A first order decomposition reaction takes 40 minutes for 30% of decomposition. Calculate t_1/2 value for it.

Solution

For the first order reaction

K = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{40} \log \frac{100}{70} = \frac{2.303}{40} \log 100 - \log 70 = \frac{2.303}{40} (2 - 1.8808) = \frac{2.303}{40} \times 0.1192 = 0.0068 \min^{- 1}

(ii) 0.0068 = \frac{2.303}{t} \log \frac{100}{50} 0.0068 = \frac{2.303}{t} \log 2 0.0068 = \frac{2.303}{t} \times 0.3010 t = \frac{2.303 \times 0.3010}{0.0068} = 101.94 \min .

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