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Two solid X and Y dissociate into gaseou...

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`.

Text Solution

Verified by Experts

`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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