Based on the given equilibrium constants:
\( \begin{array}{lc} \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 & \mathrm{~K}_1 \\ \mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} & \mathrm{K}_2 \\ \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{O} & \mathrm{K}_3 \end{array} \)
The equilibrium constant (K) of the following reaction will be :
\(2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \stackrel{\mathrm{~K}}{\rightleftharpoons} 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}\)
1. | \( \dfrac{K_{2}K_{3}^{3}}{K_{1}}\) | 2. | \( \dfrac{K_{2}K_{3}}{K_{1}}\) |
3. | \( \dfrac{K_{2}^{3}K_{3}}{K_{1}}\) | 4. | \( \dfrac{K_{3}^{3}K_{1}}{K_{2}}\) |
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)
1. | 10 litre | 2. | 2 litre |
3. | 4 litre | 4. | 5 litre |