For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,$K_p = 0.492 \ atm$ at $300 \ K$. $K_c$ for the reaction at same temperature is . . . . . . $\times 10^{-2}$. (Given: $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

The value of $K_{p}$ for the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is $3.0$ at $1000 \ K$. If initially $P_{CO_{2}} = 0.48 \ bar$ and $P_{CO} = 0 \ bar$ and pure graphite is present,calculate the equilibrium partial pressures of $CO$ and $CO_{2}$.

For a reversible reaction $A \rightleftharpoons B$,the half-life periods of the forward and backward reactions are $15 \ s$ and $18 \ s$,respectively. If both reactions follow first-order kinetics,what is the equilibrium constant $(K_c)$ of the reaction?

Consider the equilibria $(i)$ and $(ii)$ with equilibrium constants $K_1$ and $K_2$,respectively.
$SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)} ..... (i)$
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)} ..... (ii)$
$K_1$ and $K_2$ are related as

At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.