IIT JEE past question Kinetics 8

at 380 degree C, the half life period for the first order decomposition of H2O2 is 360 min.

The energy of activation of reaction is 200 kJ/mol.

Calculate the time required for 75% decomposition at 450 degree C

1 Answers

t_1/2=0.693/k

k₃₈₀=0.693/(360)

now
k₁/k₂=Ea/RT (1/T1-1/T2)

FROM HERE WE CAN FIND K₄₅₀ AS EVERY INFO IS PROVIDED

k₄₅₀=1/t 2.303 log [a]₀/[a]

k₄₅₀=1/t 2.303 log4

solving this equation we can get the time period

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