If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes)
(Take: log 2 = 0.30; log 2.5 = 0.40 )
If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes)
(Take: log 2 = 0.30; log 2.5 = 0.40 )
Asked by Shiksha User
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1 Answer
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first order reaction
K = (2.303/t) log (a? / (a? - x)
K = (2.303/90) log (a? / 0.25a? ) = 0.0154
t = 60% = (2.303/K) log (a? / a? ) . (2)
= (2.303 / 0.0154) x (1 - 0.602) = 59.51 mins ≈ 60
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