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:

Write a relation between $\Delta G$ and $Q$ and define the meaning of each term and answer the following:
$(a)$ Why a reaction proceeds forward when $Q < K$ and no net reaction occurs when $Q = K$.
$(b)$ Explain the effect of increase in pressure in terms of reaction quotient $Q$.
For the reaction: $CO_{(g)} + 3H_{2(g)} \rightleftharpoons CH_{4(g)} + H_{2}O_{(g)}$

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 $20$.

The equilibrium constant at $298 \ K$ for a reaction $A + B \rightleftharpoons C + D$ is $100$. If the initial concentration of all the four species were $1 \ M$ each,then the equilibrium concentration of $D$ (in $mol \ L^{-1}$) will be:

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are taken in a $4 \ L$ closed vessel at a constant temperature. The reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. If $0.25\%$ of $N_2$ is converted into ammonia,calculate $K_c$ for this reaction and the $K_c'$ for the reaction $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$.

At $T \ K$,$K_{c}$ for the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$ is $16$. If initially one mole each of all the four gases are taken in a $1 \ L$ vessel,the equilibrium concentrations of $SO_{3(g)}$ and $SO_{2(g)}$ in $mol \ L^{-1}$ respectively are: