For the reaction X hArr 2Y and Z hArr P+Q occuring at two different pressure P_(1) and P_(2), respectively. The ratio of the two pressure is 1:3. What will be the ratio of equilibrium constant, if degree of dissociation of X and Z are equal.

For the reaction X hArr 2Y and Z hArr P+Q occuring at two different pr

1.	For the reaction X hArr 2Y and Z hArr P+Q occuring at two different pressure P_(1) and P_(2), respectively. The ratio of the two pressure is 1:3. What will be the ratio of equilibrium constant, if degree of dissociation of X and Z are equal.
Answer» `1:36` `1:12` `1:9` `2:3` SOLUTION :`{:("for reaction",X,hArr,2Y),("Initial",1,,0),("At equilibrium",1-alpha,,2alpha):}` Total "moles" `=1-alpha+2alpha=1+alpha` `P_(X)=((1-alpha)/(1+alpha))P_(1), P_(Y)=((2alpha)/(1+alpha))P_(1)` `:. K_(p_(1))=((P_(Y))^(2))/(P_(X))=([((2alpha)/(1+alpha))P_(1)]^(2))/(((1-alpha)/(1+alpha))P_(1))=(4alpha^(2)P_(1))/((1-alpha^(2)))` ...(i) `{:("Similarly for",X,hArr,P,+,Q),("Initial",1,,0,,0),("At equilibrium",1-alpha,,2alpha,,alpha):}` Total moles `=1-alpha+alpha+alpha=1+alpha` `P_(Z)=((10alpha)/(1+alpha))P_(2), P_(Q)=(alpha/(1+alpha))P_(2), P_(P)=(alpha/(1+alpha))P_(2)` `:. K_(p_(2))=(P_(P)xxP_(Q))/P_(Z)=((alpha/(1+alpha))P_(2)xx(alpha/(1+alpha))P_(2))/(((1-alpha)/(1+alpha))P_(2))` `=(alpha^(2)P_(2))/((1-alpha^(2)))` DIVIDING equations (i) and (ii), we get, `K_(p_(1))/K_(p_(2))=(4alpha^(2)P_(1))/((1-a^(2)))xx((1-alpha^(2))/(alpha^(2)P_(2)))=(4P_(1))/P_(2)rArr (4P_(1))/P_(2)=1/3` (given) `:. P_(1)/P_(2)=1/12`