Monday, June 10, 2013

Question 4.23: The rate constant for the decomposition of hydrocarbons is 2.418 × 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.

Question 4.23: The rate constant for the decomposition of hydrocarbons is 2.418 × 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.
Answer
Given that
Activation energy , Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1
Use the formula of Arrhenius equation,

Plug the values in this formula we get
Log A = (0.3835 − 5) + 17.2082 = 12.5917
Take antilog both sides we get
A = 3.9 × 1012 s−1 (approx)

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