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\heading{E4211: Chemical potential: the law of mass action} 
\auname{Shay Izbitski}


{\bf The problem:}

\Dn


(1) Consider ideal gases of atoms ${A}$, atoms ${B}$ and atoms ${C}$
undergoing the reaction
${\nu C  \leftrightarrow A+B}$       (${\nu}$ is an integer).
${n_{A}, n_{B}}$ and ${n_{C}}$ denote the respective densities.
determine the values of the exponents ${a,b}$ and ${c}$ in the low of mass action
\[n^{a}_{A}n^{b}_{B}n^{c}_{C}=K(T)
\]
The quantity ${K\left(T\right)}$ is known as the \emph{equilibrium constant}  of the reaction.

(2) Write explicit expression for ${K(T)}$ for the reaction ${H_{2}+D_{2}\leftrightarrow  2 HD}$
when the masses $m_H, \, m_D$ and  $\omega_0$ the vibrational
frequency of ${HD}$ are given. assume the temperature is high enough to allow
classical approximation of the rotational motion.

(3) What is ${K(\infty)}$?\\

\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
We first realize that the chemical potentials must obey the reaction by

\[\mu_{A}+\mu_{B}=\nu\mu_{C}
\]

We now take the chemical potential of an ideal gas ${\mu=\tau lan(\frac{1}{\lambda^{3}_{T}n})}$ and substituting it into the former equation resulting with

\[\tau lan(\frac{1}{\lambda^{3}_{TA}n_A})+\tau lan(\frac{1}{\lambda^{3}_{TB}n_B})=\nu\tau lan(\frac{1}{\lambda^{3\nu}_{TC}n^{\nu}_{C}})
\]

From here we derive the law of mass action and we now can determine the values of ${a,b}$ and ${c}$

\[n^{-1}_{A}n^{-1}_{B}n^{\nu}_{C}=V^{2-\nu}\frac{z^{\nu}_{1C}}{z_{1A}z_{1B}}=\frac{\lambda^{-3\nu}_{TC}}{\lambda^{-3}_{TA}\lambda^{-3}_{TB}}
\]


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\Dn

(2)
We'll look at the hamiltonian of a diatomic molecule with vibrational and rotational degrees of freedom

\[H=\frac{p^2}{2m}+\hbar\omega_0(n+\frac{1}{2})+\frac{l(l+1)\hbar^2}{2I}
\]

Now we'll calculate a single diatomic hetronuclear molecule partition function

\[z_1=\frac{V}{\lambda^{3}_{T}}\sum_{n}e^{-\beta\hbar\omega_0(n+\frac{1}{2})}\sum_{l}(2l+1)e^{-\beta\frac{l(l+1)\hbar^2}{2I}}
\]

The first sum can be summarized easily while the second one can be summarized under the approximation of the classical limit

\[\sum_{l}(2l+1)e^{-\beta\frac{l(l+1)\hbar^2}{2I}}\cong\int^{\infty}_{0}e^{-\beta\frac{l(l+1)\hbar^2}{2I}}d[l(l+1)]=\frac{2I}{\beta\hbar^2}
\]

We end up with

\[z_1=\frac{V}{\lambda^{3}_{T}}\cdot\frac{e^{-\frac{1}{2}\beta\hbar\omega_0}}{1-e^{-\beta\hbar\omega_0}}\cdot\frac{2I}{\beta\hbar^2}
\]

This calculation allows us to find ${z_{1HD}}$.${z_{1HH}}$ and ${z_{1DD}}$ are single diatomic homonuclear molecule partition functions, they differ from the ${z_{1HD}}$ case in the amount of degenerated states of the rotation. the amount of degenerated states are ${l+1}$ instead of ${2l+1}$
resulting in a factor ${\frac{1}{2}}$, substituting everything together in the law of mass action gives rise to the \emph{equilibrium constant}  of the reaction.


%%%%%%%%%%%%%%
\Dn

(3)
Under the approximation of a very high temperature we can expand the vibrational term into ${\frac{1}{\beta\hbar\omega_0}}$, in addition we recall that ${\frac{1}{\lambda^{3}_{T}}=(\frac{m}{2\pi\beta\hbar^2})^{\frac{3}{2}}}$. that leads us into the next 
expression for the single diatomic hetronuclear molecule partition function

\[z_1=const\cdot m^\frac{3}{2}\frac{I}{\omega_0}
\]

The kinetic term is related with the translational mass ${m_t=m_A+m_B}$ and the vibrational and
rotational term is related with the reduced mass ${m_r=\frac{m_Am_B}{m_A+m_B}}$. we recall on the relation between ${I}$ and ${\omega_0}$ and the mass, ${I=m_rr^2}$ and ${\omega_0=\sqrt{\frac{k}{m_r}}}$ and get

\[z_1=const\cdot m^{\frac{3}{2}}_{t}m^{\frac{3}{2}}_{r}=const\cdot (m_Am_B)^{\frac{3}{2}}
\]

All we have left to do is substituting ${z_{1HH},z_{1DD}}$ and ${z_{1HD}}$ into the law of mass action
and get

\[K(\infty)=\frac{z^{2}_{1HD}}{z_{1HH}z_{1DD}}=\frac{const^2\cdot (m_Hm_D)^3}{const\cdot (m_Hm_H)^{\frac{3}{2}}\cdot\frac{1}{2}const\cdot (m_Dm_D)^{\frac{3}{2}}\cdot\frac{1}{2}}=4
\]


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\end{document} 
