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

Answer

15

Explanation

`3A_2(g)hArrA_6(g) " " K_P=1.6 atm^(-2)`
`K_P=1.6=P'_(A_2)/(P_(A_2))^3`
`P_(A_2)=root(3)(0.2/1.6)=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_2C(g)),(t=0,2P,P,0),(At eq.,2P-P^1-0.6,P-P^1,P^1),("Also for",3A_2(g)hArr,A_6(g), ),(t=0,2P,0,),("At eq.",2P-P^1-0.6,0.2,):}`
`2P-P^1-0.6=0.5" "("since"P_(A_2)"at eq. is 0.5 for similataneous equilibria")`
Also pressure of `A_2+C+A_2C+A_6`
`=(2P-P^1-0.6)+(P-P^1)+P^1+0.2`
=1.4
0.5+P+0.2=1.4
P=0.7 atm
`:. 2P-P^1-0.6=0.5`
`:. " " P^1=2xx0.7-0.6-0.5`
`P^1=0.3` atm
`:. P_(A_2)=0.5 atm , P_(C )=0.7-0.3=0.4 atm , P_(A_2C)=0.3` atm
for `K_P` for `A_2(g)+C(g)hArr A_2C(g)`
`K_P=P_(A_2C)/(P_(A_2))xxP_C=0.3/(0.5xx0.4)=1.5 "atm"^(-1)`