NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
At 380 ° C , the half life period for the first order decomposition of H 2 O 2 is 360 m i n . The energy of activation of the reaction is 200 k J m o l - 1 . Calculate the time in minutes required for 75 % decomposition at 450 ° C . [Report your answer by rounding it upto nearset whole number]
Correct answer
20.00
Step-by-step solution
K 1 = 0.693 / 360 min - 1 at 653 K = 1 . 93 × 10 - 3 m i n - 1 E a = 200 × 10 3 J ; K 2 = ? at 723 K , R = 8 . 314 J 2.303 log K 2 K 1 = E a R 1 T 1 − 1 T 2 ∴ 2.302 log K 2 1 . 93 × 10 - 3 = 200 × 10 3 8 . 314 723 - 653 723 × 653 K 2 = 0.068 m i n - 1 Now, t = 2 . 303 0.068 l o g 100 25 = 20 . 39 m i n