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  • The equilibrium constants    for the reaction <img alt="X\rightleftharpoons 2

The equilibrium constants K_{p_1} \: and\: K_{p_2}   for the reaction X\rightleftharpoons 2Y\: and \: Z\rightleftharpoons P+Q ,respectively are in the ratio of 1 : 9. If degree of dissociation of X \: and\:Z  be equal then the ratio of total pressures at these equilibria is

Option: 1

1:9


Option: 2

1:36


Option: 3

1:1


Option: 4

1:3


Answers (1)

best_answer

K_{p_{1}}=\frac{p_{y}^{2}}{P_{X}}=\frac{\left ( \frac{2\alpha }{1+\alpha }P_{1} \right )^{2}}{\left ( \frac{1-\alpha }{1+\alpha }P_{1} \right )}

K_{p_{2}}=\frac{P_{p}P_{Q}}{P_{Z}}=\frac{\left ( \frac{\alpha }{1+\alpha }P_{2} \right )\left ( \frac{\alpha }{1+\alpha }P_{2} \right )}{\left ( \frac{1-\alpha }{1+\alpha }P_{2} \right )}

\Rightarrow \; \; K_{p_{1}}=\frac{4\alpha ^{2}P_{1}}{1-\alpha ^{2}}........(i)

\Rightarrow \; \; K_{p_{1}}=\frac{4\alpha ^{2}P_{1}}{1-\alpha ^{2}}........(ii)

Given is  \frac{K_{p_{1}}}{K_{p_{2}}}=\frac{1}{9}\; ........(iii)

Substituting values of from equation (i)  &  (ii)  into  (iii)  , we get

\frac{\frac{4\alpha ^{2}P_{1}}{1-\alpha ^{2}}}{\frac{\alpha ^{2}P_{2}}{1-\alpha ^{2}}}=\frac{1}{9}\Rightarrow \frac{4P_{1}}{P_{2}}=\frac{1}{9}\Rightarrow \frac{P_{1}}{P_{2}}=\frac{1}{36}

 

Posted by

Shailly goel

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