In the equilibrium $AB \rightleftharpoons A + B$; if the equilibrium concentration of $A$ is doubled,the equilibrium concentration of $B$ would become:

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:

If $Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^+$; $K_1 = 1.6 \times 10^3$ and $[Ag(NH_3)]^+ + NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$; $K_2 = 6.8 \times 10^3$. Then the formation constant of $[Ag(NH_3)_2]^+$ is:

The equilibrium constant for the reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ at $400 \ K$ is $41$. The equilibrium constant for the reaction,$\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3$ at the same temperature will be closest to

An equilibrium mixture of the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had $0.5 mol$ $H_2S$,$0.10 mol$ $H_2$,and $0.4 mol$ $S_2$ in a $1 L$ vessel. The value of the equilibrium constant $(K_c)$ in $mol L^{-1}$ is:

The following equilibria are given:
$N_2 + 3H_2 \rightleftharpoons 2NH_3 ; K_1$
$N_2 + O_2 \rightleftharpoons 2NO ; K_2$
$H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O ; K_3$
The equilibrium constant of the reaction $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$ and $K_3$ is: