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.
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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1 Answer
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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, 75% is remaining t’ = (2.303/k’) × log ( [R]0/0.75 [R]0)
t’ = (2.303/k’) × log (1/0.75) t’ = 2.2877 / k'
But, t = t’
0.1054 / k = 2.2877 / k' / k = 2.7296
From Arrhenius equation, we obtain log k2/k1 = (Ea / 2.303 R) × (T2 - T1) / T1T2
Substituting the val
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