The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34  1.25 x 10^4 K/T.
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 x 10^{4} K/T.
Calculate E_{a} for this reaction and at what temperature will its halfperiod be 256 minutes?
Given, Arrhenius equation is,
k = Ae^{Ea/RT}
⇒ ln k = ln A \(\frac{E_a}{RT}\)
⇒ ln k = log A \(\frac{E_a}{RT}\)
⇒ log k = log A \(\frac{E_a}{2.303 \ RT}\) ....(i)
The given equation is
log k = 14.34  1.25 x 10^{4} K/T ....(ii)
From equation (i) and (ii), we obtain
\(\frac{E_a}{2.303 \ RT}\) = \(\frac{1.25 \times 10^4 K}{T}\)
⇒ E_{a} = 1.25 x 10^{4} K x 2.303 x R
= 1.25 × 10^{4} K × 2.303 × 8.314 J K^{1 }mol^{1}
= 239339.3 J mol^{1} (approximately)
= 239.34 kJ mol^{1}
Also, when t_{1/2} = 256 minutes,
k = \(\frac{0.693}{t_{1/2}}\)
= \(\frac{0.693}{256}\)
= 2.707 × 10^{3} min^{1}
= 4.51 × 10^{5} s^{1}
It is also given that,
log k = 14.34  1.25 × 10^{4 }K/T
⇒ log(4.51 x 10^{5}) = 14.34  \(\frac{1.25 \times 10^4 \ K}{T}\)
⇒ log(0.654  05) = 14.34  \(\frac{1.25 \times 10^4 \ K}{T}\)
⇒ \(\frac{1.25 \times 10^4 \ K}{T}\) = 18.686
⇒ T = \(\frac{1.25 \times 10^4 \ K}{18.686}\)
= 668.95 K
= 669 K (approximately)

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