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The values of K_(p) and K_(p_(2)) for the reaction X hArr Y+Z ...(1) and A hArr 2B...(2) Are in the ratio 9: 1 . It the degree of dissociationof X and A be equal , calculate the ratio of the total prssures of (1) and (2) at equilibrium. |
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Answer» Solution :Suppose total pressure at EQUILIBRIUM for reactions (1) and (2) are `P_(1)and P_(2)` respectively . Then ` {:(,X,hArr,Y,+,Z),("Intial.",1"MOLE",,0,,0),("At EQM.",1-alpha,,alpha,alpha,"Total"1+alpha):}` ` 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)) = ((apha)/(1+alpha) P_(1))^(2)/((1-alpha)/(1+alpha)P_(1))=(alpha^(2)P_(1))/(1- alpha^(2) )CONG alpha^(2) P_(1) ` `{:(,A,hArr,2B,),("Intial",1"mole",,0,),("At eqm.",1-alpha,,2 alpha,"Total"=1=alpha):}` ` p_(A) = (1-alpha)/(1 +alpha)p^(2),p^(B) = (2alpha)/(1 + alpha) P^(2)` ` K_(p_(2)) = ((2alpha)/(1+alpha)P_(2))^(2)/((1-alpha)/(1+alpha)P^(2))= (4 alpha^(2))/(1-alpha^(2)) P_(2) = 4 alpha ^(2) P_(2) ` ` (K_(p_(1)))/(K_(p_(2)))= (alpha^(2) P_(1))/(4alpha^(2)P_(2)) = (P_(1))/(4 P_(2))= 9/1 ("Given") or (P_(1))/(P_(2))= 36 /1 = 36 :1 ` |
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