उत्तर: (12)

समाधान:

सूत्र: वियोजन के मात्रा, $K_p$ की गणना

$ x(g) \leftrightharpoons y(g)+z(g) \quad k _{p_1}=3 $

प्रारंभिक मोल $\qquad\mathrm{n} \quad-\quad-$

साम्य पर $n-\alpha n \quad \alpha n \quad \alpha n$

$ \begin{aligned} & k _{p_1}=\frac{(\frac{\alpha}{1+\alpha} \times p_1)^{2}}{\frac{1-\alpha}{1+\alpha} p_1} \\ & 3=\frac{\alpha^{2} \times p_1}{1-\alpha^{2}} \\ & A(g) \leftrightharpoons 2 B(g) k _{p_2}=1 \end{aligned} $

$\begin{array}{lccl}\text { Initial mole } & \mathrm{n} & - & \ \text { at equilibrium } & \mathrm{x}-\alpha \mathrm{n} & 2 \alpha \mathrm{n} & \mathrm{p}_{\text {total }}=\mathrm{p}_2\end{array}$

$p _{\text{total }}=p_2$

$k _{p_2}=\frac{(\frac{2 \alpha}{1+\alpha} \times p_2)^{2}}{\frac{1-\alpha}{1+\alpha} \times p_2}$

$1=\frac{4 \alpha^{2} \times p_2}{1-\alpha^{2}}$

$\frac{k _{p_1}}{k _{p_2}}=\frac{p_1}{4 p_2}$ $\frac{3}{1}=\frac{p_1}{4 p_2}$ $\therefore p_1: p_2=12: 1$ $x=12$