For the equilibrium $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)} + Cl_{2(g)}$,what is the temperature at which $\frac{K_p}{K_c} = \frac{1}{3}$? (Given $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$)

In which of the following equilibria,the value of $K_p$ is less than $K_c$?

If $K_C$ for the equilibrium reaction $2 ABC_{(g)} \rightleftharpoons 2 AB_{(g)} + C_{2(g)}$ is $X$ at $T \ K$,its $K_P$ at the same temperature is:

$XY_2$ dissociates as $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$. When the initial pressure of $XY_2$ is $600 \ mm \ Hg$,the total equilibrium pressure is $800 \ mm \ Hg$. Calculate $K_p$ for the reaction,assuming the volume of the system remains unchanged.

$2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$ has $K_p = 0.15 \, atm$ at $298 \, K$. Calculate $K_p$ in $torr$ and $K_c$ in $mol \, L^{-1}$. ($1 \, atm = 760 \, torr$; $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$).

At $1000 \ K$,a vessel contains $CO_2$ at a pressure of $0.5 \ atm$. Some $CO_2$ is converted into $CO$ by the addition of graphite. If the total pressure at equilibrium is $0.8 \ atm$,what is the value of $K_p$ in $atm$?