250+ TOP MCQs on Design Equations for Flow Reactors and Answers

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Design Equations for Flow Reactors”.

1. The differential form of the plug flow reactor design equation is ______
A. FA0(frac{dX}{dV}) = -r_A
B. (frac{dX}{dV}) = -r_A
C. V(frac{dX}{dV}) = -r_A
D. (frac{dX}{dV}) = -r_A F_A0

Answer: A
Explanation: Molar flowrate and conversion are related in the mole balance equation to get the above equation.

2. The design equation for a plug flow reactor is ___________
A. V = (int_0^Xfrac{dX}{-rA} )
B. V = F_A0 (int_0^Xfrac{dX}{-rA} )
C. (frac{V}{-rA} = int_0^Xfrac{dX}{FA0} )
D. F_A0 = -r_A(int_0^Xfrac{dX}{V} )

Answer: B
Explanation: To determine the volume of PFR required to obtain a certain conversion, the mole balance equation in terms of molar flowrate and conversion is written and integrated to get the above equation.

3. The design equation of PBR to determine the weight of the catalyst is ________
A. (frac{W}{FA0} = int_0^X )-rA dX
B. (frac{W}{FA0} = int_0^X frac{dX}{dV} )
C. W = F_A0 (int_0^X frac{dX}{-rA} )
D. W = (int_0^X )FA0 dX

Answer: C
Explanation: The general mole balance for flow reactors is modified to get the above equation. The reaction rate is expressed in mol/s.g of catalyst in this equation.

4. In CSTR, the rate of reaction is evaluated at the exit conditions.
A. True
B. False

Answer: A
Explanation: The composition at exit is identical to the composition in the reactor. Hence, reaction rate is measured at the exit conditions.

5. A space velocity of 7 hr^-1 means ___________
A. 7 reactor volumes of feed is fed into the reactor per hour
B. Feed is fed at 7 hrs interval
C. Takes 7 hours to treat one reactor volume of feed
D. Reaction time is 7 hours

Answer: A
Explanation: Number of reactor volumes of feed that can be treated in unit time is called space velocity.

6. A space time of 10 minutes means __________
A. 10 reactor volumes of feed is fed into the reactor per minute
B. Feed is fed at 10 mins interval
C. Takes 10 mins to treat one reactor volume of feed
D. Reaction time is 10 hrs

Answer: C
Explanation: Time required to process one reactor volume of feed is called Space time.

7. Space time in terms of molar federate and initial concentration is _________
A. τ = C_A0 V₀
B. τ = (frac{CA0}{FA0} )
C. τ = C_A0 V
D. τ = (frac{VCA0}{FA0} )

Answer: D
Explanation: By definition, Space time = (frac{Volume , of , the , reactor}{Volumetric , feed , rate} = frac{V}{V0} ) and F_A0 = C_A0 V₀,
Therefore τ = (frac{VCA0}{FA0}. )

8. The performance equation for CSTR is_________
A. (frac{τ}{CA0} = frac{XA}{-rA} )
B. τ = (frac{XA}{-rA} )
C. τ = C_A0 -r_A
D. τ = (frac{XA}{CA0(-rA.} )

Answer: A
Explanation: General design equation of CSTR is modified and written in terms of initial concentration and space time to get the above equation.

9. Calculate the volume of CSTR in litre required to achieve 70% conversion. The feed contains 10 molA/litre with feed rate of 3 lit/min and rate constant 0.15 for the reaction -r_A = k(C_A)^1.5.
A. 3
B. 9
C. 18
D. 27

10. For a gaseous reaction carried out in 25 litre PFR with 5molA/lit and space time 3 mins. Calculate the conversion attained if reaction kinetics is given as follows
A → 2.5P, -r_A = (2 min^-1) C_A .
A. 0.45
B. 0.65
C. 0.95
D. 0.98

Answer: C
Explanation: For gaseous phase reaction
ε_A = (frac{2.5-1}{1}) = 1.5
For first order reaction in PFR by integral method
kτ = (1+ ε_A) ln((frac{1}{1-XA} )) – ε_AX_A
2*3 = (1+1.5)ln((frac{1}{1-XA} )) – 1.5X_A
6 = 2.5ln ((frac{1}{1-XA} )) – 1.5X_A
X_A = 0.95.