An equilibrium mixture taken in 2 litre vessel of the reaction: $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ has 4 moles of $\mathrm{SO}_2, 3$ moles of $\mathrm{O}_2$ and 6 moles of $\mathrm{SO}_3$ then the value of equilibrium constant $\left(\mathrm{K}_c\right)$ will be:

Consider the gaseous equilibrium

$2 \mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{AB}_{(\mathrm{g})}+\mathrm{B}_{2(\mathrm{~g})}$

The expression relating the degree of dissociation ( $\alpha$ ) and equilibrium constant ( $K_p$ ) and total pressure $P$ is:

If the equilibrium constant for $\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}$ is 49 The equilibrium constant for the reaction $N O_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ is $\_\_\_\_$

For a reaction,

$$A+B \rightleftharpoons 2 C$$

1.0 mole of $A, 1.5$ mole of $B$ and 0.5 mole of $C$ were taken in a 1 L vessel. At equilibrium, the concentration of $C$ was $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$. The equilibrium constant for the reaction is $x / 15$. The value of ' $x$ ' is: