Chemical Reaction Engineering Questions and Answers – Kinetics of Homogeneous Reactions – Comparison of Theories – 2


This set of Chemical Reaction Engineering Interview Questions and Answers for freshers focuses on “Kinetics of Homogeneous Reactions – Comparison of Theories – 2”.

1. Which of the following pairing is incorrect? (Where A is the frequency factor)
a) Arrhenius equation; k = A\(e^{\frac{-Ea}{RT}} \)
b) Transition state theory; k = AT\(e^{\frac{-Ea}{RT}} \)
c) Collision theory; k = AT0.5\(e^{\frac{-Ea}{RT}} \)
d) Arrhenius equation; k = AT2\(e^{\frac{-Ea}{RT}} \)
View Answer

Answer: d
Explanation: Arrhenius equation: k = A\(e^{\frac{-Ea}{RT}} \).
By collision theory, k α T0.5
By transition state theory, k α T

2. 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
View Answer

Answer: a
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.

3. 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
View Answer

Answer: b
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.

4. 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
View Answer

Answer: c
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.

5. 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
View Answer

Answer: a
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.

Sanfoundry Global Education & Learning Series – Chemical Reaction Engineering.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter