Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\text{NH}}_{\text{3}}\text{(g)}\rightleftharpoons \frac{\text{1}}{\text{2}}{\text{N}}_{\text{2}}\text{(g)}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\text{(g);}{\text{K}}_{\text{C}}\text{=}\text{4}$
${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{N}{\mathrm{H}}_3\left(\mathrm{g}\right)$$\mathrm{\Delta }{\mathrm{n}}_{\mathrm{g}}=2-4=-2$
So${\mathrm{K}}_{\mathrm{C}}$for the following reaction will be equal to$\frac{1}{4^2}$
${\mathrm{K}}_{\mathrm{P}}={\mathrm{K}}_{\mathrm{C}}(\mathrm{R}\mathrm{T}{)}^{\mathrm{\Delta }{\mathrm{n}}_{\mathrm{g}}}$
$\text{=}\frac{\text{1}}{{\text{4}}^{\text{2}}}\text{×}{\text{(800}\text{×}\text{R)}}^{-\text{2}}$
$={\left[\frac{1}{4\times 800\text{R}}\right]}^2$