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?

The variation of equilibrium constant with temperature is given below:

$T_{1} = 25^{\circ}C$	$K_{1} = 100$
$T_{2} = 100^{\circ}C$	$K_{2} = 100$

The values of $\Delta H^{\circ}$,$\Delta G^{\circ}$ at $T_{1}$ and $\Delta G^{\circ}$ at $T_{2}$ (in $kJ \ mol^{-1}$) respectively,are close to: [Use $R = 8.314 \ J \ K^{-1} \ mol^{-1}$]

For the reaction $CuSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CuSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$,which of the following is a correct representation?

In a $0.25 \ L$ tube,$4 \ mol$ of $NO$ undergoes dissociation. If the degree of dissociation is $10\%$,then the value of $K_c$ for the reaction $2NO \rightleftharpoons N_2 + O_2$ will be:

Consider the following reaction equilibrium:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initially,$1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$ are taken in a $2 \ L$ flask. At equilibrium,if the number of moles of $N_2$ is $0.6$,what is the total number of moles of all gases present in the flask?

For the reaction $X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$,the reaction is studied at a fixed temperature. Initially,$1 \ mol$ of $X_2$ is taken in a $1 \ L$ flask and $2 \ mol$ of $Y_2$ is taken in a $2 \ L$ flask. If the flasks are connected,what are the equilibrium concentrations of $X_2$ and $Y_2$? (Given: Equilibrium concentration of $XY = 0.6 \ mol/L$)