Question

Ethylene (C2H4) gas enters a well-insulated reactor and reacts completely with 400% of theoretical air, each at 25°C, 2 atm. The products exit the reactor at 2 atm. For operation at steady state and ignoring kinetic and potential energy effects, determine (a) the balanced reaction equation, (b) the temperature, in K, at which the products exit, (c) the rate of entropy production, in kJ/K per kmol of ethylene.

Answer 1

Answer:see explanation.

Explanation:

(a).The equation of reaction is;

C2H4 + 3(O2 + 3.76N2) -----> 2CO2 + 2H2O + 11.28 N2.

Theoretical air is the amount of air that will allow the complete combustion of the fuel. Air contains 21% of oxygen,79% of Nitrogen. Thus, the molecular mass of air = 29 [Kg/Kmol] . The number of oxygen needed to oxidize the hydrocarbon(C2H4) is 79/21= 3.76 moles of Nitrogen.

(b). N(total) =2+2+11.28 =15.28 kmoles of product

15.28[h(T) - h°] of air

= 15.28(Mass of air) (Cp,1000k)

T(adiabatic) = 802,310 kJ/k. Mole.

When mass of air= 29 Kg/ k.mol

Cp,1000k = 1.142 kJ.k which is the specific heat capacity of air.

T(adiabatic) = 1275k -----> assuming all the products are air.

(C). 2{∆h} co2 + 2{∆h} H20 +11.28 {∆h} N2

After series of calculations and checking of tables:

S(o2) = 320.173 - 8.314 ln 0.12= 337.46 kJ/ K.mol.k

S(H2O)= 273.986- 8.314 ln 0.1406 = 290.30 KJ/ kmol. K

S(N2) = 258.503- 8.314 ln 0.7344= 261.07 kJ/kmol.k

= 2(337.46) + 2(290.30)+11.28(261.07)-360.79-3(218.01)-11.28(193.46)

=1003.3378 kJ/kmol. K