91Ó°ÊÓ

We have given,

the reaction${\mathrm{H}}_2→2\mathrm{H}$

Where the reaction constant is given by,

localid="1650201323491" $\mathrm{K}=\frac{{\mathrm{P}}_{\mathrm{H}}^2}{{\mathrm{P}}^0{\mathrm{P}}_{{\mathrm{H}}_2}}$

We have to calculate this K value at different temperature.

The chemical potential for the given equilibrium reactions

${\mathrm{μ}}_{{\mathrm{H}}_2}=2{\mathrm{μ}}_{\mathrm{H}}$

Here we have considered that molecular rest energy is zero.

But the rest energy for the hydrogen molecules is actually not zero, even it is less than the hydrogen.

${\mathrm{μ}}_{{\mathrm{H}}_2}-\mathrm{Î\mu }=2{\mathrm{μ}}_{\mathrm{H}}$

here $\mathrm{Î\mu }$is defined as dissociate energy.

We know that chemical potential is defined as

$$\begin{gathered} {\mathrm{μ}}_{{\mathrm{H}}_2}=-\mathrm{KT}\ln \left(\frac{\mathrm{V}{\mathrm{Z}}_{{\mathrm{inH}}_2}}{{\mathrm{N}}_{{\mathrm{H}}_2}{\mathrm{v}}_{\mathrm{Q},{\mathrm{H}}_2}}\right) \\ {\mathrm{μ}}_{\mathrm{H}}=-\mathrm{KT}\ln \left(\frac{{\mathrm{VZ}}_{\mathrm{int}\mathrm{H}}}{{\mathrm{N}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right) \end{gathered}$$

substitute in above equation,

$$\begin{gathered} -\mathrm{KT}\ln \left(\frac{\mathrm{V}{\mathrm{Z}}_{{\mathrm{inH}}_2}}{{\mathrm{N}}_{{\mathrm{H}}_2}{\mathrm{v}}_{\mathrm{Q},{\mathrm{H}}_2}}\right)-\mathrm{Î\mu }=-2\mathrm{KT}\ln \left(\frac{{\mathrm{VZ}}_{\mathrm{int}\mathrm{H}}}{{\mathrm{N}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right) \\ \ln \left(\frac{\mathrm{V}{\mathrm{Z}}_{{\mathrm{inH}}_2}}{{\mathrm{N}}_{{\mathrm{H}}_2}{\mathrm{v}}_{\mathrm{Q},{\mathrm{H}}_2}}\right)+\frac{\mathrm{Î\mu }}{\mathrm{KT}}=2\ln \left(\frac{{\mathrm{VZ}}_{\mathrm{int}\mathrm{H}}}{{\mathrm{N}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right) \\ \left(\frac{\mathrm{V}{\mathrm{Z}}_{{\mathrm{inH}}_2}}{{\mathrm{N}}_{{\mathrm{H}}_2}{\mathrm{v}}_{\mathrm{Q},{\mathrm{H}}_2}}\right){\mathrm{e}}^{\mathrm{Î\mu }/\mathrm{KT}}={\left(\frac{{\mathrm{VZ}}_{\mathrm{int}\mathrm{H}}}{{\mathrm{N}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right)}^2.....................\left(1\right) \end{gathered}$$

we can write partition function for any molecules by considering there rotational and vibrational factors.

So, The hydrogen molecules has partition function as

${\mathrm{Z}}_{\mathrm{int},{\mathrm{H}}_2}=4{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}.............................\left(2\right)$

But the partition function for the hydrogen atom will be due to electron and nucleons spin is given by

${\mathrm{Z}}_{\mathrm{int},\mathrm{H}}=\left(2\right)\left(2\right)=4$......................................(3)

We know that here the quantum volume ${\mathrm{v}}_{\mathrm{Q}}$will be proportional to mass of hydrogen atom by power of minus of $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2.$}\right.$

Then,

localid="1650200789978" $$\begin{gathered} \frac{{\mathrm{v}}_{\mathrm{Q},{\mathrm{H}}_2}}{{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}={\left(\frac{2m}{m}\right)}^{-\frac{3}{2}} \\ {v}_{Q,H_2}=2^{\frac{-3}{2}}{v}_{Q,H}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}\mathit{4}\mathit{\right)} \end{gathered}$$

Substitute this all the value of partition function (2),(3)and quantum volume (4) in above equation (1)

