Question

For reaction : $SO _2( g )+\frac{1}{2} O _2( g )= SO _3( g )$ $K _{ p }=2 \times 10^{12}$ at $27^{\circ} C$ and 1 atm pressure. The $K _{ c }$ for the same reaction is ____$\times 10^{13}$. (Nearest integer) (Given $R =0.082 \,L \,atm \,K ^{-1} \,mol ^{-1}$ )

Answer 1

$SO _{2( g )}+\frac{1}{2} O _{2( g )} \rightleftharpoons SO _{3( g )} $
$ K _{ p }=2 \times 10^{12} \text { at } 300 K $
$ K _{ p }= K _{ C } \times( RT )^{\Delta n _{ g }} $
$ 2 \times 10^{12}= K _{ c } \times(0.082 \times 300)^{-1 / 2} $
$ K _{ C }=9.92 \times 10^{12} $
$ K _{ C }=0.992 \times 10^{13}$
$= 1$