4.32 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s – 1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
4.32 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s – 1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
-
1 Answer
-
4.32 Given,
k = 2.418 * 10-5 s-1
T = 546 K
Ea = 179.9 kJ mol-1 = 179.9 * 103J mol-1
The Arrhenius equation is given by k = Ae-Ea/RT Taking natural log on both sides,
Ln k = ln A- (Ea/RT) Substituting the values,
ln (2.418 * 10-5 ) = ln A-179.9/ (8.314 * 546)
ln A = 12.5917
A = 3.9 * 1012 s-1 (approximately)
Similar Questions for you
Kindly go through the solution
Ea = 216.164kJ/mol 216
Reaction rate is used to measure how fast or slow reactions occur per unit time. The rate constant is a proportionality factor that remains constant for every reaction.
Yes, in elementary reactions, order and molecularity can be the same, but this is not always the case because order is an experimental quantity, and molecularity is a theoretical concept.
Reaction Kinetics, also known as chemical kinetics, is the study of the rate of chemical reaction and the factors affecting the reaction rate, such as temperature, concentration, and catalyst.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers