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 $(R = \frac{25}{3} \ J \ K^{-1} \ mol^{-1}, \ln 2 = 0.7, \ln 3 = 1.1)$.

For the ideal gas reaction,$X + Y \rightleftharpoons Z$,a mixture with $n_{X} = 1 \, mol$,$n_{Y} = 3 \, mol$ and $n_{Z} = 2 \, mol$ is at equilibrium at $300 \, K$ and $1 \, bar$. If the pressure is isothermally increased to $2 \, bar$,the number of moles of $X$ in the new equilibrium is closest to $......$

Using the data provided,find the value of the equilibrium constant for the following reaction at $298 \ K$ and $1 \ atm$ pressure: $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
$\Delta_{f} H^0(NO_{(g)}) = 90.4 \ kJ \cdot mol^{-1}$
$\Delta_{f} H^0(NO_{2(g)}) = 32.48 \ kJ \cdot mol^{-1}$
$\Delta S^{\circ} = -70.8 \ J \cdot K^{-1} \cdot mol^{-1}$
$\text{antilog}(6.4) = 2.51 \times 10^6$ (Note: Calculation based on standard thermodynamic relations)

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 $R = 8.314 \ J \ K^{-1} \ mol^{-1}$]

Two equilibria,$AB \rightleftharpoons A^{+} + B^{-}$ and $AB + B^{-} \rightleftharpoons AB_2^-$,are simultaneously maintained in a solution with equilibrium constants $K_1$ and $K_2$ respectively. The ratio of $[A^{+}]$ to $[AB_2^-]$ in the solution is

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$56 \ g$ of nitrogen and $8 \ g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \ g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen and ammonia are respectively: