$2 \ mol$ $PCl_5$ is heated in a $4 \ L$ closed vessel at a definite temperature. At equilibrium,$55\%$ of $PCl_5$ remains undissociated. Find $K_c$. Reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

For a gaseous reversible reaction,the enthalpy of reaction at constant pressure is $1.8 \ Kcal/mol$ greater than that at constant volume at $300 \ K$. The value of $\left( \frac{K_P}{K_C} \right)$ for the reaction at $T = \left( \frac{1}{0.00821} \right) \ K$ is:

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,$K_p = 0.492 \ atm$ at $300 \ K$. $K_c$ for the reaction at same temperature is . . . . . . $\times 10^{-2}$. (Given: $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

Find the value of $\frac{P}{K_p}$ for the reaction at a certain temperature: $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$,where $P$ is the total pressure of gases at equilibrium and $P_{Br_2} = \frac{P}{9}$.

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: $A(g) \rightleftharpoons B(g) + C(g)$.

$1/T \text{ (K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)

For the elementary reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,the rate of the forward reaction is given by $r_f = 1.7 \times 10^{-18} [A_2][B_2]$. If the rate of decomposition of gaseous $AB$ into $A_2$ and $B_2$ is given by $r_r = 2.4 \times 10^{-21} [AB]^2$,then the equilibrium constant for the formation of $AB$ from $A_2$ and $B_2$ will be ...