$\left(R=0.083 \,L \operatorname{bar} \,{K}^{-1} \,{~mol}^{-1}\right)$
\({~K}_{{c}}=\frac{{K}_{{p}}}{{RT}}=\frac{47.9}{0.083 \times 288}=2\)
$\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} \text {. }$
(આપેલ : R = $\left.8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
$Fe _{2} N ( s )+\frac{3}{2} H _{2}( g ) \rightleftharpoons 2 Fe ( s )+ NH _{3}( g )$