If 87.5% of a given substance disintegrates i | vedclass

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(D) For a first-order reaction,the amount remaining after $40 	ext{ min}$ is $100\% - 87.5\% = 12.5\%$.Using the formula $k = \frac{2.303}{t} \log \f

Vedclass · Accepted Answer

$13 	ext{ min } 20 	ext{ sec}$

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(D) For a first-order reaction,the amount remaining after $40 	ext{ min}$ is $100\% - 87.5\% = 12.5\%$.Using the formula $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$:$k = \frac{2.303}{40} \log \frac{100}{12.5} = \frac{2.303}{40} \log 8 = \frac{2.303 	imes 3 	imes 0.3010}{40} = \frac{0.693 	imes 3}{40} 	ext{ min}^{-1}$.The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{k}$.$t_{1/2} = \frac{0.693 	imes 40}{0.693 	imes 3} = \frac{40}{3} 	ext{ min} = 13 	ext{ min } 20 	ext{ sec}$.

If $87.5\%$ of a given substance disintegrates in $40 \text{ minutes}$,what is the half-life of the radioactive substance?

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