For the decomposition of the compound,represented as $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,the $K_p = 2.9 \times 10^{-5} \ atm^3$. If the reaction is started with $1 \ mol$ of the compound,the total pressure at equilibrium would be............ $\times 10^{-2} \ atm$.

The partial pressures of $CH_3OH$,$CO$ and $H_2$ in the equilibrium mixture for the reaction $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$ at $427 \ ^oC$ are $2.0 \ atm$,$1.0 \ atm$ and $0.1 \ atm$ respectively. The value of $K_P$ for the decomposition of $CH_3OH$ to $CO$ and $H_2$ is:

For the reactions $(1)$ and $(2)$ :
$A \rightleftharpoons B + C \dots (1)$
$D \rightleftharpoons 2E \dots (2)$
Given $K_{P_1} : K_{P_2} = 9 : 1$.
If the degree of dissociation of $A$ and $D$ is the same,then the total pressure at equilibria $(1)$ and $(2)$ are in the ratio (Assume reactions are started with equal number of moles of $A$ and $D$). (in $: 1$)

For the following three equilibria $X, Y$,and $Z$ in the gaseous state at $300 \ K$,the increasing order of the ratio of $K_p$ to $K_c$ is:
$X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$
$Y: PCl_5 \rightleftharpoons PCl_3 + Cl_2$
$Z: 2HI \rightleftharpoons H_2 + I_2$

The standard equilibrium constant,$K_p$ at $298 \, K$ for the reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is $5.8 \times 10^5$. The value of standard equilibrium constant,if the concentration of gases is expressed in terms of $mol/L$,will be:
[Given : $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$]

$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$. The correct relationship between $K_P$,$\alpha$,and equilibrium pressure $P$ is: