For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$0.4 \ mol$ of $H_2$ and $I_2$ each are taken in a $2 \ L$ vessel. If $0.5 \ mol$ of $HI$ is formed at equilibrium,calculate the equilibrium constant $K_c$ and $K_p$.

$(i)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$ at $298 \ K$ has $\Delta G^{\Theta} = -16.5 \ kJ \ mol^{-1}$. Find $K_p$.
$(ii)$ At $298 \ K$,for $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,calculate $K_p$ and $\Delta G^{\Theta}$.

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 the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the value of $K_p = 41$ at $400 \ K$. Find out the value of $K_p$ for the following reaction at the same temperature: $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_p/K_c$ is .....