# Chemical Reaction Engineering Questions and Answers – Conversion and Reactor Sizing – Problems on Ideal Reactors

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This set of Chemical Reaction Engineering Multiple Choice Questions & Answers (MCQs) focuses 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: –$$\frac{dC_A}{dt}$$ = 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

Explanation: 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

Explanation: 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

Explanation: $$\frac{0.75}{X_A} = \frac{0.25}{2(1-XA)}$$
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

Explanation: 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

Explanation: For a CSTR, the performance equation is FA0XA = (-rA)V
t = $$\frac{C_{A0}- CA}{(-rA)}$$
rA = $$\frac{10-5}{100}$$ = 0.05.

10. A reaction of first order takes 10 minutes to achieve 50% conversion. The conversion achieved after 20 minutes is ____
a) 0.5
b) 0.75
c) 0.87
d) 0.9 