Chemical Reaction Engineering Questions and Answers – Reaction Kinetics Elements – Modelling the Rate Coefficient – 2

This set of Chemical Reaction Engineering Questions and Answers for Freshers focuses on “Reaction Kinetics Elements – Modelling the Rate Coefficient – 2”.

1. The reaction rate constant is related to reaction rate as ____
a) k α (-rA)
b) k α \(\frac{1}{(-r_A)} \)
c) k α e(-rA)
d) k is independent of (-rA)
View Answer

Answer: a
Explanation: The rate of reaction, (-rA) = kCAn
Where, n is the reaction order. The rate increases as the value of k increases.

2. State true or false.
The Arrhenius law plot of lnk vs 1/ T gives a straight line with large slope for large activation energy.
a) True
b) False
View Answer

Answer: a
Explanation: The plot of lnk vs 1/ T gives a straight line of slope –\(\frac{E_a}{(R×T)}\). As Ea increases, slope increases.

3. The unit of activation energy is ____
a) mol
b) mol.K
c) K
d) J/ mol
View Answer

Answer: d
Explanation: Activation energy is expressed in J/ mol or kJ/ mol. Activation energy is the energy required for a chemical reaction to occur.
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4. State true or false.
The value of frequency factor affects the temperature sensitivity of reaction.
a) True
b) False
View Answer

Answer: b
Explanation: Frequency factor in Arrhenius equation does not affect the temperature sensitivity of the reaction. Frequency factor is independent of temperature. It takes the units of rate constant.

5. The rate constant and half life for reactions are related as ____
a) k α t0.5
b) k α \(\frac{1}{(t_{0.5})} \)
c) k α e(-t0.5)
d) k is independent of t0.5
View Answer

Answer: b
Explanation: For 1st order reaction, t0.5 = \(\frac{0.6931}{k}.\) For all reactions of order other than 1, t0.5 = \(\frac{(0.5^{(1-n)}-1)×C_{A0}^{1-n}}{k(1-n)}. \)
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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