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NEET Class neet Answered

Please answer the following question with explanation
question image
Asked by deepakudgiri29 | 25 Dec, 2018, 05:19: PM
answered-by-expert Expert Answer
We have 27°C = 300 K,
In the activated region 3.8×10-16 of the reactant molecules exists
i.e. N = 3.8×10-18 N0
 
As we know, 
 
space straight N equals space straight N subscript 0 space straight e to the power of negative straight E subscript straight a divided by RT end exponent

straight E subscript straight a space equals space RT space log space straight e space straight N subscript 0 over straight N space equals space 8.314 space cross times 300 space log space straight e fraction numerator straight N subscript 0 over denominator 3.8 cross times 10 to the power of negative 18 end exponent straight N subscript 0 end fraction

straight E subscript straight a space equals space 100 space kJ divided by mol
Answered by Ramandeep | 07 Jan, 2019, 12:45: PM
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