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Chemical Equilibrium question

2008 · Q117
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Chemical Equilibrium question

2008 · Q117

NEETChemistryChemical EquilibriumMCQ+4 / −1
The values of for the reactions,

X ⇌\rightleftharpoons⇌ Y + Z      . . . .(i)
A ⇌\rightleftharpoons⇌ 2B       . . . .(ii)

are in the ratio 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (i) and (ii) are in the ratio
  1. A
    36 : 1
  2. B
    1 : 1
  3. C
    3 : 1
  4. D
    1 : 9
View written solutionFree

Correct answer: A

Given

X ⇌\rightleftharpoons⇌ Y + Z      . . . .(i)

A ⇌\rightleftharpoons⇌ 2B       . . . .(ii)

Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then

KP1KP2=91{{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {9 \over 1}KP2​​KP1​​​=19​

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X⇌ Y+Z
Initial mole100
At equilibrium1 - $\alpha $$\alpha $$\alpha $


Total number of moles at equilibrium

= 1 - $\alpha $ + $\alpha $ + $\alpha $ = 1 + $\alpha $

$ \therefore $ KP1 = $${{{P_Y} \times {P_Z}} \over {{P_X}}}$$ = $${{{\alpha \over {1 + \alpha }} \times {P_1} \times {\alpha \over {1 + \alpha }} \times {P_1}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_1}}}$$

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A⇌2B
Initial mole10
At equilibrium1 - $\alpha $2$\alpha $


Total number of moles at equilibrium

= 1 - $\alpha $ + 2$\alpha $ = 1 + $\alpha $

$ \therefore $ KP2 = $${{{{\left( {{P_B}} \right)}^2}} \over {{P_A}}}$$ = $${{{{\left( {{{2\alpha } \over {1 + \alpha }} \times {P_2}} \right)}^2}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_2}}}$$

$ \therefore $ $${{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {{{P_1}} \over {4{P_2}}}$$

$ \Rightarrow $ $${{{P_1}} \over {4{P_2}}} = {9 \over 1}$$

$ \Rightarrow $ $${{{P_1}} \over {{P_2}}} = {{36} \over 1}$$
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