SELECT * FROM question_mgmt as q WHERE id=1848 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 time required for 10 completion of a first or

Question:

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 1010 s-1. Calculate k at 318 K and Ea.

Answer:

For a first order reaction,

t = 2.303 / k log a  / a - x

At 298 K,

t = 2.303 / k log 100  / 90

= 0.1054 / k

At 308 K,

t' = 2.303 / k' log 100  / 75

= 2.2877 / k'

According to the question,

t = t'

⇒ 0.1054 / k  =  2.2877 / k'

⇒ k' / k  = 2.7296

From Arrhenius equation,we obtain

 To calculate k at 318 K,

It is given that, A = 4 x 1010 s-1, T = 318K

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10-2 s -1

    


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