^{‡}S=R\left(\ln \frac{A}{B}-2\right)\\ =\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\left(\ln \frac{0.93\text{x}{\text{10}}^{\text{8}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}}{\text{1}\text{.56}\text{x}{\text{10}}^{\text{11}}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}}-2\right)\\ =\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\left(-9.425\right)\\ =-78.35{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

The entropy of activation during collision is calculated as $-78.35{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.