The equilibrium $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ shows that $K_P$ (in $atm$) is double the value of $K_C$ (in $mol/L$) at a particular temperature $T$. Then,$T$ is $...... \ K$.

The equilibrium constant $(K_p)$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{-4}$ and at $127^{\circ} C$ is $0.60 \times 10^{-4}$. Calculate the mean heat of formation of ammonia per mole in this temperature range. (in $cal$)

At $1100 \ K$ temperature,$CaCO_{3(s)}$ and $CaO_{(s)}$ are in equilibrium. The pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.
Find the equilibrium constant $(K_p)$ for the reaction: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

Consider the equilibria $(i)$ and $(ii)$ with equilibrium constants $K_1$ and $K_2$,respectively.
$SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)} ..... (i)$
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)} ..... (ii)$
$K_1$ and $K_2$ are related as

For the reversible reaction in equilibrium:
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.

For the equilibrium $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)} + Cl_{2(g)}$,what is the temperature at which $\frac{K_p}{K_c} = \frac{1}{3}$? (Given $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$)