The value of $K_{C}$ is $64$ at $800 \ K$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
The value of $K_{C}$ for the following reaction is:
$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$

The equilibrium constant $K_C$ of the reaction,$2 A \rightleftharpoons B + C$ is $0.5$ at $25^{\circ} C$. The reaction will proceed in the backward direction,when concentrations $[A], [B]$ and $[C]$ are,respectively:

At $490\,^{\circ}C$,the equilibrium constant for the synthesis of $HI$ is $50$. The value of $K$ for the dissociation of $HI$ will be:

At $T \ K$,the equilibrium constant for the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$ is $K_c$. If the reaction takes place in the following form $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,its equilibrium constant is $K_c^{\prime}$. The correct relationship between $K_c$ and $K_c^{\prime}$ is

The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.

At a definite temperature,the equilibrium constant $K_{c}$ is given by the following equation: $K_{c} = \frac{[I_{2}][H_{5}IO_{6}]^{5}}{[IO_{3}^{-}]^{7}[H_{2}O]^{9}[H^{+}]^{7}}$. Write the balanced chemical equilibrium equation.