Correct Answer - C
`K_(p)=K_(c )(RT)^(Deltan_(g))`
or `(K_(p))/(K_(c ))=(RT)^(Deltan_(g))`
Now, `K_(p) gt K_(c) ( `given)
`implies (K_(p))/(K_(c)) gt 1`
`(RT)^(Deltan_(g)) gt1`
At `T gt 15K`
`RT gt 0.0821 xx 15 L ` atm `mol^(-1)`
`RT gt 1.22 L` atm `mol^(-1)`
`implies Delta n_(g) gt 0`
`implies ` No. of moles of gaseous products `gt `No. of moles of gaseous reatants
Such a reaction in the forward direction will favoured by low pressure.
