$4$ moles of $A$ are mixed with $4$ moles of $B$. At equilibrium for the reaction $A + B \rightleftharpoons C + D$,$2$ moles of $C$ and $D$ are formed. The equilibrium constant for the reaction will be

The equilibrium constant of the following reaction is $0.5$. $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$. At equilibrium,$[CO] = 0.18 \ mol \ L^{-1}$ and $[H_2] = 0.22 \ mol \ L^{-1}$. Calculate the concentration of $CH_3OH$.

At $780 \ K$ and $10 \ atm$ pressure,the equilibrium constant for the reaction $2 \ A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$ is $3.52$. At the same temperature and $7.04 \ atm$ pressure,the equilibrium constant for the same reaction is:

$A$ tenfold increase in pressure on the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at equilibrium,makes $K_p$

At $T \ K$,the equilibrium constants for the following two reactions are given below:
$2 A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)} ; K_1 = 16$
$2 B_{(g)} + C_{(g)} \rightleftharpoons 2 D_{(g)} ; K_2 = 25$
What is the value of equilibrium constant $(K)$ for the reaction given below at $T \ K$?
$A_{(g)} + \frac{1}{2} B_{(g)} \rightleftharpoons D_{(g)}$

If for ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$ and ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$ the equilibrium constants are $K_1$ and $K_2$ respectively,the reaction ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$ would have equilibrium constant