If the equilibrium constant for the reaction,$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $64$ at $500 \ K$,then the equilibrium constant for the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ at the same temperature is

If the equilibrium constant for the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$ is $K$,then what is the equilibrium constant for the reaction $\frac{1}{2}{N_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rightleftharpoons NO_{(g)}$?

The dissociation of $SO_{3(g)}$ into $SO_{2(g)}$ and $O_{2(g)}$ is carried out in a closed container at a constant temperature $T$. The equilibrium constant for the reaction is $K_P = x \ atm$. The partial pressure variation for the three gases is as shown in the graph. The value of $x$ is

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:

Write the expressions for $K_{P}$ and $K_{C}$ for the reaction: $3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_{4(s)} + 4H_{2(g)}$.

Explain the calculation of equilibrium concentrations given the value of the equilibrium constant.