InterviewSolution
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InterviewSolution
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The values of `K_(p1)` and `K_(p2)` for the two equilibrium reactions `X hArr +Z and A hArr 2B` are in the ratio 9, 1,. If degree of dissociation of X and A be equal , calculate the ratio of the total pressure of the equilibrium mixture in the two cases. |
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Answer» Let the degree of disscoiation in the two cases be `alpha` (i) Calculattion of `K_(p1)` `{:("For equilibrium", X,hArr, Y, +,Z),("initial moles" : ,1,,0,,0),("moles at eqm. point",(1:alpha),,alpha,,alpha):}` Total no. of ,moles `=1 -alpha +alpha+alpha=(1+alpha)` `K_(p1) = (P_(y)xxP_(z))/(P_(x)) = (((alpha)/(1+alpha)P_(1))xx((alpha)/(1+alpha)P_(1)))/(((1-alpha)/(1+alpha)P_(1)))` `=(P_(1)^(2)alpha^(2))/((1+alpha)^(2))xx((1+alpha)/((1-alpha)P_(1)))=(P_(1)alpha)^(2)/((1+alpha)(1-alpha))` (II) Calculation of `K_(p2)`. `{:("For equilibrium",A, hArr, 2B),("initial moles",1,,0),("Moles at eqm. point",(1-alpha),,2alpha):}` Total no. of moes `=1 alpha +2alpha =(1+alpha)` `K_(p2) = (pB)^(2)/(pA)=((2alpha)/(1+alpha)P_2)^(2)/(((1-alpha)/(1+alpha)P_(2)))` `P_(2)^(2) xx (4alpha^(2))/(1+alpha)^(2) xx ((1+alpha))/((1-alpha)P_(2))= (P_(2)xx4alpha^(2))/((1+alpha)(1-alpha))` Dividing eqn. (i) with eqn. (ii), `(K_p1)/(K_p2) = ((P_(1)alpha)^(2))/((1-alpha^(2)))xx ((1-alpha^(2)))/(P_(2)xx4alpha^(2)) = P_(1)/(4_(p_(2))` `" Now " " "(K_(p1))/(K_(p2))=9/1 ("given") :. P_(1)/(4P_(2))=9/1 or P_(1)/P_(2) = (36)/(1)` |
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