For the reaction ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ at $721 \ K$,the value of the equilibrium constant $({K_c})$ is $50$. When the equilibrium concentration of both is $0.5 \ M$,the value of ${K_p}$ under the same conditions will be:

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 $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

If ${K_c}$ is the equilibrium constant for the formation of $NH_3$,the dissociation constant of ammonia under the same temperature will be

$2 \ mol$ $PCl_5$ is heated in a $4 \ L$ closed vessel at a definite temperature. At equilibrium,$55\%$ of $PCl_5$ remains undissociated. Find $K_c$. Reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

Equilibrium constant for the reaction $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$ is $81$. If the velocity constant of the forward reaction is $162 \ L \ mol^{-1} \ s^{-1}$,what is the velocity constant (in $L \ mol^{-1} \ s^{-1}$) for the backward reaction?

The gas phase reaction $2A_{(g)} \rightleftharpoons A_{2(g)}$ at $400 \ K$ has $\Delta G^{\circ} = +25.2 \ kJ \ mol^{-1}$. The equilibrium constant $K_{C}$ for this reaction is $...... \times 10^{-2}$. (Round off to the nearest integer) $[$Use: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$\ln 10 = 2.3$,$\log_{10} 2 = 0.30$,$1 \ atm = 1 \ bar]$ $[$antilog $(-3.3) = 5.01 \times 10^{-4}]$