The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation: log k = 14.34 – 1.25 × 10^{4}K/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 × 10^{4}K/T → eqn 2

Comparing 2 equations,

E_{a}/2.303R = 1.25 × 10^{4}K

E_{a} = 1.25 × 10^{4}K × 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^{-5}s^{–1}

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

log 4.51 × 10^{-5} s^{–1} = 14.34 – 1.25 × 10^{4}K/T

log(0.654-5) = 14.34– 1.25 × 10^{4}K/T

T = 1.25 × 10^{4}/[ 14.34- log(0.654-5)]

T = 668.9K or T = 669 K

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