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 $PCl_3$?

Identify the incorrect statement.

In a one-litre flask,$6$ moles of $A$ undergoes the reaction $A_{(g)} \rightleftharpoons P_{(g)}$. The progress of product formation at two temperatures (in Kelvin),$T_1$ and $T_2$,is shown in the figure:
If $T_1=2 T_2$ and $(\Delta G_2^{\Theta}-\Delta G_1^{\Theta})=R T_2 \ln x$,then the value of $x$ is. . . . .
$[\Delta G_1^{\Theta}$ and $\Delta G_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$,respectively.]

At $500 \ K$,for a reversible reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2 AB_{(g)}$ in a closed container,$K_C = 2 \times 10^{-5}$. In the presence of a catalyst,the equilibrium is attained $10$ times faster. The equilibrium constant $K_C$ in the presence of a catalyst at the same temperature is:

$3.00 \ mol$ of $PCl_5$ kept in $1 \ L$ closed reaction vessel was allowed to attain equilibrium at $380 \ K$. If $1.59 \ mol$ of reactant was converted into the product at equilibrium,then $K_c$ is:

$5 \, \text{moles}$ of $PCl_5$ are heated in a closed vessel of $5 \, \text{L}$ capacity. At equilibrium,$40\%$ of $PCl_5$ is found to be dissociated. What is the value of $K_c$ (in $, \text{M}$)?