Carbon monoxide at 50 degree F is completely burned at 1 atm pressure with 50 %
ID: 499780 • Letter: C
Question
Carbon monoxide at 50 degree F is completely burned at 1 atm pressure with 50 % excess air which is at 1000 degree F. The products of combustion leave the combustion chamber at 800 degree F. Calculate the heat evolved from the combustion chamber expressed as British thermal units per pound of CO entering. (remember: T(degree F) = 1.8 times T(degree C) + 32, therefore 25 degree C is ... degree F) Data: Standard heat of combustion of CO = -67, 636 cal/g-mol = 121, 745 Btu/lb-mol Enthalpy (Btu/lb-mol); reference is 32 degree F: Adapted from Himmelblau "Basic Principles and Calculations in Chemical Engineering" - Application of the energy balance to a reactionExplanation / Answer
Basis : 1 lb of CO, the reactino is CO+0.5O2--àCO2
Mass balances :
Moles of CO= mass/molar mass= 1/28=0.036, As per the stoichiometry of the reaction, moles of oygen required= 0.5 per moles. hence moles of oxygen required =0.036*0.5= 0.018, moles of air to be supplied ( since air contains 21% O2 and 79%N2) = 0.018/0.21=0.086, moles of air supplied= 1.5*0.086= 0.129 moles
Products ( moles) : CO2= 0.036, N2=0.129*0.79= 0.102, O2= 0.129*0.21-0.018= 0.009
Energy balance: since the referecne temperature is 25 deg.c (77 deg.F), enthalpy will have to be calculated with reference to this temperature.
Enthalpy = (Enthalpy at which the reactant is entering or produc is levaing- enthalpy at 77 deg.F)* no of moles of that particular chemicals
Reactants (Btu) : (313.3-125.2) btu/lbmole*0.036=6.8, Air = (6984-312.7)*0.129=861
Enthalpy of reactants = 6.8+861= 867.8 btu
Products :Enthalpy ( Btu): CO2=(8026-392.2)*0.036=275, N2= (5443-312.2)*0.102=523, O2=(5690-315.1)*0.009=48
Enthapy of products= 275+523+48=846 btu
Standard heat of combustion = -121745 Btu/lb mol*0.036=-4382 Btu
Heat of reaction= enthalpy of products+ standard heat of reaction- enthalpy of reactants
=846-4382-867.8=-4403.8 Btu