For a gaseous phase reaction `A+2B hArr AB_(2), K_(c)=0.3475 L^(2) "mole"^(-2)` at `200^(@)C`. When `2` moles of B are mixed with one "mole" of A, what total pressure is required to convert `60%` of A in `AB_(2)`?

Answer

A

Explanation

`{:(,A,+,B,hArr,AB_(2)),("Initial moles",1,,2,,0),("moles at",(1-x),,(2-2x),,x),("equilibrium",,,,,):}`
Total moles at equilibrium `=1-x+2-2x+x=3-2x`
Let pressure at equilibrium br P,
Now,
`P'_(AB_(2))=[x/(3-2x)]P, P'_(A)=[(1-x)/(3-2x)]P, P'_(B)=[(2-2x)/(3-2x)]P`
`:. K_(p)=(x.P)/((3-2x).P ((1-x))/((3-2x)).P^(2) ((2-2x)^(2))/((3-2x)^(2)))`
`K_(P)=(x.(3-2x)^(2))/(P^(2)(1-x)(2-2x)^(2))` ...(i)
Alternate to derive `K_(p)` or equation (i),
`:' K_(p)=n_(AB_(2))/(n_(A)xx(n_(B))^(2))xx(P/(Sigman))^(Deltan)`
`:. K_(p)=(x)/((1-x)(2-2x)^(2))xx[P/((3-2x))]^(-2)`
`=(x(3-2x)^(2))/((1-x)(2-2x)^(2).p^(2))` ...(i)
Given that, `x=0.6` and `Deltan=-2`
`:' K_(P)=K_(C)(RT)^(Deltan)=0.3475xx(0.0821xx473)^(-2)` ...(ii)
by equations (1) and (2),
`:. 0.3475xx(0.0821xx473)^(-2)=(0.6(3-1.2)^(2))/(P^(2)(1-0.6)(2-1.2)^(2))`
`=(0.6xx(1.8)^(2))/(P^(2)(0.4)(0.8)^(2))`
`:. P=181.5 "atm"`