Answer
c) 0.14
Work Step by Step
$K_{c} = 1.1\times10^{-3}$
$R = \frac{0.08206L.atm}{K.mol}$
T = (1280+273) K = 1553 K
$ \Delta n = 2-1 =1 $
Using the relation, $K_{p} = K_{c}(RT)^{\Delta n}$
We get, $K_{p} = 1.1\times10^{-3} (\frac{0.08206L.atm}{K.mol} \times 1553 K) \approx 0.14$
