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  4. The half-life period of a first order reaction is 1 hour. What is the time/in hour taken for 87.5 % completion of the reaction?

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Q. The half-life period of a first order reaction is $1$ hour. What is the time/in hour taken for $87.5 \%$ completion of the reaction?

Chemical Kinetics

Solution:

For first order reaction

$\frac{t_{1 / 2}}{t_{x \%}}=\frac{0.3}{\log \left(\frac{100}{100-x}\right)} \Rightarrow \frac{t_{1 / 2}}{t_{87.5 \%}}=\frac{0.3}{\log \left(\frac{100}{100-87.5}\right)}=0.3$

$\Rightarrow t_{87.5}=\frac{t_{1 / 2}}{0.3}=\frac{1}{0.3}=3.33 hr \approx 3 hr$