Answer
$3.5\times10^{-7}$
Work Step by Step
$K_{p} = 1.8\times10^{-5}$
T = (350+273) K = 623 K
$\Delta n$ = (2+1)-2 = 1
$K_{p} = K_{c} (RT)^{\Delta n}$
Rearranging the equation,
$K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{1.8\times10^{-5}}{0.08206L.atm.K^{-1}mol^{-1}\times 623 K} = 3.5\times10^{-7}$
