The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?


We know, The Arrhenius equation is given by k = Ae-Ea/RT


Taking natural log on both sides,


Ln k = ln A-(Ea/RT)


Thus, log k = log A -(Ea/2.303RT) eqn 1


The given equation is log k = 14.34 – 1.25 × 104K/T eqn 2


Comparing 2 equations,


Ea/2.303R = 1.25 × 104K


Ea = 1.25 × 104K × 2.303 × 8.314


Ea = 239339.3 J mol-1 (approximately)


Ea = 239.34 kJ mol-1


Also, when t1/2 = 256 minutes,


k = 0.693 / t1/2


= 0.693 / 256


= 2.707 × 10-3 min-1


k = 4.51 × 10-5s–1


Substitute k = 4.51 × 10-5s–1 in eqn 2,


log 4.51 × 10-5 s–1 = 14.34 – 1.25 × 104K/T


log(0.654-5) = 14.34– 1.25 × 104K/T


T = 1.25 × 104/[ 14.34- log(0.654-5)]


T = 668.9K or T = 669 K


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