The equilibrium of three reactions with their respective equilibrium constants are given below,

\( \begin{array}{lc} \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 & \mathrm{~K}_1 \\ \mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} & \mathrm{K}_2 \\ \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{O} & \mathrm{K}_3 \end{array} \)

The equilibrium constant (K) of the following reaction will be : 
$2{\mathrm{NH}}_3+\frac{5}{2}{\mathrm{O}}_2$ $\stackrel{\mathrm{K}}{\rightleftarrows }$ $2\mathrm{NO}+3{\mathrm{H}}_2\mathrm{O}$

1. \( \frac{K_{2}K_{3}^{3}}{K_{1}}\)
2. \( \frac{K_{2}K_{3}}{K_{1}}\)
3. \( \frac{K_{2}^{3}K_{3}}{K_{1}}\)
4. \( \frac{K_{3}^{3}K_{1}}{K_{2}}\)

A 20-litre container at 400 K contains CO₂ (g) at pressure 0.4 atm and an excess of SrO (neglecting the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container.
The maximum volume of the container, when the pressure of CO₂ attains its maximum value will be:
(Given that: SrCO₃(s) ⇋ SrO(s) + CO₂(g) , Kp = 1.6 atm)