In the figure shown below,reactant $A$ (represented by square) is in equilibrium with product $B$ (represented by circle). The equilibrium constant is:

At $200^{\circ} C$,nitric oxide reacts with oxygen to form nitrogen dioxide as follows: $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$,$K_C = 3 \times 10^6$. In a mixture of the three species at equilibrium,we can accurately predict that:

In the reaction $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$,one mole of $AB$ reacts with one mole of $CD$. When equilibrium is established,$3/4$ mole of each $AB$ and $CD$ are converted into $AD$ and $CB$. If the volume remains constant,the equilibrium constant for the reaction 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

For the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be

For the system $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)}$,the expression for equilibrium constant $K$ is