Chemical Reaction Engineering Multiple Choice Questions on “Conversion and Reactor Sizing – Problems on Ideal Reactors”.
1. If the rate of a reaction is given as (-rA) = (frac{k_1+C_A^2}{k_2×C_A} frac{mol}{m^3 min} ), then the units of k1 and k2 respectively are ____
a) (frac{mol^2}{m^6} ) and min
b) min and (frac{mol^2}{m^6} )
c) mol and min
d) m3 and mol
Answer: a
Clarification: Two quantities of the same dimensions can only be added. As the unit of concentration is (frac{mol^2}{m^3} ) , k1 has the units (frac{mol^2}{m^6} ).
Equating the units of all known quantities on both sides, (frac{mol}{m^3 min} = frac{mol^2}{m^6} × frac{1}{k_2×(frac{mol}{m^3})}. )
Hence, k2 has the unit of min.
2. State true or false.
For pure gaseous reactants, the best reactor suited for carrying out reaction is CSTR.
a) True
b) False
Answer: a
Clarification: For pure gaseous reactants, (frac{C_p}{-∆H_r} ) is small, where Cp is the specific heat capacity of the gas and ∆Hr is the enthalpy change of reaction. CSTR gives a high conversion for small values of (frac{C_p}{-∆H_r}. )
3. State true or false.
For a first order reaction, the conversion of the reaction is a function of CA0.
a) True
b) False
Answer: b
Clarification: For a first order reaction, –(frac{dC_A}{dt} ) = kCA0
CA = CA0(1 – XA)
Differentiating on either sides, dCA = -CA0dXA
-CA0 (frac{dXA}{dt} ) = kCA0(1 – XA)
(frac{dXA}{(1 – XA)} ) = kdt
-ln(1-XA) = kt
Hence, conversion is not a function of CA0.
4. For a first order reaction, if the rate constant is 1 min-1 and the reaction time is 2 min, then the ratio of final to initial concentration is ____
a) 0.167
b) 0.135
c) 0.198
d) 0.764
Answer: b
Clarification: –(frac{dC_A}{dt}) = kCA
Integrating between initial and final concentrations,
CA = CA0e-kt
(frac{C_A}{C_{A0}} ) = e-2
(frac{C_A}{C_{A0}} ) = 0.135.
5. The residence time (in min) for the reactants at a flow rate of 5 (frac{litre}{min}) inside a reactor of volume 10 litre is ____
a) 5
b) 1
c) 10
d) 2
Answer: d
Clarification: Residence time = (frac{volume}{volumetric , flow , rate} = frac{10}{5} ) = 2.
6. A second order reaction occurs in a CSTR. Determine the ratio of volumes required for 45% conversion to 90% conversion, provided initial concentration, flow rate and other parameters are constant.
a) 1.64
b) 60.5
c) 0.35
d) 0.0165
Answer: b
Clarification: For a second order reaction occurring in a CSTR,
XA1 = (1-XA1)2 V1
(frac{0.45}{0.90} = frac{(1-0.45)^2 V1}{(1-0.9)^2 V2} )
(frac{V_1}{V_2} ) = 0.0165.
7. The present CSTR is replaced by another CSTR of double the volume. A first order liquid phase reaction occurs in both the reactors. If the conversion achieved in the old reactor is 75%, what is the conversion achieved in the new reactor?
a) 0.857
b) 0.675
c) 0.765
d) 0.99
Answer: a
Clarification: (frac{0.75}{X_A} = frac{0.25}{2(1-XA)} )
XA = 0.857.
8. A liquid phase reaction occurring in a CSTR follows first order dynamics. The feed rate is 500m3/min. Initial concentration of the feed is 50mol/m3. If the reaction conversion is 50% and the reaction is carried out in a 0.5 m3 CSTR, what is the value of rate constant?
a) 40
b) 20
c) 10
d) 5
Answer: b
Clarification: FA0 XA = (-rA)V
(-rA) = kCA0(1 – XA)
500 × 0.5 = k × 50 × (1-0.5) × 0.5
k = 20.
9. For a first order liquid phase reaction occurring in a CSTR, the initial concentration is 10 (frac{mol}{m^3} ) and the final concentration is 5 (frac{mol}{m^3} ). If the residence time is 100 minutes, then the value of reaction rate is ____
a) 0.75
b) 0.33
c) 0.4
d) 0.05