Chapter 12, Problem 12.14P

Interpretation Introduction

Interpretation:

The equilibrium constant for the dissociation of $I_2$ at 500K has to be calculated.

Concept Introduction:

The dissociation of $I_2$ is given below.

  $I_2\to 2I$

The equilibrium constant for the dissociation of $I_2$ can be calculated using the following formula.

  $K=\frac{{\left(\frac{q_{I,m}^{\Theta }}{N_A}\right)}^2}{\left(\frac{q_{I_2,m}^{\Theta }}{N_A}\right)}{e}^{\frac{-E_{dis}}{RT}}$

Where,

    $\begin{array}{c}K=equilibriumcons\tan t\\ q_{I,m}^{\Theta }=partitionco-efficientofI\\ q_{I_2,m}^{\Theta }=partitionco-efficientof{I}_2\\ N_A=avogadronumber\\ E_{dis}=dissociationenergy\\ R=universalgascons\tan t\\ T=temperature\end{array}$

The expression for the translational partition function is given below.

  $q^T=\frac{{\left(2\pi mkT\right)}^{\frac{3}{2}}V}{h^3}$

Where,

    $\begin{array}{c}{q}^T=translationalpartitionfunction\\ m=molecularmass\\ T=temperature\\ h=plank\text{'}scons\tan t\\ V=volumeats\tan dardconditions\\ k=boltzmanncons\tan t\end{array}$

The expression for rotational partition function is given below.

  $q^R=\frac{kT}{\sigma hB}$

Where,

    $\begin{array}{c}{q}^R=rotationalpartitionfunction\\ k=boltzmanncons\tan t\\ T=temperature\\ h=plank\text{'}scons\tan t\\ \sigma =symmetrynumber\\ B=rotationalcons\tan t\end{array}$

The expression for vibrational partition function is given below.

  $q^V=\frac{1}{1-e^{\frac{-h\nu }{kT}}}$

Where,

    $\begin{array}{c}{q}^V=vibrationalpartitionfunction\\ h=plank\text{'}scons\tan t\\ k=boltzmanncons\tan t\\ T=temperature\\ \nu =vibrationalfrequency\end{array}$

Answer to Problem 12.14P

The equilibrium constant for the dissociation of $I_2$ at 500K is calculated as $1.615\times 10^{-20}.$

Explanation of Solution

The equilibrium constant for the dissociation of $I_2$ can be calculated using the following formula.

  $K=\frac{{\left(\frac{q_{I,m}^{\Theta }}{N_A}\right)}^2}{\left(\frac{q_{I_2,m}^{\Theta }}{N_A}\right)}{e}^{\frac{-E_{dis}}{RT}}$

The partition functions of $I$ and $I_2$ can be written as the product of partition function for each mode of vibration.

The translational partition function for $I$ can be calculated as follows,

Given,

    $\begin{array}{c}m=127\times 1.6606\times {10}^{-27}kg\\ T=500K\\ h=6.626\times {10}^{-34}Js\\ V=volumeats\tan dardconditions\\ k=1.38\times {10}^{-23}J/K\end{array}$

  $\begin{array}{c}{q}_I^T{}^{}=\frac{{\left(2\pi mkT\right)}^{\frac{3}{2}}V}{h^3}\\ \\ =\frac{{\left\{2\times 3.14\times \left(127\times 1.6606\times {10}^{-27}kg\right)\times 1.38\times {10}^{-23}J/K\times 500K\right\}}^{\frac{3}{2}}V}{{\left(6.626\times {10}^{-34}Js\right)}^3}\\ \\ =\frac{{\left\{2\times 3.14\times \left(127\times 1.6606\times {10}^{-27}kg\right)\times 1.38\times {10}^{-23}J/K\times 500K\right\}}^{\frac{3}{2}}V}{{\left(6.626\times {10}^{-34}Js\right)}^3}\\ \\ =V\times 2.97\times 10^{27}\end{array}$

The translational partition function for $I_2$ can be calculated as follows,

Given,

    $\begin{array}{c}m=2\times 127\times 1.6606\times {10}^{-27}kg\\ T=500K\\ h=6.626\times {10}^{-34}Js\\ V=volumeats\tan dardconditions\\ k=1.38\times {10}^{-23}J/K\end{array}$

  $\begin{array}{c}{q}_{I_2}^T{}^{}=\frac{{\left(2\pi mkT\right)}^{\frac{3}{2}}V}{h^3}\\ \\ =\frac{{\left\{2\times 3.14\times \left(2\times 127\times 1.6606\times {10}^{-27}kg\right)\times 1.38\times {10}^{-23}J/K\times 500K\right\}}^{\frac{3}{2}}V}{{\left(6.626\times {10}^{-34}Js\right)}^3}\\ \\ =\frac{{\left\{2\times 3.14\times \left(127\times 1.6606\times {10}^{-27}kg\right)\times 1.38\times {10}^{-23}J/K\times 500K\right\}}^{\frac{3}{2}}V}{{\left(6.626\times {10}^{-34}Js\right)}^3}\\ \\ =V\times 8.42\times 10^{27}\end{array}$

Rotational partition function for $I_2$ can be calculated using following formula.

