The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
k= 2.418 × 10-5 s-1
T= 546 K
Ea= 179.9 kJ mol - 1 = 179.9 × 103J mol - 1
According to the Arrhenius equation,
= (0.3835 - 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 1012 s - 1(approximately)
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 23: The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy ....
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