Two moles of $NH_3$ when put into a previously evacuated vessel $(1 \ L)$,partially dissociate into $N_2$ and $H_2$. If at equilibrium one mole of $NH_3$ is present,the equilibrium constant is

$PCl_5$,$PCl_3$,and $Cl_2$ are at equilibrium at $500 \ K$ with concentrations $[PCl_3] = 1.59 \ M$,$[Cl_2] = 1.59 \ M$,and $[PCl_5] = 1.41 \ M$. Calculate $K_c$ for the reaction:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; K_1$
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_2$
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightleftharpoons NH_{3(g)}; K_3$
$2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}; K_4$
If $K_1 = K_2^x = K_3^y = K_4^z$,then the correct values of $x, y,$ and $z$ are respectively:

The equilibrium concentrations of $HI$,$I_2$,and $H_2$ are $0.7 \ M$,$0.1 \ M$,and $0.1 \ M$ respectively. The equilibrium constant for the reaction $I_2 + H_2 \rightleftharpoons 2HI$ is:

In which of the following reactions will the concentration of the product be greater than the concentration of the reactant at equilibrium? ($k$ = equilibrium constant)

For the system $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}$,the equilibrium concentrations are $[A] = 0.06 \ mol/L$,$[B] = 0.12 \ mol/L$,and $[C] = 0.216 \ mol/L$. The $K_{eq}$ for the reaction is: