For the reaction $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$,$2 \ mol$ each of $A$ and $B$ are taken in a flask. Which of the following will always be true when the system reaches equilibrium?

For which of the following reactions will $K_p < K_c$?

$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Hence,the resultant pressure is: (in $atm$)

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

An equilibrium shifts towards reactants at higher temperatures. Find the correct graph.

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)