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[Post New]7 May 2008 21:25:09 IST
    Subject: Sir,please explain this question of nikhilesh
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golu1234 (0)

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The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 /s at 546 K.If the energy of activation is 179.9 KJ/mol. Calculate the value of pre-exponential factor.
    
[Post New]7 May 2008 21:30:24 IST
    Subject: Re:Sir,please explain this question of nikhilesh
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donnowatnametochoose (34)

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..here find A.
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[Post New]7 May 2008 21:31:28 IST
    Subject: Re:Sir,please explain this question of nikhilesh
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golu1234 (0)

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Please calculate yaaaaaaaaaaar . ans is not coming
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[Post New]7 May 2008 23:00:10 IST
    Subject: Sir,please explain this question of nikhilesh
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krishna.gopal (2399)

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I am getting A = 3.9*10^12. What is the answer. Theapproach is right. Either their is some calculation mistake on your part or there is something wrong with answer
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi
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[Post New]7 May 2008 23:01:46 IST
    Subject: Re:Sir,please explain this question of nikhilesh
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arpan1 (665)

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hey ans is 3.9 x 10 ^ 13


 


 


 


krishna.gopal u have commited some calculation mistake


this one is easy and direct application of formula
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