1. The activation energy for a reaction is 122137.5 J/ mol. If the rate constant at 330K is 0.5 min-1, then the value of rate constant (in min-1) at 360K by Arrhenius law is ____
a) 20.42
b) 18.65
c) 30.12
d) 12.32
Explanation: ln(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) \)
ln(\(\frac{k_2}{0.5}) = -\frac{E}{8.314} (\frac{1}{360}- \frac{1}{330}) \)
k2 = 20.42 min-1.
2. If the rate constant of a reaction at 275K is 1 min-1 and the rate constant at 300K is 2 min-1, what is the activation energy (in J/ mol) as obtained by Arrhenius law?
a) 24655
b) 19019.14
c) 366543.2
d) 18989.32
Explanation: ln(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) \)
ln(2) = –\(\frac{E}{8.314} (\frac{1}{360}- \frac{1}{275}) \)
E = 19019.14 J/ mol.
3. If the rate constant of a reaction at 600K is 100 times the rate constant at 500K, then the value of the activation energy obtained by Transition state theory is ____
a) 120987.12
b) 167435.15
c) 110319.28
d) 156435.54
Explanation: By Transition state theory, ln(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) + ln(\frac{T_2}{T_1}) \)
ln(\(\frac{100k_1}{k_1}) = -\frac{E}{8.314} (\frac{1}{600} – \frac{1}{500}) + ln(\frac{600}{500}) \)
E = 110319.28.
4. The activation energy of a reaction is 155326 J/ mol. The rate constant of the reaction at and 300K as a function of rate constant at 400K, obtained by the Collision theory is ____
a) k1 = 1.5 × 10-7k2
b) k1 = 1.2 × 10-6 k2
c) k1 = 1.5 × 10-6 k2
d) k1 = 1.2 × 10-7 k2
Explanation: By collision theory, ln(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) + 0.5ln(\frac{T_2}{T_1}) \)
ln(\(\frac{k_2}{k_1}) = -\frac{155326}{8.314}(\frac{1}{400} – \frac{1}{300}) + 0.5ln(\frac{400}{300}) \)
Hence, k1 = 1.5 × 10-7k2.
5. The combination of ideal reactors among the following is ____
a) Plug flow reactor and batch reactor
b) Batch reactor and mixed flow reactor
c) Plug flow reactor and mixed flow reactor
d) Batch reactor only
Explanation: Batch reactor is an unsteady state reactor. Plug flow reactor and mixed flow reactors operate under steady state conditions.
6.
There is no axial mixing in a plug flow reactor.
a) True
b) False
Explanation: There is no mixing of early and later entering fluid in a plug flow reactor. There is no overtaking between the fluid molecules.
7.
The average concentration of product is low inside a CSTR.
a) True
b) False
Explanation: There is continuous mixing of all fluid molecules inside a CSTR. As there is continuous recycle, the average concentration is low in the reactor.
8. Which of the following is true for a reaction occurring in a batch reactor?
a) The accumulation term is zero
b) Only the component input term is zero
c) Only the component output term is zero
d) Both the component input and output terms are zero
Explanation: Reaction is carried out in batches. As and when a new batch of reactant is fed to the reactor, the preceding batch product is withdrawn.
9. Which one of the following combinations gives the highest conversion for second order reaction?
a) PFR followed by smaller CSTR followed by a bigger CSTR
b) PFR followed by bigger CSTR followed by a bigger CSTR
c) Smaller CSTR followed by PFR followed by a bigger CSTR
d) Bigger CSTR followed by a PFR followed by smaller CSTR
Explanation: For reactions of order greater than one, higher conversion is achieved in a PFR. Smaller the volume of a CSTR, it tends to behave as a PFR. Hence, for higher conversions, the reactors are arranged in the order PFR followed by smaller CSTR followed by a bigger CSTR.
10. For a PFR, the area under the curve \(\frac{1}{-r_A}\) vs XA gives ____
a) \(\frac{F_{A_0}}{dV} \)
b) \(\frac{dV}{F_{A_0}} \)
c) dVCA0
d) dVFA0
Explanation: For a PFR, FA0dXA = (-rA)dV. \(\frac{dXA}{(-rA)}\) gives . Area under the curve gives \(\frac{dV}{FA0}. \)