One mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:

For the formation of ammonia from its constituent elements ($1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$) in a closed vessel of volume $V \ L$,the value of $K_C$ is [units of $K_C = mol^{-2} \ L^2$].

In the synthesis of $HI$,the amounts of $H_{2(g)}$,$I_{2(g)}$,and $HI_{(g)}$ at equilibrium were found to be $0.8 \ mol$,$0.8 \ mol$,and $2.4 \ mol$ respectively in a $10 \ L$ vessel. Calculate the equilibrium constant $(K_c)$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ and the equilibrium constant for the reverse reaction.

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_p = 41$ at $400 \ K$. Calculate $K_c$ for the following reactions at $400 \ K$:
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$

For the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_c$ is $26$ at $250 \ ^oC$. The value of $K_p$ at this temperature will be ...........

For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is