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JEE Main 2008
Exam
JEE Main
Year
2008
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2008 · Chemistry · Chemical Equilibrium

The equilibrium constants K_ P_1 and K_ P_2 for the reactions X...

The equilibrium constants KP1K_{P_1} and KP2K_{P_2} for the reactions X⇌2YX \rightleftharpoons 2 Y and Z⇌P+QZ \rightleftharpoons P+Q, respectively are in the ratio of 1:91: 9. If the degree of dissociation of XX and ZZ be equal then the ratio of total pressure at these equilibria is

Options

  1. A1:361 : 36
  2. B1:11 : 1
  3. C1:31 : 3
  4. D1:91 : 9

Correct answer

A. 1:361 : 36

Step-by-step solution

X⇌2Y10(1−x)2xkp1=(2x)2(1−x)(P11+x)1Z\rightleftharpoonsP+Q100(1−x)xxkp2=x2(1−x)(P21+x)14×P1P2=19⇒P1P2=136\begin{aligned}& X \rightleftharpoons2 Y \\ & 10 \\ & (1-x) \quad 2 x \\ & k_{p_1}=\frac{(2 x)^2}{(1-x)}\left(\frac{P_1}{1+x}\right)^1 \\ & Z \rightleftharpoonsP+Q \\ & 100 \\ & (1-x) \quad x \quad x \\ & \mathrm{k}_{\mathrm{p}_2}=\frac{\mathrm{x}^2}{(1-\mathrm{x})}\left(\frac{\mathrm{P}_2}{1+\mathrm{x}}\right)^1 \\ & \frac{4 \times \mathrm{P}_1}{\mathrm{P}_2}=\frac{1}{9} \Rightarrow \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{1}{36}\\ & \end{aligned}
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