The decomposition of a compound is found to follow a first-order rate law. If it takes 15 minutes for 20 percent of original material to react.
Calculate :

(i) the specific rate constant
(ii) the time at which 10 percent of the original material remains unreacted
(iii) the time it takes for the next 20 percent of the reactant left to react after the first 15 minutes.

Solution

According to the first order rate -law

(i) k = \frac{2.303}{t} \log \frac{a}{a - x} k = \frac{2.303}{t} \log \frac{100}{80} k = 0.154 \log 1.25 k = 0.154 \times 0.0969 k = 1.50 \times 10^{- 1} \min^{- 1}

Time at which 10 percent of the original material remains unreacted is ,

(ii) t = \frac{2.303}{1.50 \times 10^{- 2}} \times \log \frac{100}{10} t = 1.54 \times 10^{2} \min = 154 \min .

Time for the next 20 percent reaction

(iii) t = \frac{2.303}{1.50 \times 10^{- 2}} \times \log \frac{100}{60} t = 1.54 \times 10^{2} [\log 10 - \log 6] = 154 \times [1.0 - 0.7782] = 34.15 \min .

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