Consider
(i) `C(s)+O_(2)hArrCO_(2)(g) K_(P_(2)=(7)/(8)`
(ii) `2C(s)+O_(2)hArr2CO(g) K_(P_(2)=12.5atm`
As `100 L` of air (`80% N_(2),20% O_(2)` by volume) is pased over excess heated coke to establish these equilibrium the equilibrium mixture is found to measure `105 L` at constnat temperature & pressure (`105atm`). Assuming no other reaction, find the sum of partial pressure of `CO "and" CO_(2)` in the final equilibrium mixture.

Answer

15

Explanation

Since (1) causes no change in volume due to reaction change from `100 L` is due to (II) only.
`therefore V(CO)`finally`=10L`
`P(O)` finally`=((10)/(105))xx105=10L`
`therefore P(O_(2))=P_(CO)^(2)//K_(P_(2))=(10^(2)atm^(2))/(125atm)=`solutions
Also, `K_(P_(1))=(P_(CO_(2)))/(P_(O_(2))) therefore P(CO_(2))=KP_(1.)P(O_(2))`
`=(7)/(8)xx"solution"=7atm`
`therefore P(CO)+P(CO_(2))` finally is `8+7=15atm`