Given,

    $\begin{array}{c}k=1.38\times {10}^{-23}J/K\\ T=500K\\ h=6.626\times {10}^{-34}Js\\ \sigma =2forsymmetricallinearrotor\\ B=3.73m^{-1}\times 3\times {10}^{8}m/s\end{array}$

$\begin{array}{c}{q}_{I_2}^R=\frac{kT}{\sigma hB}\\ \\ =\frac{\left(1.38\times {10}^{-23}J/K\right)\times 500K}{2\times \left(6.626\times {10}^{-34}Js\right)\times \left(3.73m^{-1}\times 3\times {10}^{8}m/s\right)}\\ \\ =4.63\times 10^3\end{array}$

Vibrational partition function for $I_2$ can be calculated using following formula.

Given,

    $\begin{array}{c}h=6.626\times {10}^{-34}Js\\ k=1.38\times {10}^{-23}J/K\\ T=500K\\ \nu =\frac{1}{\lambda }c=21.4m^{-1}\times 3\times {10}^{8}m/s\end{array}$

  $\begin{array}{c}{q}^V=\frac{1}{1-e^{\frac{-\left(6.626\times {10}^{-34}Js\right).\left(21.4m^{-1}\times 3\times {10}^{8}m/s\right)}{\left(1.38\times {10}^{-23}J/K\right)\left(500K\right)}}}\\ \\ =1.622\times 10^3\end{array}$

Therefore the equilibrium constant can be calculated by substituting these valus in the equation as follows,

Given,

    $\begin{array}{c}\text{Dissociation}\text{energy}\text{=}{E}_{dis}=\text{151}\text{kJ/mol}\text{=1}{\text{.51×10}}^{\text{5}}\text{J/mol}\\ \\ {\text{q}}_{\text{I,m}}^{\text{Θ}}{\text{=q}}_{\text{I}}^{\text{T}}\text{=V×2}{\text{.97×10}}^{\text{27}}\\ \\ {\text{q}}_{{\text{I}}_{\text{2}}\text{,m}}^{\text{Θ}}{\text{=q}}_{{\text{I}}_{\text{2}}}^{\text{T}}\text{×}{\text{q}}_{{\text{I}}_{\text{2}}}^{\text{R}}\text{×}{\text{q}}^{\text{V}}\text{=}\left(\text{V×8}{\text{.42×10}}^{\text{27}}\right)\text{×}\left(\text{4}{\text{.63×10}}^{\text{3}}\right)\text{×}\left(\text{1}{\text{.622×10}}^{\text{3}}\right)=V\times 6.32{\text{×10}}^{33}\\ \\ V=\frac{RT}{P}\\ \\ P={10}^{5}Pa\\ \\ R=8.314J/K\end{array}$

  $\begin{array}{c}K=\frac{{\left(\frac{q_{I,m}^{\Theta }}{N_A}\right)}^2}{\left(\frac{q_{I_2,m}^{\Theta }}{N_A}\right)}{e}^{\frac{-E_{dis}}{RT}}=\frac{{\left(q_{I,m}^{\Theta }\right)}^2}{N_A\left(q_{I_2,m}^{\Theta }\right)}{e}^{\frac{-E_{dis}}{RT}}\\ \\ =\frac{{\left(\text{V×2}{\text{.97×10}}^{\text{27}}\right)}^2}{N_A\times \left(V\times 6.32{\text{×10}}^{33}\right)}{e}^{\frac{-E_{dis}}{RT}}\\ \\ =\frac{RT{\left(\text{2}{\text{.97×10}}^{\text{27}}\right)}^2}{P\times N_A\times 6.32{\text{×10}}^{33}}{e}^{\frac{-E_{dis}}{RT}}\\ \\ =\frac{\left(8.314J/K\right)\times \left(500K\right)\text{×}{\left(\text{2}{\text{.97×10}}^{\text{27}}\right)}^2}{{10}^{5}Pa\times \left(6.022\times {10}^{23}\right)\times 6.32{\text{×10}}^{33}}{e}^{\frac{-\text{1}{\text{.51×10}}^{\text{5}}\text{J/mol}}{\left(8.314J/K\right)(500K)}}\\ \\ =1.615\times 10^{-20}\end{array}$

The equilibrium constant for the dissociation of $I_2$ at 500K is calculated as $1.615\times 10^{-20}.$