For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,at equilibrium,the mole fraction of $PCl_5$ is $0.4$ and the mole fraction of $Cl_2$ is $0.3$. What will be the mole fraction of $PCl_3$?

$A$ mixture of $1.57 \ mol$ of $N_2$,$1.92 \ mol$ of $H_2$ and $8.13 \ mol$ of $NH_3$ is introduced into a $20 \ L$ reaction vessel at $500 \ K$. At this temperature,the equilibrium constant,$K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ is $1.7 \times 10^2$. Is the reaction mixture at equilibrium? If not,what is the direction of the net reaction?

The equilibrium constants of the following are
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is:

$2$ moles of $N_2$ are mixed with $6$ moles of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,find the value of $K_c$ for the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

For the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$,the equilibrium constant $K_c$ is $16$. If $1 \ mol$ of each gas is taken in a $1 \ dm^3$ vessel,the equilibrium concentration of $NO$ will be ....

The equilibrium composition for the reaction $PCl_3 + Cl_2 \rightleftharpoons PCl_5$ at $298 \, K$ is given below.
$[PCl_3]_{eq} = 0.2 \, mol \, L^{-1}$
$[Cl_2]_{eq} = 0.1 \, mol \, L^{-1}$
$[PCl_5]_{eq} = 0.40 \, mol \, L^{-1}$
If $0.2 \, mol$ of $Cl_2$ is added at the same temperature,the equilibrium concentration of $PCl_5$ is $.... \times 10^{-2} \, mol \, L^{-1}$. Given: $K_c$ for the reaction at $298 \, K$ is $20$.