The equilibrium $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ shows that $K_P$ (in $atm$) is double the value of $K_C$ (in $mol/L$) at a particular temperature $T$. Then,$T$ is $...... \ K$.

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$1 mol$ of $PCl_5$ is taken at $5 atm$ pressure. If $50\%$ of $PCl_5$ dissociates at equilibrium,calculate $K_p$.

At $3000 \ K$ the equilibrium pressures of $CO_2$,$CO$ and $O_2$ are $0.6 \ atm$,$0.4 \ atm$ and $0.2 \ atm$ respectively. $K_p$ for the reaction,$2CO_2 \rightleftharpoons 2CO + O_2$ is

For the given equilibrium reaction,$2 A(g) \rightleftharpoons 2 B(g) + C(g)$,the equilibrium constant $(K_c)$ at $1000 \ K$ is $4 \times 10^{-4}$. Calculate $K_p$ for the reaction at $800 \ K$ temperature.

$XY_2$ 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.

Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of $CO$ and $H_2$. In second stage,$CO$ formed in first stage is reacted with more steam in water gas shift reaction,
$CO_{(g)} + H_2O_{(g)} \longleftrightarrow CO_{2(g)} + H_{2(g)}$
If a reaction vessel at $400^{\circ}C$ is charged with an equimolar mixture of $CO$ and steam such that $P_{CO} = P_{H_2O} = 4.0 \ bar,$ what will be the partial pressure of $H_2$ at equilibrium? $K_p = 10.1$ at $400^{\circ}C$