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 (-r_{A}) = \(\frac{k_1+C_A^2}{k_2×C_A} \frac{mol}{m^3 min} \), then the units of k_{1} and k_{2} respectively are ____

a) \(\frac{mol^2}{m^6} \) and min

b) min and \(\frac{mol^2}{m^6} \)

c) mol and min

d) m^{3} and mol

View Answer

Explanation: Two quantities of the same dimensions can only be added. As the unit of concentration is \(\frac{mol^2}{m^3} \) , k

_{1}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, k

_{2}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

View Answer

Explanation: For pure gaseous reactants, \(\frac{C_p}{-∆H_r} \) is small, where C

_{p}is the specific heat capacity of the gas and ∆H

_{r}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 C_{A0}.

a) True

b) False

View Answer

Explanation: For a first order reaction, –\(\frac{dC_A}{dt} \) = kC

_{A0}

C

_{A}= C

_{A0}(1 – X

_{A})

Differentiating on either sides, dC

_{A}= -C

_{A0}dX

_{A}

-C

_{A0}\(\frac{dXA}{dt} \) = kC

_{A0}(1 – X

_{A})

\(\frac{dXA}{(1 – XA)} \) = kdt

-ln(1-X

_{A}) = kt

Hence, conversion is not a function of C

_{A0}.

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

View Answer

Explanation: –\(\frac{dC_A}{dt}\) = kC

_{A}

Integrating between initial and final concentrations,

C

_{A}= C

_{A0}e

^{-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

View Answer

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

View Answer

Explanation: For a second order reaction occurring in a CSTR,

X

_{A1}= (1-X

_{A1})

^{2}V

_{1}

\(\frac{0.45}{0.90} = \frac{(1-0.45)^2 V1}{(1-0.9)^2 V2} \)

\(\frac{V_1}{V_2} \) = 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

View Answer

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

X

_{A}= 0.857.

8. A liquid phase reaction occurring in a CSTR follows first order dynamics. The feed rate is 500m^{3}/min. Initial concentration of the feed is 50mol/m^{3}. If the reaction conversion is 50% and the reaction is carried out in a 0.5 m^{3} CSTR, what is the value of rate constant?

a) 40

b) 20

c) 10

d) 5

View Answer

Explanation: F

_{A0}X

_{A}= (-r

_{A})V

(-r

_{A}) = kC

_{A0}(1 – X

_{A})

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

View Answer

Explanation: For a CSTR, the performance equation is F

_{A0}X

_{A}= (-r

_{A})V

t = \(\frac{C_{A0}- CA}{(-rA)} \)

r

_{A}= \(\frac{10-5}{100} \) = 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

View Answer

Explanation: -ln(1-X

_{A}) = kt

X

_{A}= 0.75.

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