In the thermal dissociation of $PCl_5$,the total pressure in the gaseous equilibrium mixture is $1.0 \ atm$ when half of $PCl_5$ is found to dissociate. The equilibrium constant of the reaction $(K_p)$ in atmosphere is

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 $1073 \ K$

$5.1 \ g$ $NH_4SH$ is introduced in a $3.0 \ L$ evacuated flask at $327 \ ^\circ C$. $30\%$ of the solid $NH_4SH$ decomposes into $NH_3$ and $H_2S$ gases. The $K_p$ of the reaction at $327 \ ^\circ C$ is ($R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,molar mass of $S = 32 \ g \ mol^{-1}$,molar mass of $N = 14 \ g \ mol^{-1}$)

In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,if at equilibrium there are $2 \ mol$ each of $PCl_5$,$PCl_3$,and $Cl_2$,and the total pressure is $3 \ atm$,then the value of the equilibrium constant $K_p$ is ....... $atm$.

$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
The reaction was started with some amount of $NH_4HS$. The equilibrium pressure at $25^{\circ}C$ is $0.5 \ atm$. What is $K_p$ for the reaction (in $atm^2$)?

The ratio $\frac{K_p}{K_C}$ for the reaction: $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$ is: