At $600\ ^{\circ}C$,$NH_4COONH_2\ (s) \rightleftharpoons 2NH_3\ (g) + CO_2\ (g)$ has an equilibrium constant $K_p = 3.2 \times 10^2\ bar^3$. Calculate $K_c$. $\left( R = 0.082\ L\ atm\ K^{-1}\ mol^{-1} \right)$

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ (by mass) of water reacted with $CO_{(g)}$ as follows: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2_{(g)}} + CO_{2_{(g)}}$. The value of $K_p$ is:

For the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$,if the initial concentration of $[H_2] = [CO_2] = 1 \ M$ and $x \ mol/L$ of hydrogen is consumed at equilibrium,the correct expression for $K_c$ is:

For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,$AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P$. Therefore,$P$ is correctly related to $K_p$ by which of the following options?

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 $400^{\circ}C$

At a certain temperature,only $50\%$ $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is: