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(D) For a first-order reaction,the half-life is given as $t_{1/2} = 120 \, min$.The rate constant $k$ is calculated as $k = \frac{0.693}{t_{1/2}} = \f

Vedclass · Accepted Answer

Nearly $400 \, minutes$

Vedclass · Answer

(D) For a first-order reaction,the half-life is given as $t_{1/2} = 120 \, min$.The rate constant $k$ is calculated as $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120} = 5.775 	imes 10^{-3} \, min^{-1}$.For $90\%$ decomposition,the amount remaining is $(a - x) = 100 - 90 = 10$.Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log_{10} \frac{a}{a - x}$.Substituting the values: $t = \frac{2.303}{5.775 	imes 10^{-3}} \log_{10} \frac{100}{10} = \frac{2.303}{5.775 	imes 10^{-3}} 	imes 1 = 398.78 \, min$.Rounding to the nearest value,we get $t \approx 400 \, minutes$.

The thermal decomposition of a compound is of first order. If a sample of the compound decomposes $50\%$ in $120 \, minutes$,in what time will it undergo $90\%$ decomposition?

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