`K_c" for "N_2O_4 (g) Leftrightarrow 2NO_2 (g)` is 0.00466 at 298K. If a one-litre container initially contained 0.8 mole of `N_2O_4`, what would be the concentrations of `N_2O_4 and NO_2` at equilibrium? Also, calculate the equilibrium concentration of `N_2O_4` and `NO_2`, if the volume is halved is halved at the same temperature.

Explanation

To solve the problem, we need to find the equilibrium concentrations of \( N_2O_4 \) and \( NO_2 \) in two scenarios: first, when the initial concentration of \( N_2O_4 \) is 0.8 moles in a 1-litre container, and second, when the volume is halved.

### Step 1: Write the equilibrium expression

The equilibrium expression for the reaction:

\[ N_2O_4 (g) \leftrightarrow 2 NO_2 (g) \]

...