Final answer:
The rotational partition function for SO2 at 298 K is calculated using the rotational constants for the molecule, which are 2.03, 0.344, and 0.293 cm−¹, and the temperature of 298 K. The resulting function describes the distribution of rotational energy levels at thermal equilibrium.
Step-by-step explanation:
To calculate the rotational partition function for SO2 at 298 K, we use the rotational temperatures, θr, for a non-linear molecule which is given by:
θr = βhc/k
where β is the rotational constant in cm−1, h is Planck's constant, c is the speed of light, and k is Boltzmann's constant.
The rotational partition function, Ξrot, for a non-linear molecule is approximated by:
Ξrot = π^(3/2)/(ΣβAβBβC)^(1/2) * T^3
Substituting the rotational constants for SO2: βA = 2.03 cm−1, βB = 0.344 cm−1, βC = 0.293 cm−1, and the temperature (298 K), we get:
Ξrot = π^(3/2)/((2.03)*(0.344)*(0.293))^(1/2) * (298)^3
Calculating the above expression will yield the rotational partition function for SO2 at 298 K.