For the reaction,$A \rightleftharpoons n B$,the concentration of $A$ decreases from $0.06 \ mol \ L^{-1}$ to $0.03 \ mol \ L^{-1}$ and that of $B$ rises from $0$ to $0.06 \ mol \ L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction,respectively,are

Match the items in List-$X$ with List-$Y$ and select the correct option.

List-$X$	List-$Y$
$(A)$ $A_{(g)} \rightleftharpoons B_{(g)} + \text{Heat}$	$(i)$ Equilibrium constant
$(B)$ $r_b/r_f$	$(ii)$ Favored at low temperature
$(C)$ $r_f/r_b$	$(iii)$ [Equilibrium constant]$^{-1}$
$(D)$ $2A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$	$(iv)$ $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
$(E)$ Effect of pressure	$(V)$ $\Delta n < 0$

For the reaction $A + B \rightleftharpoons C + D$ at $250^\circ C$ in a $1 \ L$ vessel,the initial concentration of $A$ is $3$ and $B$ is $n$. If the equilibrium concentration of $C$ is equal to the equilibrium concentration of $B$,what is the equilibrium concentration of $D$?

Consider the reaction,$P(aq) \rightleftharpoons Q(aq)$ with an equilibrium constant $K=1.5$. The reaction is started in a vessel with a concentration of $[P]$ of $2 \ M$ and concentration of $[Q]=0$. When the equilibrium is established,half the amount of $P$ is removed,and the reaction is allowed to re-equilibrate. The concentration of $Q$ in the vessel (in $M$) is closest to

Given the equilibrium constants for the following three reactions:
$(1) N_2 + 3H_2 \rightleftharpoons 2NH_3; K_1$
$(2) N_2 + O_2 \rightleftharpoons 2NO; K_2$
$(3) H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O; K_3$
The equilibrium constant for the reaction of $NH_3$ with oxygen to form $NO$ and $H_2O$ is:

Attainment of the equilibrium $A_{(g)} \rightleftharpoons 3C_{(g)} + 2B_{(g)}$ gave the following graph. Find the correct option. $(\text{Percentage dissociation} = \text{fraction dissociated} \times 100)$