For the following reaction in gaseous phase $CO + 1/2 O_2 \to CO_2$,the ratio $K_p/K_c$ is:

One mole of $PCl_{5(g)}$ was heated in a $1 \ L$ closed flask at $500 \ K$. At equilibrium,$0.1 \ mole$ of $Cl_{2(g)}$ was formed. What is its $K_{p}$ (in $atm$)? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

Given $K_p$ for the reaction $\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$ at a fixed temperature is $0.25 \ atm$. Find the $K_p$ for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$ at the same temperature.

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the equilibrium constant $K_c = 1.8 \times 10^{-6}$ at $184 \, ^\circ C$. Comparing $K_p$ and $K_c$ at $184 \, ^\circ C$,we find that:

For a reversible reaction $A \rightleftharpoons B$,the pre-exponential factor is the same for both the forward and backward reactions and has a value of $20 \ s^{-1}$. If the enthalpy change for the forward reaction is $-41.5 \ kJ \ mol^{-1}$,the value of the equilibrium constant at $500 \ K$ is:

For a reaction,if the graph of $\ln K_{eq}$ versus $1/T$ is plotted as shown,the reaction must be: