SELECT * FROM question_mgmt as q WHERE id=1850 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='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Chemical Kinetics the activation energy for the reaction 2hi g rar

Question:

The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Answer:

In the given case:

Ea = 209.5 kJ mol - 1 = 209500 J mol - 1

T = 581 K

R = 8.314 JK - 1 mol - 1

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as

 


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