How do you estimate the average kinetic energy of #"2 mols"# of #"O"_2# gas at #"298 K"#?

We need to consider what kinds of kinetic energy contributions there are... and once we do, we should get #"12.4 kJ"#.

And the average kinetic energy at the high-temperature limit is given by the equipartition theorem:

#<< kappa >> = << kappa >>_(tr) + << kappa >>_(rot) + << kappa >>_(vib) + . . . #

#= K/n = N/2 RT#

where:

• #<< kappa >>_(k)# is the average kinetic energy contribution by the #k#th type of motion. This energy is in #bb"J/mol"#.
• #K# is the average kinetic energy in #"J"#.
• #N# is the number of degrees of freedom for that kind of motion.
• #R = "8.314472 J/mol"cdot"K"# is the universal gas constant.
• #T# is the temperature in #"K"#.

The high-temperature limit is simply the temperature where the molecule is actually accessing certain energy levels that allow it to move a certain way.

Breaking it down...

• For translation, #"O"_2# molecules move in #3# dimensions. Thus, #N_(tr) = 3#. The high-temperature limit pretty much always applies here.
• For rotation, you need two angles to define a linear rotating molecule, so #N_(rot) = 2#. The high-temperature limit applies above approximately #"10 K"#.
• For vibration, we don't have to worry about it, and we will show why. Call it #N_(vib) ~~ 0# at #"298 K"#.

We can verify above what temperatures we can use #N/2RT# by using the following:

#Theta_(rot) = (tildeB_e)/(k_B) stackrel(?)(" << ") T#

#Theta_(vib) = (tildeomega_e)/(k_B) stackrel(?)(" << ") T#

where these are the rotational and vibrational temperatures, respectively.

• #tildeB_e# is the rotational constant.
• #omega_e# is the fundamental frequency of vibration.
• #k_B ~~ "0.695 cm"^(-1)"/K"# is the Boltzmann constant.

These data are readily available on NIST for diatomic molecules:

#tildeB_e = "1.4376766 cm"^(-1)#
#tildeomega_e = "1580.193 cm"^(-1)#

These temperatures are then:

#Theta_(rot) = ("1.4376766 cm"^(-1))/("0.695 cm"^(-1)"/K") = "2.07 K"#

#Theta_(vib) = "1580.193 cm"^(-1)/("0.695 cm"^(-1)"/K") = "2273.7 K"#

Since we are well above #"2.07 K"# but well below #"2273.7 K"#, we are to include #N_(rot) = 2# and ignore #N_(vib)# completely.

Therefore, the average kinetic energy at #"298 K"# is around:

#color(green)(<< kappa >>) = N/2RT#

#= N_(tr)/2RT + N_(rot)/2RT#

#= 3/2RT + 2/2RT = color(green)(5/2RT)# #" J/mol"#

And we then get:

#color(green)(<< kappa >>) = 5/2 cdot "8.314472 J/mol"cdot"K" cdot "298 K"#

#=# #"6194.3 J/mol"#

#=# #color(green)("6.19 kJ/mol")#

Since you want #"2 mols"# of #"O"_2#...

#color(blue)(K) = n<< kappa >>#

#= "2 mols" cdot "6.19 kJ/mol"#

#=# #color(blue)("12.4 kJ")#