For the reversible reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$, the unit of the equilibrium constant $K_c$ is:

The equilibrium constant of the following reaction is $0.5$. $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$. At equilibrium,$[CO] = 0.18 \ mol \ L^{-1}$ and $[H_2] = 0.22 \ mol \ L^{-1}$. Calculate the concentration of $CH_3OH$.

The equilibrium constant for the equilibrium $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$ is $1 \times 10^{-5}$. What will be the equilibrium concentration of $HX$ if the equilibrium concentrations for $H_2$ and $X_2$ are $1.2 \times 10^{-3} \ M$ and $1.2 \times 10^{-4} \ M$ respectively?

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_p$ changes with:

For the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be

At $400 \ K$ for the reaction $2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}$,the concentration of $NO_{2}$ is $0.710 \ M$ and $N_{2}O_{4}$ is $0.145 \ M$. Find the equilibrium constant $K_{c}$. (in $M^{-1}$)