$8 \ mol$ of $AB_{3(g)}$ are introduced into a $1.0 \ dm^3$ vessel. If it dissociates as $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. At equilibrium,$2 \ mol$ of $A_2$ are found to be present. The equilibrium constant of this reaction is

$A$ $1 \, M$ solution of glucose reaches dissociation equilibrium according to the equation $C_6H_{12}O_6 \rightleftharpoons 6HCHO$. What is the concentration of $HCHO$ at equilibrium if the equilibrium constant $K_c$ for the formation of glucose from formaldehyde is $6 \times 10^{22}$?

At $600 \ K$,ammonium carbamate decomposes in a closed vessel: $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. At equilibrium,the total pressure is $3 \ bar$. Calculate $K_p$. (in $bar^3$)

From the given data of equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
Wait,the provided question text has a typo in the reaction equations. Assuming the standard problem format where we relate equilibrium constants for reverse or combined reactions,if the target reaction is the same as reaction $(1)$,the answer is $K_1$. However,based on the options provided,this is likely a question asking for the relationship between $K_1$ and $K_2$ where reaction $(2)$ is the reverse of reaction $(1)$. If reaction $(2)$ is the reverse of reaction $(1)$,then $K_2 = \frac{1}{K_1}$. Given the options,please re-verify the input. Assuming the question asks for the equilibrium constant of a reaction derived from these,if the target reaction is $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$,the answer is $K_1^{-1}$. Given the options,if we assume the target reaction is the reverse of reaction $(1)$,then $K = \frac{1}{K_1}$.

At high temperature,$2 \, \text{mol}$ of $NH_3$ is placed in a $500 \, \text{mL}$ vessel. For the decomposition reaction $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$,if $1 \, \text{mol}$ of $NH_3$ remains at equilibrium,then $K_c$ is equal to:

At $67\,^{\circ}C$ and $1\ bar$ pressure,dinitrogen tetraoxide is $50\%$ dissociated into nitrogen dioxide. $\Delta G^{\circ}$ for the process $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is $(R = \frac{25}{3} \ J \ K^{-1} \ mol^{-1}, \ln 2 = 0.7, \ln 3 = 1.1)$.