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(D) For a $1^{st}$ order reaction,the half-life $t_{1/2}$ is given as $1 \, h$. This means $50 \, \%$ of the reaction is completed in $1 \, h$.For $87

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$3.00$

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(D) For a $1^{st}$ order reaction,the half-life $t_{1/2}$ is given as $1 \, h$. This means $50 \, \%$ of the reaction is completed in $1 \, h$.For $87.5 \, \%$ completion,the amount remaining is $100 \, \% - 87.5 \, \% = 12.5 \, \%$.We can relate the number of half-lives $(n)$ to the fraction remaining: $\frac{[A]_t}{[A]_0} = (\frac{1}{2})^n$.Here,$\frac{12.5}{100} = \frac{1}{8} = (\frac{1}{2})^3$.Thus,$n = 3$ half-lives are required.Total time $t = n 	imes t_{1/2} = 3 	imes 1 \, h = 3 \, h$.

It takes $1 \, h$ for a first order reaction to go to $50 \, \%$ completion. The total time required for the same reaction to reach $87.5 \, \%$ completion will be $..... \, h$.

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