---A---

vanvleck 2093$ calc
* let kb= 8.6173e-5
* let T=25+273
* ? exp(-.37/(kb*T))
    .5527555774189063E-06
* ? exp(-.027/(kb*T))
    .3494411447887067    

HCl will not be vibrationally active at room temperature; I2 will be

---B---

Sum[Exp[-e n/(k T)],{n,0,Infinity}]
avgE=Sum[e n Exp[-e n/(k T)],{n,0,Infinity}]/Sum[Exp[-e n/(k T)],{n,0,Infinity}]
c=D[avgE,T]
na = 6.0221 10^23
e=.01  1.6022  10^-19
k=1.3807 10^-23

Plot[na avgE ,{T,0,300}]
Export["MB_avgE.eps",%]

Plot[na c ,{T,0,300}]
Export["MB_c.eps",%]

----C---

ClearAll[e,k,avgE,c]
z=1+Exp[-e/(k T)]
avgE=e Exp[-e/(k T)]/z

e=.01  1.6022  10^-19
k=1.3807 10^-23


LogLinearPlot[na avgE ,{T,10,1000}]
Export["MBC_avgE.eps",%]

c=D[avgE,T]
LogLinearPlot[na c ,{T,10,1000}]
Export["MBC_c.eps",%]

if e<<kT then the argument of Boltzmann is small and Exp[0]=1, so
both states equally occupied
so avgE=e/2
In[32]:= na*e/2                                                                                                                                   

Out[32]= 482.43

all depends on e<<kT , i.e., T>>e/k