Consider the reaction $C{O}_2(g)+H_2(g)harrCO(g)+H_{2}O(g)$ at $1$ bar and $900K,$ as well as the following data:

The partition functions of $H_2$ and $CO$ :

$q_{\text{r}\text{o}\text{t}}(H_2)=5\mathrm{.}1397$;$q_{\text{r}\text{o}\text{t}}(CO)=323\mathrm{.}90$

The rotational constants of $C{O}_2($linear$)$ and $H_{2}O$ :

$B(C{O}_2)=0\mathrm{.}3902cm-1$

$A(H_{2}O)=27\mathrm{.}877c{m}^{-1}$;$B(H_{2}O)=14\mathrm{.}512c{m}^{-1}$;$C(H_{2}O)=9\mathrm{.}285c{m}^{-1}$

The difference in electronic energy:

$E_r=D_e(C{O}_2)+D_e(H_2)-D_e(CO)-D_e(H_{2}O)=-29\mathrm{.}89k\frac{J}{m}ol$

a$.$ Calculate the rotational contribution to the equilibrium constant $K_c.$

b$.$ Compute the electronic contribution to the equilibrium constant $K_c$

please show the calculations and the results. maybe handwritten