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# The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 10 ^ 4K/T

4.27   The rate constant for the first order decomposition of $H_2O_2$ is given by the following equation:
$\log k = 14.34 - 1.25 \times 10 ^ 4K/T$ .

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

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The Arrhenius equation is given by
$k=Ae^{-E_{a}/RT}$

taking log on both sides,

$\log k = \log A -\frac{E_{a}}{2.303RT}$....................(i)
given equation,

$\log k = 14.34 - 1.25 \times 10 ^ 4K/T$.....................(ii)

On comparing both equation we get,

$E_{a}/2.303R=1.25 \times 10^{4}$

activation energy
$\\=1.25 \times 10^{4} \times 2.303 \times 8.314\\ =239.34\ KJ/mol$

half life ($t_{1/2}$)  = 256 min

k = 0.693/256
$k = 4.51\times 10^{-5} s^{-1}$

With the help of equation (ii),

$\log4.51\times 10^{-5} s^{-1} = 14.34-\frac{1.25\times 10^{4}}{T}$

$\frac{1.25\times 10^{4}}{T} = 18.686$
T = $\frac{1.25\times 10^{4}}{18.686}$

= 669 (approx)

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