Given $K_p$ for the reaction $\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$ at a fixed temperature is $0.25 \ atm$. Find the $K_p$ for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$ at the same temperature.

The reaction $A(g) \rightleftharpoons B(g) + C(g)$ was initiated with the amount $a$ of $A(g)$. At equilibrium,it is found that the amount of $A(g)$ remaining is $(a-x)$ at a total pressure of $p$. The equilibrium constant $K_{p}$ of the reaction can be calculated from the expression:

$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?

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 $CO_{2}$.

If ${K_c}$ is the equilibrium constant for the formation of $NH_3$,the dissociation constant of ammonia under the same temperature will be

For the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_p$ at $250 \ ^oC$ is $0.61 \ atm^{-1}$. The value of $K_c$ at this temperature will be .... $(mol \ L^{-1})$.