The equilibrium constant $K_c$ for the following equilibrium:
$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$
at $563 \ K$ is $100$. At equilibrium,the number of moles of $SO_3$ in the $10 \ L$ flask is twice the number of moles of $SO_2$. Calculate the number of moles of oxygen.

The variation of equilibrium constant with temperature is given below:

$T_{1} = 25^{\circ}C$	$K_{1} = 100$
$T_{2} = 100^{\circ}C$	$K_{2} = 100$

The values of $\Delta H^{\circ}$,$\Delta G^{\circ}$ at $T_{1}$ and $\Delta G^{\circ}$ at $T_{2}$ (in $kJ \ mol^{-1}$) respectively,are close to: [Use $R = 8.314 \ J \ K^{-1} \ mol^{-1}$]

At $473 \ K$, equilibrium constant $K_{c}$ for decomposition of phosphorus pentachloride, $PCl_{5}$, is $8.3 \times 10^{-3}$. If decomposition is depicted as,
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}, \Delta_{r}H^{\Theta} = 124.0 \ kJ \ mol^{-1}$
$(a)$ Write an expression for $K_{c}$ for the reaction.
$(b)$ What is the value of $K_{c}$ for the reverse reaction at the same temperature?
$(c)$ What would be the effect on $K_{c}$ if $(i)$ more $PCl_{5}$ is added $(ii)$ pressure is increased $(iii)$ the temperature is increased?

$PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2} \quad K_{c} = 1.844$
$3.0 \ \text{moles}$ of $PCl_{5}$ is introduced in a $1 \ \text{L}$ closed reaction vessel at $380 \ \text{K}$. The number of moles of $PCl_{5}$ at equilibrium is $..... \times 10^{-3}$. (Round off to the Nearest Integer)

In a reaction,$A + B \rightleftharpoons C + D$,$40 \%$ of $B$ has reacted at equilibrium,when $1 \ mol$ of $A$ was heated with $1 \ mol$ of $B$ in a $10 \ L$ closed vessel. The value of $K_C$ is

For the gas phase reaction; $C_2H_{4(g)} + H_{2(g)} \rightleftharpoons C_2H_{6(g)}$,$\Delta H = -32.7 \ kcal$ carried out in a vessel,the equilibrium concentration of $C_2H_4$ can be increased by: