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
Truong-Son N.
Apr 15, 2017

Consider a general diatomic gas, #AB(g)#, in the following dissociation reaction:

#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 #AB#, it has six DOFs. Hence, #AB(g)# has #K_(avg) = 1/2nRT# for each DOF, totalling #K_(avg) = 3nRT#.

For monatomic gases, they only have translational DOFs, so their kinetic energies are assumed to be #3/2nRT#.

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 #AB(g)# were properly conserved and transferred into the individual atomic gases #A(g)# and #B(g)# as the diatomic molecular gas #AB(g)# dissociated.

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