The rate constant for the first order decomposition of $H_{2}O_{2}$ is given by the following equation:
$\log k = 14.34 - 1.25 \times 10^{4} \, K / T$
Calculate $E_{a}$ for this reaction and at what temperature will its half-period be $256 \, min$?

The Arrhenius equation is given by:
$k = Ae^{-E_{a} / RT}$
Taking the logarithm on both sides:
$\log k = \log A - \frac{E_{a}}{2.303 \, RT} \quad (i)$
The given equation is:
$\log k = 14.34 - \frac{1.25 \times 10^{4} \, K}{T} \quad (ii)$
Comparing $(i)$ and $(ii)$:
$\frac{E_{a}}{2.303 \, R} = 1.25 \times 10^{4} \, K$
$E_{a} = 1.25 \times 10^{4} \times 2.303 \times 8.314 \, J \, mol^{-1} \approx 239.34 \, kJ \, mol^{-1}$
For a first-order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ as:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{256 \, min} \approx 2.707 \times 10^{-3} \, min^{-1} \approx 4.51 \times 10^{-5} \, s^{-1}$
Substituting $k$ into the given equation:
$\log(4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^{4}}{T}$
$-4.346 = 14.34 - \frac{1.25 \times 10^{4}}{T}$
$\frac{1.25 \times 10^{4}}{T} = 18.686$
$T = \frac{1.25 \times 10^{4}}{18.686} \approx 669 \, K$