InterviewSolution
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InterviewSolution
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Two solid `X` and `Y` dissociate into gaseous products at a certain temperature as followas: `X(s)hArrA(g)+C(g)` , and `Y(s)hArrB(g)+C(g)` At a given temperature, the pressure over excess solid `X` is `40 mm` and total pressure over solid `Y` is `80 mm`. Calculate a. The value of `K_(p)` for two reactions. b. The ratio of moles of A and B in the vapour state over a mixture of `X` and `Y`. c. The total pressure of gases over a mixture of `X` and `Y`. |
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Answer» `X(S)hArrA(g)+C(g)` At wquilibrium, `A` and `C` are in equal proportions, so their pressures will br same. `p_(A)=p_(C )` Also `p_(A)+p_(C )=(20)^(2)=400 mm^(2)` `Y(s)hArrB(g)+C(g)` `p_(B)=p_(C )=40 mm(p_(B)+p_(C )=80)` rArr `K_(p)=p_(B).p_(C )=40^(2)=1600 mm^(2)` b. Now for a mixture of `X` and `Y`, we will have to consider both the equilibrium simultaneously. `X(s)hArrA(g)+C(g)` `Y(s)hArrB(g)+C(g)` Let `p_(A)=a mm, p_(B)=b mm` Note that the pressure of `C` due to dissociation of `X` will also be a mm and similarly the pressure of `C` due to dissociation of `Y` will also be `b mm`. `rArr p_(C )=(a+b) mm` `K_(p)(for X)=p_(A).p_(C )=a(a+b)=400 ...(i)` `K_(p)(for Y)=p_(B).p_(C )=(a+b)=1600 ...(ii)` From (i) and (ii), we get: `a/b=1/4` as volume and temperature are constant, the "mole" ratio will be same as the pressure ratio. c. The total pressure `=P_(T)=p_(A)+p_(B)+p_(C )` `=a+b+(a+b)=2(a+b)` Adding (i) and (ii) `a+b=sqrt(K_(PX)+K_(PY))=sqrt(2000)=20sqrt(5)mm` rArr Total pressure `=2(a+b)=89.44 mm` |
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