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Following two equilibria are established...

Following two equilibria are established on mixing two gases `A_(2)` and `C`.
i. `3A_(2)(g) hArr A_(6)(g) " " K_(p)=1.6 atm^(-2)`
ii. `A_(2)(g)+C(g) hArr A_(2)C(g)`
If `A_(2)` and `C` mixed in `2:1` molar, ratio calculate the equilibrium partial pressure of `A_(2)`, C, `A_(2)C` and `K_(p)` for the reaction (ii). Given that the total pressure to be `1.4` atm and partial pressure of `A_(6)` to be `0.2` atm at equilibrium

Text Solution

Verified by Experts

`3A_(2)(g) hArr A_(6)(g)`
`K_(p)=1.6=(P'_(A_(6)))/((P_(A_(2)))^(3))`
`P_(A_(2))=root(3)(0.2/16)=0.5 "atm"`
Also pressure of `A_(2)` used for the formation of `A_(6)=0.6 "atm"`
`{:("For",A_(2)(g),+,C(g),hArr,A_(2)C(g)),(t=0,2P,,P,,0),("At Eq",2P-P'-0.6,,P-P',,P'),("Also for",3A_(2)(g),,hArr,A_(6)(g),),(t=0,2P,,,0,),("At Eq",2P-P'-0.6,,,0.2,):}`
(since `P_(A_(2))` at equilibrium is `0.5` for simultaneous equilibria)
Also pressure of `A_(2)+C+A_(2)C+A_(6)`
`=(2P-P'-0.6)+(P-P')+P'+0.2=1.4`
`0.5+P+0.2=1.4`
`P=0.7 "atm"`
`:. 2P-P'-0.6=0.5`
`:. P'=2xx0.7-0.6-0.5`
`P'=0.3 "atm"`
`:. P_(A_(2))=0.5 "atm", P_(C)=0.7-0.3=0.4 "atm", P_(A_(2)C)=0.3 "atm"`
Also `K_(p)` for `A_(2)(g)+C(g) hArr A_(2)C(g)`
`K_(p)=(P_(A_(2)C))/(P_(A_(2))xxP_(C))=0.3/(0.5xx0.4)=1.5 "atm"^(-1)`
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