The dissociation of molecular hydrogen into atomic hydrogen, ${\mathrm{H}}_2\to 2\mathrm{H}$can be treated as an ideal gas reaction using the techniques of Section 5.6. The equilibrium constant K for this reaction is defined as

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

where${\mathrm{P}}^0$is a reference pressure conventionally taken to be$1\mathrm{bar},$and the other P's are the partial pressures of the two species at equilibrium. Now, using the methods of Boltzmann statistics developed in this chapter, you are ready to calculate K from first principles. Do so. That is, derive a formula for K in terms of more basic quantities such as the energy needed to dissociate one molecule (see Problem 1.53) and the internal partition function for molecular hydrogen. This internal partition function is a product of rotational and vibrational contributions, which you can estimate using the methods and data in Section 6.2. (An${\mathrm{H}}_2$ molecule doesn't have any electronic spin degeneracy, but an H atom does-the electron can be in two different spin states. Neglect electronic excited states, which are important only at very high temperatures. The degeneracy due to nuclear spin alignments cancels, but include it if you wish.) Calculate K numerically at$\mathrm{T}=300\mathrm{K},1000\mathrm{K},3000\mathrm{K},\mathrm{and}6000\mathrm{K}.$ Discuss the implications, working out a couple of numerical examples to show when hydrogen is mostly dissociated and when it is not.

We have given,

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

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

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

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.

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

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

We know that chemical potential is defined as

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

substitute in above equation,

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

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)

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

Since using thermodynamics we know that,

Substitute it above

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

Since in the question is given

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

Here

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

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

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

But we know that

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

then,

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

The partition function for the the vibrational factor is given by

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

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

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

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

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

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,

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

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.