The reaction quotient $Q$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is given by $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$. The reaction will proceed from right to left when:

For the reaction $3A + 2B \rightleftharpoons C$,the expression for the equilibrium constant $K_c$ is:

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

$PCl_5$,$PCl_3$,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:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:
$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

The equilibrium constant $K_{c}$ at $298 \ K$ for the reaction $A + B \rightleftharpoons C + D$ is $100$. Starting with an equimolar solution with concentrations of $A$,$B$,$C$ and $D$ all equal to $1 \ M$,the equilibrium concentration of $D$ is $....... \times 10^{-2} \ M$. (Nearest integer)