Two solid compounds A and B dissociate into gaseous products at `20^(@)C` as
a. `A(s) hArr A'(s)+H_(2)S(g)`
b. `B(s) hArr B'(g)+H_(2)S(g)`
At `20^(@)C` pressure over excess solid A is `50` mm and that over excess solid B is `68` mm. Find:
a. The dissociation constant of A and B
b. Relative number of moles of A' and B' in the vapour phase over a mixture of the solids A and B.
c. Show that the total pressure of gas over the solid mixture would be `84.4` mm.

Answer

A, B, C, D

Explanation

a. For dissociation of
`A(s) hArr A'(g)+H_(2)S(g)`
`:' P'_(A)+P'_(H_(2)S_((g)))=50`
`:. P'_(A)=25` mm and `P'_(H_(2)S_((g)))=25 mm [ :' ("Given" 1:1 "ratio")]`
`:. K_(p_(1))=P'_(A)xxP'_(H_(2)S_((g)))=25xx25=625 (mm)^(2)`
Similarly, for
`B(s) hArr B'(g)+H_(2)S(g)`
`K_(p_(2))=P'_(B)xxP'_(H_(2)S)=34xx34=1156 (mm)^(2)`
b. For mixture showing an equilibrium simultaneousely,
`{:(A(s),hArr,A'(g),+,H_(2)S,,B(s),hArr,B'(g),+,H_(2)S),(,,x,,x,,,y,,y),(,,,,,,,,,"Pressure after dissociation"):}`
`:. K_(P_(1))=x(x+y)` and `K_(P_(2))=y(x+y)`
`:. K_(P_(1))/K_(P_(2))=x/y :. x/y=625/1156`
`:.` V and T are same and thus, ratio of pressure is ratio of "moles", i.e.,
`n_(A')/n_(B')=625/1156=0.5406`
c. `:' x/y=625/1156` and `x(x+y)=625`
`:. x=14.81 mm` and `y=27.38 mm`
`:. P_(T)=P'_(A)+P'_(B)+P'_(H_(2)S)=x+y+(x+y)`
`=2(x+y)=2(14.81+27.38)`
`P_(T)=84.38 mm`