The half-life period of a first order reactio | vedclass

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(C) Given half-life $t_{1/2} = 15 \ minutes$. Total time $T = 1 \ hr = 60 \ minutes$. The number of half-lives $n$ is calculated as $n = \frac{T}{t_{1

Vedclass · Accepted Answer

$\frac{1}{16}$ of the original amount

Vedclass · Answer

(C) Given half-life $t_{1/2} = 15 \ minutes$. Total time $T = 1 \ hr = 60 \ minutes$. The number of half-lives $n$ is calculated as $n = \frac{T}{t_{1/2}} = \frac{60}{15} = 4$. The amount of substance remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$. Substituting the value of $n$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$. Therefore,the amount of substance left after $1 \ hour$ is $\frac{1}{16}$ of the original amount.

The half-life period of a first order reaction is $15 \ minutes$. The amount of substance left after one hour will be

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