Question 10.2: The reaction A(ν, J) + M ←→ A(ν, J - ΔJ) + M describes rotat...
The reaction A(ν, J) + M \rightleftarrows A(ν, J – ΔJ) + M describes rotational relaxation for a generic diatomic molecule from rotational level, J, to rotational level, J − ΔJ, upon collision with the nonreactive species, M. Employing the simplex model, show that the equilibrium constant based on number density for this energy-transfer reaction is given by
K_{c}=\left(1-\frac{ 2\Delta J}{g_{J}} \right) exp\left[\frac{ \theta_{r}\Delta J}{T}\left(g_{J}-\Delta J\right) \right] ,
where the rotational degeneracy is g_{J} = 2J + 1.
From Eq. (10.59), the equilibrium constant for this energy-transfer reaction is
K_{c} = \prod\limits_{i}{n^{\nu i}} = \prod\limits_{i}{\phi^{\nu_{i}}_{i}}\exp \left(\frac{\sum_{i}\nu_{i}D_{\circ i}}{kT}\right). (10.59)
K_{c} = \frac{n(\nu,J-\Delta J)}{n(\nu,J)}.
Now, from Eq. (4.14), the population ratio between the number density associated specifically with (ν, J) and the total number density describing all possible rotational and vibrational levels is
\frac{N_{j}}{N} = \frac{g_{j}e^{-\varepsilon_{j}/kT}}{Z}, (4.14)
\frac{n(\nu,J)}{n} = \frac{g_{J}}{Z_{rot}Z_{vib}}\exp \left[-\frac{J(J+1)\theta_{r}+\nu\theta_{\nu}}{T}\right].
Similarly, for the number density associated specifically with (ν, J − ΔJ ) , we have
\frac{n(\nu,J-\Delta J)}{n} = \frac{g_{J-\Delta J}}{Z_{rot}Z_{vib}}\exp\left[-\frac{(J-\Delta J)(J-\Delta J+1 )\theta_{r}+\nu\theta_{\nu}}{T}\right].
Dividing the latter by the former, we obtain for the equilibrium constant
K_{c} = \frac{n(\nu,J-\Delta J)}{n(\nu,J)} = \frac{2(J-\Delta J)+1}{2J+1}\times\exp\left\{\frac{\left[J(J+1)-(J-\Delta J)(J-\Delta J +1)\right]\theta_{r}}{T}\right\}
so that
K_{c} = \frac{2J+1-2\Delta J}{2J+1}\exp\left[\frac{(2J+1-\Delta J)\Delta J\theta_{r}}{T} \right].
On this basis, we find that
K_{c} = \left(1-\frac{2\Delta J}{g_{J}}\right)\exp\left[\frac{\theta_{r}\Delta J}{T}(g_{J}-\Delta J )\right],
which verifies the given expression for the equilibrium constant.