## Chemistry (4th Edition)

$K_{p} = 0.105$ $K_{c} = 2.05\times10^{-3}$
Given, $K_{p}$ = 0.105 T = (350+273) K = 623 K $\Delta n = 1+1-1 = 1$ We know, $K_{p} = K_{c}(RT)^{\Delta n}$ Or $K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{0.105}{\frac{0.08206L.atm}{K.mol}\times623 K} = 2.05\times10^{-3}$