Question

Pure sulfur is burned with 63% excess air according to the following reaction.
S + O2 = SO2 The furnace gases formed enter a reactor under a pressure of 1 atm at 500 ° C.
80% of SO2 is converted into SO3 by SO2 +1/2 O2 =SO3 reaction. According to this,
a) Calculate the equilibrium constant Kp of the SO2 +1/2 O2 = SO3 reaction. (10 points)
b) What should be the reactor pressure to increase the conversion rate of SO2 to 95% at the same temperature?

Answer 1

the problem is based on the concepts of thermodynamics and stoichiometry.

we know that for a reversible reaction

the equilibrium constant is written as:

these are concentration

and for the gaseous reaction the Kp is given as,

here, ng = moles of gaseous products - moles gaseous reactants

the ideal gas law in terms of concentration is given as,

here,

P = pressure

C =concentration

T = temperature

R = 8.314J/mol.K

these equations will be used to solve the problem.

the data given to us in the problem is:

1. excess oxygen take is 63%

2. conversion of SO2 in reactor = X = 80%

3. Pressure of reactor inlet = P = 1 atm * 101325PA/1atm = 101325Pa

4. temperature = 500 C+ 273 = 773 K.

we are given that pure S is reacted with 63% oxygen to produce SO2, the reaction is :

so, let 'Y' mol/m3 concentration os sulfur is taken initially.

so, from the stoichiometry, 1 mol/m3 od S reacts with 1 mol/m^3 of O2, so

S mol/m3 of sulfur reacts with = S mol/m3 of O2.

moles of O2 coming in with reactant = S + 63%of S = S + 0.63S = 1.63S mol/m3

hence, in the outlet stream from the furnace the components will be ;

S = 0 mol/m3

O2 = moles of O2 inlet - moles of O2 reacted = 1.63S - S = 0.63S mol/m3

SO2 = S mol/m^3

so, the total concentration of the gas = SO2 + O2 = S + 0.63S mol/m3 = 1.63S mol/m3

now we use the ideal gas equation,

we get

C = 15.766 mol/m3.

this total concentration is equal to the total concentration os gases in the furnace outlet gas,

so,

1.63S mol/m3 = 15.766 mol/m3

we get

S = 9.67 mol/m3 ... (1)

now ,

S = 0 mol/m3

O2 = 0.63S mol/m3 = 0.63 * 9.67 mol/m3 = 6.09 mol/m3

SO2 = S mol/m^3 = 9.67 mol/m3

now the second reaction is SO2 reacting with O2 in a reactor to give SO3,

we have the inlet concentration of SO2 and O2,

we are given that SO2 conversion is 80%,

here the limiting reagent is O2 because less moles of O2 is taking part in reaction than required.

so,

in the outlet stream of the reactor after 80% conversion of SO2 will be, let 'x' be conversion of O2,

O2 = 6.09*(1-x)

SO2 = 9.67 - 2*6.09*x

SO3 = 2*6.09*x

so, initial SO2 = 9.67 mol/m3

converted SO2 = 0.8*9.67 mol/m3 = 7.736 mol/m3

SO coming out = 9.67 mol/m3 - 7.736 mol/m3 = 1.934 mol/m3

so,

9.67 - 2*6.09*x = 1.934

we get

x = (9.67 - 1.934)/(2*6.09) = 0.6351

so,

O2 = 6.09*(1-x) = 6.09 ( 1 -0.6351) = 2.22 mol/m3

SO2 = 9.67 - 2*6.09*x = 1.934 mol/m3

SO3 = 2*6.09*x = 7.735 mol/m3

so we have the concentration in outlet gas, so,

so, now using ,

for the reaction,

so,

.... {ans.}

hence the Kp value is 0.00112 1/Pa.

b)

to increase the conversion of SO2 to 95 %. we will have to increase the temperature. hence the Kc value will remains the same when temperature is kept same.

so,

let final pressure is 'P' Pa.

so, using the equation

now we know that total concentration of gases from the furnace will be 1.63S.

so,

so,

....... (2)

now,

this S is concentration os sulfur taken hence this will be same as the initial one.

so,

SO2 produced = S mol/m3

O2 in the outlet gas from furnace = 0.63S

so, this gas will go to reactor at the P, and 773K,

so,

the conversion of SO2 is 95%.;

at 95% conversion SO2, the concentration will be:

SO2 = S (1 -0.95) = 0.05S

O2 = 0.63S - 0.5*S*0.95 = 0.63S - 0.475S = 0.155S

SO3 = S*0.95 = 0.95X

so,

so, equating it with Kc calculated,

solving for S we get

S = 44.864 mol/m3

so,

we get

so , it atm unit,

.. {ans.}

hence the final pressure is to be increase till 4.638 atm from 1 atm at constant temperature.