Sulfur dioxide is oxidized to sulfur trioxide in a continuous flow reactor. SO_2

Question

Sulfur dioxide is oxidized to sulfur trioxide in a continuous flow reactor. SO_2 and 100% excess air are fed to the reactor at 450 degree C. The reaction proceeds to a 65% SO_2 conversion, and the product stream emerge from the reactor at 550 deree C. The production rate of SO_3 is 1.00 times 10^2 kg middod min^-1. The reactor is surrounded by a water jacket into which water at 25 degree C is fed. Calculate the feed rates of the SO_2 and air feed streams. Calculate the heat (kW) that must be transferred from the reactor to the cooling water.

Explanation / Answer

The reaction is SO2+0.5O2--àSO3

1 mole of Sulfur requires 0.5 mloles of oxygen and produces 1 mole of SO3.

Production of SO3 is 100 kg/min

Moles of SO3 produced = mass/molar mass = 100*1000/80 =1250 gmoles/min

SO2 required as per the reaction = 1250 gmoles/min

But conversion of SO2 is only 65%. Hence moles of SO2 required =1250/0.65=1923 gmoles/min

Moles of oxygen required = 1923/2, but oxygen supplied is 100% excess. Moles of oxygen supplied =

(1923/2)*(1+1)= 1923 moles/min, since air contains 21%O2 and 79%N2, moles of air supplied = 1923/0.21= 9157 gmoles/min= 9.157 kgmoles/min

Feed rates : SO2= 1.923 kgmoles/min = 1.923*64 kg/min =123 kg/min and Air =9.157*29 kg/min=265.5 kg/min

2. Enthalpy of reactants

Average specific heats : SO2= 4.796*8.314 J/mole.K =39.88 J/mole.K and Air = 3.509*8.314 joules/mole.K= 29.2 J/mole.K

the reactants need to e brought to 25 deg.c to supply standard heat of reaction

hence enthalpy of reactants = moles of reactant* specific heat* temperature difference=

1.923*1000*39.88*(25-450)+ 9.157*1000*29.2*(25-450) =-1.46*108 J/min = -1.46*105 Kj/min

Heat of reaction for the reaction SO2+0.5O2-àSO3 is -396 Kj/mole

Heat of reaction = 1250 moles/min*(-396 KJ/mole) =-495000 KJ/min =-4.95*105 Kj/min

Products are at 550 deg.c. They contains

N2 = 9157*0.79 gmoles/min=7234 gmoles/min , O2= O2 supplied-O2 converted = 1923- 1250/2 = 1298 moles/min, SO2= 1923-1250 = 673 gmoles/min, SO3= 1250 gmoles/min

CP data for Products : N2= 3.502*8.314 J/mole.K= 29.12 J/mole.K, O2= 3.535*8.314=29.38 Joules/mole.K, SO2= 39.88 J/mole.K, SO3= 6.092*8.314 Joules/mole.K= 50.65 J/mole.K

Enthalpy of products = (7234*29.12+1298*29.38+673*39.88+1250*50.65)*(550-25) joules/min=1.78*108 Joules/min = 1.78*105 Kj/min

Q= heat transferred = enthalpy of products+ heat of reaction + enthalpy of products

= (-1.46-4.95+1.78)*105 Kj/min=-4.63*105 KJ/min

Heat need to be transferred to cooling water= 4.63*105/60 Kj/sec =7717 Kj/sec= 7717 Kw

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