Equilibrium constant K _ C for the following reaction at 800 K is,...

{\text{NH}}_{\text{3}} \text{(g)} ⇌ \dfrac{\text{1}}{\text{2}} {\text{N}}_{\text{2}} \text{(g)}\text{+}\dfrac{\text{3}}{\text{2}} {\text{H}}_{\text{2}} \text{(g);}{\text{K}}_{\text{C}} \text{=}\text{4}$<br />${N}_{2} ( g ) + 3{H}_{2} ( g ) ⇌ 2N{H}_{3} ( g ) \Delta{n}_{g} = 2 - 4 = - 2$<br />So${K}_{C}$for the following reaction will be equal to$\dfrac{1}{{4}^{2}}$<br />${K}_{P} = {K}_{C} ( RT{)}^{\Delta{n}_{g}}$<br />$\text{=}\dfrac{\text{1}}{{\text{4}}^{\text{2}}} \text{\times}{\text{(800}\text{\times}\text{R)}}^{- \text{2}}$<br />$= {\left[\dfrac{1}{4 \times 800\text{R}}\right]}^{2}