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.

Consider (i) C(s)+O_(2)hArrCO_(2)(g) K_(P_(2)=(7)/(8) (ii) 2C(s)+O_(

1.	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» Solution :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`