NEET UG 2008 — Chemistry Organic Chemistry

From equation. $\begin{array}{lcc} & X & \rightleftharpoons & Y & + & Z \\ \text {Initial mole } & 10 & & 0 & & 0 \\ \text {mole at equilibrium } & (1-a) & & a & & a \end{array}$Initial mole mole at equilibrium ​X10(1−a)​⇌​Y0a​+​Z0a​ $$\begin{aligned} & K_{p_1}=\frac{p_Y \times p_z}{p_X} \\ & =\frac{\left[\frac{\alpha \times p_1}{1+\alpha}\right]\left[\frac{\alpha \times p_1}{1+\alpha}\right]}{\left[\frac{1-\alpha}{1+\alpha}\right] p_1} \\ & K_{P_1}=\frac{\alpha^2 p_1}{1-\alpha^2} \end{aligned}$$​Kp1​​=pX​pY​×pz​​=[1+α1−α​]p1​[1+αα×p1​​][1+αα×p1​​]​KP1​​=1−α2α2p1​​​ From equation $\begin{array}{lcc} & A & \rightleftharpoons & 2 B \\ \text {Initial mole } & 10 & & 0 \\ \text {mole at equilibrium } & (1-a) & & 2 a \end{array}$Initial mole mole at equilibrium ​A10(1−a)​⇌​2B02a​ $$\begin{aligned} & K_{p_2}=\frac{\left[\frac{2 \alpha}{1+\alpha} \cdot p_2\right]^2}{\left[\frac{1-\alpha}{1+\alpha}\right] p_2}=\frac{4 \alpha^2 p_2}{1-\alpha^2} \end{aligned}$$​Kp2​​=[1+α1−α​]p2​[1+α2α​⋅p2​]2​=1−α24α2p2​​​ From Eqs (i) and (ii) $\frac{K_{P_1}}{K_{P_2}}=\frac{p_1}{4 p_2}$KP2​​KP1​​​=4p2​p1​​ $$\begin{array}{ll} \Rightarrow & \frac{9}{1}=\frac{p_1}{4 p_2} \\ \Rightarrow & \therefore \frac{p_1}{p_2}=\frac{36}{1} \end{array}$$⇒⇒​19​=4p2​p1​​∴p2​p1​​=136​​