A batch reactor is charged with 7.00 kmol of sulfur dioxide and 4.30 kmol oxygen

Question

A batch reactor is charged with 7.00 kmol of sulfur dioxide and 4.30 kmol oxygen, which react reversibly to form sulfur trioxide. The equilibrium constant for this reaction at the pressure and temperature of the reactor is provided below. (K=8.50)

SO2 + ½ O2 ? SO3

a) What is the limiting reactant? How do you know?

b) What is the maximum possible percent conversion of the limiting reactant? Calculate the composition of the outlet stream for this case.

c) Increasing the amount of oxygen fed to the reactor should increase the conversion of sulfur dioxide. What initial input of oxygen is required to achieve 90% conversion of sulfur dioxide?

d) What is the maximum possible sulfur dioxide conversion that can be obtained by increasing the amount of oxygen charged to the reactor?

e) If the reactor is initially charged with 7.00 kmol sulfur dioxide and 1.60 kmol oxygen, what is the final composition?

f) Repeat the previous problem, using elements rather than compounds as components in the material balance equation.

Explanation / Answer

(a) According to the reaction stoichiometry,

1 mole SO2 reacts with 0.5 mole of O2.

Thus, 7 kmol of SO2 will require 3.5 kmol of O2.

Since O2 is present more than this required amount, it means it is in excess, thus SO2 is the limiting reagent.

(b)

                SO2 + 0.5 O2 ---> SO3

Initial         7            4.3            0

Eqb.         7-2x        4.3-x       2x

Thus, for the above reaction, K = 2x/[(7-2x)*(4.3-x)1/2] = 8.5

Solving for x, we get : x = 3.153

Thus, SO2 consumed = 2*x = 6.306 kmol

This is the max. possible consumption of SO2 at the given conditions

Thus the maximum possible % conversion for this is : 6.306/7*100 = 90.08%

In this case as per the reaction equation, amount of SO3 formed = amount of SO2 consumed = 6.306 kmol SO3

Remaining O2 is : Initial - Consumed = 4.3 - x = 1.147 kmol O2

Remaining SO2 is : Initial - Consumed = 7 - 6.306 = 0.694 kmol O2

(c)

As calculated above, for 90% conversion of SO2, amount of O2 required = 3.153 kmol

(d) The maximum possible conversion can be brought very close to 100%, around 99% by increasing O2 largely.

(e)

In this case, the equation for x becomes :

2x/[(7-2x)*(4.3-x)1/2] = 8.5

Solving we get :

x = 1.59

Thus, SO3 present = 2*x = 3.18 kmol

SO2 left = 7-3.18 = 3.82 kmol

O2 left = 0.01 kmol

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