InterviewSolution
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InterviewSolution
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The values of `K_(p)` and `Kp_(2)` fot the reactions `XhArrY+Z` , (a) and `A hArr 2B` , (b) are in the ration of `9:1`. If the degree of dissociation of X and A is equal, then the total pressure at equilibriums (a) and (b) is in the rationA. `3:1`B. `1:9`C. `36:1`D. `1:1` |
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Answer» Correct Answer - C Let us assume that `P_(1)` and `P_(2)` are the total pressures at equilibrium for reactions (a) and (b), respectively. First consider reaction (a): `{:(,X hArr,Y+,Z),("Initial moles",1,0,0),("Equilibrium moles",1-alpha,alpha,alpha):}` where `alpha` is the degree of dissociation of X. `K_(p_(1))= (P_(y)P_(z))/(P_(x))` Partial pressure = (Mole fraction) x (Total pressure) `:. P_(x)= (1-alpha)/(1+alpha)P_(1)`, `P_(Y)=(alpha)/(1+alpha)P_(1)`, `P_(Z)=(alpha)/(1+alpha)P_(1)` `:. K_(P_(1))=(((alpha)/(1+alpha)P_(1))^(2))/((1-alpha)/(1+alpha)P_(1)) = (alpha^(2))/((1+alpha)(1-alpha))P_(1)` `= (alpha^(2))/(1-alpha^(2))P_(1)` Now consider reaction (b): `{:(,A hArr,2B),("Initial moles",1,0),("Equilibrium moles",1-alpha,2alpha):}` where `alpha` is the degree of dissociation of A. `P_(A) = (1-alpha)/(1+alpha)P_(2)`, `P_(B)= (2alpha)/(1+alpha)P_(2)` `K_(P_(2))=(P_(B)^(2))/(P_(A))=(((2alpha)/(1+alpha)P_(2))^(2))/(((1-alpha)/(1+alpha)P_(2)))` `= (4alpha^(2))/(1-alpha^(2))P_(2)` Let us take the ratio of `K_(P_(1))` and `K_(P_(2))` : `(K_(P_(1)))/(K_(P_(2)))=((alpha^(2))/(1-alpha^(2))P_(1))/((4alpha^(2))/(1-alpha^(2))P_(2))` `= (P_(1))/(4P_(2))` (as `alpha` of X and A are equal) Since `(K_(P_(1)))/(K_(P_(2)))=(9)/(1)`, we have `(P_(1))/(4P_(2))=(9)/(1)` or `(P_(1))/(P_(2))=(36)/(1)=36 : 1` |
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