# 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?

##### 1 Answer

Consider a general diatomic gas,

#AB(g) -> A(g) + B(g)#

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

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

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

#K_(avg) = N/2nRT# ,where

#N# is the degrees of freedom,#n# is#"mol"# s, and#R# and#T# are as usual from the ideal gas law.

In total, for

For monatomic gases, they only have translational DOFs, so their kinetic energies are assumed to be

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

#DeltaK_(avg) = K_(avg,AB) - [K_(avg,A) + K_(avg,B)]#

#= K_(avg,AB) - 2K_(avg,A)#

#= 3nRT - 2*3/2nRT#

Therefore:

#color(blue)((DeltaK_(avg))/n) = (3-3)RT = color(blue)("0 J/mol")#

This is saying that all the energy that was partitioned into the translational, rotational, and vibrational motions of