For the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_p$ at $250 \ ^oC$ is $0.61 \ atm^{-1}$. The value of $K_c$ at this temperature will be .... $(mol \ L^{-1})$.

For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$ at $495 \ K$,$\Delta_{r}G^{\circ} = -9.478 \ kJ \ mol^{-1}$. If we start the reaction in a closed container at $495 \ K$ with $22 \ mmol$ of $A$,the amount of $B$ in the equilibrium mixture is $x \ mmol$. Find $x$ (Round off to the nearest integer). $[R = 8.314 \ J \ mol^{-1} \ K^{-1}; \ln 10 = 2.303]$

At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

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 ...

The equilibrium constant of $NH_4COONH_2$ in a closed vessel at $400 \ K$ temperature is $600 \ bar^3$. What will be the total pressure at equilibrium (in $bar$)?
$NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$

For the reversible reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $500 \ ^oC$,the value of $K_P$ is $1.44 \times 10^{-5}$ when partial pressure is measured in atmospheres. The corresponding value of $K_c$ with concentration in $\text{mol L}^{-1}$ is: