Chemistry - Chemical Equilibrium Question with Solution | TestHub

$2\mathrm{\text{N}}{\mathrm{\text{O}}}_3\left(\mathrm{\text{g}}\right)\rightleftharpoons 2\mathrm{\text{N}}\mathrm{\text{O}}\left(\mathrm{\text{g}}\right)+{\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)$

${\text{K}}_{\text{c}}=1.8\times {10}^{-6}\mathrm{\text{a}}\mathrm{\text{t}}184\mathrm{\text{℃}}(=457\mathrm{\text{K}})$

$\text{R}=0.00831\mathrm{\text{k}}\mathrm{\text{J}}\mathrm{\text{m}}\mathrm{\text{o}}{\mathrm{\text{l}}}^{-1}{\mathrm{\text{K}}}^{-1}$

${\text{K}}_{\text{p}}={\text{K}}_{\text{c}}{\left(\text{R}\text{T}\right)}^{∆{\text{n}}_{\mathrm{\text{g}}}}$

Where,

$∆{\text{n}}_{\mathrm{\text{g}}}=$(gaseous products - gaseous reactants)

=3-2=1

$\therefore {\text{K}}_{\text{p}}=1.8\times {10}^{-6}\times 0.00831\times 457$

$=6.836\times {10}^{-6}>1.8\times {10}^{-6}$

Thus,${\text{K}}_{\text{p}}>{\text{K}}_{\text{c}}$