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

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8 months ago

4.38 We know, time t = (2.303/k) * log ( [R]0/ [R])

Where, k- rate constant

[R]0-Initial concentration

[R]-Concentration at time 't'

At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) * log ( [R]0/0.9 [R]0)

t = (2.303/k) * log (1/0.9) t = 0.1054 / k

At temperature 308K, 25% is completed

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