At $444 \, ^\circ C$,the equilibrium constant for the reaction $HI \rightleftharpoons 1/2 H_2 + 1/2 I_2$ is $64$. What will be the equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

In a chemical reaction,the rate constant for the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant for the forward reaction will be .......

Assertion : For reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the unit of $K_C$ is $L^2 \, mol^{-2}$.
Reason : For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}$.

$5.1 \ g$ $NH_4SH$ is introduced in a $3.0 \ L$ evacuated flask at $327 \ ^\circ C$. $30\%$ of the solid $NH_4SH$ decomposes into $NH_3$ and $H_2S$ gases. The $K_p$ of the reaction at $327 \ ^\circ C$ is ($R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,molar mass of $S = 32 \ g \ mol^{-1}$,molar mass of $N = 14 \ g \ mol^{-1}$)

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

For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2}O_{2(g)}$,the equilibrium constant $K_C = 4.9 \times 10^{-2}$. Calculate the $K_C$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.