$A + B \rightleftharpoons C + D$. If the final equilibrium concentrations of $A$ and $B$ are equal,and the equilibrium concentration of $D$ is twice that of $A$,what is the equilibrium constant $(K_c)$ of the reaction?

Equilibrium constants for the following reactions at $1200 \ K$ are given:
$2 \ H_2O_{(g)} \rightleftharpoons 2 \ H_{2(g)} + O_{2(g)}$
$K_1 = 6.4 \times 10^{-8}$
$2 \ CO_{2(g)} \rightleftharpoons 2 \ CO_{(g)} + O_{2(g)}$
$K_2 = 1.6 \times 10^{-6}$
The equilibrium constant for the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
at $1200 \ K$ will be

Find $\Delta n_g$ when $1 \text{ mol}$ of each $NH_{3(g)}$ and $HCl_{(g)}$ reacts to form solid $NH_4Cl_{(s)}$.

Partial pressures of $A$,$B$,$C$,and $D$ for the gaseous system $A + 2B \rightleftharpoons C + 3D$ are $A = 0.20 \ atm$,$B = 0.10 \ atm$,$C = 0.30 \ atm$,and $D = 0.50 \ atm$. The numerical value of the equilibrium constant $(K_p)$ is:

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

Equilibrium constant for the reaction $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$ is $81$. If the velocity constant of the forward reaction is $162 \ L \ mol^{-1} \ s^{-1}$,what is the velocity constant (in $L \ mol^{-1} \ s^{-1}$) for the backward reaction?