## Chemistry (4th Edition)

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