The time required for 10% completion of a first a order reaction at 298K is equal to that required for its 25% completion at 308 K. If pre-exponential factor for the reaction is 3.56 × 109 s–1. Calculate its rate constant at 318 K and also E a .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. t = 
log 
= 
log 
⇒ 
= 
= 
= 
= 
= 
= 2.7511
Now log 
= 
⇒ log 2.7511 = 
× 
⇒ 0.4395 = Ea × 5.7 × 10 –6 ⇒ Ea = 0.0771 × 10
6 J/mol
= 7.71 × 10 4 J/mol
Now from log k = log A – 
log k = log (3.56 × 10 9 ) – 
log k = log (3.56 × 10 9 ) – 12.663
= 9 + log 3.56 – 12.663
= – 3.663 + 0.5514
= – 3.1116
∴ k =Anti log (–3.1116)= 9.22×10 –4 sec –1
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