In the gas phase reaction,$C_2H_4 + H_2 \rightleftharpoons C_2H_6$,the equilibrium constant can be expressed in units of

$A + B \rightleftharpoons C + D$. If the final equilibrium concentrations of $A$ and $B$ are equal,and the equilibrium concentration of $D$ is twice that of $A$,what is the equilibrium constant $(K_c)$ of the reaction?

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

$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
The reaction was started with some amount of $NH_4HS$. The equilibrium pressure at $25^{\circ}C$ is $0.5 \ atm$. What is $K_p$ for the reaction (in $atm^2$)?

For which of the following reactions is $K_p = K_c$?

At $1000 \ K$,the value of $K_c$ for the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$ is $10 \ mol \ L^{-1}$. The value of $K_p$ (in $atm$) is: (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)