# Carbon monoxide, (CO) at 300°C is burned under atmospheric pressure with dry air at the same temperature in 80% excess of that theoretically required. The products of combustion leave the reaction chamber at 800°C. Calculate the heat evolved in the reaction chamber in kcal per kg-mol of CO burned, assuming complete combustion. (Take the values of molal heat capacities at the intermediate temperatures).

Question-AnswerCategory: Material And Energy BalanceCarbon monoxide, (CO) at 300°C is burned under atmospheric pressure with dry air at the same temperature in 80% excess of that theoretically required. The products of combustion leave the reaction chamber at 800°C. Calculate the heat evolved in the reaction chamber in kcal per kg-mol of CO burned, assuming complete combustion. (Take the values of molal heat capacities at the intermediate temperatures).

Carbon monoxide, (CO) at 300°C is burned under atmospheric pressure with dry air at the same temperature in 80% excess of that theoretically required. The products of combustion leave the reaction chamber at 800°C. Calculate the heat evolved in the reaction chamber in kcal per kg-mol of CO burned, assuming complete combustion. (Take the values of molal heat capacities at the intermediate temperatures).

lets write the equation as:
CO + 0.5 O2 –> CO2
entering at 300C
used O2 = 1.8*0.5 = 0.9;
balancing on 1 mol basis:
CO + 0.9O2 + 3.386 N2 –> CO2 + 0.4O2 + 3.386 N2
enthalpy of input stream = heat of formation of CO + heat till 573 K of all other components.
= -110.53 KJ/mole + 1*30.3*275 + 0.9*31.83*275 + 3.386*29.93*275
= -110.53KJ +8332.5 +7877.93 + 27869.32
= -66450.25 J
enthalpy of output stream: (298 to 1073)
-393.15KJ + 1*49.27*775+ 0.4*32.85*775 + 3.386*30.675*775
= -393150 + 38184.25 +10183.5 + 80495.8
= -264286.45 J
enthalpy change = out – in = -264286.45 + 66450.25 = -197836.2 J
= -197.836 KJ /mole
= -47.33 Kcal/mol
for Kmol multiply numerator and denominator by 1000
thus answer is -47329.18 Kcal/Kg mol