In a closed vessel,$PCl_{5(g)}$ is obtained by the chemical reaction between $PCl_{3(g)}$ and $Cl_{2(g)}$. If the equilibrium concentrations in this vessel of $PCl_3$,$Cl_2$,and $PCl_5$ at $500 \ K$ are $1.59 \ M$,$1.59 \ M$,and $1.41 \ M$ respectively,then find the equilibrium constant $K_c$ for the reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

If the equilibrium constant for $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ is $K,$ the equilibrium constant for $\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{(g)}$ will be

When $4 \ mol$ of $A$ is mixed with $4 \ mol$ of $B$,$2 \ mol$ of $C$ is formed at equilibrium. For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant is:

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 $CH_3OH$.

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 $3.90$.

$PCl_5$,$PCl_3$,and $Cl_2$ are at equilibrium at $500 \ K$ with concentrations $[PCl_3] = 1.59 \ M$,$[Cl_2] = 1.59 \ M$,and $[PCl_5] = 1.41 \ M$. Calculate $K_c$ for the reaction:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$