According to Le Chatelier's principle, in the reaction $\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$, the formation of methane is favoured by

(a) Increasing the concentration of CO

(b) Increasing the concentration of $\mathrm{H}_2 \mathrm{O}$

(c) Decreasing the concentration of $\mathrm{CH}_4$

(d) Decreasing the concentration of $\mathrm{H}_2$

The equilibrium constant at 298 K for the reaction $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$ is 100 . If the initial concentrations of all the four species were 1 M each, then equilibrium concentration of D (in $\mathrm{mol} \mathrm{L}^{-1}$ ) will be

At $$500 \mathrm{~K}$$, for a reversible reaction $$A_2(g)+B_2(g) \rightleftharpoons 2 A B(g)$$ in a closed container, $$K_C=2 \times 10^{-5}$$. In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant $$K_C$$ in the presence of catalyst at the same temperature is

1 mole of $$\mathrm{HI}$$ is heated in a closed container of capacity of $$2 \mathrm{~L}$$. At equilibrium half a mole of $$\mathrm{HI}$$ is dissociated. The equilibrium constant of the reaction is