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Question

The values of $${K}_{{p}_{1}}$$ and $${K}_{{p}_{2}}$$ for the reactions:

(i)  $$X \rightleftharpoons Y + Z$$                   
(ii) $$A \rightleftharpoons 2B$$
They are in the ratio $$9 : 1$$. Assuming degree of dissociation of $$X$$ and $$A$$ be same, the dissociation pressure at equilibrium (i) and (ii) are in the ratio :


A
36:1
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B
1:1
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C
3:1
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D
1:9
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Solution

The correct option is C $$36 : 1$$
                         $$X   \ \ \   \rightleftharpoons  \ \ \  Y \ \   +   \ \ \ Z$$
                     $$1 - \alpha $$           $$  \alpha $$          $$ \alpha$$

For (i) $${K}_{{p}_{1}} = \displaystyle\frac{{n}_{Y}\cdot {n}_{Z}}{{n}_{X}} \displaystyle\frac{{p}_{1}}{{\left[\sum{n}\right]}_{1}} = \displaystyle\frac{{\alpha}^{2}}{\left(1 - \alpha\right)}\cdot \displaystyle\frac{{P}_{1}}{\left(1 + \alpha\right)}$$

                         $$A\ \ \    \rightleftharpoons  \ \ \ \ 2B$$

                      $$1 - \alpha $$           $$ 2\alpha$$

For (ii) $${K}_{{p}_{2}} = \displaystyle\frac{{\left({n}_{B}\right)}^{2}}{\left({n}_{A}\right)}\times \displaystyle\frac{{P}_{2}}{{\left[\sum{n}\right]}_{2}} = \displaystyle\frac{4{\alpha}^{2}}{\left(1 - \alpha\right)}\cdot \displaystyle\frac{{P}_{2}}{\left(1 + \alpha\right)}$$

$$             \displaystyle\frac{{K}_{{p}_{1}}}{{K}_{{p}_{2}}} = \displaystyle\frac{{\alpha}^{2}\times {P}_{1}}{\left(1 - \alpha\right)\times \left(1 + \alpha\right)}\times \displaystyle\frac{\left(1 - \alpha\right)\left(1 + \alpha\right)}{4{\alpha}^{2}\times {P}_{2}}$$
$$           \displaystyle\frac{{P}_{1}}{{P}_{2}} = \displaystyle\frac{4\times 9}{1} = \displaystyle\frac{36}{1}$$

Chemistry

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