SELECT * FROM question_mgmt as q WHERE id=1842 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=59 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Chemical Kinetics the rate constant for the decomposition of hydroca

Question:

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.

Answer:

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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