Only N_(2) and CO_(2) gases remain after 15.5 g of carbon is treated with 25 litres of air at 25^(@)C and 5.5 atm pressure. Assume air of composition: O_(2)-19%,N_(2)-80% and CO_(2)-1% (by volume). Calculate the heat evolved under constant pressure Given, C+O_(2) to CO_(2),DeltaH=-94.05"kcal/mole" C+(1)/(2)O_(2) to CO,DeltaH=-26.41"kcal/mole"

Only N_(2) and CO_(2) gases remain after 15.5 g of carbon is treated w

1.	Only N_(2) and CO_(2) gases remain after 15.5 g of carbon is treated with 25 litres of air at 25^(@)C and 5.5 atm pressure. Assume air of composition: O_(2)-19%,N_(2)-80% and CO_(2)-1% (by volume). Calculate the heat evolved under constant pressure Given, C+O_(2) to CO_(2),DeltaH=-94.05"kcal/mole" C+(1)/(2)O_(2) to CO,DeltaH=-26.41"kcal/mole"
Answer» SOLUTION :Moles of `C=(15.5)/(12)=1.292` Moles of `O_(2)=(pV)/(RT)=(5.5xx(0.19xx25))/(0.0821xx298)=1.068` `C+O_(2) to CO_(2)` x moles (say) x moles `C+(1)/(2)O_(2) to CO` `(1.292-x)"moles" (1.292-x)"moles"` As `O_(2)` is fully consumed, number of moles of O before reaction = number of moles of O after reaction =moles of O in `CO_(2)+"moles"` of O in CO or `2xx1.068=2x+(1.292-x)` or `x=0.844` MOLE of `CO_(2)=0.844` Mole of `CO=1.292-0.844=0.448`. Total HEAT evolved `=0.844(-94.05)+0.448(-26.41)=-91.2"kcal"`