For the reaction at $25\,^{\circ}C$,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,if $\Delta G^{\circ}_f$ for $N_2O_4$ and $NO_2$ are $23.49 \, kcal$ and $12.39 \, kcal$ respectively,then $K_p$ for the reaction is:

At a certain temperature in a $5\,L$ vessel,$2\,moles$ of carbon monoxide and $3\,moles$ of chlorine were allowed to reach equilibrium according to the reaction,$CO + Cl_2 \rightleftharpoons COCl_2$. At equilibrium,if $1\,mole$ of $CO$ is present,then the equilibrium constant $(K_c)$ for the reaction is:

Calculate $K_{C}$ for the reversible process given below if $K_{P}=167$ and $T=800^{\circ}C$.
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

For the reaction,$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_c$ at $250 \ ^oC$ is $26$. The value of $K_p$ at this temperature will be:

In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant is $10$. If the rate constant for the forward reaction is $203$,what will be the rate constant for the backward reaction?