$\left(\frac{4\mathrm{V}{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}}{{\mathrm{N}}_{{\mathrm{H}}_2}{2}^{\frac{-3}{2}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right){\mathrm{e}}^{\mathrm{Î\mu }/\mathrm{KT}}={\left(\frac{4\mathrm{V}}{{\mathrm{N}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right)}^2$

Since using thermodynamics we know that,

Substitute it above

$$\begin{gathered} \left(\frac{4\mathrm{KT}{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}}{{\mathrm{P}}_{{\mathrm{H}}_2}{2}^{\frac{-3}{2}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right){\mathrm{e}}^{\mathrm{Î\mu }/\mathrm{KT}}={\left(\frac{4\mathrm{KT}}{{\mathrm{P}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right)}^2 \\ \left(\frac{{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}}{{\mathrm{P}}_{{\mathrm{H}}_2}{2}^{\frac{-3}{2}}}\right){\mathrm{e}}^{\mathrm{Î\mu }/\mathrm{KT}}=\left(\frac{4\mathrm{KT}}{{\mathrm{P}}_{\mathrm{H}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}\right) \\ \frac{{\mathrm{P}}_{\mathrm{H}}^2}{{\mathrm{P}}_{{\mathrm{H}}_2}}=\frac{\sqrt{2}{\mathrm{KTe}}^{-\frac{\mathrm{Î\mu }}{\mathrm{KT}}}}{{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}}...........................\left(5\right) \end{gathered}$$

Since in the question is given

$$\begin{gathered} \mathrm{K}=\frac{{\mathrm{P}}_{\mathrm{H}}^2}{{\mathrm{P}}^0{\mathrm{P}}_{{\mathrm{H}}_2}} \\ \mathrm{K}=\frac{\sqrt{2}{\mathrm{KTe}}^{\frac{-\mathrm{Î\mu }}{\mathrm{kT}}}}{{\mathrm{P}}^0{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}} \\ \mathrm{K}={\mathrm{K}}_0\left(\mathrm{T}\right)\frac{{\mathrm{e}}^{-\raisebox{1ex}{$\mathrm{Î\mu }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kT}$}\right.}}{{\mathrm{Z}}_{\mathrm{rot}}{\mathrm{Z}}_{\mathrm{vib}}} \end{gathered}$$

Here

localid="1650201682594" $$\begin{gathered} {\mathrm{K}}_0\left(\mathrm{T}\right)=\frac{\sqrt{2}\mathrm{kT}}{{\mathrm{P}}^0{\mathrm{v}}_{\mathrm{Q},\mathrm{H}}} \\ {\mathrm{v}}_{\mathrm{Q}}={\left(\frac{{\mathrm{h}}^2}{2\mathrm{Ï€³¾°ì°Õ}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\mathrm{K}}_0\left(\mathrm{T}\right)=\frac{\sqrt{2}{\mathrm{kT}}^{{\displaystyle \frac{5}{2}}}}{{\mathrm{P}}^0}{\left(\frac{{\mathrm{h}}^2}{2\mathrm{Ï€³¾°ì°Õ}}\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\mathrm{K}}_0\left(\mathrm{T}\right)=\frac{\sqrt{2}k{T}^{\frac{5}{2}}}{{\mathrm{P}}^0}{\left(\frac{2\mathrm{Ï€³¾}}{{\mathrm{h}}^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Since we know the mass of hydrogen atom , then the value of this equation will be,

$$\begin{gathered} {\mathrm{K}}_0\left(\mathrm{T}\right)=\frac{\sqrt{2}(1.38×{10}^{-23}\mathrm{J}/\mathrm{K})(300\mathrm{K}){\mathrm{t}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{10}^5\mathrm{pascal}}{\left(\frac{2\mathrm{π}(1.66×{10}^{-27}\mathrm{kg})}{{(6.62×{10}^{-34}\mathrm{J}.\mathrm{s})}^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\mathrm{K}}_0\left(\mathrm{T}\right)=57170{\mathrm{t}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

But we know that

$\mathrm{t}=\frac{\mathrm{T}}{300\mathrm{K}}$

then,

$$\begin{gathered} \mathrm{K}=\left(571707{\mathrm{t}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\frac{{\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{Î\mu }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{KT}$}\right.}}{{\mathrm{Z}}_{\mathrm{vib}}{\mathrm{Z}}_{\mathrm{rot}}} \\ \mathrm{K}=\left(571707{\mathrm{t}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\frac{{\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{Î\mu }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{KT}$}\right.}}{{\mathrm{Z}}_{\mathrm{vib}}{\displaystyle \frac{\mathrm{kT}}{2\mathrm{Î\mu }}}} \\ \mathrm{K}=\left(571707{\left(\mathrm{t}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\frac{\mathrm{e}\raisebox{1ex}{$-\mathrm{Î\mu }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{KT}$}\right.}{\mathrm{Z}} \end{gathered}$$_vib(1.7t)K=(33592t32)e-εKTZvib

The partition function for the the vibrational factor is given by

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{vib}}=\frac{1}{1-{\mathrm{e}}^{{\displaystyle \raisebox{1ex}{$-\mathrm{Î\mu }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kT}$}\right.}}} \\ {\mathrm{Z}}_{\mathrm{vib}}=\frac{1}{1-{\mathrm{e}}^{{\displaystyle \raisebox{1ex}{$-0.44\mathrm{eV}$}\!\left/ \!\raisebox{-1ex}{$(8.02×{10}^{-5}\mathrm{eV}/\mathrm{K})(300\mathrm{K})\mathrm{t}$}\right.}}} \\ {\mathrm{Z}}_{\mathrm{vib}}=\frac{1}{1-{\mathrm{e}}^{-{\displaystyle \raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.}}}....................\left(6\right) \end{gathered}$$

Then the K value will becomes

$\mathrm{K}=(333592.{\mathrm{t}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})(1-{\mathrm{e}}^{\raisebox{1ex}{$-17$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.}){\mathrm{e}}^{\raisebox{1ex}{$-173.6$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.}$

For temperature$300\mathrm{K}$, the K value will be

localid="1650202838856" $$\begin{gathered} \mathrm{K}=(333592.1^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})(1-{\mathrm{e}}^{\raisebox{1ex}{$-17$}\!\left/ \!\raisebox{-1ex}{$1$}\right.}){\mathrm{e}}^{\raisebox{1ex}{$-173.6$}\!\left/ \!\raisebox{-1ex}{$1$}\right.} \\ {\mathrm{K}}_{300}=1.357×{10}^{-71} \end{gathered}$$

Similarly for temperature $1000\mathrm{K}$,

$\mathrm{t}=\frac{1000\mathrm{k}}{300\mathrm{k}}=\frac{10}{3}$

$$\begin{gathered} \mathrm{K}=(333592.{\left(\frac{10}{3}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})(1-{\mathrm{e}}^{\raisebox{1ex}{$-17×3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}){\mathrm{e}}^{\raisebox{1ex}{$-173.6×3$}\!\left/ \!\raisebox{-1ex}{$1$}\right.0} \\ {\mathrm{K}}_{1000}=4.87×{10}^{-18} \end{gathered}$$

For temperature $300\mathrm{K}$,

localid="1650203019179" $$\begin{gathered} \mathrm{t}=\frac{3000\mathrm{K}}{300\mathrm{K}}=10 \\ \mathrm{K}=(333592.{\left(10\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})(1-{\mathrm{e}}^{\raisebox{1ex}{$-17$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}){\mathrm{e}}^{\raisebox{1ex}{$-173.6$}\!\left/ \!\raisebox{-1ex}{$1$}\right.0} \\ {\mathrm{K}}_{3000}=0.025 \end{gathered}$$

for temperature $6000\mathrm{K}$,

$$\begin{gathered} \mathrm{t}=\frac{6000\mathrm{K}}{300\mathrm{K}}=20 \\ \mathrm{K}=(333592.{\left(20\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})(1-{\mathrm{e}}^{\raisebox{1ex}{$-17$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}){\mathrm{e}}^{\raisebox{1ex}{$-173.6$}\!\left/ \!\raisebox{-1ex}{$2$}\right.0} \\ {\mathrm{K}}_{6000}=292.5 \end{gathered}$$

Every molecule of hydrogen dissociates into two atoms only when an energy equal to or greater than the dissociation energy is supplied. if the temperature and pressure are not such that so that it can increases its energy in sufficient amount then it will not dissociate.

There are some example where the hydrogen mostly dissociate is given by,

${\mathrm{H}}_2+{\mathrm{I}}_2→2\mathrm{HI}$

where hydrogen is dissociate at high temperature and pressure.

whenever $\mathrm{HCl}$ will dissolved into water to form hydrochloric acid, most of its molecules of hydrogen and chlorine dissociate.