# What is the change in molar kinetic energy for the dissociation of the diatomic gas "AB"(g) into "A"(g) and "B"(g) in the high temperature limit?

Apr 15, 2017

Consider a general diatomic gas, $A B \left(g\right)$, in the following dissociation reaction:

$A B \left(g\right) \to A \left(g\right) + B \left(g\right)$

We say that the diatomic gas has certain degrees of freedom (DOFs):

• $1$ per dimension of translational motion: $x , y , z \implies$ $3$ DOFs
• $1$ per rotational angle: $\theta$ and $\phi$ $\implies$ $2$ DOFs
• $1$ per vibrational motion (there is only one: symmetrical stretch). $\implies$ $1$ DOF

By the equipartition theorem for average kinetic energy, we have:

${K}_{a v g} = \frac{N}{2} n R T$,

where $N$ is the degrees of freedom, $n$ is $\text{mol}$s, and $R$ and $T$ are as usual from the ideal gas law.

In total, for $A B$, it has six DOFs. Hence, $A B \left(g\right)$ has ${K}_{a v g} = \frac{1}{2} n R T$ for each DOF, totalling ${K}_{a v g} = 3 n R T$.

For monatomic gases, they only have translational DOFs, so their kinetic energies are assumed to be $\frac{3}{2} n R T$.

When the temperature does not change, we can write the change in average kinetic energy as:

$\Delta {K}_{a v g} = {K}_{a v g , A B} - \left[{K}_{a v g , A} + {K}_{a v g , B}\right]$

$= {K}_{a v g , A B} - 2 {K}_{a v g , A}$

$= 3 n R T - 2 \cdot \frac{3}{2} n R T$

Therefore:

$\textcolor{b l u e}{\frac{\Delta {K}_{a v g}}{n}} = \left(3 - 3\right) R T = \textcolor{b l u e}{\text{0 J/mol}}$

This is saying that all the energy that was partitioned into the translational, rotational, and vibrational motions of $A B \left(g\right)$ were properly conserved and transferred into the individual atomic gases $A \left(g\right)$ and $B \left(g\right)$ as the diatomic molecular gas $A B \left(g\right)$ dissociated.