One mole of $H_2O$ and one mole of $CO$ are taken in a $10 \ L$ vessel and heated to $725 \ K$. At equilibrium,$40 \%$ of water (by mass) reacts with $CO$ according to the equation:
$H_2O_{(g)} + CO_{(g)} \longleftrightarrow H_{2(g)} + CO_{2(g)}$
Calculate the equilibrium constant $(K_c)$ for the reaction.

The value of $K_{c} = 4.24$ at $800 \, K$ for the reaction,
$CO_{(g)} + H_{2}O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$
Calculate equilibrium concentrations of $CO_{2}$,$H_{2}$,$CO$ and $H_{2}O$ at $800 \, K$,if only $CO$ and $H_{2}O$ are present initially at concentrations of $0.10 \, M$ each.

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 $bar^3$)

The following reaction is performed at $298 \, K$.
$2 NO_{(g)} + O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$
The standard free energy of formation of $NO_{(g)}$ is $86.6 \, kJ/mol$ at $298 \, K$. What is the standard free energy of formation of $NO_{2(g)}$ at $298 \, K$? $(K_p = 1.6 \times 10^{12})$

$37.8 \ g$ $N_2O_5$ was taken in a $1 \ L$ reaction vessel and allowed to undergo the following reaction at $500 \ K$:
$2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$
The total pressure at equilibrium was found to be $18.65 \ bar$. Then,$K_p = \text{ . . . . . . } \times 10^{-2}$ [nearest integer].
Assume $N_2O_5$ to behave ideally under these conditions.
Given: $R = 0.082 \ bar \ L \ mol^{-1} \ K^{-1}$

$\Delta _f G^o$ at $500 \, K$ for substance '$S$' in liquid state and gaseous state are $+100.7 \, kcal \, mol^{-1}$ and $+103 \, kcal \, mol^{-1}$,respectively. The vapour pressure of liquid '$S$' at $500 \, K$ is approximately equal to $(R = 2 \, cal \, K^{-1} \, mol^{-1}) \dots \dots \text{atm}$.