${\mathrm{K}}_{\mathrm{p}} \mathrm{is} \mathrm{less} \mathrm{than} {\mathrm{K}}_{\mathrm{C}} \mathrm{in}$

We know that the relation between ${\mathrm{K}}_{\mathrm{p}} \mathrm{and} {\mathrm{K}}_{\mathrm{C}} \mathrm{is} {\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆\mathrm{n}}$

here, $∆\mathrm{n}$ represents the difference in the no. of moles of the gaseous products and reactants.

For a reaction , ${\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}\to 2{\mathrm{NO}}_{2\left(\mathrm{g}\right)}$

$∆\mathrm{n}=$no. moles of gas products- no. moles of gas reactants

$∆\mathrm{n}=2-1=1$

$2{\mathrm{HI}}_{\left(\mathrm{g}\right)}\to {\mathrm{H}}_{2\left(\mathrm{g}\right)}+{\mathrm{I}}_{2\left(\mathrm{g}\right)}$

For a reaction, $2{\mathrm{SO}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{SO}}_{3\left(\mathrm{g}\right)}$

$∆\mathrm{n}=2-(2+1)=-1$

For a reaction, ${\mathrm{N}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{NO}}_{\left(\mathrm{g}\right)}$

$∆\mathrm{n}=2-(1+1)=0$

if $∆\mathrm{n}=-\mathrm{ve}; \mathrm{then} {\mathrm{K}}_{\mathrm{p}}<{\mathrm{K}}_{\mathrm{C}}$

For the reaction in the formation of ${\mathrm{SO}}_3,$we will get $∆\mathrm{n}=-\mathrm{ve}$ 

Therefore, the correct answer is option “C”.