For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

$A$ reaction is $A + B \rightleftharpoons C + D$. Initially,we start with equal concentrations of $A$ and $B$. At equilibrium,the number of moles of $C$ is two times that of $A$. What is the equilibrium constant $(K_c)$ of the reaction?

Given the reaction between $2$ gases represented by $A_2$ and $B_2$ to give the compound $AB_{(g)}$:
$A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
At equilibrium,the concentrations are $[A_2] = 3.0 \times 10^{-3} \, M$,$[B_2] = 4.2 \times 10^{-3} \, M$,and $[AB] = 2.8 \times 10^{-3} \, M$.
If the reaction takes place in a sealed vessel at $527^{\circ}C$,then the value of $K_c$ will be:

The equilibrium constant $K$ for the synthesis of $HI$ is $50$. The equilibrium constant $K$ for the dissociation of $HI$ is:

The reaction,$CO_{(g)} + 3H_{2(g)} \longleftrightarrow CH_{4(g)} + H_{2}O_{(g)}$ is at equilibrium at $1300 \, K$ in a $1 \, L$ flask. It also contains $0.30 \, mol$ of $CO$,$0.10 \, mol$ of $H_{2}$,and $0.02 \, mol$ of $H_{2}O$ and an unknown amount of $CH_{4}$ in the flask. Determine the concentration of $CH_{4}$ in the mixture. The equilibrium constant,$K_{c}$ for the reaction at the given temperature is $3.90$.

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[SO_{2}] = 0.60 \, M, [O_{2}] = 0.82 \, M$ and $[SO_{3}] = 1.90 \, M?$
$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$