The value of $K_c$ for the following equilibrium is
$ \mathrm{CaCO}_{3(s)} \rightleftharpoons \math

Question

The value of $K_c$ for the following equilibrium is 
$ \mathrm{CaCO}_{3(s)} \rightleftharpoons \mathrm{CaO}_{(s)}+\mathrm{CO}_{2(g)} . $
Given : $K_p=167$ bar at 1073 K .

Professor PC Thomas Classes · Answer

$\begin{aligned} & (\mathrm{a}): K_p=K_c(R T)^{\Delta n_g} ; \Delta n_g=1 \ & K_p=167 \mathrm{bar}, \ & K_c=\frac{167 \mathrm{bar}}{0.0831 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} 	imes 1073 \mathrm{~K}}=1.873 \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$