If the equilibrium constant for the reaction $2AB \rightleftharpoons A_2 + B_2$ is $49$,then the equilibrium constant for the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$ will be:

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

The equilibrium constant expression for a gas reaction is,
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.

In a chemical equilibrium $A + B \rightleftharpoons C + D$,when $1 \ mol$ each of the two reactants are mixed,$0.6 \ mol$ each of the products are formed. The equilibrium constant calculated is

If the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ at $720 \ K$ has $K = 48$,then for the reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$,find its equilibrium constant.

The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be