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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The equilibrium constants K_ P 1 and K_ P 2 for the reaction (X...

The equilibrium constants KP1K_{P 1} and KP2K_{P 2} for the reaction (X \rightleftharpoons2 Y) and (Z ⇌\rightleftharpoonsP+Q), respectively are in the ratio of 1:1: 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(g)⇌2Y(g)\text{X}( \text{g} ) ⇌ \text{2Y}( \text{g} )

& & & & & & & & & & & 1 & & & & & & & & 0&

At equilibrium & & 1−x1 - \text{x} & & & & 2x\text{2x}

Total no. of moles = 1+x1 + \text{x}

{K}_{{p}_{1}} \text{=}\dfrac{{\left[\dfrac{\text{2x}}{\text{1+x}} {\text{.p}_{1}\right]}^{2}}{\left(\dfrac{\text{1} - \text{x}}{\text{1+x}}\right){\text{.p}_{1}} \text{=}\dfrac{{\text{4x}^{2}}{\left(\text{1} - \text{x}\right)} \dfrac{{p}_{1}}{\left(\text{1+x}\right)} \text{=}\dfrac{{\text{4x}^{2} {p}_{1}}{\left(\text{1} - x^{2}\right)}

Z(g)⇌P(g)+Q(g)\text{Z}( \text{g} ) ⇌ \text{P}( \text{g} )\text{+Q}( \text{g} )

& & & & & & & & & & & 1 & & & & & & & & & & & & & & 0 & & & & & & 0

At equilibrium & 1−x1 - \text{x} & & & & & & x\text{x} & & x\text{x}

Similarly, Kp2=x2(1−x)p2(1+x)=x2p2(1−x2){K}_{{p}_{2}} \text{=}\dfrac{x^{2}}{\left(\text{1} - \text{x}\right)} \dfrac{{p}_{2}}{\left(\text{1+x}\right)} \text{=}\dfrac{x^{2} {p}_{2}}{\left(\text{1} - x^{2}\right)}

∴Kp1Kp2=19∴ \dfrac{{K}_{\text{p1}}}{{K}_{\text{p2}}} \text{=}\dfrac{1}{9}

So, 4\timesp1p2=19\dfrac{\text{4}\text{\times}{p}_{1}}{{p}_{2}} \text{=}\dfrac{1}{9}

⇒p1p2=136⇒ \dfrac{{p}_{1}}{{p}_{2}} \text{=}\dfrac{1}{\text{36}}
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