The rate constant for the first order decomposition of
Question:

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

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

Solution:

Arrhenius equation is given by,

$k=\mathrm{Ae}^{-E_{\alpha} / \mathrm{R} T}$

$\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$

$\Rightarrow \ln k=\log \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$ 

$\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{RT}}$    …(i)

The given equation is

$\log k=14.34-1.25 \times 10^{4} \mathrm{~K} / T$   …(ii)

From equation (i) and (ii), we obtain

$\frac{E_{a}}{2.303 \mathrm{RT}}=\frac{1.25 \times 10^{4} \mathrm{~K}}{T}$

$\Rightarrow E_{o}=1.25 \times 10^{4} \mathrm{~K} \times 2.303 \times \mathrm{R}$

= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1

= 239339.3 J mol1 (approximately)

= 239.34 kJ mol−1

 

Also, when t1/2 = 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 × 104 K/T

$\Rightarrow \log \left(4.51 \times 10^{-5}\right)=14.34-\frac{1.25 \times 10^{4} \mathrm{~K}}{T}$

$\Rightarrow \log (0.654-05)=14.34-\frac{1.25 \times 10^{4} \mathrm{~K}}{T}

$\Rightarrow T=\frac{1.25 \times 10^{4} \mathrm{~K}}{18.686}$

= 668.95 K

= 669 K (approximately)

 

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