Question

At equilibrium conditions, three gas-phase reactions carried out in a 1.5m3 batch reactor are found to have an overall conversion of A of 80% with individual reaction selectivity’s of 60% and 40% for reactions 1 and 2, respectively. Note that selectivity has the traditional definition (amount of A reacted in reaction i/ total amount reacted)
A + 2B -> C + D    (1)
A + D ->B             (2)
C + B -> 2E + A   (3)
Assuming that only A and B are fed to the reactor, determine the following if 5 moles of A remain in the reactor at equilibrium:
a) Determine the number of moles of A generated through reaction 3 if 4 moles of C are present in the reactor at equilibrium.

b) Determine the pressure in the reactor at equilibrium if the initial pressure was 150 kPa and the reactor operated isothermally at 100°C. Assume that B is only fed in the quantity needed for stoichiometry.

Answer 1

(a)

Data given:

v=1.5 m³

X_A= 0.8

n_A = 5 moles

Selectivity of A for reaction (1) is 0.6

Selectivity of A for reaction (2) is 0.4

Now we know that

final concentration of A is C_A = n_A /V =5 mole/1.5*10³ (lit) = 0.0033 mol/lit

Now from conversion definition

So solving this equation we get

C_A0 = 0.0165 mol/lit

So initial moles n_A0 = C_A0*V=0.0165*1.5*10³=24.75 moles

Now According to stoichiometry if 1 mol of C reacted then it produces 1 mol of A.

So 4 moles of C reacted then it produces 4 moles of A.

(b)

Now P_A0 = 150 kPa= 150000 Pa

Now n_A0 = 24.75 moles

So for reaction (1) selectivity is 0.6

no of moles of A reacted in reaction(1) =0.6*24.75=14.85 moles

So according to stoichiometry no of moles of B reacted = 2*14.85= 29.7 moles

Now only 5 moles of A remain in reactor

R=8.314 J/molK

T=100 ⁰C=100+273.15 =373.15 K

Now initial moles 24.75 moles

Now

initial pressure P₀ = 150 kPa=150000 Pa given

Now for total pressure at equilibrium

Now stochiometry coefficient of A is a=1 ,∆v=-1

So

P=109152.1 pa =109.152 kPa