For the reaction,$A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$ at $25 \, ^oC$,$2 \, moles$ of $A$,$1 \, mole$ of $B$ and $1 \, mole$ of $C$ are present in a $1 \, L$ vessel. If $K_c$ for the reaction is $2$,then the reaction will proceed in:

$15$ moles of $H_2$ and $5.2$ moles of $I_2$ are mixed and allowed to attain equilibrium at $500 \, ^oC$. At equilibrium,the concentration of $HI$ is found to be $10$ moles. The equilibrium constant for the formation of $HI$ is

For the reaction ${N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$,the equilibrium constant ${K_p = 35}$ at a given temperature. Calculate the values of ${K_p}$ for the following reactions at the same temperature:
$(i) \ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$
$(ii) \ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$

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

Equilibrium constant,$K_{c}$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ at $500 \, K$ is $0.061$. At a particular time,the analysis shows that the composition of the reaction mixture is $[N_{2}] = 3.0 \, mol \, L^{-1}$,$[H_{2}] = 2.0 \, mol \, L^{-1}$,and $[NH_{3}] = 0.5 \, mol \, L^{-1}$. Is the reaction at equilibrium? If not,in which direction does the reaction tend to proceed to reach equilibrium?

For the chemical equilibrium $A + B \rightleftharpoons C + D$,when one mole of each reactant is mixed,$0.4 \ mol$ of each product is formed. The equilibrium constant $K_c$ is: