Find out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$; $K_{p} = 1.8 \times 10^{-2}$ at $500 \ K$
$(ii)$ $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$; $K_{p} = 167$ at $1073 \ K$

(N/A) The relation between $K_{p}$ and $K_{c}$ is given by the formula:
$K_{p} = K_{c}(RT)^{\Delta n}$
$(i)$ For the reaction $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$:
$\Delta n = (2 + 1) - 2 = 1$
$R = 0.0831 \ \text{bar L K}^{-1} \text{mol}^{-1}$
$T = 500 \ K$
$K_{p} = 1.8 \times 10^{-2}$
$1.8 \times 10^{-2} = K_{c}(0.0831 \times 500)^{1}$
$K_{c} = \frac{1.8 \times 10^{-2}}{41.55} = 4.33 \times 10^{-4}$
$(ii)$ For the reaction $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$:
$\Delta n = 1 - 0 = 1$
$R = 0.0831 \ \text{bar L K}^{-1} \text{mol}^{-1}$
$T = 1073 \ K$
$K_{p} = 167$
$167 = K_{c}(0.0831 \times 1073)^{1}$
$K_{c} = \frac{167}{89.1663} \approx 1.87$