Answer
$1.76\times10^{20}$
Work Step by Step
$K_{c} = 2.24\times10^{22}$
R = $\frac{0.0821 L.atm}{K.mol}$ (Universal gas constant)
T = (1273+273) K = 1546 K
$\Delta n$ = 2 - 2 - 1 = -1
Substituting the values in the relation,
$K_{p} = K_{c}(RT)^{\Delta n} = (2.24\times10^{22})(\frac{0.0821L.atm}{K.mol}\times 1546 K)^{-1} = 1.76\times10^{20}$
