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The dissociation of $CO_2$ is represented as $2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)$. If $2 \ mol$ of $CO_2$ are taken initially and $40\%$ of $CO_2$ dissociates,what will be the total number of moles at equilibrium?

At $717 \ K$,$3.2 \ mol$ of $HI$ is heated in a closed tube. $20\%$ of $HI$ decomposes at equilibrium according to the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$. Find $K_c$ and the moles of $HI$,$H_2$,and $I_2$ at equilibrium.

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When the reaction $A + 2B \rightleftharpoons 2C + D$ was studied,it was observed that the initial concentration of $B$ was $1.5$ times that of $A$,and the equilibrium concentrations of $A$ and $C$ were equal. Then $K_C$ for the given equilibrium is equal to

For the reaction $Br_2 \rightleftharpoons 2Br$,the equilibrium constants at $327\,^oC$ and $527\,^oC$ are $6.1 \times 10^{-12}$ and $1.0 \times 10^{-7}$ respectively. The reaction is ...........

$5 \ \text{moles}$ of $SO_2$ and $5 \ \text{moles}$ of $O_2$ are allowed to react to form $SO_3$ in a closed vessel. At the equilibrium stage,$60\%$ of $SO_2$ is used up. The total number of moles of $SO_2$,$O_2$,and $SO_3$ in the vessel now is:

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