The equilibrium for the reaction

The equilibrium for the reaction

Question:

Numerical
Constants: 
a.m.u. = 1.660 *10-27 kg
kB = 1.3806 *10−23 JK−1=0.69503 cm−1K−1
ħ= 1.05457 *10−34 Js
c = 2.9979 *108ms−1
NA = 6.022 *1023
R = 8.314 J mol−1 K−1
Consider the equilibrium for the reaction H2 (g) +D2 (g) ↔2HD (g). Here D indicates Deuterium 
(nuclear spin =1) where we will use mD=mH, mH=1 a.m.u. The nuclear spin of H=1/2. We
consider the equilibrium at room temperature (T=300K) and you can use the high temperature
formula for the rotational partition function. Assume that the rigid rotor and harmonic
approximations can be used. The equilibrium distance, R=0.74*10-10m for each species is the
same while also the force-constant and the electronic energy is the same for each species. The
vibration frequency (or better wave number) for H2 is ωH2=4400cm-1.
(a)
(b)
(c)
(d)
(e)
(f)
(g) use the pieces under (c) to (f), what is the numerical value of the equilibrium constant for the
reaction at T=300K?