At 380°C, the half-life period for the first-order decomposition of H₂O₂ is 360 min. The energy of activation of the reaction is 200 kJ mol^-1. Calculate the time required for 75% decomposition at 450°C if the half-life for decomposition of H₂O₂ is 10.17 min at 450°C:

20.4 min

The correct answer is option 1. 20.4 min.

At 450ºC

\(k =\frac{0.693}{t_{1/2}}\)

\(k = \frac{0.693}{10.17}\)

\(k = 0.0681 \text{ min}^{-1}\)

Now, for 75%decomposition,

\(x= 75\)% of a \( = 0.75 a\)

\(∴ t =\frac{2.303}{0.0681} log\frac{a}{a − 0.75 a}\)

\(⇒ t = \frac{2.303}{0.0681}log \frac{1}{0.25}\)

\(⇒ t = \frac{2.303}{0.0681} × log 4\)

\(⇒ t = \frac{2.303}{0.0681} × 0.602\)

\(⇒ t = 20.37 \text{ min}\)

\(⇒ t ≈ 20.4 \text{ min}\)