Question 10.3: Calculate the equilibrium constant based on pressure for the...
Calculate the equilibrium constant based on pressure for the reaction, N_{2} + M \rightleftarrows N + N + M, at 3000 K.
From Eq. (10.64), the equilibrium constant is
K_{p} = \frac{(\pi m_{A})^{{3}/{2}}}{h^{3}P_{\circ}}(kT)^{{5}/{2}}\left(\frac{Z^{2}_{A,int}}{Z_{A_{2},int}}\right)e^{-D_{\circ}/kT}. (10.64)
K_{p} = \frac{(\pi m_{N})^{{3}/{2}}}{h^{3}P_{\circ}}(kT)^{{5}/{2}}\left(\frac{Z^{2}_{N,int}}{Z_{N_{2},int}}\right)e^{-D_{\circ}/kT}.
Now, from Examples 9.2 and 9.5, the internal contributions to the partition function at 3000 K are
Z_{N,int} = Z_{N,el} = 4.0010
Z_{N_{2},int} = Z_{el}Z^{\circ}_{R-V}Z_{corr} = (1.0000)(778.80) (1.0139) = 789.63.
From Appendix K.1, the dissociation energy for molecular nitrogen is
D_{\circ} = 9.759 eV (1.602 × 10^{-19}J/eV) = 1.5634 × 10^{-18}J.
Hence, we have
\left(\frac{Z^{2}_{N,int}}{Z_{N_{2},int}}\right)e^{-D_{\circ}/kT} = \frac{(4.0010)^{2}}{(789.63)}\exp\left[-\frac{1.5634\times10^{-18}}{(1.3807\times10^{-23})(3000)}\right] = 8.2211\times10^{-19}.
Evaluating the equilibrium constant, we thus obtain
K_{p} = \frac{\left[\pi (14.0067)\left(1.6605\times10^{-27}\right)\right]^{{3}/{2}}}{\left(6.6261\times10^{-34}\right)^{3}(1.0\times 10^{5})}\left[(1.3807\times10^{-23})(3000)\right]^{{5}/{2}}\times\left(8.2211\times10^{-19}\right) = 1.949\times10^{-10}.