Chemical Reaction Engineering Interview Questions and Answers for freshers 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}} )
Answer: d
Clarification: 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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: a
Clarification: 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.