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 E_a 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-(E_a/RT)

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

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

Comparing 2 equations,

E_a/2.303R = 1.25 × 10⁴K

E_a = 1.25 × 10⁴K × 2.303 × 8.314

E_a = 239339.3 J mol^-1 (approximately)

E_a = 239.34 kJ mol^-1

Also, when t_1/2 = 256 minutes,

k = 0.693 / t_1/2

= 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